Algebra questions

Question and mark-scheme extracts © Pearson Education Ltd. Reproduced for educational use; the original paper is linked on every question. Contact hello@oscarsongtutoring.co.uk for removal.

• Edexcel 1MA1 · November 2024 · Paper 1F · Q2 · 1 mark

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  In this question learners were presented with a list of the first four even numbers and asked to find the 10th even number. This was answered well by the majority of learners, however where errors were seen these included giving an answer of 10 (the next term in the sequence) or an answer of 20th.

  Text version

  Here are the first four even numbers: 2, 4, 6, 8.

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• Edexcel 1MA1 · November 2024 · Paper 1F · Q8 · 3 marks

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  In part (a) of this question learners were asked to give the coordinates of the point marked. The majority of learners were able to correctly identify these as (3, 1). The most common incorrect answer was (1, 3) where the x coordinate and y coordinate were reversed. Part (b) required learners to plot the point (3, −4) on the grid. This was again correctly completed by the vast majority of learners. The most common incorrect response was to plot the point (−4, 3), although some learners also plotted points that were close to the required point but out by one grid square either horizontally or vertically. Occasionally learners plotted both (3, −4) and (−4, 3) giving a choice of answers which was not awarded the mark. Part (c) required learners to write down the coordinate of the midpoint of AC. This was answered correctly by slightly fewer learners than part (a) and part (b), but was still answered correctly by the majority. It was common to see learners using the diagram to find the required midpoint, often by drawing a line between A & C, and successfully giving the answer (0, 2). Incorrect answers included reversal of the order of the x and y coordinates. A minority of learners had attempted to find the midpoint of AC on the diagram but had not done so successfully, others had incorrectly attempted to find the midpoint of AB. It was unusual to see a response that suggested that the learner did not understand what a midpoint was.

  Text version

  The coordinate grid runs from −5 to 5 on both axes. The plotted points are shown below; each character position is one unit in x. x: −5 −4 −3 −2 −1 0 1 2 3 4 5 y= 5 · · · · · · · · · · · y= 4 · · · · · · · · · · · y= 3 · · C · · · · · · · · y= 2 · · · · · · · · · · · y= 1 · · · · · · · · A · · y= 0 · · · · · O · · · · ·

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• Edexcel 1MA1 · November 2024 · Paper 1F · Q13 · 4 marks

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  In part (a) of the question learners were asked to find the output of the number machine when the input was 11. The majority of learners were able to do this correctly often showing their working to achieve the answer of 8. Only a minority of learners made errors in arithmetic in otherwise correct working. A common incorrect answer was 13.5 which arose from learners applying order of operations rules to the two separate steps within the function machine. Another incorrect approach was was putting the output as 11 then calculating either × 2 – 5 = 17 or   + 5 = 10.5. Part (b) of this question required learners to complete the number machine by adding the missing operation. This was answered well by the majority of learners, although caused slightly more difficulty than part (a). The answer of −3 was much more common than × 0.7 Where learners were not able to find either of the possible answers they were often able to gain 1 mark for finding 7 as an intermediate step or dividing 28 by 4. Some learners did not transfer their answer over to the function machine, leaving a choice of methods and answers. Candidates need to ensure they are identifying their answer and not just leaving it in the body of their working.

  Text version

  A number machine adds 5 and then divides by 2. A different machine applies one operation and then multiplies by 4.

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• Edexcel 1MA1 · November 2024 · Paper 1F · Q15 · 4 marks

  4 marks

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  The factorisation of 6 + 15 (part (a) of this question) was answered correctly by around half of the learners. Common incorrect answers included attempts at factorisation that would not expand to give the initial expression, for example 3( + 5), 3(2 + 5) and 3(2 + 3). In part (b) learners were asked to solve 4(3 + 1) = 28. Around half of the learners were able to solve this successfully with the more popular approach being to expand the bracket before solving the equation. There were also a significant minority of learners who were able to gain partial credit often for expanding the bracket correctly or for correctly isolating terms in y following an incorrect attempt at expanding the brackets. Common errors included use of addition rather than multiplication when attempting to remove the bracket, use of subtraction rather than division when attempting to divide through to remove the bracket or attempts including collecting together terms which did not simplify in this way. Another common error was to forget to multiply the 4 and +1 leading to 12 + 1 = 28. Quite a number of students used no algebraic solution, and instead approached this via a trial and improvement method.

  Text version

  Algebraic manipulation.

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• Edexcel 1MA1 · November 2024 · Paper 1F · Q16 · 5 marks

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  This question presented learners with information about the costs of two different combinations of teas and coffees. They were asked to find the cost for a third combination. This proved to be challenging for learners with the majority not being able to make progress towards a correct solution. Around a quarter of learners were able to correctly find the required cost. In attempting this question the majority of learners chose to work with the information provided without converting to simultaneous equations. This proved to be successful for those learners who could recognise how the two separate pieces of information could be used to determine the cost of just some teas (more commonly) or just some coffees (as was seen occasionally). Most that were awarded one mark for finding some teas, also successfully found the cost for one tea. Many learners, however, struggled to see how to find the cost of only one type of drink. The most common incorrect approach was assuming that the cost of a cup of tea and a cup of coffee were equal (£2.25). Others attempted to guess the cost of one of the two drinks and see what the cost would be for the second drink type. This approach had the potential to be successful if these costs were checked using both of the initial sets of information, but often the values only worked for one set of cost conditions and this was not recognised. It was common to see extensive working where these trial and improvement approaches had been tried. Some candidates recognised that you could multiply the first cost scenario by three and then add on a cup of tea to gain the correct answer of 15.5(0).

  Text version

  One cup of tea and one cup of coffee cost £4.50. Three cups of tea and one cup of coffee cost £8.50.

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• Edexcel 1MA1 · November 2024 · Paper 1F · Q23 · 4 marks

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  This question gave learners information about the interior angles of a pentagon and asked them to find the size of one of the missing angles. There were a minority of learners who could give a fully correct answer which often came from trial and improvement. only a very small minority were able to use algebraic methods to find c. Some learners gained 1 or 2 marks for the interior angle sum of a pentagon and/or for the start to a process of giving each angle in a common form. A common error here was to work with the wrong interior angle sum, often 360°, or to assume that angle d or e was the same size as angle b based on looking at the diagram. Quite a few did their angle sum – 155 but then divided by 4 forgetting that the other angles were not all the same size.

  Text version

  A pentagon has interior angles a, b, c, d and e. Angle a = angle c, b = 155°, d = 3c and e = 2c.

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• Edexcel 1MA1 · November 2024 · Paper 1F · Q25 · 2 marks

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  This question proved very challenging with only a minority of learners able to correctly change the subject of the formula. It was common to see the unknowns being incorrectly combined in a variety of different ways such as 13# or 30#. It was also common to see errors in attempts to rearrange when incorrect inverse operations were used. There were a few examples where ‘# =’ was omitted giving just $% " on the answer line. Many candidates made an incorrect first step, by either incorrectly subtracting 3, attempted to divide by three but did not divide 10 by 3 too, or subtracted 10 incorrectly. Many just swapped the y and w.

  Text version

  Make w the subject of the formula y = 3w - 10.

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• Edexcel 1MA1 · November 2024 · Paper 1F · Q26 · 3 marks

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  This question was not well answered. Only a small minority were able to make progress with factorising the quadratic and it was uncommon to see attempts at using the quadratic formula. Many learners attempted to solve the quadratic using methods based on those for linear equations. A number of learners found one root, often  = 5, through use of trial and improvement, but did not find the second root and therefore did not gain marks. Of those learners who were able factorise the quadratic, the majority did not progress on to find both solutions. Summary Based on the performance on this paper, learners / centres should work on:  improving numeracy skills, including multiplication and division skills when working with integers, decimals, and fractions (including mixed numbers), including writing calculations in the correct order and checking that their answers have a sensible value,  the interpretation of problems involving ratio, particularly combined ratios,  considering the context of the problem they are solving and checking if their answer would be feasible,  knowing, stating and applying angle facts, particularly those relating to angles in parallel lines,  solving equations and inequalities,  processes for estimation (with emphasis of rounding a value to 1 significant figure) and identification of whether the chosen rounding leads to an overestimate or an underestimate of the answer and giving reasons for their decision,  encouraging learners to set out their working using logical steps,  encouraging learners to read the question carefully so that they understand what they are required to do and what format the answer should be stated in,  interpretation of questions involving context and multiple step processes,  understanding of real-life graphs including interpreting intercepts and gradients in context.

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  Solve x² - 2x - 15 = 0.

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• Edexcel 1MA1 · November 2024 · Paper 1H · Q6 · 4 marks

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  It was pleasing that many learners gained full marks for finding the size of angle a. Those that gained the first mark for using (n – 2) × 180 to find the sum of the interior angles of the pentagon or for stating that the sum of the interior angles is 540 often went on to make further progress. Starting a process to give each angle in a common form earned the second mark and many did this by writing d = 3c or e = 2c. Some wrote the angles in terms of a different variable such as x. This was often followed by an equation, e.g. c + c + 3c + 2c + 155 = 540, which earned the third mark. Common errors at this stage were to omit one ‘c’ or two ‘c’s from the equation or failing to change the given ‘a’ to ‘c’ thereby keeping two variables in the equation and not earning the third mark. Many of those with a correct equation went on to give a correct final answer but arithmetic errors, particularly when subtracting 155 from 540 or dividing 385 by 7, were quite common. Learners making these errors could still be credited with the process marks if their intended calculations were shown. After gaining the first mark some learners did not show a process to write each angle in a common form but resorted to a trial and improvement approach and gained no more marks. Those who relied on memory for the sum of the interior angles of a pentagon and stated an incorrect figure were still able to earn the second and third marks.

  Text version

  A pentagon has angles a,b,c,d,e where a=c, b=155°, d=3c and e=2c.

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• Edexcel 1MA1 · November 2024 · Paper 1H · Q8 · 3 marks

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  Many learners made a good start, most often dividing 720 by 80 to find the area of the base of the rectangle and this earned them the first process mark. To complete the solution, they needed to divide the area of the base by 2 to find the length of the rectangle. A good proportion managed to work out the length as 4.5 cm and scored full marks. Some lost the accuracy mark due to an arithmetic error, most commonly 720 ÷ 80 = 90, but they could still gain the second process mark for dividing by 2. After finding the area of the base many learners made no further correct progress and some gave 9 as the final answer or multiplied by 2 instead of dividing by 2. It was very common to see b × h ÷ 2 being used in an attempt to find the length of the base. Presumably this was prompted by the triangular cross section of the prism. Some learners used the pressure formula incorrectly, working out 720 × 80, and gained no marks.

  Text version

  A triangular prism rests on a rectangular face of width 2 m. It exerts 720 N and pressure is 80 N/m². pressure = force/area.

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• Edexcel 1MA1 · November 2024 · Paper 1H · Q11 · 5 marks

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  In part (a) it was pleasing that many learners were able to explain what is wrong with Kate’s statement. Correct explanations often stated that the sum of a and b should be 5 and the product of a and b should be 6 and some explained that Kate got the numbers the wrong way around which was also acceptable. Explanations that mentioned just one of the mistakes were accepted but it was not acceptable to simply state that the sum is not 6 and the product is not 5. A common error was to give the correct factorisation and not explain what is wrong with Kate’s statement. A relatively small proportion of learners gained both marks for a fully correct factorisation in part (b) . Many gained one mark for a correct partial factorisation, most commonly 2(m2 – 1) but sometimes this was (2m – 2)(m + 1) or (m – 1)(2m + 2). A common incorrect factorisation that gained no marks was 2m(m – 1). More fully correct factorisations were seen in part (c) than in part (b). Correct partial factorisations such as x(a + b) – y(a + b) gained one mark. The incorrect partial factorisation x(a + b) – y(a – b) was quite common.

  Text version

  Factorisation.

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• Edexcel 1MA1 · November 2024 · Paper 1H · Q13 · 5 marks

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  There was quite a lot to do in order to work out the value of x so it was pleasing to see some well presented and fully correct solutions. Overall, though, this question was not answered particularly well and many learners failed to score any marks at all. The first mark was awarded for setting up an equation using volumes which could be the first step or it could occur later in the process. This should have been a straightforward mark but many learners failed to set up a correct equation and ‘+ 142’ or ‘– 142’ often appeared on the wrong side of the equation. Some learners only worked with cuboid A and it was not uncommon to see the volume of cuboid A equated to 142 cm3. Learners were more successful in gaining the second mark which was awarded for a process to find an expanded expression for the area of one face, where one incorrect term in the expansion of two brackets was condoned. Some went on to earn the third mark for a complete process to find a fully expanded expression for the volume of one cuboid. For this mark the expression could be unsimplified but it had to be correct so errors in their expanding cost some learners this mark. Some of those who had earned the first three marks were able to earn the fourth mark for correctly rearranging the expanded terms in their equation to get a 3-term quadratic.

  Text version

  Cuboid A dimensions are (x+2), (2x-1), (x-1). Cuboid B dimensions are (x+3), 2x, (x-3). A's volume is 142 cm³ greater than B's.

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• Edexcel 1MA1 · November 2024 · Paper 1H · Q18 · 4 marks

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  Those with some knowledge of inverse functions usually answered part (a) by rearranging x y   to make x the subject or rearranging y x   to make y the subject. A correct first step in the rearrangement earned the method mark. Many of the learners that made a correct first step went on to give a correct answer but errors in the rearranging cost some the accuracy mark. Following 4x = 5y – 3 with 4x – 3 = 5y was quite a common error. Some completed the rearrangement correctly but gave the answer in terms of y and lost the accuracy mark. A common incorrect answer was x  because f−1 (x) was interpreted by some as the reciprocal of f(x). In part (b), the most common approach used to work out the value of gh(5) was to work out h(5) = –9 and then work out g(–9). A clear intention to substitute –9 correctly earned the method mark. A surprising number of learners made hard work of evaluating (–9 – 1)2. It was common to see this being expanded to give four terms which frequently did not lead to an answer of 100 or to see –102 = –100 and these responses earned one mark only. Some found gh(x) = (1 – 2x – 1)2, which earned the method mark, and then substituted x = 5. Misunderstanding of composite functions was evident in many answers. Many learners found both g(5) and h(5) and then multiplied them or added one to the other.

  Text version

  f(x)=(5x-3)/4. Also g(x)=(x-1)² and h(x)=1-2x.

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• Edexcel 1MA1 · November 2024 · Paper 1H · Q20 · 4 marks

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  Very few learners showed any understanding of what was required to answer this question. There were some excellent responses from the most able learners taking this paper but these were few and far between. Some made a successful start by substituting into x2 + y2 = 4 to find the value of p and earned the first mark but often they made no further correct progress. Some earned the second mark for finding the gradient of the normal/radius or the gradient of the tangent. Mistakes were often made at this stage. Instead of finding the value of p a few learners chose to start by finding the gradient of the tangent in terms of p and then substituted x = p, y = 1 and the gradient into y = mx + c. Finding the value of p then enabled them to complete the solution. Learners should be encouraged to sketch the circle and tangent in questions like this as showing the key values could help them find the gradients of the radius and the tangent. Summary Based on their performance on this paper, learners should: ● practise their arithmetic skills, particularly division by a decimal number and negative number arithmetic ● consider whether or not an answer to a calculation is of a sensible size ● practise finding missing angles using angles on parallel lines and giving correctly worded reasons ● know how to calculate the sum of the interior angles of a polygon ● practise using the relationships between lengths, areas and volumes in similar figures ● practise finding angles using the circle theorems and stating the circle theorems used

  Text version

  Circle C has equation x²+y²=4. Consider the point (p,1) on C where p>0.

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• Edexcel 1MA1 · November 2024 · Paper 2F · Q11 · 3 marks

  3 marks

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  Simplifying the algebraic expression involving multiplication in part (a) was well done by a large majority of the cohort. Success in part (b) was more varied though, with scores of 0, 1 and 2 marks being awarded in equal proportions. It was common for learners to become confused with the minus sign, and one of the most common wrong answers was 5d – 5e but such answers were awarded 1 mark for correctly showing a partial simplification and 1d + 5e due to dealing with the signs incorrectly. Answers of 5d + − 3e scored only 1 mark as they were not fully simplified, but this was rarely seen.

  Text version

  Simplify algebraic expressions.

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• Edexcel 1MA1 · November 2024 · Paper 2F · Q14 · 3 marks

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  This question assessed the ability to interpret and complete a travel graph, with responses to part (a) typically being correct to gain a mark. A large number of good responses were seen in part (b), with many learners able to draw a horizontal line of the correct length, followed by a correct straight line back down to the time axis. Sometimes only one of the two was correct, which gained just 1 mark. In responses where the horizontal part was incorrect, it was usually either the wrong length or completely missing, rather than being drawn at an angle. For the part representing the journey home, some learners incorrectly drew an upwards- sloping line, and others drew lines to the incorrect time on the x-axis or drew lines which were either not a single line or were curved rather than straight, which were not acceptable.

  Text version

  A travel graph shows Amy reaches a friend's house 3 km from home, then reaches the skate park 5 km from home at 3 pm.

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• Edexcel 1MA1 · November 2024 · Paper 2F · Q19 · 4 marks

  4 marks

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  Fully factorising the expression in part (a) proved challenging for many learners and whilst some learners knew what to do, the correct response was not seen regularly. Of the learners gaining credit, the majority gained both of the marks available for a complete factorisation shown. The award of 1 mark was awarded less often than full marks for any partially correct factorisation, with the common factor of 5 or w being the most frequently seen. Some learners showed incorrect use of algebra, sometimes attempting to write the expression as a single term such as 10w2 or by including w in a factorisation where w had already been removed as a common factor which resulted in no marks being awarded. In part (b) there were a great number of fully correct responses with 2 marks being frequently awarded. Some learners were confused about which circle should be open or closed, but these usually gained 1 mark for showing the line from −2 to 4. Alternatively, just one correct circle drawn open at −2 or closed at 4 was also given 1 mark to learners. Crosses or circling of the correct numbers on the number line were not acceptable in place of the endpoint circles, and nor were arrows. Similarly, the use of the number line to join correct endpoint circles is not acceptable for full marks but often gained the partial mark.

  Text version

  Factorisation and inequality representation.

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• Edexcel 1MA1 · November 2024 · Paper 2F · Q24 · 4 marks

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  A standard and familiar topic, this question discriminated well, with a range of different scores seen. Many learners were able to gain 2 marks of the 4 available for either completing the table correctly in part (a) or finding at least 2 correct values and beginning to plot points. The most common error when completing the table predominantly involved the negative x value or when x = 0. Another commonly seen wrong answer was an attempt at an arithmetic sequence to give 1, (0), −1, −2, (−3), −4, −5 which gained no marks. When plotting the points in part (b), most learners could do so accurately and gained one mark for plotting. Most attempted to join them with a smooth curve but where points were joined with clear line segments between points, a maximum of 1 mark could be awarded. Pleasingly, this was less frequently seen compared to previous series.

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  For y=x²-2x-3, x values are -2,-1,0,1,2,3,4.

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• Edexcel 1MA1 · November 2024 · Paper 2F · Q26 · 5 marks

  5 marks

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  This 5 mark multi-step problem required learners to use a variety of topics that learners find more challenging such as forming and solving equations and using Pythagoras’ Theorem. Whilst many attempts to begin to solve the problem were shown, there was often not enough working to gain credit, such as forming a suitable expression for the perimeter but not setting this equal to 72 to form an equation for example. Of the learners who were able to gain the first mark for forming a suitable equation, many then usually managed to get the second mark for correctly isolating terms in x. After finding a value for x, many correctly substituted to find the length of all 3 sides of the triangle but unfortunately, did not know what to do next and it was very rare to see any attempt to use Pythagoras’ Theorem. The most commonly seen next step was either attempting to find the area by treating one of the slant lengths as the perpendicular height, or simply stopping. The layout of working was often unstructured and multiple processes, both correct and incorrect, were attempted which again often led to fewer marks being awarded due to not clearly identifying the work they wanted to be marked. Learners should be reminded that when providing a choice of methods, the appropriate marks for the lowest scoring method will be awarded unless one of those methods is crossed out or leads to the answer on the answer line.

  Text version

  Isosceles triangle ABC has AC=BC. AC=2x+11, BC=4x-4, AB=2x+5 cm; perimeter 72 cm.

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• Edexcel 1MA1 · November 2024 · Paper 2H · Q5 · 4 marks

  4 marks

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  Another familiar question for learners and one they continue to do well on. Many learners gained full marks for a correct table of values and a correct graph. Those who were less successful often miscalculated the value for x . This was almost certainly down to missing a bracket around    when inputting on their calculator. In this case learners normally gained 1 mark in part (a) and then the B1 in part (b) for plotting. Those who gained no marks typically tried to get the table to fit a linear pattern and had no values correct. Unlike the graph in summer, the minimum point was plotted here, rather than it being between 2 points, and learners do better in this case as we do not get a ‘flat bottom’ to the curve. The only real case of marks being dropped on the graph was either carelessness causing them to clearly miss one of the points plotted, or the use of line segments.

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  For y=x²-2x-3, x=-2,-1,0,1,2,3,4.

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• Edexcel 1MA1 · November 2024 · Paper 2H · Q7 · 5 marks

  5 marks

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  Another problem-solving question, and this one worth five marks. The first is for forming an appropriate equation using the information on the diagram. Either summing all sides to 72, or for equating the two equal sides. The second for the process to isolate the terms in x. Many learners were able to gain both marks. It was more common to see the longer equation set to 72, but most were able to then manipulate the algebra and typically solve to find the value of x. The next two marks were then for a process to find the area of the triangle. This could be done in two main ways. The first, and most common, was to use Pythagoras’ theorem to find the perpendicular height of the triangle and then use the standard formula for the area of a triangle. The second method was to use either standard trigonometry or the cosine rule to find one of the angles in the triangle and then 1 sin 2 ab C to find the area. These marks were less commonly earned than the previous two. Of those who attempted Pythagoras, many forgot to halve the base of the large triangle when finding the height. Even if was the case, the final mark could still be awarded if what they believed to be the height was clearly stated, and not one of the other lengths in the question. Those who attempted some forms of trigonometry were often the more accomplished mathematicians and as such typically went on to complete the question successfully. Overall, the full range of marks was awarded and the majority scored at least 2.

  Text version

  Isosceles triangle has sides 2x+11,4x-4,2x+5; equal first two; perimeter 72.

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• Edexcel 1MA1 · November 2024 · Paper 2H · Q10 · 4 marks

  4 marks

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  A standard simultaneous equation question and one that learners are familiar with. Centres, and therefore learners, have learned from previous series that clear algebraic working is required for full credit here and it was rare to award the SCB1 or SCB2 for correct answers without working. Although we saw the full range of marks scored, those who gained M1 for a method to isolate normally got the second M1 for substitution at the least. Those who gained the first M1A1 for full method to find one value also then normally went on to gain full marks. There were exceptions to this, but not many. As is ever the case with solving linear simultaneous equations, the major misconception comes in knowing whether to add or subtract the equations to eliminate, and a significant number of learners made the same mistake in this series.

  Text version

  Solve 4x-5y=20 and 6x+7y=-57 algebraically.

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• Edexcel 1MA1 · November 2024 · Paper 2H · Q15 · 6 marks

  6 marks

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  Part (a) of question 15 was looking for an expression for the nth term of a quadratic sequence. This is a standard question with a standard 3-mark scheme. The first mark is for finding a constant second difference and recognising it is an expression in n . The second mark comes from working with 5n and finding the difference between that and the original sequence, or for getting as far as n n  . There was then an accuracy mark for the final expression. The alternative method working with the first and second difference along with the expression '2 ...' a  '3 ...' a b   and ' ...' a b c    was seen and gained appropriate credit, although seen less frequently than in previous series. There was some success with 1 mark awarded, and a reasonable proportion going onto score 2 or even 3. Part (b) was found less accessible and few understood how to substitute into the iterative formula. Some learners gained a mark, normally for 4 9k k   but then didn’t know how to solve to find k, which meant that further marks could not be scored. Those who did find the value of k, often then went on to score both further marks. This was typically not the case only when they got mixed up with the numbering and gave the value of 3u in place of 4u .

  Text version

  Quadratic sequence 3,20,47,84,131. Separately u(n+1)=ku(n)+k, u1=9,u2=4.

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• Edexcel 1MA1 · November 2024 · Paper 2H · Q17 · 3 marks

  3 marks

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  This was the stage of the paper where the level of demand poses a challenge to many learners. That said, most did at least make a good attempt at question 17, and we saw plenty achieving the first mark for writing at least two of the fractions correctly with a common denominator. The second mark proved more challenging as they had to have all fractions with a common denominator and numerators expanded. There was normally an error here, often due to the two negatives in the final fraction. We saw very few examples of learners being able to complete the algebra to the required result of 2 y .

  Text version

  Show that (6x-y)/(10xy)+1/(2x)-(2y-7x)/(5xy) simplifies to k/y for integer k.

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• Edexcel 1MA1 · November 2024 · Paper 2H · Q21 · 5 marks

  5 marks

  21

  
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  Learners had two routes into the problem. They could work with gradients and the fact that perpendicular gradients have a product of 1  to form an equation to be solved. Alternatively, they could use Pythagoras to form an equation. This did lead to 4 expressions to each be squared, expanded and then summed leading to 12 terms. The sheer quantity of terms often leads to mistakes in manipulation and simplification. The presence of having to deal with multiple negative signs also proved a challenge. Where learners did have some success it was normally for a single correct gradient. From there most learners didn’t know how to proceed further.

  Text version

  P=(-4,5), Q=(6,-6), R=(k,k+3); angle PRQ=90°.

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• Edexcel 1MA1 · November 2024 · Paper 2H · Q22 · 5 marks

  5 marks

  22

  
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  The final question was a probability problem in which learners had to generate their own expressions without any scaffolding provided in the question. Many tried to work simply with numerical probabilities, and this scored zero. To gain the first mark a correct second probability was needed, and this did need to include algebra. The second mark was for a correct product, either for both red, both yellow, or one of each. The third mark was for forming a correct equation. Those who gained the first two, often failed to gain this mark as they only worked with one product and not two. The final two marks came from manipulating the algebra into an equation to solve (either quadratic or linear), and then the correct answer. Summary Based on the performance on this paper, learners/centres should:  Spend time learning how to describe the various regions in Venn diagrams clearly, ideally using set notation.  Practise articulating their mathematics in verbal discussions to aid them in written responses.  Learn the shapes of common graphs. In particular remembering quadratic graphs are curves and not a series of line segments.  With transformations take care to read the question carefully so they know which shape is the object and which is the image. Also, that if asked for a single transformation giving two will result in zero marks.  Continue to work on annotating working clearly, especially in extended problem questions, to ensure all marks can be awarded.  Practise further working with iterative formulae.  Whenever possible use accurate figures without prematurely rounding or truncating in calculations and only round at the final stage.  Practise working with multiple negative signs when working with algebra.

  Text version

  A box has only red/yellow counters; 3/5 are red. Two are drawn without replacement; probability same colour is 41/80.

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• Edexcel 1MA1 · November 2024 · Paper 3F · Q6 · 3 marks

  3 marks

  6

  
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  Part (a) was answered well with most learners able to give a correct answer of 5 for one mark; the most common incorrect answer seen was 24. Part (b) also saw plenty of success with most learners giving the correct answer of 17 for 1 mark; the most common incorrect answer was 3, presumably coming from truncating the answer of 24 ÷ 7. Part (c) was also answered well; common incorrect answers included 2, 4, and 6.

  Text version

  Solve three equations.

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• Edexcel 1MA1 · November 2024 · Paper 3F · Q12 · 2 marks

  2 marks

  12

  
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  This question saw mixed results. A good number of learners were able to give a correct expression. Some gave the correct expression but did incorrect further working such as ‘simplifying’ to e.g. 13xy or 13, gaining M1A0. Some also turned their expression into a formula such as T = 6x + 7y or Total = 6x + 7y, showing a lack of understanding of the definition of an expression. There were also a good number who gave expressions that were completely incorrect such as x6 + y7. Some learners used a and o instead of x and y, this could only gain credit if a and o were defined as number of packs of apples and number of bags of oranges respectively.

  Text version

  A shop sells apples in packs of 6 and oranges in bags of 7. It sells x packs of apples and y bags of oranges.

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• Edexcel 1MA1 · November 2024 · Paper 3F · Q25 · 3 marks

  3 marks

  25

  
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  Although part (a) saw mixed results, it was pleasing to see some learners pick up 2 marks at this stage of the paper. Some gained B1 as they were not able to simplify all 3 terms but managed to do so with 2 terms; the y power as 5 instead of 6 or the coefficient as 9 instead of 14 were often seen. Many learners were not able to give an answer in the correct form e.g. some gave their answer as an addition, whereas others appeared not to be able to make any sort of correct start. Some that got the three terms correct only gained B1 due to the inclusion of multiplication signs. Part (b) saw a good number of this cohort able to gain B1 with a correct answer. That being said, many were not able to simplify the powers correctly; common incorrect answers were m−1 and m5.

  Text version

  Simplify expressions with indices.

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• Edexcel 1MA1 · November 2024 · Paper 3F · Q27 · 3 marks

  3 marks

  27

  
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  It was very rare to see learners making much progress with this question. Those that gained 1 mark usually produced an equation with a value of c as 3. Many learners did not know how to start the method and simply wrote down pairs of coordinates such as x = 1.5, y = 3. Summary Based on the performance on this paper, learners should:  learn the names of special quadrilaterals and their associated properties  practise reading scales for axes on graphs  work with the given information in the question and not include further interpretation of their own  work on calculator skills, including the use of the mixed number button  understand that a frequency polygon requires the points to be joined with line segments, not a curve  work on percentage profit and finding one amount as a percentage of another in general  practise finding the equation of a straight line from a graph  practise finding an angle bisector  show complete mathematical methods e.g. 2 of 200 or 35% of 80 are not credit worthy unless seen with an accurate result  be encouraged to use rulers with graphs such as frequency polygons

  Text version

  A straight line L crosses the y-axis at 3 and passes through (2,-1).

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• Edexcel 1MA1 · November 2024 · Paper 3H · Q9 · 3 marks

  3 marks

  9

  
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  A well answered question with common errors being that learners either didn’t simplify the 2 × 7 or added rather than multiplied them. There were some who gave their final answer as a sum and hence scored B0 and the retention of multiplication signs in a final answer scored a maximum of B1.

  Text version

  Indices.

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• Edexcel 1MA1 · November 2024 · Paper 3H · Q14 · 3 marks

  3 marks

  14

  
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  A pleasing number of learners were familiar with the requirement of proving algebraically that the given recurring decimal was equal to the given fraction. Most chose to find the value of 10x and 1000x and subtract these. Where full marks were not awarded, this was invariably due to not showing the recurring nature of the given decimal and using terminating decimals throughout the working.

  Text version

  Prove recurring decimal 0.46262626... equals 229/495.

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• Edexcel 1MA1 · November 2024 · Paper 3H · Q15 · 4 marks

  4 marks

  15

  
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  Many learners knew that the first step was to multiply both sides of the equation by the denominator, however too often this was not always carried out correctly. When the first mark was awarded this was often the only mark gained due to errors in the method of isolating the terms in p on one side of the equation and factorising the expression. Those who did not know the correct first step often incorrectly cancelled terms in the fraction or attempted to add or subtract terms in p from the fraction.

  Text version

  t = 2(2p - 3)/(5 - 2p).

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• Edexcel 1MA1 · November 2024 · Paper 3H · Q16 · 5 marks

  5 marks

  16

  
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  Few fully correct solutions were seen to this question. Those who attempted the question but gained no marks often found the gradient between the given points (2, -4) and (12, -6.5) which gained no credit. Several attempts at using Pythagoras’ Theorem to find the lengths of sides were also seen. The most commonly, awarded mark was for finding the gradient of OB. Of those who realised that the gradient of OB was necessary to proceed, some of these were able to find the equation of DE and/or OD, with many correct equations leading to a correct solution. Those who just guessed at a co-ordinate gained no marks since the command that “you must show all your working” means that a correct answer with no supportive working gains 0 marks.

  Text version

  OBCD rectangle; D,C,E collinear; B=(2,-4), E=(12,-6.5).

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• Edexcel 1MA1 · November 2024 · Paper 3H · Q19 · 4 marks

  4 marks

  19

  
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  A challenging question with only a few forming an equation in terms of one variable. Of those who correctly linked the radius and the height, only a few used Pythagoras to find the slant height. Of those who were successful on this question, clear algebraic manipulation was used to obtain an expression for the curved surface, most commonly embedded within a full equation for the surface area of the cone. Most learners in this case would then simplify the equation through eliminating π first and then simplifying the fraction to obtain a correct solution.

  Text version

  Cone radius r=3/4 height h; total surface area 54π cm²; slant l.

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• Edexcel 1MA1 · November 2024 · Paper 3H · Q20 · 4 marks

  4 marks

  20

  
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  Most of those who obtained marks on this question had correctly recognised that making y, rather than x, the subject would be the most sensible approach. Of those who made a correct substitution for y2, it was invariably sign errors that cost them the most marks. Use of the quadratic formula was seen more often than use of completing the square and again it was sign errors that led to incorrect solutions. Reasons for a learner obtaining 3 rather than 4 marks was for failing to find the y values, not clearly identifying the final pairs of values (good practice being x1, y1, x2, y2) or leaving answers in surd form despite the question instructing that answers should be given to 3 significant figures.

  Text version

  Solve y²=3x²+4 and y+2x=7.

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• Edexcel 1MA1 · November 2024 · Paper 3H · Q21 · 4 marks

  4 marks

  21

  
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  In part (a), there were many blank responses, attempts at translations and reflections in the y axis. Those who reflected in the x axis often did so correctly although there were a number that had one incorrect segment. In part (b), there were various misunderstandings apparent. Some incorrectly identified point P as (90, 1), whilst others added 180 to their y coordinate of point P rather than subtracting 180 and some added both the 180 and the 4 to their y coordinate. Invariably those that obtained one mark usually did so for having (90, b) or (a, 3) rather than for identifying the coordinates of P.

  Text version

  For part (a), the graph y = f(x) consists of straight line segments joining (−3,3) to (−2,−2) to (3,3) to (6,3) on a unit grid. For part (b), a sketch shows y = sin x° for one complete cycle beginning at O. Point P is the minimum turning point in that cycle. The graph is translated to y = sin(x + 180)° + 4, and P moves to R.

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• Edexcel 1MA1 · November 2024 · Paper 3H · Q22 · 3 marks

  3 marks

  22

  
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  Various responses were seen to this question with common incorrect answers being from 30 × 92 or 54 × 10 + 76 × 10 + 92 × 10. Of those who gained just one mark it was usually for correctly finding the area of the triangle in the first section. Of those that had an appreciation of what they needed to do, many limited their marks by incorrectly reading values from the graph or not doing as instructed and using 3 strips of equal width.

  Text version

  Velocity-time graph for 0-30 s; ordinates at 0,10,20,30 s are 0,54,76,92 m/s.

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• Edexcel 1MA1 · November 2024 · Paper 3H · Q23 · 5 marks

  5 marks

  23

  
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  A challenging question with the most common mark being awarded for drawing a curve with a y intercept at –7. Of those that recognised the need to complete the square, scores were higher; common mistakes in this being to not square the 3p. Most of those who correctly identified the turning point drew it in the correct quadrant but some did not gain full marks as they incorrectly identified roots at –1 and 7 or –p and 7p. Summary Based on their performance on this paper, learners are offered the following advice:  Practise basic algebraic processes such as expanding a single bracket, factorising and collecting terms that can be used when changing the subject of a formula.  When comparing distributions, ensure that comparative words are used such as “greater” or “higher”  Ensure that recurring decimal notation, rather than terminating decimals are used when converting a recurring decimal to a fraction.  Achieve a greater understanding of equivalent proportion in capture-recapture.  Work on appropriate rounding, especially with monetary questions.

  Text version

  y=x²-6px-7, p>0.

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• Edexcel 1MA1 · June 2024 · Paper 1F · Q9 · 5 marks

  5 marks

  9

  
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  In part (a) the majority of candidates were able to correctly calculate the output of the number machine for an input of 16. Incorrect answers, where seen, tended to be the result of an arithmetic error, for example 26 −10 incorrectly evaluated as 6, or from working with only part of the number machine leading to answers of 26 or 3. Part (b) of the question was also well answered. Where candidates did not correctly find the input which would give an output of 28, they were often able to gain partial credit for a start to a method to use inverse operations, often by working out 28 + 10. The most common incorrect answers were 46 or 9 which came from the result of using an input of 28 rather than using this as the output or from subtracting 10 from 28 and then dividing by 2. Where candidates knew to add 10 and then divide by 2, dividing 38 by 2 proved problematic. There were many instances of students performing the inverse operations correctly, getting the answer 19 then writing a different number on the answer line. Other candidates obtained an incorrect answer of 36 by calculating (28 – 10)  2, therefore only partially using inverse operations. Many candidates found it useful using the number machine to write their workings, helping them to answer the questions step by step. In the final part of this question students were asked to show that there is a number for which the output of the number machine was the same as the input. Many candidates were able to correctly identify 10 as this number and the majority showed that this was the case either by indicating that 10 × 2 = 20 and 20 −10 = 10 or by showing the number machine with an input of 10 and an output of 10. Where candidates did not show that an input of 10 gave an output of 10, they were often able to gain partial credit for identifying that 10 was the value which met the requirements of the question without showing that this input gave a matching output or for a correctly evaluated trial. Partial credit was also available for candidates who had made a correct trial of a number different to 10. The most common incorrect answers seen were where candidates changed the number machine used in order to make their input and output match, often by changing the amount subtracted.

  Text version

  Here is a number machine.

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• Edexcel 1MA1 · June 2024 · Paper 1F · Q17 · 3 marks

  3 marks

  17

  
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  Many candidates were able to correctly solve the equation given, most commonly expanding the brackets as the first step in their method of solving. It was also common to see well set out working. Where candidates were not able to solve the equation completely, they were often able to gain 1 mark for correctly expanding the brackets or for correctly isolating terms in x following on from their incorrect expansion. A common incorrect first step was to try to subtract 5 from both sides without expanding the brackets first. Others expanded the brackets correctly, but went on to add 10 to the left hand side and subtract 10 from the right hand side and giving an answer of 𝑥= 1. Those who were awarded the first two of three marks had failed to divide 28 by 8 correctly and reached answers like 3.4.

  Text version

  Solve 2(4x – 5) = 18

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• Edexcel 1MA1 · June 2024 · Paper 1F · Q20 · 2 marks

  2 marks

  20

  
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  A minority of candidates were able to find the nth term for the arithmetic sequence given in the question. More candidates were able to give an answer in the form 4𝑛+ 𝑘 having identified the common difference for the sequence, but finding the incorrect constant term or no constant term (i.e. 4n). Common incorrect answers included identifying the next term in the sequence (17), stating the term-to-term rule for the sequence, +4, 𝑛+ 4 where the need for an algebraic expression was recognised but the common difference was incorrectly used, or switching the coefficient of n with the constant term −3𝑛+ 4

  Text version

  Here are the first four terms of an arithmetic sequence.

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• Edexcel 1MA1 · June 2024 · Paper 1F · Q25 · 4 marks

  4 marks

  25

  
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  In part (a) of this question candidates were asked to find the equation of the line shown. This was answered poorly by the majority of candidates and a significant minority did not attempt the question. Where candidates did attempt to answer the question it was unusual to see a fully correct equation for the line L. Many candidates drew a triangle under the line as a first step to calculating the gradient but then didn’t know how to progress from there. A greater number of responses correctly identified the y-intercept as 3 or used the correct structure for the equation of a line, 𝑦= 𝑚𝑥+ 𝑐 which were awarded 1 mark. Common incorrect answers included giving coordinates, often (2, 6) which was the closest point on the line to the letter L or (-2, 3) which was obtained from the intercepts with the axes. Where equations of lines were attempted they often made use of the values -2, 3, 2 or 6 combined with the variables x and y, for example 𝑦= 6𝑥 or 3𝑦= −2𝑥. In part (b) candidates were asked to write down the equation of a straight line parallel to 𝑦= 5𝑥. There were a minority of candidates who were able to give a correct equation. A significant proportion of candidates did not attempt the question. Where attempts were seen, common incorrect responses included 𝑦= −5𝑥, 5𝑦= 𝑥, 𝑦= 𝑥+ 5, 𝑦= 𝑘𝑥 with k ≠ 5, simply restating the original equation 𝑦= 5𝑥, a coordinate or a numerical value.

  Text version

  Straight line L passes through (-2, 0) and (0, 3).

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• Edexcel 1MA1 · June 2024 · Paper 1F · Q28 · 3 marks

  3 marks

  28

  
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  Solving the inequality 𝑥+ 11 ≤5 − 2 𝑥 proved very challenging for candidates with most unable to make progress towards a solution with a significant proportion of candidates not attempting the question at all. Where candidates did have a correct process many struggled to carry out − 1.5 as their final step or, having evaluated this to -4 gave this value rather than an inequality as their final answer. Many candidates chose to show their intended calculation step and then present the subsequent line of working, this was awarded credit where the elements of the inequality that should be acted upon by that step had changed (allowing for errors in arithmetic) and the other aspects of the inequality had not. A common incorrect approach was to incorrectly combine the terms in x with the constant terms or to attempt to multiply through by 2 and not apply this to all of the terms. Only a minority of candidates attempted trial and improvement which was generally not a successful approach. Summary Based on the performance on this paper, students/centres should work on: • improving numeracy skills, including multiplication and division skills when working with integers, decimals, and fractions (including mixed numbers), • the interpretation of problems involving ratio, particularly combined ratios, • considering the context of the problem they are solving and would their answer be feasible, • knowing, stating and applying angle facts, • describing a single transformation, • solving equations and inequalities, • working with number machines and inverse number machines, • processes for estimation (with emphasis of rounding a value to 1 significant figure) and identification of whether the chosen rounding leads to an overestimate or an underestimate of the answer, • finding the equation of a given line, • understanding of set notation, • identification of different types of percentage question – including when reversed percentage change is required and how this should be performed, • encouraging students to set out their working using logical steps, • encouraging students to read the question carefully so that they understand what they are required to do and what format the answer should be stated in, • interpretation of questions involving context and multiple step processes.

  Text version

  Solve x + 11 ≤ 5 - (1/2)x.

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• Edexcel 1MA1 · June 2024 · Paper 1H · Q1 · 2 marks

  2 marks

  1

  
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  This familiar type of question was answered well. Many of the students who recognised that 4n was needed in their expression were able to write the correct expression, 4n – 3, for the nth term of the sequence. Answers such as 4n, 4n + 1 and 4n +3 were common and these gained one of the two marks for an expression of the form 4n + k, with k ≠ – 3. The most common incorrect answers that gained no marks were n + 4 and –3n + 4.

  Text version

  Here are the first four terms of an arithmetic sequence.

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• Edexcel 1MA1 · June 2024 · Paper 1H · Q6 · 4 marks

  4 marks

  6

  
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  Part (a) was generally answered quite well. Many students were able to get one mark for being able to use y = mx + c. Some students drew an appropriate triangle on the diagram but did not know how to use it, others did use it and showed correct working or stated that the gradient is 3 2 . Incorrect gradients included − and 2 3 . Students should be encouraged to write down the formula for a gradient before applying it as this may minimise the use of incorrect values. Many went on to use the gradient correctly in an equation but some students did not know what to do with it. When the gradient was used correctly a common wrong answer was y = 3 2 x – 2 where the x intercept was used rather than the y intercept and some gave the answer as y x = . Answers such as these were awarded the two method marks. A few students, who knew what they were doing, spoiled their final answers by using L instead of y in their equation, but were awarded two of the three marks. Students who did not know how to find the gradient could gain the first method mark for demonstrating understanding of the y-intercept. A similar number of students were successful in part (b) when writing down the equation of a straight line parallel to y = 5x. It was clear that some students did not know that parallel lines have the same gradient. Common incorrect answers often included the digit 5, for example y x = − , y x = , y x = − and even y = 5x. Some students simply changed the 5 for a different digit and gave answers such as y = 4x or y = 2x.

  Text version

  Straight line L passes through (-2, 0) and (0, 3).

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• Edexcel 1MA1 · June 2024 · Paper 1H · Q10 · 4 marks

  4 marks

  10

  
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  Many of the students appeared familiar with simultaneous equations and started by multiplying both equations to make the coefficients of x or y the same. Most then subtracted one equation from the other to eliminate one of the variables and gained the first method mark. Accuracy was sometimes lost through arithmetic errors in the multiplying or subtracting. The most successful students were those who eliminated y first as this made the arithmetic easier. Eliminating x using 10x – 4y = 46 and 10x – 15y = 90 tended to lead to more subtraction errors than using 15x – 6y = 69 and 4x – 6y = 36 to eliminate y. Having carried out the necessary multiplications some students added the two equations and gained no marks. Students who used a correct method to find one value usually went on to substitute this value into an equation in order to find the other value and so gained the second method mark. A number of students who substituted x = 3 into the second equation to get 15 – 2y = 23 could not solve the equation correctly, often arriving at y = 4. Eliminating one variable by rearranging one equation and substituting into the other equation was attempted by a few students but this method rarely led to a fully correct answer. There were many responses in which students had made multiple attempts to find a solution and the working was often difficult for examiners to follow.

  Text version

  Solve the simultaneous equations

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• Edexcel 1MA1 · June 2024 · Paper 1H · Q12 · 3 marks

  3 marks

  12

  
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  Most of the students correctly identified graph H as the graph of y = x2 – 4. Less of the students identified graph F as the graph of y = –x2. Not surprisingly, the other two cubic graphs, graph C and graph E, were common incorrect responses. Even fewer students were able to identify graph J as the graph of y x = − . Common incorrect answers were A, C, F and G.

  Text version

  Nine graphs are labelled A, B, C, D, E, F, G, H and J. A is a downward-opening parabola with its vertex above the x-axis; B is a positive reciprocal graph; C is a positive-leading cubic with two turning points; D is an increasing straight line; E is y = x^3; F is y = -x^3; G is a decreasing straight line through the origin; H is an upward-opening parabola with its vertex below the x-axis; J is a negative reciprocal graph. Identify the graph that could have equation (i) y = x^2 - 4, (ii) y = -x^3, (iii) y = -5/x.

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• Edexcel 1MA1 · June 2024 · Paper 1H · Q14 · 3 marks

  3 marks

  14

  
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  This question, which required the expansion and then simplification of a product of three linear expressions, was generally answered well. Usually, students showed their working in an organised way or used a grid to show the terms in their products. Errors were usually restricted to incorrect terms or difficulties in dealing with the signs when collecting terms together, rather than a flawed strategy. Some students omitted terms from their expansion when at the stage of multiplying a quadratic expression by a linear expression and they could not be awarded the second mark because their method was not complete. For students who did not give a fully correct answer, it was commonplace to see 2 marks, or 1 mark awarded. Some lower attaining students sometimes either stopped after multiplying two linear expressions together or lacked a clear strategy and tried to multiply all three brackets together at once. Misreads of one or more of the expressions were common and there were a surprising number of transcription errors made when transferring a correct answer from the final line of working to the answer line.

  Text version

  Expand and simplify (3x - 1)(2x + 3)(x - 5).

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• Edexcel 1MA1 · June 2024 · Paper 1H · Q20 · 3 marks

  3 marks

  20

  
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  This question was not answered as well as might have been expected. The expressions for 2x and 2y contained a combination of surds and indices and many students worked with surds but not with indices and made no meaningful progress. It was common, for example, to see ( ) written as 4 2 rather than as 2 . Many of the students who did get the first mark for 2 or for y = 2.5 were unable to write 3 n as n−. Those who got the second mark for an equation such as n = − + usually went on to give a correct answer although arithmetic errors cost some students the final mark. Any errors made when converting 35 6 into a mixed number were ignored as the improper fraction is acceptable as the final answer.

  Text version

  2^x = 2^n / cuberoot(2) and 2^y = (sqrt(2))^5. Given that x + y = 8, work out n.

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• Edexcel 1MA1 · June 2024 · Paper 1H · Q21 · 5 marks

  5 marks

  21

  
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  Many students gained the first mark for dividing 300 by 20 to find the area of one face of the cuboid or for forming an equation using the volume such as 20wh = 300. Many students then gave an answer of 3 because 3 × 5 = 15 and the width is greater than the height. Those who did make further correct progress gained the second mark for forming an equation using the surface area. A common mistake at this stage was assuming that four of the faces of the cuboid have the same area. Fewer than half of the students who formed two correct equations went on to gain any more marks. Some students completed the solution by eliminating one variable and solving the resulting quadratic equation. The majority of those who found both 2.5 and 6 then chose the correct value for the height. Others simplified the two equations to wh = 15 and w + h = 8.5 and hence deduced that h = 2.5. An apparent lack of confidence with simultaneous equations meant that a significant number of students resorted to trial and improvement methods once they had equations for volume and surface area. Some tried several pairs of numbers multiplying to 15 until they found the pair that resulted in the correct total surface area. Others were not successful because they only used integer values. A common misconception was to divide the surface area by 6, assuming that the shape was a cube.

  Text version

  A solid cuboid has volume 300 cm^3 and total surface area 370 cm^2. Its length is 20 cm, and its width is greater than its height. Work out the height of the cuboid. Show all your working.

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• Edexcel 1MA1 · June 2024 · Paper 1H · Q22 · 4 marks

  4 marks

  22

  
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  In part (a) a good number of students gained the method mark for sketching a graph with the correct shape between x = 0 and x = 360. About half of these students went on to gain both marks for a fully correct sketch with labels at 1 and –1. Incorrect sketches that gained no marks were often attempts at a sine curve with an incorrect period or attempts at a cosine curve but sketches of horizontal lines were also quite common. Part (b) was not as well answered as part (a). Some students gained the method mark for stating that sin30 = , which was often seen in a table of exact trig values, or for giving one correct solution. About one third of those who gained the method mark went on to give the two correct solutions.

  Text version

  (a) Sketch y = sin(x°) for 0 ≤ x ≤ 360. [2 marks]

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• Edexcel 1MA1 · June 2024 · Paper 1H · Q23 · 5 marks

  5 marks

  23

  
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  It was pleasing to see some fully correct and well presented solutions to this grade 9 problem. Many students attempted to draw diagrams to help them structure their response. Some of those who did then started to solve the problem by using the equation of a circle and gained the first mark. Some made no further progress but a good number were able to show a correct process to find the y coordinate of Q which gained the second mark. Selecting the negative square root (because L has a positive gradient) gained the third mark. Students who used 11 rather than –11 were still able to gain the fourth mark for a process to find the y intercept. This process usually involved finding the gradient of L and if the gradient was incorrect then the mark could only be awarded if it could be seen that the gradient had come from a correct process. A significant number of students attempted to solve the problem by using only gradients and scored no marks. Summary Based on their performance on this paper, students should: ● communicate clearly when a decision and a reason are required ● develop an understanding of the purpose of rounding so that they can choose appropriate. rounded values when working out an estimate ● become more familiar with the set notation for the complement of a set ● practise applying the rules of indices in problems, including the use of negative and fractional indices ● practise solving problems involving similar triangles ● ensure that axes are correctly labelled on graphs and charts ● practise drawing and interpreting histograms

  Text version

  C is a circle with centre (0, 0)

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• Edexcel 1MA1 · June 2024 · Paper 2F · Q5 · 1 mark

  1 mark

  5

  
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  Students generally completed the statement for a simple algebraic expression correctly and were awarded the mark.

  Text version

  Complete the statement: ___ × m = 2m.

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• Edexcel 1MA1 · June 2024 · Paper 2F · Q8 · 4 marks

  4 marks

  8

  
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  Many students were able to write down the required coordinates in part (a) and a similar level of success was seen when plotting the point in part (c). When the mark wasn’t awarded in part (a), it was often due to writing the x and y coordinates round the wrong way and often led to the answer to part (c) being plotted incorrectly at (2, −4) instead. Success in part (b) was not as frequent however, still a majority gained the mark nonetheless. The most successful approaches to finding the midpoint of AB often appeared to be when the two points were joined and often led to either full marks or 1 mark being awarded. When only the method mark was awarded it was often for either transposing the x and y coordinates or for marking the midpoint on the grid. Other common errors included the exclusion of the negative sign for the y coordinate, stating x = 0 and y = −2 or giving the coordinates of point B due to not reading the question carefully.

  Text version

  Points A = (2, 1) and B = (-2, -5) are shown on a coordinate grid. (a) Write down A. [1 mark] (b) Find the midpoint of AB. [2 marks] (c) Mark C = (-4, 2). [1 mark]

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• Edexcel 1MA1 · June 2024 · Paper 2F · Q13 · 3 marks

  3 marks

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  The expression in part (a) was often simplified correctly, with over half of the cohort being awarded the mark. The most commonly seen error was to add the coefficients rather than multiply or still including multiplication signs. Success in part (b) was very pleasing with a significant proportion of students demonstrating they understood substitution of negative numbers and could also evaluate correctly to score full marks. The most common incorrect answers stemmed either from doing 3 + 2 – 4 instead of 3 + 2 × (−4), or from evaluating incorrectly as −20 due to failing to apply the correct order of operations. The use of brackets 3 + 2(−4) was sufficient for students to achieve the method mark for a correct substitution and students should be encouraged to include brackets around negative values in every topic as this would mean incorrect evaluation from poor calculator use could be avoided.

  Text version

  (a) Simplify 2c × 3d [1 marks]

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• Edexcel 1MA1 · June 2024 · Paper 2F · Q16 · 3 marks

  3 marks

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  Part (a) required an accurate reading from the conversion graph provided, and many students were able to give an answer within the acceptable range of 167 to 173 and those not gaining the mark often demonstrated they knew how to use the graph correctly but made an error when interpreting the grams scale. Part (b) however, was often poorly attempted in comparison. Successful students were those who used the graph wisely to extrapolate data and clearly showed what reading they were using and how this multiplied to make 1000g. Many students were either competent in using readings from the graph as a factor of 1000 or applied a build-up method of various readings. Of the students identifying the need to use a factor of 1000 and reading from the graph, most go on to get full marks. Some students only gained one of the two marks available as they were unable to interpret the scale for ounces, e.g. reading 100g = 3.5 ounces as 100g = 2.8 ounces, but could follow through and scale up correctly. Another common error that students made was to use 300 grams as part of a build-up, usually equating this to 10 ounces or more before multiplying by 3 and adding a reading for 100g. This approach gained no credit due to 300g being outside of the range of the graph. Extrapolation has always proved to be a challenging topic for students and in this series, although improvements can be seen, many of the students show insufficient working when the answer is outside of the specified range as incorrect readings need to be supported to gain credit.

  Text version

  A straight ounces-to-grams conversion graph passes through approximately (0 oz, 0 g) and (10 oz, 283 g).

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• Edexcel 1MA1 · June 2024 · Paper 2F · Q22 · 2 marks

  2 marks

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  This 2-mark question, on the whole, was answered very well with most students being able to substitute both of the values correctly and being able to find the difference to gain both marks. A common mistake was to substitute the incorrect values or not using the formula at all and dividing the number of bottles, redundant information, by the number of machines. It is therefore important for centres to ensure that students read the questions carefully and not assume that every number is to be used in the question.

  Text version

  The time H hours for n machines to make 5000 bottles is H = 72/n. On Monday n = 6 and on Tuesday n = 9. How much longer did Monday take?

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• Edexcel 1MA1 · June 2024 · Paper 2F · Q24 · 6 marks

  6 marks

  24

  
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  A standard and familiar question split into 3 parts and assessing knowledge of quadratic graphs. In part (a) many students achieved the full 2 marks for finding all three values correctly. However, equally as many achieved 1 mark for finding one or two correct values, with the corresponding y value for when x was negative being incorrect. Students should be encouraged to use their calculator to find the values, using brackets when substituting negative values. The majority of students achieved one mark for plotting at least 5 of their coordinates correctly, with many of these being for following through their incorrect points provided a mark in part (a) had already been awarded. Some of the students who got full marks in (a) only achieved one mark in (b) as they did not join their points, joined them using line segments or their parabola did not have a clear vertex below the x-axis. The final part of the question was not answered very well at all and was often left blank. Very few students achieved the full two marks for two solutions in range. The method mark was occasionally awarded for an answer of 2.5 to 2.7 but many students often disregarded the negative value of x and very few attempted to draw the line y = 4 or marked the solutions on the graph which would have also led to the method mark being awarded. Another common mistake leading to lost marks was writing the answer(s) embedded in coordinates.

  Text version

  (a) Complete the table for y = x^2 - x at x = -2, -1, 0, 1, 2, 3. [2 marks]

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• Edexcel 1MA1 · June 2024 · Paper 2F · Q26 · 4 marks

  4 marks

  26

  
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  The majority of students were not able to form a suitable equation involving at least two angles. When marks were awarded, students were often able to secure 3 marks. However, they missed out on the final mark due to not providing a suitable conclusion to show that ABCD was in fact a trapezium. When marks were awarded, only a small minority were awarded 1 or 2 marks and this was often due to being able to form an equation but either not recognising the need to solve; simplifying incorrectly; using an incorrect method to isolate; or isolating correctly with arithmetic errors. The main error when beginning the problem occurred when students did not recognise that a trapezium has one pair of parallel sides, and consequently, the co-interior angles on the same side of the transversal sum up to 180 degrees. Instead, they approached the problem as if dealing with a general quadrilateral, which often affected the final communication mark. Additionally, some students correctly added the four interior angles or a suitable pair of angles but did not equate the sum to 360° or 180° degrees to allow them to solve for the variable x. This oversight prevented them from securing any marks in this section. Lots of responses tried to use the formula for area of a trapezium which again scored no credit.

  Text version

  The four interior angles of a quadrilateral are labelled 4x + 15 degrees, 4x + 8 degrees, 2x + 15 degrees and 3x - 3 degrees. Show that one pair of opposite sides is parallel.

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• Edexcel 1MA1 · June 2024 · Paper 2H · Q3 · 2 marks

  2 marks

  3

  
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  On the whole, this small problem was answered well, but many were put off by the extra numbers in the question. The ‘5000’ was simply there to show the same number of bottles were made both days, but many tried to use this figure in the formula, often in place of 72. There were also cases where students wrote 72 ÷ 6 but then went on to use this in an incorrect calculation. This said, many did gain full credit for using the formula twice and subtracting to get an answer of 4.

  Text version

  The time H hours for n machines to make 5000 bottles is H = 72/n. On Monday n = 6 and on Tuesday n = 9. How much longer did Monday take?

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• Edexcel 1MA1 · June 2024 · Paper 2H · Q5 · 6 marks

  6 marks

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  Another familiar style of question, and again, many scored good marks. As it always the case in part (a) a number struggle to substitute negative numbers into quadratic equations, and so the y value when x = −1 was often incorrect. This then meant they could only score 1 at best in part (b). Those who completed the table correctly typically then scored both marks in (b). However, we are still seeing the same two errors we often see, line segments, rather than curves, and a ‘flat bottom’ to the graph. Both of these resulted in a lost mark for the graph. Part (c) is always the lowest scoring, but a good number gained marks, and often 2. Those who had an incorrect graph in (b) could still gain both marks in (c) providing the graph led to 2 solutions when y = 4. Unfortunately, some were able to find both correct values for x, but lost a mark for poor notation, normally writing the solutions as coordinates.

  Text version

  (a) Complete the table for y = x^2 - x at x = -2, -1, 0, 1, 2, 3. [2 marks]

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• Edexcel 1MA1 · June 2024 · Paper 2H · Q7 · 4 marks

  4 marks

  7

  
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  There were three main ways of completing this problem. The first, and most common, was to form an equation by summing all angles to 360°. The second was to assume it was a trapezium and set one pair of co-interior angles to 180°, and the third was to set both pairs of co-interior angles equal to each other. An equation to equate one of these was needed to score any marks. In the case of the first two approaches, to gain the final mark, substitution to show the angles were co-interior was required, as was some form of explanation. In the third approach an explanation was enough for the final mark. Many students gained 3 marks by forming a suitable equation and solving to get x = 25. However, the final mark was often harder to come by, with the most common approach showing that if x = 25 the angle sum is 360° and hence it is a trapezium. Obviously, the angles summing to 360° is not sufficient to show it is a trapezium. A notable amount of students added opposite angles equal to 180 to try to solve for x, showing they are unclear of the properties of a trapezium.

  Text version

  The four interior angles of a quadrilateral are labelled 4x + 15 degrees, 4x + 8 degrees, 2x + 15 degrees and 3x - 3 degrees. Show that one pair of opposite sides is parallel.

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• Edexcel 1MA1 · June 2024 · Paper 2H · Q10 · 3 marks

  3 marks

  10

  
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  This question started with an independent B mark for a correct expression for either sin a or tan b. It was pleasing to see so many students gaining this credit and then setting the values equal to each other to form an equation. Many did then struggle to solve this question and the process mark was a correct start to do so. To gain this, students had to have a correct process to clear both fractions or to write them with a common denominator with a correct matching numerator. A significant number formed the correct equation but didn’t gain the P mark as they only multiplied the equations by 4, thus only clearing one denominator. Those who were able to clear the denominators, often then went on to solve correctly and gave an answer of 1.25 or 4.

  Text version

  Two right-angled triangles define acute angles a and b. For angle a, the opposite side is 7 - 2x and the hypotenuse is 8. For angle b, the opposite side is 1 + x and the adjacent side is 4. Given sin(a) = tan(b), find x.

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• Edexcel 1MA1 · June 2024 · Paper 2H · Q13 · 3 marks

  3 marks

  13

  
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  Students gained the first mark for at least two correct boundaries, normally, but not always, for y = −3 and x = 2. The second mark was harder to gain, but again many did. This could be gained for drawing the remaining two boundaries or one more correct boundary and shading correct for at least two correct boundaries. The second of these was probably seen more often. A fully correct region identified was needed to gain the C1, and although not scored by a majority, it was scored by a significant minority.

  Text version

  On a coordinate grid, shade and label the region R satisfying x ≤ 2, y ≥ -3, y ≤ 2x + 1 and 3x + 2y ≤ 6.

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• Edexcel 1MA1 · June 2024 · Paper 2H · Q14 · 6 marks

  6 marks

  14

  
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  This question tested two topics that were new to 1MA1 and students are now scoring some good marks. In part (a) a tangent needed to be drawn and the gradient calculated. Many gained the first mark for drawing the tangent, and of those a significant number gained the second mark for a correct method to find the gradient. Unfortunately, many who did this, lost the third mark due to a missing negative sign in their answer. Weaker students read off at x = 5 and either multiplied or divided these two values, and thus scored zero. In this question, if no tangent was drawn, no marks could be scored. In part (b) a method to find the area under the curve was required and a pleasing number attempted this with trapezia. This was also seen in terms of triangle and rectangles, which was perfectly acceptable. To gain the first mark, a single correct area of a trapezium (or rectangle and triangle) was required, with a complete method needed for the second M mark. Those who used rectangles whose height was the point where the middle of the bar crossed the graph could also gain full credit here, and normally did. This method led to an answer of 78 compared to the 79 got from the other. Quite a few found the area under the tangent or had problems reading from the scale, but method marks could still be gained if this was limited to just one error.

  Text version

  The official velocity-time curve passes through (0,25), (2,16), (4,9), (5,6), (6,4), (8,1) and (10,0); a tangent drawn at t=5 passes approximately through (3,11) and (7,1).

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• Edexcel 1MA1 · June 2024 · Paper 2H · Q16 · 2 marks

  2 marks

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  It was pleasing to see just how many gained the first mark of this question for getting the critical values. It obviously helped that the quadratic was already factorised but pleasing nonetheless. Unfortunately, a significant number had the values of 5 and −0.75 rather than −5 and 0.75 and they gained no marks. Of those who got the critical values it was about 50 : 50 as to who would go on to gain the second mark. This mark was lost either because they were written with ‘=’ or with an incorrect inequality symbol and sometimes as −5 < 0 < Some attempted to expand the quadratic as a first step. This almost always led to a score of zero. Many tried to solve using the quadratic formula and often used in incorrectly.

  Text version

  Solve (4x – 3) (x + 5) < 0

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• Edexcel 1MA1 · June 2024 · Paper 2H · Q20 · 3 marks

  3 marks

  20

  
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  Completing the square is always a topic that many struggle with, especially, if like here, the coefficient of x2 is greater than 1. We saw a decent number gain a mark for taking out a factor of 2, and many were then able to find the (𝑥−3)2term, but very few were able to complete the process and gain the second. If they did, most gained the final mark, except for a few with the correct values but incorrect signs. Some were able to deduce the first coordinate of 𝑥= 3, which gained them 1 mark, but were then unable to find the second coordinate.

  Text version

  Find the coordinates of the turning point of the graph y = 2x^2 - 12x + 7.

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• Edexcel 1MA1 · June 2024 · Paper 2H · Q21 · 2 marks

  2 marks

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  Another topic testing the very top end, and test it did. Few scored both marks for a fully correct graph, but we did see credit given fairly often for a correct translation. Some of those who gained for the translation reflected in the x-axis rather than the y-axis. In some cases, the minimum point was transformed correctly but all the other points were slightly incorrect.

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  The graph y=g(x) is a smooth U-shaped curve with endpoints (-1,1) and (3,1), vertex (1,-2), and symmetry line x=1.

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• Edexcel 1MA1 · June 2024 · Paper 3F · Q4 · 1 mark

  1 mark

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  This question saw varied results; the majority of students were able to simplify correctly to reach an answer of 3a. Unfortunately, some commonly and incorrectly answered with “2a” and “12a” gaining no marks. Some students failed to fully simplify their answers and gave the answer of 8a – 5a.

  Text version

  Simplify 7a + a – 5a

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• Edexcel 1MA1 · June 2024 · Paper 3F · Q16 · 3 marks

  3 marks

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  Most students were able to give a correct answer for part (a); the most common incorrect answer seen was 4m. Part (b) was not answered as well, with much fewer students able to gain the full 2 marks for an answer of 5x + 2y. Of those that didn’t, some were able to gain 1 mark for either 5x or 2y or a linear expression in the form ax + by where a, b > 0. Some students gave a correct expression but included it in a formula e.g. T = 5x + 2y which gained only 1 mark as students were being assessed on their understanding of the term ‘expression’. A significant proportion of the cohort scored 0 marks and the majority of these students generally did not understand the context of the question. Common incorrect solutions did not include x or y e.g. 5 + 2 = 7 or multiplying 5 and 2 by some chosen values and finding the sum of the products.

  Text version

  (a) Simplify m × m × m × m [1 marks]

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• Edexcel 1MA1 · June 2024 · Paper 3F · Q19 · 3 marks

  3 marks

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  A minority of this cohort gained 3 marks on this question; these students produced a table of values and a correct graph from there but some were seen without a table implying they used the equation of the line, gradient and intercept. Of those that didn’t gain full marks, most gained no marks and many did not make any attempt to generate coordinates to be able to plot and join; students should be encouraged to draw the table of values, those who did nearly almost always gained 3 marks. The most common errors were from finding incorrect coordinates using the negative x-values; the majority of these students only lost 1 mark as long as they plotted their points correctly and had a correct line segment through at least 3 of the correct points. Students should be encouraged to recognise that equations in the form y = mx + c should produce graphs that are a straight line without a 'dog leg' or any other type of curve. If no marks were scored the most common incorrect response was simply plotting points at (−2, 3) and (3, −2) and joining up.

  Text version

  On the grid below, draw the graph of y = 3x – 2 for values of x from –2 to 3

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• Edexcel 1MA1 · June 2024 · Paper 3F · Q21 · 2 marks

  2 marks

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  Both parts of this question saw mixed results; most students were not able to factorise correctly. Of those who gained 0 marks, many did not seem to have an idea of what factorise meant, and instead tried an incorrect simplification of the expressions. Common incorrect responses for part (a) included −9x and 6(x − 2.5) and for part (b) it was 6m2.

  Text version

  (a) Factorise 6x - 15. [1 mark]

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• Edexcel 1MA1 · June 2024 · Paper 3F · Q29 · 4 marks

  4 marks

  29

  
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  Part (a) saw mixed results with a similar number of students awarded 0 or 1 or 2 marks. Of those that gained 1 mark, the most common error usually related to brackets around the −5, either omitting them or putting them in the wrong place, often leading to an answer of 65. Part (b) saw little success with around most students gaining 0 marks. The majority of those that were able to provide a correct first step, e.g. multiplying both sides by 3 to reach 3p = h – 5, generally went on to score 2 marks. h = (p + 5) × 3 was the most common incorrect answer along with h = p + 5 × 3. Other incorrect solutions included adding 5 first instead of multiplying by 3 and the interchange of h and p e.g. an answer of p h − = , was often seen. Summary Based on the performance on this paper, students should: • learn the names of the parts of the circle. • practise drawing pie charts from a frequency table. • check their answers, in particular when completing two-way tables. • practise drawing graphs of linear functions. • understand the geometrical rules for angles and practise using these to answer longer, multi-step problems. • practise using their calculators to deal with questions involving substitution. • use tables and diagrams to extract information from a question to enable them to navigate through a problem. • understand the difference between simple and compound interest. • come prepared with the necessary equipment, particularly a calculator of which they are familiar. • show their workings out especially when calculating percentages rather than just writing 15% of ... or 15% × ... • understand that the sum of the probabilities of all outcomes for an event is 1 and cannot be greater than 1. • become more familiar with including appropriate units to answers. • practise plans and elevations. • practise articulating their mathematics competently, using the correct mathematical language. A regular practice of mathematical discussions would help in the writing of explanations. • ensure incorrect working is properly crossed out and it is clear which response they wish the examiner to consider as their final workings/answer.

  Text version

  (a) w = 40 - t^2. Find w when t = -5. [2 marks]

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• Edexcel 1MA1 · June 2024 · Paper 3H · Q10 · 3 marks

  3 marks

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  This question was a good discriminator of students understanding of algebraic concepts and proved to be challenging for many. Those that gained all 3 marks for this question usually went down the route of finding the number of yellow counters or the total number of counters in bag A in terms of x rather than for starting to work with ratio using algebra eg 3y , 5y . Too often students incorrectly had the number of red counters as 3x instead of x and hence the number of yellow counters as 5x instead of 5 x . This incorrect method led to the total number of counters in bag B being 4x rather than the total number of counters in bag A and bag B being 4x and hence gained no marks. Another common error was for students to think that replacing x with a numerical value was an algebraic approach, as such this gained no marks in the same way that substituting in values for n when trying to prove 2 + n n is always even would gain no marks. There were a few students who lost out on gaining the C mark as rather than using fractions they worked with decimals but did not have these as recurring decimals. The variable ‘x’ was frequently lost on the total number of counters for bag B meaning marks were deducted.

  Text version

  There are only red counters and yellow counters in bag A.

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• Edexcel 1MA1 · June 2024 · Paper 3H · Q15 · 6 marks

  6 marks

  15

  
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  Part (a) was well answered with a high proportion of students gaining the mark. When the mark was not awarded it was usually as the student had expanded the brackets or for incorrectly simplifying the − a . In part (b) most students were able to gain at least one of the available marks. Some were awarded one mark for making an error with the signs or for giving two pairs of brackets which when expanded gave two out of three terms correct. Students should be encouraged to multiply out their brackets as a final check - there was little evidence of this taking place. A popular error that resulted in no marks being awarded was to factorise the first two terms and give k(3k + 11) − 4 as the answer. Some students went on to equate the expression to zero and solve the equation, which on this occasion was condoned and didn’t result in loss of marks. It was surprising how many students used the quadratic formula to find the solutions when the expression was equal to zero, this often resulted in no marks being awarded as the students either thought that this was the requirement of the question or didn’t know how to use these solutions to help them factorise. A surprising number gave a solution of 3(k-1/3)(k+4) which also scored 1 mark. Students should be aware that factors within brackets should be integer values. Part (c) was a good discriminator and many students were able to show that to divide the algebraic fractions, they needed to multiply the first fraction by the reciprocal of the second fraction. Fewer students realised the need to factorise the numerator and denominator of the first fraction and many spent time expanding the brackets for no gain and could not then simplify the complex expression generated. Of those that realised the need to factorise many struggled to factorise the 4−x often factorising this to ( 2)( 2) + − x x instead of (2 )(2 ) + − x x as sign errors were condoned in one bracket, these students were able to gain 2 of the 3 marks. Almost all of those who correctly factorised went on to simplify the fraction correctly and gain full marks.

  Text version

  (a) Simplify (a - 3)^2 / (5(a - 3)). [1 mark]

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• Edexcel 1MA1 · June 2024 · Paper 3H · Q16 · 5 marks

  5 marks

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  Part (a) was well answered with the majority of students giving the correct answer of –6. In part (b) the most successful approach was to find g(1) and then substitute this answer into f(x). However many also found f(1) and then multiplied these two together. As this showed a lack of understanding of composite functions this did not gain the method mark. Less common was to find an expression for fg(x) and then to evaluate fg(1). This approach was rarely successful due to students having difficulty dealing with the –3x. Part (c) was not well answered with the most common error being to substitute = x into g( ) x and some incorrectly interpreted g ( ) −x as the reciprocal of g( )x . It was common for the students who recognised that they could proceed on this question by finding the inverse function but often made a sign error and so arrived at − x or equivalent which prevented them from gaining the method mark. Less students approached the problem by equating the 5 – 3x to 4 and solving for x but when this method was chosen, it was often done successfully.

  Text version

  f(x) = 12/(x + 1) and g(x) = 5 - 3x. (a) Find f(-3). [1 mark] (b) Find fg(1). [2 marks] (c) Find g^-1(4). [2 marks]

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• Edexcel 1MA1 · June 2024 · Paper 3H · Q17 · 3 marks

  3 marks

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  This question was answered well and it was clear that a good number of students understood the process of iteration. Although a method using powers was rare - most preferring to find the height after each bounce. The students who gained 2 of the 3 marks on this question usually carried out extra iterations as they failed to realise that they only needed to carry out the iteration three times. Some could only show a method for the first bounce and went no further. Less successful responses seemed to be confused by the notation of the subscript and tried to have this as a power or add 1 as a first step or multiplying by 0.55 and then adding 1. Or they demonstrated little understanding of the context (a bouncing ball) by showing values that increased eg by writing 8 =0.55 × h n+1 with the ball getting higher after each bounce.

  Text version

  A ball is thrown upwards and reaches a maximum height.

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• Edexcel 1MA1 · June 2024 · Paper 3H · Q20 · 7 marks

  7 marks

  20

  
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  There were a fair number of incorrect responses to both parts of this question with students not understanding what an arithmetic or a geometric sequence was. Many students added or multiplied the first and second terms and equated this to the third term or they added or multiplied all three terms together and equated this to zero. It was evident that those who added the first two terms and equated this to the third were confusing the sequences with a Fibonacci sequence. In part (a) the most successful students worked out the differences of the adjacent terms and equated these. There were too many students who missed off brackets and this often resulted in no marks being awarded; it is important that students take care with algebraic notation and ensure that brackets are used when necessary. Although the vast majority of students who successfully equated the common differences then went on to achieve full marks there were a few basic algebraic manipulation errors seen with the solving. Part (b) was designed to challenge the more able students and provide them with the opportunity to demonstrate an understanding of the structure of a geometric sequence. Unfortunately, some treated the sequence as quadratic and despite being tested in part (a) attempts at arithmetic sequence approaches were not uncommon. Some students proceeded straight to + + = − + y y y y and usually went on to gain at least the first three marks. A number of students made simple errors when expanding brackets and sign errors when collecting terms, from this point no further marks could be awarded. Some students used a different approach and rearranged their equations to make y the subject and in terms of r but this route was often unsuccessful due to frequent errors.

  Text version

  x – 4, x + 2 and 3x + 1 are three consecutive terms of an arithmetic sequence.

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• Edexcel 1MA1 · November 2023 · Paper 1F · Q3 · 1 mark

  1 mark

  3

  
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  This simplifying question saw mixed results. The majority of students were able to give a correct answer of 12a but there were a variety of incorrect answers seen. Students should note that this type of simplifying question requires no multiplication signs in the answer; 12 × a and 3 × 4a were often seen and gained no marks.

  Text version

  Simplify 3 x a x 4.

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• Edexcel 1MA1 · November 2023 · Paper 1F · Q12 · 6 marks

  6 marks

  12

  
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  Part (a)(i) was answered well with most students able to gain 2 marks for an answer of 26. Of those that didn’t, it was common to see incorrect algebraic expressions such as 6g + 20h and 23 + 45 was often seen. Part (a)(ii) saw less success with many students unable to make any progress with the substitution. Many were unsure as to which letters should be substituted for which values and substituting g for 38 was often seen. For those that did manage to show a correct substitution, some were able to rearrange correctly to reach an answer of 13. One commonly seen incorrect answer was g = 26. Students should be advised to write the correct answer on the answer, on occasions 13 could be seen embedded in a calculation but the wrong answer given on the answer line, leading to 0 marks. Part (b) saw mixed results as less than half of students in this cohort gained 2 marks. Of those that didn’t, many gained 1 mark, usually in one of two ways; successfully working out 3 × −3 as −9 but being unable to accurately subtract 2 or for a full substitution; common incorrect answers included −7, 7 and 11. Some students wrote notes saying 2 negatives make a positive, leading to the misconception that the negative for the 9 and the subtraction sign meant the two values needed to be added, leading to an answer of 11.

  Text version

  P = 2g + 4h and V = 3r - q.

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• Edexcel 1MA1 · November 2023 · Paper 1F · Q14 · 3 marks

  3 marks

  14

  
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  This question saw the full range of marks awarded. Of those students who managed to plot the correct 5 points it was usual to see this accompanied by a straight line through the points. Some students gained 2 marks, usually for a line segment through 3 correct points; students should be able to recognise that an equation of the form y = mx + c produces a straight line and therefore if one point is not in line with the others it will likely be incorrect. Some students gained 1 mark for producing 2 correct points either plotted or stated in the workspace but there were difficulties in calculating y when x was negative. More than half of the cohort gained 0 marks so this is an area of the specification that needs to be worked on.

  Text version

  The equation is y = 4x - 1 for -2 <= x <= 2.

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• Edexcel 1MA1 · November 2023 · Paper 1F · Q24 · 5 marks

  5 marks

  24

  
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  This was another challenging multi-step question for this cohort; the majority gained 0 marks. Some were able to make a correct start and set up an equation in x, although some failed to include the x from shelf A. Progress from there required a correct process to solve their equation and substitute the value for x into one of the expressions for shelves B or C, any students that did this successfully often gained the 3rd P mark as well for 7500 ÷ “25”. It was rare to see any students get as far as the 4th P mark but those that did usually went on to gain full marks assuming their arithmetic had been correct. There were many completely incorrect solutions seen such as expanding (3x + 1)(2x − 5) or setting 3x + 1 equal to 7500 and going from there.

  Text version

  Shelf A has x books, shelf B has 3x + 1 books and shelf C has 2x - 5 books. There are 44 books in total. All books have equal mass, and the books on shelf B have mass 7500 g.

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• Edexcel 1MA1 · November 2023 · Paper 1F · Q27 · 3 marks

  3 marks

  27

  
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  For part (a) there were many good attempts to expand the brackets but these attempts rarely resulted in credit being awarded. Common errors included 6x or 5x instead of 6x2 and incorrect signs being used. Those students using the grid method to expand the quadratic were usually successful at gaining at least 1 mark. In part (b) a small number of students were able to give a correct factorisation. Some students who gained B0 were able to give a pair of brackets with the product of the constants being −16, such as (x + 2)(x – 8). Summary Based on the performance on this paper, students/centres should work on: • measuring angles accurately. • improving numeracy skills, including division skills for integers and decimals, multiplication skills for decimals. • the interpretation of problems involving ratio. • the formation of algebraic equations and solving them along with subsequent substitution to apply to problems in context. • drawing a straight line graph by correctly calculating a table of values. • using equilateral, isosceles and right-angled triangles to solve angle questions. • encouraging students to show a full method for percentage calculation, noting that before method marks can be awarded a full explanation of each step has to be shown. If the answer is just stated then process marks will be lost if there are any arithmetic errors. • encouraging students to set out their working using logical steps. • encouraging students to read the question carefully so that they understand what they are required to do and what format the answer should be stated in. • encouraging students to use approximations to check if their answers are sensible. • encouraging students to set their work out in an orderly fashion. • improving the expertise and accuracy required in geometric constructions. and • in questions requiring finding the prime factors of a number, encourage a more structured approach of dividing the number by the smallest prime numbers successively until the result was a prime number. • ensure students know common metric equivalents from the specification, from one unit to another. • practise arithmetic processes involving negative numbers.

  Text version

  Algebraic manipulation.

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• Edexcel 1MA1 · November 2023 · Paper 1H · Q7 · 5 marks

  5 marks

  7

  
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  There were a good number of well presented and fully correct solutions. After scoring the first mark for forming a correct equation some students made no further progress but many went on to gain full marks. Some students got as far as finding that the mass of one book is 300 grams but gave this as the final answer and scored 3 of the 5 marks. Many attempts failed at the first hurdle when students could not use the information given to form a correct equation. A common incorrect equation was 3x + 1 = 7500.

  Text version

  Mano has three shelves of books. There are x books on shelf A. There are (3x + 1) books on shelf B. There are (2x − 5) books on shelf C. There is a total of 44 books on the three shelves. All the books have the same mass. The books on shelf B have a total mass of 7500 g.

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• Edexcel 1MA1 · November 2023 · Paper 1H · Q12 · 3 marks

  3 marks

  12

  
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  Some students made a successful start by finding the gradient of the straight line L and were then able to find that the gradient of a line perpendicular to L is − . Having found the gradient of the perpendicular line some made no further correct progress, but many went on to substitute x = 6 and y = –5 in y = − x + c to work out the value of c. Arithmetic slips were quite common. Substituting − , 6 and –5 into y − y1= m(x − x1) gave some students a quick route to an answer. Some students did not take sufficient care over finding the gradient of the straight line L and assumed it was 3. These students could gain one mark for showing an otherwise correct method to get to an answer of y = − x – 3.

  Text version

  The straight line L has equation 2y = 3x − 7

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• Edexcel 1MA1 · November 2023 · Paper 1H · Q17 · 4 marks

  4 marks

  17

  
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  Those students who knew how to factorise a quadratic expression often gained both available marks in part (a). Some made an error with the signs, giving an answer of (2x – 1)(3x + 4), and were awarded one mark. Some were awarded one mark for giving two pairs of brackets which when expanded gave two out of three terms correct. Students could be encouraged to multiply out their brackets as a final check – there was little evidence of this taking place. A popular error was to factorise the first two terms and give x(6x – 5) – 4 as the answer. In part (b) many of the students who gained one mark for giving two critical values with at least one correct then gave the critical values as their final answer. Disappointingly few students gave a fully correct answer. Students who drew a sketch of the curve and identified the part which was < 0 were able to give the correct inequality symbols in their final answer. Many students did not realise that they could use their answer from part (a) and started again – often with no success.

  Text version

  This question is about the quadratic expression 6x² − 5x − 4

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• Edexcel 1MA1 · November 2023 · Paper 1H · Q22 · 4 marks

  4 marks

  22

  
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  This proved to be a challenging question with very few students knowing how to work out the value of the common ratio of the sequence. Some of the students who were able to use the 2nd term and 3rd term to write down the common ratio 13 9 2 2 2 + + made no further progress but it was pleasing that there were others who showed a complete process and went on to gain full marks. When attempting to rationalise the denominator errors were sometimes made in expanding terms, most commonly in the numerator. One error was condoned for the third mark. Many students did not know how to use the 2nd term and 3rd term to find the common ratio. Summary Based on their performance on this paper, students should: ● practise their arithmetic skills, particularly division and operations with decimals ● consider whether or not an answer to a calculation is of a sensible size ● take care when interpreting the scales on the axes of graphs ● practise using the standard ruler and compass constructions ● understand the difference between standard and reverse percentage ● practise using the relationships between lengths, areas and volumes in similar figures

  Text version

  The 2nd term of a geometric sequence is 3 + 2√2. The 3rd term of the sequence is 13 + 9√2

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• Edexcel 1MA1 · November 2023 · Paper 2F · Q20 · 4 marks

  4 marks

  20

  
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  A large number of students assumed the graph would be a straight line and completed the table to suit their line with no calculations using the given equation being shown. Some students gained 1 mark by correctly evaluating the y coordinate for the positive values of x only, although very few gained both marks for part (a) as they struggled evaluating the y coordinates for the negative values of x, indicating difficulty in carrying out calculations with negative numbers. Similarly, many students achieved the method mark in part (b) for correctly plotting 4 of their points correctly but plotted the points with negative x and/or y values incorrectly. Very few students produced a fully correct curve, with a very small number who evaluated and plotted all points correctly losing the accuracy mark as they joined all points with line segments rather than a curve or did not show the minimum point of with a clear curve.

  Text version

  The function is y = x^2 - x - 2 for x = -2, -1, 0, 1, 2 and 3.

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• Edexcel 1MA1 · November 2023 · Paper 2F · Q22 · 6 marks

  6 marks

  22

  
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  Success in expanding and simplifying an expression in part (a) varied considerably. Many fully correct answers were seen, and of those not gaining full marks, a mark was often awarded for either expanding one bracket correctly or collecting like terms for one variable, often y, correctly. Many expanded the brackets correctly but then did not know how to simplify either ignoring the signs or adding all terms together to give a single term as an answer. The importance of adding / subtracting negative numbers correctly in algebra questions appears to be challenging for Foundation tier students. A small number of students confused this type of expansion with multiplying 2 brackets resulting in having a term in y2 but this was less common than in previous similar style questions. Part (b) was answered with moderate success with a pleasing number of students able to factorise the terms to identify the HCF of 3x in order to obtain their answer of 3x(2x+5). However, a large number of students appear to not know what the word ‘factorise’ means; with several students trying to add the terms together or attempting to solve and find a value for x. There were a few students who knew what factorising meant, but due to the x squared term, thought this was a two bracket factorisation and subsequently failed to achieve any marks. Rearranging a formula proved to be challenging for the majority, with very few students being able to rearrange fully or partially. The most common errors included simply transposing the f and g in the original equation or stating the correct process but failing to apply this correctly to one or both sides, eg stating −11 and/or ÷3 but not applying this to both sides of the equals sign.

  Text version

  Algebra.

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• Edexcel 1MA1 · November 2023 · Paper 2F · Q25 · 3 marks

  3 marks

  25

  
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  The level of success when solving this algebraic equation with the unknown variable on both sides varied greatly. Students appeared to find this question challenging with a very small number not attempting at all. A correct 1st step of adding x or adding 14 was often indicated as the intention but incorrectly executed leading to 4x = 38 or similar. A common error was to subtract x from both sides instead of adding. Elimination of −14 was more commonly carried out correctly. Most students attempted to answer the question algebraically, but a larger number than expected attempted an entirely numerical approach. Trial and improvement methods usually led to no marks, but some obtained the correct answer of 11 and therefore gained full marks.

  Text version

  Solve 5x - 14 = 52 - x.

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• Edexcel 1MA1 · November 2023 · Paper 2F · Q30 · 3 marks

  3 marks

  30

  
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  This final question appeared to be challenging for a great number of students. Of those students who tried to create a multiple of one or both equations and identified the need to subtract the equations to eliminate a variable, most struggled with the negative numbers or forgot to multiply all parts of the equation. A negative sign was frequently lost in the working and calculating with negative values caused a large number of arithmetic errors. When students successfully showed a correct method and were awarded the first mark, the majority often went on to substitute successfully to earn the second method mark. It was rare to see a fully correct response. Some students used checks of their answers, with many then able to correct any earlier arithmetic errors. When students gained no credit at all, it was often due to either trying to combine the initial equations without first altering either equation or attempting to use a trial and improvement method which scored no marks unless correct values for x and y were found. Summary Based on their performance on this paper, students should: • ensure they have access to a suitable calculator and have a good working knowledge of how to use it • read questions carefully, including after reaching a final answer to check whether the magnitude is sensible, units are appropriate, and the level of accuracy required is shown • practice questions involving proportion and manipulating algebraic expressions • be able to recall formulae and angle facts to solve problems and support working shown.

  Text version

  The simultaneous equations are 3x + y = -4.5 and 4x + 3y = -3.5.

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• Edexcel 1MA1 · November 2023 · Paper 2H · Q1 · 6 marks

  6 marks

  1

  
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  A very familiar 3-part algebra question to start the paper, was accessed by all, and with significant success. Part (a) was an expand and simplify, with almost all gaining at least one of the two marks on offer, normally for a correct expansion of one of the brackets and with a large number gaining both for the simplification. In part (b) students were required to factorise an expression. Again, many gained at least one mark for a partial factorisation, but it was more common to award both marks. The third part of the question was a relatively simple rearrangement which was answered well. The first mark was awarded for a correct first step of rearrangement, normally seen in the form of subtracting 11 from both sides. A significant number gained both marks. It is worth noting that a small proportion, lost the second mark as they failed to give their answer in the form “g =”.

  Text version

  This question is about algebraic manipulation.

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• Edexcel 1MA1 · November 2023 · Paper 2H · Q4 · 3 marks

  3 marks

  4

  
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  This question was a straightforward solving of an equation with the unknown on both sides. A good proportion of students gained marks. Many were able to gain at least one for carrying out one of the two processes correctly, normally dealing with the constant term. The fact that the variable was a negative, caused many students problems, with a significant number trying to subtract x rather than adding it. This is an area that students would benefit from extra practice.

  Text version

  Solve 5x − 14 = 52 − x

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• Edexcel 1MA1 · November 2023 · Paper 2H · Q9 · 3 marks

  3 marks

  9

  
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  This now familiar topic was answered, overall, very well. A majority were able to gain at least one mark for a correct expansion of a pair of brackets. Many were then able to continue and gain all the marks. Some however, when expanding the final bracket, failed to gain enough terms, and the method mark requires a complete method to find all terms, so those who only had 5 or 7 lost this mark.

  Text version

  Expand and simplify (x + 7)(x − 2)(x + 3)

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• Edexcel 1MA1 · November 2023 · Paper 2H · Q17 · 6 marks

  6 marks

  17

  
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  Another familiar style question, but performance suggests it is not one that many students had spent much time preparing for. Part (a) saw some awarded 1 mark, but very few were able to gain 2 marks as they didn’t complete their explanation, normally missing out a statement about a change of signs. The rearrangement in part (b) only carried one mark, but again it seems many have little experience of this when factorisation is required, and poor incorrect algebra was commonplace. Part (c) only required substitution and evaluation, and it was not surprising to see that more students were able to gain marks in this part compared to the others. Many scoring 1, 2 or 3 marks. Those who gained 2 normally carried out extra iterations and therefore should be encouraged to read the question carefully to ensure they complete the correct number.

  Text version

  This question is about the equation x³ + 2x − 6 = 0

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• Edexcel 1MA1 · November 2023 · Paper 2H · Q19 · 4 marks

  4 marks

  19

  
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  This was one of the better answered questions at this level, with a good number able to gain 2 marks in part (a). This was normally done by two steps of substitution, rather than forming the composite function. However, some that attempted the two steps of substitution, completed the first step correctly, but then instead of substituting their found value into g(x), they re-substituted −3, and unfortunately gained no credit. In part (b) it was pleasing to see so many understand the concept of an inverse function, and many were able to rearrange to find this.

  Text version

  The functions f and g are such that f(x) = (2x + 3)² and g(x) = 2x − 1

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• Edexcel 1MA1 · November 2023 · Paper 2H · Q20 · 4 marks

  4 marks

  20

  
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  This hard algebra question required students to deal with 3 different aspects. There was division by a fraction, factorisation of a harder quadratic, and then the addition of fractions with different denominators. A reasonable number of students were able to gain one or two marks, normally for the first two steps. However, many struggled with the factorising, and often lost a factor of 2. This appears to come from their method for solving quadratics, they can divide through by a common factor, and they attempted to do the same here. Those who made a mistake in the factorisation step, then were not able to gain further credit as any fractions they had, would not be correct.

  Text version

  Write 14/(3x − 21) + [(x + 4) ÷ (2x² − 6x − 56)/(2x + 3)] in the form (ax + b)/(cx + d) where a, b, c and d are integers.

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• Edexcel 1MA1 · November 2023 · Paper 3F · Q8 · 4 marks

  4 marks

  8

  
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  Most candidates were able to articulate that the sequence was decreasing by 6, by saying exactly that or in an alternative way such as; "subtract 6", "decreasing by 6", "minus 6" or "go down 6". Some spoiled their answers by not quantifying the reduction. (b) The majority of candidates were able to continue the sequence, often next to the number sequence given. They then went on to either work out the difference between the 5th and 7th term by completing a subtraction. There were two main methods seen; 73 − 61 or it was sometime shown in two different steps: 73 − 6 = 67 and 67 – 6 = 61. (c) Candidates always find response type questions more difficult to answer and often do not give enough detail to explain their reasoning, but on this occasion this question was well answered. Answers such as "subtracting 6 will not get to 52" or it goes past 52 or similar were not enough to be credited with the mark without further supportive evidence of their statement. Most candidates chose to either comment: "the sequence is odd" and/or "52 is an even number" or continued the sequence far enough to show that the 8th term was 55 and/or the 9th term was 49.

  Text version

  The first four terms are 97, 91, 85, 79.

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• Edexcel 1MA1 · November 2023 · Paper 3F · Q13 · 2 marks

  2 marks

  13

  
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  In part (a) man candidates correctly simplified the coefficient to 20 but there were numerous errors with the power of h, especially giving the power as 9 or 6. Quite a few candidates showed 22h6 – 2h3in the working box but their final answer was often correct, suggesting that they were adding and subtracting both the coefficient and the power. Many times the candidate was seen to be applying the power to the co-efficient, leaving out the h altogether before popping it back on to their answer at the end. In part (b) Most candidates used order of operations correctly and got the correct answer. Some chose to divide each term in the bracket by 3 and sadly failed to simplify their answer and left as 3y + 4y. It was also common to see the y dropped and the answer of 7 just given.

  Text version

  Simplify algebraic expressions.

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• Edexcel 1MA1 · November 2023 · Paper 3F · Q16 · 2 marks

  2 marks

  16

  
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  Generally, this was not well answered. If the candidate chose to multiply through by 7 as their first step, they were often unsuccessful at gaining method marks because they omitted to multiply the 9 as well as the other terms. With an incorrect first step no further marks were possible. Whilst those candidates who showed the intention to subtract the 9 from both sides as a first step were more successful at gaining method marks, they would often fail to carry out the subtraction on one or both sides. They had difficulty in subtracting a larger number from a smaller one often ending up with a positive solution demonstrating their struggles when working with negative numbers. If this step was completed successfully, they would often go on to divide by 7 rather than multiply. Another common error was to dispense with the divide by 7 altogether and to rewrite the equation as 7x + 9 = 4 as they are more familiar with this type of equation.

  Text version

  Solve x/7 + 9 = 4.

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• Edexcel 1MA1 · November 2023 · Paper 3F · Q18 · 5 marks

  5 marks

  18

  
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  There were four distinct steps to this problem: find the missing length of the triangle base (6.4), calculate the area of the triangle (19.2), calculate the area of the rectangle (67.2) and thus calculate the length of the rectangle (4.8). Where there were earlier mistakes in the base length or in the calculation to find the area of a triangle it was still possible for candidates to be awarded marks for correct application of later processes using their figures, under certain circumstances. The likely misconceptions were seen which included incorrectly assuming the base was 6 or 3.8 and forgetting to correctly apply the formula for a triangle by halving the product of "6.4 and 6". Centres are encouraged to teach candidates to structure working with labels as to what they are doing; e.g. "area of a triangle = ..." That way more marks would be yielded from further working from previous mistakes as examiners are in a better position to understand what the candidates is doing, and can apply marks whenever possible.. In most cases where candidates scored the first three marks most went on to finish this problem.

  Text version

  A symmetric compound shape consists of a rectangle of length 14 cm and width w, plus a centred triangle. The triangle's base lies along the 14 cm side leaving 3.8 cm at each end, and its perpendicular height is 6 cm. The rectangle's area is 3.5 times the triangle's area.

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• Edexcel 1MA1 · November 2023 · Paper 3F · Q23 · 4 marks

  4 marks

  23

  
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  In general part (a) of this question was well answered with most candidates picking up at least one mark for stating that the inequality should have been drawn from – 3 and/or that the circle at 4 should have been shaded/coloured. Again, as usual, sometimes responses lacked detail and were not precise enough which meant that the mark was not awarded. Examples included: "the circle was not shaded" or "the line was not long enough" or "the circle was not drawn at the correct number". Some candidates chose to correct the diagram and re-drew the correct inequality, with or without an explanation. In both these cases the marks were awarded when it was clear and when there was no contradiction. In part (b) few candidates scored full marks. The most common responses scored M1 for finding the critical value of 4.6. A few candidates who got this far then stated 5 instead of 4 as the greatest integer. A common error seen was where candidates subtracted 7 rather than adding (scoring 0). There were fewer embedded answers seen this year, showing candidates were using more formal methods.

  Text version

  The inequality is -3 < x <= 4. Jenna draws a line segment from -2 to 4 with an open circle at each endpoint.

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• Edexcel 1MA1 · November 2023 · Paper 3F · Q28 · 3 marks

  3 marks

  28

  
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  Examiner report

  By this stage in the paper many candidates were struggling. It was common to see a table of values generated by taking the points from the straight line usually followed by an attempt to find an equation by trial and error. Candidates who had some knowledge of this skill knew to highlight the y-intercept at (0, 3), drawing a gradient triangle on the graph. However, many were unable to take the gradient triangle further and were unable to show correct working and/or that the gradient was "−2". A method mark was awarded for the few correct gradient workings seen and for arriving at "−2". Candidates who got this far nearly always went on to find the correct solution when using this method. It was more likely to see a solution of the form y = mx+3 which demonstrated understanding of the y-intercept. A few candidates, who knew what they were doing, spoiled their final solutions by using L instead of y, but were awarded the two method marks.

  Text version

  Line L passes through (-1,5), (0,3), (1,1), (2,-1) and (3,-3).

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• Edexcel 1MA1 · November 2023 · Paper 3H · Q4 · 4 marks

  4 marks

  4

  
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  Examiner report

  Part (a) of this question was very well answered with the great majority of students giving clear and concise descriptions of the two mistakes on the diagram. On occasion the vagueness of a statement cost students a mark, for example when they said that one of the circles should be shaded but did not specify which circle. Part (b) was also quite well answered and most students worked accurately to find the critical value. This was more than enough to score the first mark. A surprising number of students left their answer as 4.6 so had not addressed the requirement stated in the question to give an integer. Other students solved the inequality and wrote y < 4.6 on the answer line. Students are reminded to read questions carefully and take on board any specific requirements. The overwhelming majority of students who did give an integer answer gave the correct integer, 4.

  Text version

  Jenna is asked to show the inequality -3 < x <= 4 on a number line. In her answer she draws a line segment from -2 to 4, with an open (empty) circle at -2 and an open (empty) circle at 4.

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• Edexcel 1MA1 · November 2023 · Paper 3H · Q9 · 3 marks

  3 marks

  9

  
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  Examiner report

  This question produced a good spread of marks. Most students who attempted the question scored at least one of the three marks for forming an equation of the form y = mx + c with 3 in place of c or for substituting their gradient for m. Many students scored full marks for a complete and correct equation. Students scoring 2 of the 3 marks available usually gave the answer y = 2x + 3 and had demonstrated a correct method to find the gradient of the line. Examiners did not see equations written in other forms though these would have been acceptable.

  Text version

  The straight line L passes through the points (0, 3) and (2, -1).

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• Edexcel 1MA1 · November 2023 · Paper 3H · Q10 · 3 marks

  3 marks

  10

  
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  Students generally started their method to write m as the subject of the formula in one of two ways, either subtracting p from both sides or multiplying each side by 5. Students who did the former were generally much more successful than those who did the latter. Those students who started by multiplying each side by 5 often failed to multiply the p on the right hand side of the equals sign by 5. Students who started by subtracting p usually gave a correct answer in one of the forms m = 5(k – p) , m = 5k – 5p or m = 2.5(k – p).

  Text version

  Rearranging a formula.

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• Edexcel 1MA1 · November 2023 · Paper 3H · Q14 · 2 marks

  2 marks

  14

  
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  Examiner report

  A minority of students scored full marks for their responses to this question. However, many students were able to gain at least one mark and the question was a good discriminator between students who could identify the coefficient of n², those who could not get this far and those who went on to get a fully correct expression.

  Text version

  Here are the first six terms of a quadratic sequence. 5 11 21 35 53 75

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• Edexcel 1MA1 · November 2023 · Paper 3H · Q17 · 7 marks

  7 marks

  17

  
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  This question also discriminated well between students targeting the higher grades. In part (a) of the question, most students working at this level were able to make an attempt to complete the square. They were often successful and where they did not give a fully correct response, they were often able to get as far as including (x − 4)2 as part of their answer. (x − 8)2 was an error commonly seen by examiners. An encouraging number of students were able to use their completed square form to write down the coordinates of the turning point on the graph. However, confusion over signs prevented some students from its accurate identification from their completed square form. In part (b), examiners were surprised that only a minority of students realised that they should use the quadratic formula. The question requested that students give their solutions correct to 3 significant figures and this provided a hint that the equation could not be factorised. Many students did attempt to factorise the quadratic or manipulate the equation in some other way. Despite being given the quadratic formula on the formula sheet, some students used an incorrect version in their working, for example –b ± √b2− 4ac 2a . Some other students replaced the “±” in the formula with “+” and so only found one of the two solutions. There were a good number of fully correct solutions, with most students giving their solutions correct to 3 significant figures. A few students lost a mark as they gave one solution as 0.45 rather than 0.449 The identification of the mistake in Alex’s working for part (c) of the question was usually clearly expressed in one way or another.

  Text version

  This question is about quadratic expressions and equations.

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• Edexcel 1MA1 · November 2023 · Paper 3H · Q20 · 1 mark

  1 mark

  20

  
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  Those students aiming to achieve a high grade were usually able to write down the coordinates of the turning point in part (a) of this question. There were many different incorrect pairs of coordinates seen, with perhaps, not surprisingly (–10, 6) being the most common. Some candidates had apparently not read the question carefully and wrote down the coordinates of the graph given for part (b) of the question. Part (b) was less well done. Those students who appeared to recognise the transformations needed sometimes carried them out with such inaccuracy that it was not possible for examiners to award any credit. Students are advised in questions like this to consider what happens to specific points before trying to sketch the whole graph. In this question examiners wanted to see the image of the maximum point and endpoints of the graph correctly shown as part of the transformed graph. A very common error was for students to reflect the graph in the x-axis instead of the y-axis.

  Text version

  The turning point on the graph of y = g(x) has coordinates (-3, 6).

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• Edexcel 1MA1 · November 2023 · Paper 3H · Q23 · 4 marks

  4 marks

  23

  
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  There was a good number of complete, clear and concise solutions to this question. The great majority of students who attempted this question started by eliminating the fractions and were given due credit for this first step. Most of the students following this approach were able to isolate terms in x and y accurately to get as far as 54y2 = 150x2 or equivalent, then went on to get 3y = 5x. A small number of students getting this far made an error when interpreting their linear relationship to get a ratio for x : y and wrote 5 : 3 as their answer. The few students who used an approach involving simultaneous equations were usually successful.

  Text version

  Given that (2x^2 + y^2)/(4x^2 - y^2) = 43/11, where x > 0 and y > 0,

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• Edexcel 1MA1 · June 2023 · Paper 1F · Q5 · 1 mark

  1 mark

  5

  
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  It was not uncommon to see an answer of 1 coming from (3 – 2), although many correct answers were seen.

  Text version

  Solve p - 2 = 3.

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• Edexcel 1MA1 · June 2023 · Paper 1F · Q9 · 4 marks

  4 marks

  9

  
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  Parts (a), (b) and (c) were usually answered correctly. Failure tended to be when students reversed the x and y coordinates. Some candidates drew multiple points and were not awarded any marks in (b). They need to be reminded to label points when requested. Part (d) was poorly done, many just labelling the point (0, −4) or drawing the line x = −4, thinking it had to be parallel to the y-axis. Weaker students often joined the point (0, −4) to A or B or their C.

  Text version

  On a coordinate grid, A = (-3, 4) and B = (5, 2).

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• Edexcel 1MA1 · June 2023 · Paper 1F · Q13 · 4 marks

  4 marks

  13

  
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  In part (a), answers of 5A and 5𝑎 1 were sometimes seen and they were acceptable as alternatives for 5a for the award of the mark. This was answered well but incorrect answers of 5, 45, 3a, 5a/3 and 45a were often seen. Many students found difficulty in dealing with negative expressions in part (b). 12b (5b + 7b) and 3c (4c – c) were very common errors made. Some wrote 19 + - 2b + 5c for their answer failing to simplify the two signs. This lost the accuracy mark. A significant number of students correctly collected the like terms but then tried to simplify their answers further, usually giving 17b + 5c or 22bc as their final answer losing the accuracy mark. Factorisation in part (c) was not well done. The most common incorrect answer was 2d.

  Text version

  Simplify and factorise the following expressions.

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• Edexcel 1MA1 · June 2023 · Paper 1F · Q18 · 3 marks

  3 marks

  18

  
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  Understanding of algebraic formulae was very poor indeed, with the most common incorrect answer being T = b + c but 33 = 5 + 28 and 33T=5b + 28c were also regularly seen. Often when students did find the correct formula, they tried to simplify it giving T = 33bc as their final answer. Weaker students struggled to gain any credit at all in this question.

  Text version

  Naomi has b bags of apples with 5 apples per bag and c crates with 28 apples per crate. She has T apples in total.

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• Edexcel 1MA1 · June 2023 · Paper 1F · Q19 · 2 marks

  2 marks

  19

  
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  Many students correctly found the difference of 8 between terms but were unable to offer a complete expression for the nth term, n + 8 and 8n, the latter gaining one mark, were common answers. After finding 8n many candidates were unable to find the correct number to subtract. Other common incorrect responses were 13n – 8 and –13n + 8

  Text version

  The first five terms of an arithmetic sequence are -5, 3, 11, 19, 27.

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• Edexcel 1MA1 · June 2023 · Paper 1F · Q28 · 1 mark

  1 mark

  28

  
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  Very few realised that the solution of the simultaneous equations was determined by the point of intersection of the two lines. Many spent far too much time trying to solve the simultaneous equations algebraically usually without any success.

  Text version

  The graphs of 2 - 2y = x and 2y = 3x - 22 are drawn on the same coordinate grid.

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• Edexcel 1MA1 · June 2023 · Paper 1H · Q10 · 4 marks

  4 marks

  10

  
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  It was pleasing that many students gained full marks for finding the size of angle AED. Those who gained the first mark for using (n – 2) × 180 to find the sum of the interior angles of the pentagon or for stating that the sum of the interior angles is 540 usually went on to make further progress. After subtracting 365 from 540 to find the sum of the two unknown angles some students divided the result by 4 instead of by 5. Some who did divide by 5 then decided that angle AED = 35 and lost the final mark. Arithmetic errors, particularly when subtracting 365 from 540 or dividing 175 by 5, were quite common. However, students making these errors could still be credited with the process marks if their intended calculations were shown. Some students relied on memory to give the sum of the interior angles of a pentagon and stated an incorrect figure such as 520, 560, 580 without working and gained no marks. A small number of students used an exterior angle approach and subtracted the three exterior angles from 360 to leave 185 to gain two marks but very few went beyond that.

  Text version

  ABCDE is a pentagon in which angle EAB = 120°, angle BCD = 110° and angle CDE = 135°. Angle AED = 4 × angle ABC.

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• Edexcel 1MA1 · June 2023 · Paper 1H · Q11 · 3 marks

  3 marks

  11

  
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  Many students gained the first mark for simplifying either the numerator or the denominator with at least two of three terms correct in a product. At this stage the most common mistakes were 6x10y6 or 36x25y9 or ax7y5 instead of 36x10y6 in the numerator and 12x2y4 instead of 12x3y4 in the denominator. A good proportion of students completed the algebra correctly and arrived at an answer of 3x7y2. Some students made one error in the numerator or denominator and gained two marks for an answer of the form axbyc with two of a, b and c correct (of these the error was often an incorrect coefficient). A small number of students used a different approach and instead of simplifying the numerator and the denominator they chose to start by simplifying 6x5y3 ÷ 3x2y7 and 6x5y3 ÷ 4xy–3. A common error was for students to introduce plus signs into the numerator and then attempt to expand the two brackets, leading to more than one product for the numerator.

  Text version

  Write (6x⁵y³)² / (3x²y⁷ × 4xy⁻³) in the form ax^b y^c where a, b and c are integers.

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• Edexcel 1MA1 · June 2023 · Paper 1H · Q15 · 3 marks

  3 marks

  15

  
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  This question was not answered as well as might have been expected and a surprising number of students failed to gain any marks at all. After gaining the first mark for making the y term the subject of 6y + kx – 12 = 0 or for finding that the gradient of a line perpendicular to L1 is − some students made no further progress. Many of those that did get both k y x = − + and the gradient of − could not make the final step of equating the gradients to find the value of k and it was common to see an answer of 1 12 from multiplying − by k − . The double minus of k − and − caused lots of issues and a significant number of students gave the final answer as –3 instead of 3. An answer of 3 with no supportive working gained no marks because the question required students to show all their working. Common incorrect methods included trying to solve the equations simultaneously.

  Text version

  The equation of line L1 is y = 2x − 5. The equation of line L2 is 6y + kx − 12 = 0. L1 is perpendicular to L2.

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• Edexcel 1MA1 · June 2023 · Paper 1H · Q17 · 4 marks

  4 marks

  17

  
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  More than half of the students who gained the first mark for multiplying both sides of the formula by (5x + 3) to clear the fraction went on to get full marks. Many students, though, did not know how to deal with the fact that x appeared on both sides of the equation or how to deal with the xy term. Attempts to isolate the terms in x were often not successful with sign errors quite common and only some of those who did isolate the terms in x went on to factorise correctly. It was common to see incorrect steps such as 5xy – 8x = –3y – 28 followed by 5x – 8x = 3 y y − − . Students who made a mistake when expanding 4(2x – 7) were still able to gain two of the four marks. Some attempts to clear the fraction failed because students did not include brackets and this resulted in them not multiplying both 5x and 3 by y. Instead of clearing the fraction as a first step many students worked with the right-hand side of the formula, incorrectly simplifying 8 x x − + to get 3x – 25 or 3x – 31 for example, and gained no marks.

  Text version

  y = 4(2x − 7) / (5x + 3)

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• Edexcel 1MA1 · June 2023 · Paper 1H · Q18 · 4 marks

  4 marks

  18

  
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  Many students struggled to find a strategy that would enable them to solve this problem and there were many solutions that had calculations dotted around the page making the working very difficult for examiners to follow. Many of the algebraic methods broke down at an early stage. After gaining the first mark for writing an equation such as 7c + 5t = 480 many students were unable to use the ratio 5 : 9 correctly to write down a second equation. Writing 5c = 9t instead of 9c = 5t was a common mistake and the incorrect equations 5c + 9t = 14 and c + t = 14 were frequently seen. Nevertheless, it was pleasing to see some successful solutions from students using an algebraic method. The majority of the correct answers came from students adopting a numerical approach. The most efficient of these saw students working with the ratio of total costs, finding 7 × 5 = 35 and 5 × 9 = 45, often as 35 : 45, and then dividing 480 by 80 to get 6. Multiplying 5 by 6 and 9 by 6 completed the solution. Some students worked out the total cost of the tomatoes and the total cost of the carrots and gained the first two marks but did not find the cost of 1 kg of each. Many of the incorrect strategies were based on dividing 480 by 12 or by 14.

  Text version

  7 kg of carrots and 5 kg of tomatoes cost a total of 480p. cost of 1 kg of carrots : cost of 1 kg of tomatoes = 5 : 9

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• Edexcel 1MA1 · June 2023 · Paper 1H · Q20 · 5 marks

  5 marks

  20

  
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  In part (a) the students who knew about inverse functions usually tried to find g ( ) x − by rearranging x y + = to make x the subject or rearranging y x + = to make y the subject. Many got as far as 5 y x = + or 5 x y = + and gained the first mark. A good proportion of the students who made a correct first step went on to give a correct answer. Those who completed the rearrangement correctly sometimes gave the answer in terms of y and lost the accuracy mark. Some unwisely attempted to expand (5x – 2)2 and if they did so incorrectly they lost the accuracy mark. A common error in the final step of the rearrangement was to follow 5 y x − = with 5 y x − = . Some students attempted to find the inverse function by using a flow diagram but this approach appeared to be less successful. Incorrect expressions for g−1(x) such as 5(x2 – 2) or 5x2 – 2 were often the result of applying the inverse operations in the wrong order. Unfortunately, many students seemed unaware of the concept of an inverse function and some mistook the notation to mean the reciprocal. In part (b), many students started by finding an expression for gf(x) and gained the first mark. Instead of equating gf(x) with 3 a common error was to substitute x = 3 into gf(x) and students doing this gained no more marks. Those who did equate gf(x) with 3 often went on to solve the equation and score full marks. Some students made mistakes when rearranging the equation and some made arithmetic errors when squaring 13 or when dividing 165 by 3, with 53 (from 165 ÷ 3) being a common incorrect answer.

  Text version

  For x ≥ 0, the functions f and g are such that f(x) = 3x + 4 and g(x) = (√x + 2) / 5

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• Edexcel 1MA1 · June 2023 · Paper 1H · Q23 · 4 marks

  4 marks

  23

  
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  This question involved writing two fractions with a common denominator, rationalising and simplifying surds. There were many valiant attempts achieving varying degrees of success and not many blank responses. Many students started with a method to write the two fractions with a common denominator and gained the first mark. A common mistake, which prevented the award of the second mark, was incorrectly simplifying 2 3 4 3 4 3 − − − − to 1 2 3 4 3 − − . Some students were unable to continue the solution any further and gave their single fraction as the final answer. A good number of students though went on to gain the third mark for a method to rationalise the denominator with some giving a fully correct solution. The alternative approach used by a good proportion of students was to start by rationalising the denominator of each fraction before writing the fractions with a common denominator. The main obstacle to a successful outcome for these students was making errors when manipulating the surds with common mistakes being √3 × √3 = 9 and √3(3√3) = 3√3 + 3. Those who used this approach were more likely to give a fully correct answer. Some students used a hybrid approach, rationalising one fraction before trying to achieve a common denominator and then rationalising again. This route had comparable success with the other routes.

  Text version

  Write 3√3/(4 − √3) − 2/√3 in the form (a√3 + b)/c where a, b and c are integers.

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• Edexcel 1MA1 · June 2023 · Paper 1H · Q24 · 5 marks

  5 marks

  24

  
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  It was pleasing to see so many students attempting the final question on the paper. Higher attaining students usually made a good start by showing a correct method to find the critical values for at least one of the inequalities and often for both. Many factorised 4x2 – 25 and gained the first mark for (2x + 5)(2x – 5). Those who rearranged it to x2 < 25/4 often failed to complete the method by finding both the positive and negative square root. Inevitably, some errors (mostly sign errors) were made in the factorisation of 12 – 5x – 3x2 and those who rearranged the inequality to 3x2 + 5x – 12 < 0 tended to make fewer mistakes. After finding the critical values many students were not able to use them correctly to solve the inequalities. It was very common to see x < –2.5 and x < 2.5 for 4x2 – 25 < 0 and x > –3 and x > 4 3 for 12 – 5x – 3x2 > 0 with students using < or > to match the original inequality. Students were more successful in writing the solution to the first inequality, –2.5 < x < 2.5, than the solution to the second inequality. The latter was often given as x < –3 and x > 4 3 rather than as –3 < x < 4 3 . Drawing a sketch of the curve helped some students but there were others who drew a correct sketch and were still unable to identify the relevant region. Furthermore, students who drew sketches often drew y = 12 – 5x – 3x2 as if it were a positive quadratic which led them to deduce an incorrect critical region. Students who solved both inequalities correctly were often able to identify the set of possible values as –2.5 < x < 4 3 and gain full marks. Some students gave the final answer as –2, –1, 0, 1 or used ≤ instead of < and lost the accuracy mark. Summary Based on their performance on this paper, students should: ● practise their arithmetic skills, particularly division and operations with decimals ● take care when carrying out arithmetic operations and use logical checking processes to make sure that their answer is sensible ● become more familiar with the set notation for the complement of a set and how to identify it on a Venn diagram ● know the difference between stating the type of correlation and describing a relationship and should not confuse positive correlation with direct proportion ● practise finding angles using the circle theorems and stating the circle theorems used ● practise algebraic manipulation including rearranging formulae, using index laws and solving equations

  Text version

  Find the set of possible values of x for which 4x² − 25 < 0 and 12 − 5x − 3x² > 0. You must show all your working.

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• Edexcel 1MA1 · June 2023 · Paper 2F · Q4 · 1 mark

  1 mark

  4

  
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  Simplifying a simple algebraic expression seemed to cause very few problems for most students. Common incorrect responses included 82t or 7t from not applying the correct operator.

  Text version

  Simplify 3 x 4t.

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• Edexcel 1MA1 · June 2023 · Paper 2F · Q9 · 3 marks

  3 marks

  9

  
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  Almost all students could continue the sequence correctly to gain the mark in part (a). Part (b) performed less successfully. Whilst many were generally able to write the correct values using ratio notation, not all continued to write their ratio in the simplest form and although many achieved the correct simplified ratio, some went on to do further incorrect simplification such as 4:9 leading to 2:3 which lost the accuracy mark. However correct simplification to 1:2.25 (1: n) did not lose the accuracy mark and was occasionally seen. When no marks were awarded, it was often due to not understanding how to write the appropriate values using ratio notation, such as simply writing as a fraction instead. Other commonly seen responses that gained no credit included choosing the incorrect terms from the sequence or selecting three terms e.g. 8:23 or 8:13:18 and a small minority attempted to find the nth term.

  Text version

  The first five terms of a sequence are 3, 8, 13, 18, 23.

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• Edexcel 1MA1 · June 2023 · Paper 2F · Q10 · 4 marks

  4 marks

  10

  
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  Part (a) was usually correct. Rare mistakes arose from misreading the scale or reading off from the incorrect axes. Part (b) required the use of the conversion graph combined with working with time and was attempted by nearly all students. Very few students drew lines on the graph from £9 to 6 hours and this sometimes led to the graph not being read correctly. However, very many responses gained full or partial marks even if the graph showed no indication of use. The most common error occurred in the use of correct time notation when writing the final answer, with a significant number of students losing the final accuracy mark. Occasionally a correct answer in 24-hour notation was given on the answer line along with an incorrectly converted 12-hour time. There was some confusion when adding on 6 hours, with students counting 08:00 as 1, 09:00 as 2 etc and arriving at a final answer of 1:00. Where a list of times appeared in the working, they were able to gain the method marks for a correct reading from the graph and attempting to add to 8 am. Less commonly seen errors were: misreading the Cost axis and mistaking £9 for £8.50 resulting in a time in hours and minutes that they did not know how to deal with, and inaccurate lines drawn by hand on the graph giving wrong ‘hour’ values.

  Text version

  A parking-cost graph is a straight line from (0 hours, £0) to (12 hours, £18), with linear scales on both axes.

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• Edexcel 1MA1 · June 2023 · Paper 2F · Q23 · 6 marks

  6 marks

  23

  
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  Mixed responses were seen in part (a). Some students understood the concept but didn't write down an integer and gave responses such as 4.9. In part (b), most students were able to demonstrate some knowledge of inequalities, but the accuracy mark was often lost due to not knowing how to interpret the inequality symbols correctly by use of open and closed circles. Just giving a line was the most common response by those students who attempted this part of the question. However, some students drew a line from −4 to 0 or −4 to −1, indicating that the inequalities were not being interpreted correctly. If only one circle was correct it was more often the closed circle at −4. Of the students scoring no marks, common answers included: a cross at −1½, lines from −3 to 0 or from −4 to 0, lines that start correctly at one end but end with an arrow at the other, and lines which extend beyond one of the endpoints. A common misconception noted was the students treating it as two separate inequalities by drawing two circles and two lines with direction. Some students tended to use two arrows to help with the direction of each part of the inequality, again not realising that it was one inequality. To improve performance of this familiar skill, centres should emphasise the need for a single line connecting both circles rather than two separate lines. Centres could also focus on stating integer values represented by the inequality as this would help students appreciate that the number lies between these two values and should not be represented by two inequalities. The final part of this question was not answered well, with most scoring no marks. The fraction part of the inequality seemed to confuse many students. Those who started trying to reverse this part of the process were usually unsuccessful due to either changing the fraction to 10 from multiplying the numerator and denominator or multiplying by 5 but omitting the 2 from the next stage of their calculations. However, converting to 0.4 was seen often and being a calculator paper, by following algebraic processes and using the calculator for calculations, students should have been able to reach 25, but a surprising number who used 0.4 did not. The intention to use a correct first step was often carried out incorrectly, contradicted or carried out on one side only. For those who chose to begin by correctly adding 4 to both sides, many were then unable to continue correctly and deal with the fractional element. Finally, for those who had replaced the ‘<’ with an ‘=’ in the working, most remembered to return it for the answer. However 25 alone was seen on the answer line more often than the correct answer when marks were awarded.

  Text version

  Work with the following inequalities.

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• Edexcel 1MA1 · June 2023 · Paper 2F · Q24 · 4 marks

  4 marks

  24

  
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  This question proved to be challenging to most students and was not answered as well as expected. Some students were able to begin to write an expression for the area of the triangle, but many did not divide by 2. Of those who attempted to write an expression for the area of the rectangle, many gained the mark by writing 5 × 4x − 1 but this was often without using brackets which then caused difficulty in simplifying the expression later. Often 4x − 1 was simplified as 3x or the area as 19x. This demonstrated a weak understanding of creating expressions to represent the areas and collecting like terms. Whilst a small number of students were able to write a correct expression for both shapes, it was rare for any of them to include the additional 10 cm2 correctly in either of their expressions or in an equation. Multiplying or dividing by 10 or 100 was often seen instead. Some students omitted algebra altogether and wrote either ‘48’ or ‘24’ for the area of the triangle. Trial and improvement methods were often seen and, when used, were quite often unsuccessful. Another common mistake was to calculate the perimeter of the shapes and attempts to use Pythagoras’ theorem by some students were seen. Centres are encouraged to remind students that if only a method of trial and improvement is seen then they will score no marks unless a fully correct answer is given. This approach should not be encouraged.

  Text version

  A triangle has base 8 cm and perpendicular height 6x cm. A rectangle measures 5 cm by (4x - 1) cm. The triangle's area is 10 cm^2 greater than the rectangle's.

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• Edexcel 1MA1 · June 2023 · Paper 2F · Q28 · 3 marks

  3 marks

  28

  
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  Student’s responses to this last question were very mixed, with some not attempting it at all. Very few students successfully matched all the graphs to a correct equation, although a few scored 1 mark for two graphs correctly assigned. Summary Based on their performance on this paper, students should: • read questions carefully, including after reaching a final answer to check whether the magnitude is sensible, units are appropriate, and the level of accuracy required is shown • practice questions involving proportion and algebraic expressions • practice using measuring equipment such as rulers and, more particularly, protractors • give clear and succinct explanations when a written answer is required • use a calculator rather than relying on paper methods, particularly when working with percentages • avoid using inconsistent units or rounded or truncated figures in calculations so that their final answer is inaccurate

  Text version

  Five graph sketches are described: A is a decreasing straight line with positive y-intercept; B is an upward-opening parabola through the origin with a second positive root; C is a reciprocal hyperbola in quadrants I and III; D is an increasing straight line with positive y-intercept; E is an increasing cubic shifted downward.

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• Edexcel 1MA1 · June 2023 · Paper 2H · Q4 · 4 marks

  4 marks

  4

  
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  In part (a) many candidates gained the mark for a correct answer, although some are still unclear on the meaning of the inequality symbols; something that is surprising at Higher level. Part (b) was generally answered well, but we saw the typical error of the shading being the incorrect way round on the circles. In this case candidates gained one of the two marks. A number of candidates drew their line on the number line which made marking trickier. Where partial marks were awarded, in most cases, it was for a correct line drawn with incorrect end points. In part (c), most were able to gain some credit with many getting at least 2 for finding the critical value. However, the fraction in the inequality caused problems for a significant proportion of learners. Those who attempted to work with the fraction first, typically scored no marks as they failed to multiply or divide all 3 terms or failed to deal with the inverse of multiplying by . Those who started with the −4 first often fared better. It was pleasing however to see fewer candidates losing marks for using ‘=’ rather than ‘<’.

  Text version

  −2 ≤ n < 5. n is an integer.

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• Edexcel 1MA1 · June 2023 · Paper 2H · Q5 · 4 marks

  4 marks

  5

  
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  This was the first problem solving question on the paper and almost all candidates gained some credit. There were three steps in setting up an equation to solve. Finding an expression for the area of each shape, assigning the ‘10’ as a plus or minus to one of the terms, and putting an equation together. Typically, candidates got 2 of these 3 parts correct. Normally either failing to half for the triangle or adding 10 to the triangle rather than subtracting. The mark scheme this time was generous here and allowed candidates to miss the brackets from the expression for the area of the rectangle for all three method marks, meaning many candidates were able to score well. Those who formed the correct equation, did then gain the final mark for the correct answer.

  Text version

  Here is a triangle and a rectangle. In triangle ABC: AB = 6x, BC = 8, and angle ABC is a right angle. In rectangle PQRS: PQ = 5 and PS = 4x − 1. All measurements are in centimetres. The area of the triangle is 10 cm² greater than the area of the rectangle.

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• Edexcel 1MA1 · June 2023 · Paper 2H · Q11 · 4 marks

  4 marks

  11

  
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  This is a question often seen on paper 1, but it was answered quite well by many, with it not being uncommon to award 4 marks. With it appearing on a calculator paper though, it meant some candidates used their scientific calculators to solve. Unfortunately, these candidates had not paid attention to the demand of the question which asked for algebra. Without supporting algebra, the correct answers were awarded a Special case B2. It is vital that candidates read the demand carefully to ensure they are answering the questions as they are intended to. Of those who did use algebra, it was typical to award at least 2 marks, for a correct method to eliminate (often condoning a single arithmetic error) and for substituting correctly to find the second value. The main issue with the substitution method was subtracting a negative e.g. 18y – (–8y) = 10y was a very common error.

  Text version

  Use algebra to solve the simultaneous equations: 2x + 6y = 5 and 3x − 4y = −12

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• Edexcel 1MA1 · June 2023 · Paper 2H · Q14 · 3 marks

  3 marks

  14

  
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  A straightforward algebra question that allowed many to score some credit. Even some of the weaker candidates were able to gain one mark for correctly factorising the numerator of the fraction, the coefficient of x2 being more than 1 caused more issues but still a significant number went on to gain all 3 marks. Unfortunately, quite a few candidates got their signs wrong on the numerator, eg which often then led to an incorrectly factorised denominator as they were looking for a common factor.

  Text version

  Show that (x² − x − 6)/(2x² − 5x − 3) can be written in the form (ax + b)/(cx + d) where a, b, c and d are integers.

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• Edexcel 1MA1 · June 2023 · Paper 2H · Q15 · 3 marks

  3 marks

  15

  
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  Quadratic sequences have become a familiar topic now, and it is pleasing to see so many candidates being able to gain marks. This year we have seen a significant number of candidates using the formula approach to find the values of a, b and c in the expression compared to previous years. This did often bring success, but it was also evident that some had not fully understood this approach. It was common to award 2 marks for candidates working with n2 and getting as far as the sequence 2, 5, 8, … or seeing n2 + 3n. There was a lot more variety in responses where students used the n squared term to find the difference between that and the sequence. Often students found the difference as -2, -5, -8 resulting in an incorrect nth term. We did also see an increase in candidates working with different variables, normally x, which was able to gain full credit. However, a surprising number worked in two variables, such as , and these candidates were unable to get full marks.

  Text version

  Here are the first four terms of a quadratic sequence: 3, 9, 17, 27

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• Edexcel 1MA1 · June 2023 · Paper 2H · Q17 · 4 marks

  4 marks

  17

  
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  Clear algebraic working was lacking for a good number of candidates in part (a) and was the cause of them not gaining the mark, particularly not showing x4 = x2 + 5. Those that clearly showed each step in the rearrangement were rewarded. In part (b) many were able to gain some credit for use of the iteration formula, but many went beyond the requested value of x3 and lost the accuracy mark. Some candidates also found the values of x1, x2, and x3, and then found the mean of these 3 values and gave this as their answer, losing the accuracy mark. In terms of substitution, it is worth noting here, like with use of the quadratic formula, that ensuring the lid of the square root sign extends fully is important to show correct substitution.

  Text version

  This question is about solving x⁴ − x² − 5 = 0 by iteration.

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• Edexcel 1MA1 · June 2023 · Paper 3F · Q3 · 1 mark

  1 mark

  3

  
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  This question was well answered. Most candidates were able to state the correct answer of 4m. Weaker students wrote the answer incorrectly as m4.

  Text version

  Simplify m + m + m + m.

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• Edexcel 1MA1 · June 2023 · Paper 3F · Q11 · 3 marks

  3 marks

  11

  
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  Overall an excellent response from the majority of students correctly answering the question. Most students answered by subtracting 7 then dividing by 5. The use of inverse operations was a common approach to this question with candidates choosing to work numerically rather than algebraically. Very little algebra was seen here, though there were some trial and improvement approaches. A common mistake was to put the answer 65 from the execution of just one inverse operation rather than two. Candidates sometimes carried out the calculation on their calculator without showing their working but checked their solution and showed it, proving that their answer was correct. Occasionally a function machine approach was used but this was not as successful as the other methods shown. Sometimes candidates lost the final mark for only showing their answer embedded in the expression.

  Text version

  Gabriel thinks of a number, multiplies it by 5, adds 7 and gets 72.

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• Edexcel 1MA1 · June 2023 · Paper 3F · Q18 · 3 marks

  3 marks

  18

  
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  Candidates who were confident with their algebra often gained full marks. The most common approach involved expanding brackets whilst initial division by 4 was rare but successful when used. Candidates who expanded the brackets correctly, generally went on to answer the question well, although some mistakenly subtracted 12 from both sides, rather than adding. Other approaches such as trial and error or dividing both sides by 4 were also successful although less commonly used. When candidates failed to expand the bracket, they often failed to score any marks at all, choosing to guess an answer or to incorrectly remove the −3 first. A common mistake was to see 8x – 12 = 20 turning into 8x = 8 and then x = 1 given as the answer. There were cases of embedded answers shown within algebraic expressions.

  Text version

  Solve 4(2x - 3) = 20.

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• Edexcel 1MA1 · June 2023 · Paper 3F · Q20 · 4 marks

  4 marks

  20

  
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  In parts (a) and (b), most candidates had learnt and understood the index laws and were able to apply these and arrive at the correct solutions. The common misconceptions of adding for part (a) or multiplying for part (b) were observed. It was also common to see variations of these with multiple variable terms such as 2m or 2x followed by an index. In part (c) candidates that were able to multiply out the brackets and arrive at one correct term were rewarded with a method mark. Singular mistakes or responses that were spoiled by adding the two terms together gained no further credit. However, those that were able to multiply correctly through both terms and state these two terms secured the accuracy mark. No credit was given for simply multiplying by ‘4’.

  Text version

  Simplify and expand expressions using indices.

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• Edexcel 1MA1 · June 2023 · Paper 3F · Q26 · 2 marks

  2 marks

  26

  
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  Most candidates answered part (a) correctly. Some wrote the x and y coordinates the wrong way round, and a few listed the roots, which should be the answer for part (b) Part (b) was poorly answered with most candidates leaving the question blank or writing different numbers between −2 and 3 that were not within the required range. Common mistakes included writing in co-ordinate form. even when they had the correct figure. Some students wrote −2, the intercept on the y axis, or tried to isolate x and solve the equation using algebra, not noticing they are asked to estimate by using the graph. A common incorrect answer was (−2,2). The proportion of blank answers shows that many candidates were poorly prepared and did not understand how to find roots from a quadratic graph.

  Text version

  The graph of y = x^2 - 2x - 2 is shown on a grid.

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• Edexcel 1MA1 · June 2023 · Paper 3H · Q1 · 4 marks

  4 marks

  1

  
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  This question provided a good start to the paper. Most students gained full marks. Nearly all students scored the marks available in parts (a) and (b) though examiners did see 𝑚5 and 𝑥40 from some lower attaining students. 2𝑥13 was also seen occasionally as an answer in part (b). Part (c) was also done well. However, a significant number of students gave the correct answer in the working space but then misguidedly tried to “simplify” the expression by combining the two terms to give an expression such as 16𝑝5 for their final answer.

  Text version

  This question is about simplifying and expanding algebraic expressions.

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• Edexcel 1MA1 · June 2023 · Paper 3H · Q12 · 3 marks

  3 marks

  12

  
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  Students usually scored well in this question involving the expansion and then simplification of a product of three linear expressions. They generally showed their working in an organized way or used a grid to show the terms in their products. Errors were usually restricted to incorrect terms or difficulties in dealing with the signs when collecting terms together, rather than a flawed strategy although some students omitted terms from their expansion when at the stage of multiplying a quadratic expression by a linear expression. For students who did not give a fully correct answer, it was commonplace to see 2 marks or 1 mark awarded.

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  Show that (x − 1)(x + 3)(x − 5) can be written in the form ax^3 + bx^2 + cx + d where a, b, c and d are integers.

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• Edexcel 1MA1 · June 2023 · Paper 3H · Q13 · 3 marks

  3 marks

  13

  
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  A minority of students scored full marks for their responses to this question and there were more students than average that did not attempt the question. Weaker responses include those where a student had substituted a value or values for n, but of course, this could not serve as a proof and so could not be given any marks. Students who did approach the question more generally and attempted to write down an expression for the next triangular number term often wrote down incorrect expressions such as 𝑛(𝑛+2) , 𝑛+1(𝑛+2) or 𝑛(𝑛+1) + 1. Students writing down the expression 𝑛+1(𝑛+2) sometimes recovered their error in omitting brackets and went on to give a correct proof. Not all students who obtained the expression n2 + 2n + 1 could factorise it to prove it was a square number.

  Text version

  An expression for the nth term of the sequence of triangular numbers is n(n + 1)/2. Prove that the sum of any two consecutive triangular numbers is a square number.

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• Edexcel 1MA1 · June 2023 · Paper 3H · Q15 · 3 marks

  3 marks

  15

  
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  Part (a) of this question was not well answered with under a half of students taking the paper giving a correct factorisation. Very few students took the hint from part (a) about how to make progress with part (b) of the question. Those who did realise there was a connection and factorised the numerical expression were usually able to reason why this represented a product of two consecutive odd numbers. Some students wrote down (2n + 1)(2n − 1) indicating that they knew this was a general expression for the product of two consecutive odd numbers. Unfortunately, it was rare for students to securely link this expression with 240 − 1. A small number of students evaluated each of 220 − 1 and 220 + 1. This was accepted by examiners. Centres are encouraged to complete work with students where results in algebra, for example the difference of two squares, can be used with number.

  Text version

  This question is about the difference of two squares.

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• Edexcel 1MA1 · June 2023 · Paper 3H · Q21 · 4 marks

  4 marks

  21

  
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  Most students attempted this question testing algebraic techniques and it discriminated well between higher attaining students. The most concise method of multiplying throughout the equation by (x + 4) and (2 − 2x) as a first step was not used by many students. Instead, most students tried to combine the two algebraic fractions on the left- hand side of the equation first. A good proportion of students were able to deal with the fractions to write an equation of the form (2 − 2x) + 3(x + 4) = (2 − 2x)(x + 4) or equivalent. A much lower proportion of students could work with accuracy to write the equation with all terms on one side to get 2x2 + 7x + 6 = 0 or −2x2 − 7x − 6 = 0. Students who did get this far, often went on to get correct values for x, though examiners did see a significant number of incorrect factorisations and errors in using the quadratic formula. Errors in using the quadratic formula were often the result of substituting in three negative values for a, b and c. Centres could encourage students to multiply through an equation by −1 in such cases before using the formula.

  Text version

  Solve 1/(x + 4) + 3/(2 − 2x) = 1

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• Edexcel 1MA1 · June 2023 · Paper 3H · Q23 · 4 marks

  4 marks

  23

  
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  This question was accessible to and discriminated well between the highest attaining students sitting this paper. There were a number of routes through the question available to students. The most concise route was for students to find the gradient of the tangent at the point (−3, 4) on the circle and compare it with the gradient of the line joining the point (−8, 0) to the point (−3, 4). Few students chose this route, instead opting to find the equation of the tangent at the point (−3, 4), then show that the tangent would not pass through the point with coordinates (−8, 0) but through one of the points (− 3 , 0) or (−8, 4). Though this was the most frequently seen route, a significant number of students did not get as far as a correct conclusion supported by correct reasoning and necessary values. A number of students opted to approach the problem by using Pythagoras’ rule to show the triangle with vertices (−8, 0),(−3, 4) and (0, 0) was not right-angled. It was rare to find a complete solution from students using this strategy.

  Text version

  A circle has equation x^2 + y^2 = 25. The point P with coordinates (−3, 4) lies on the circle. Alex says that the tangent to the circle at P crosses the x-axis at the point (−8, 0). Is Alex correct? You must show how you get your answer.

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• Edexcel 1MA1 · June 2023 · Paper 3H · Q24 · 5 marks

  5 marks

  24

  
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  This question also discriminated well between high attaining students. Most students who made a good attempt at the question recognised the non-replacement nature of the problem and were able to make a start by forming expressions for probabilities in terms of x and/or y. A good proportion of these students were also able to use the ratio given in the question to make a substitution in order to get their expressions in terms of x only or y only. However, a significant number of students used “y = 4x” instead of the correct “y = 3x”. Invariably and sensibly, the overwhelming majority of students opted for getting their expressions in terms of x only. Those students who attempted substituting 3x for y into their probabilities before combining their products had more success than those who substituted later as the algebra became more complex. A significant number of students obtained correct expressions for the probability of Freda taking two pink counters and the probability of her taking two blue counters but gave the probability of her taking two green counters as 2𝑥−5 3𝑥 × 2𝑥−4 3𝑥−1 instead of 2𝑥−5 3𝑥 × 2𝑥−6 3𝑥−1 . This had clearly arisen from an error with signs earlier on in a student’s working. It was encouraging to see a good number of fully correct solutions from students working at the highest level. Summary Based on their performance on this paper, students are offered the following advice: • carry out a check to make sure you have not miscopied any numbers given in a question before you complete the question. • check that you have not miscopied algebraic expressions from a previous line in your working. • practise solving equations involving algebraic fractions. • compare the medians and either the ranges or the interquartile ranges in context when asked to compare two distributions represented by box plots. • ensure you can write and manipulate algebraic expressions accurately particularly with regard to the use of brackets and the signs of terms. • practise solving real life problems involving inverse proportionality. • carry out common sense checks where possible.

  Text version

  There is a total of y counters in a box. There are x pink counters and 5 blue counters in the box. The rest of the counters are green. x : y = 1 : 3. Freda takes at random two counters from the box. Find, in terms of x, an expression for the probability that Freda takes two counters of the same colour. Give your answer as a fraction in the form (ax^2 + bx + c)/(dx^2 + ex) where a, b, c, d and e are integers.

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• Edexcel 1MA1 · November 2022 · Paper 1F · Q5 · 1 mark

  1 mark

  5

  
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  Many students failed to score in this question by not simplifying the given expression fully. 3w × 5t and 15 × wt were popular partial simplifications gaining no marks.

  Text version

  Simplify 3 × w × 5 × t

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• Edexcel 1MA1 · November 2022 · Paper 1F · Q8 · 3 marks

  3 marks

  8

  
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  Part (a) was answered well, with very few failing to score the mark. Candidates performed less well in part (b) with many not collecting the like terms of 3n and n together as a first step. Many tried to subtract the n from both sides of the equation either resulting in 2n = 24 (giving n = 12) or 3n = 24 (giving n = 8). In this question, students usually scored no marks or full marks. Some used an embedded solution, with mixed success. Many candidates who achieved full marks did not attempt the algebraic approach and instead simply stated that 3 × 6 + 6 = 24, with a correct answer on the answer line.

  Text version

  Complete all parts of this question.

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• Edexcel 1MA1 · November 2022 · Paper 1F · Q15 · 2 marks

  2 marks

  15

  
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  The diagram proved very useful in getting one correct coordinate, for which one mark was awarded, usually y = 1, but rarely was a fully correct answer seen. Whilst many students did use the diagram, drawing the correct line segment, it was also rare to see the correct positioning of the required midpoint. This was often seen placed on the y-axis. There were very few instances of candidates using written methods to find the midpoint.

  Text version

  On a coordinate grid, P = (−3, −2) and Q = (2, 4).

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• Edexcel 1MA1 · November 2022 · Paper 1F · Q16 · 4 marks

  4 marks

  16

  
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  Those students who attempted an algebraic solution often picked up good marks, 2 or more, for a correct expression for the perimeter and then equating it to the given value of 52. Simplifying their algebraic expression often proved more difficult and it was not uncommon to see quadratic terms creeping in. Some students assumed all the lengths to be the same and calculated 52/4 gaining no marks. Some students correctly found the value of x and then failed to answer the question by finding 2x, the length of DC. Poor arithmetic let many students down in being unable to divide 57 by 6 and hence not being able to achieve the accuracy mark. Many students did however adopt a trial and improvement approach which usually cost them a great deal of time for little reward.

  Text version

  Quadrilateral ABCD has side lengths AB = x + 1 cm, BC = x − 1 cm, CD = 2x cm and DA = 2x − 5 cm. Its perimeter is 52 cm.

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• Edexcel 1MA1 · November 2022 · Paper 1F · Q25 · 2 marks

  2 marks

  25

  
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  This was very poorly answered. The formula and the force both being given meant there was only the need to work out the area by multiplying 4 by 2. Some students still think that when working with area, they need to square the values first before using them to divide into 10 000. Those who correctly worked out the area usually tried to divide 10 000 by 8, for one mark, but often then failed, again through poor arithmetic. Some stated the area as 6 (4 + 2) and therefore gained no credit.

  Text version

  pressure = force area A storage tank exerts a force of 10 000 newtons on the ground. The base of the tank in contact with the ground is a 4 m by 2 m rectangle. Work out the pressure on the ground due to the tank.

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• Edexcel 1MA1 · November 2022 · Paper 1F · Q26 · 5 marks

  5 marks

  26

  
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  There were many trial and improvement attempts to answer part (a) of this question, the majority of which were unsuccessful, however a few did throw up the critical value of 6. Those following an algebraic approach were able to score one mark initially for a correct process to start a solution. Subtracting 3 from both sides was usually successful but multiplying both sides by 2 was often carried out incorrectly by omitting to multiply the 3 by 2 resulting in 5x + 3 > 36 or 5x + 3 = 36. A correct first step was often followed by an incorrect second step or no second step at all. Replacing the inequality by an equality is not penalised but students must remember to replace the inequality for the final answer. Far too often an answer of x = 6 was given. In part (b), the ability to factorise a polynomial was sketchy. Factorisation of the first two terms [eg. x(x + 10) + 9] was often seen but this gained no credit. (x + 10)(x – 1) was also a common incorrect answer seen. Common incorrect answers included (x + 1x)(x + 9) or (x + 9x)(x + 1) Summary On the evidence of performance on this paper, students need to: ● take greater care when reading the questions. Performance in questions 6, 10b, 12 and 18 in particular was severely affected by this. ● take care when carrying out arithmetic operations and check their working to avoid careless errors. ● write clearly so that correct values quoted are not altered in subsequent working. ● show clear methods of working particularly in basic calculations such as finding fractions or percentages of quantities. ● set their working out clearly, crossing out working that is being replaced. ● be able to convert using metric units of distance. ● be clear on the specific wording required when giving geometric reasoning

  Text version

  Complete all parts of this question.

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• Edexcel 1MA1 · November 2022 · Paper 1H · Q9 · 4 marks

  4 marks

  9

  
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  This question was answered quite well by the majority of students. The table in part (a) was often completed correctly. Most errors occurred with the substitution of the negative values of x into the equation and evaluating y as 4 when x = –2 was a common error. In part (b) the plotting of the points was usually accurate. Most students realised that a curve was needed to join the points and it was pleasing to see many freehand curves drawn that passed through all the points. When the points were joined with line segments a maximum of one mark could be awarded. Students who had gained just one mark in part (a) generally went on to gain one mark in part (b) for plotting at least five of the points from their table correctly. Students should be encouraged to make sure that their curve passes through all of the points and doesn’t consist of more than one curve between any two points.

  Text version

  For y = 6x − x³, the table uses x = −3, −2, −1, 0, 1, 2, 3 and already gives y = 9 at x = −3, y = 4 at x = 2 and y = −9 at x = 3.

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• Edexcel 1MA1 · November 2022 · Paper 1H · Q12 · 4 marks

  4 marks

  12

  
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  Many of the students appeared familiar with simultaneous equations and started by multiplying both equations to make the coefficients of x or y the same. Most then subtracted one equation from the other to eliminate one of the variables and gained the first method mark. Accuracy was sometimes lost through arithmetic errors in the multiplying or subtracting. Some students got to 7y = –14 or –7y = 14 and then wrote y = 2. Students who used a correct method to find one value usually went on to substitute this value into an equation in order to find the other value and gained the second method mark. Eliminating one variable by rearranging one equation and substituting into the other equation was attempted by a few students but this method rarely led to a fully correct answer. Responses which scored no marks tended to involve an immediate addition or subtraction of the equations before balancing coefficients.

  Text version

  Solve the simultaneous equations   5x + 2y = 11 4x + 3y = 6 x = y =

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• Edexcel 1MA1 · November 2022 · Paper 1H · Q16 · 4 marks

  4 marks

  16

  
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  In part (a) few students used the difference of two squares to prove that (2m + 1)2 – (2n – 1)2 = 4(m + n)(m – n + 1) but those that did were usually able to complete the proof without any errors. The majority of students started by attempting to expand (2m + 1)2 and (2n – 1)2 and many were able to gain the first mark. Some students gained the first mark by correctly expanding the right-hand side. For the second mark it was necessary to see a correct expression for the left-hand side after expansion. Many of those who attempted to subtract 4n2 – 4n + 1 from 4m2 + 4m + 1 were not able to write a correct expression with or without brackets - their omission was usually followed by sign errors - and it was common to see the subtraction written as 4m2 + 4m + 1 – 4n2 – 4n + 1. After showing that the left-hand side can be simplified to 4m2 + 4m – 4n2 + 4n students were still left with the task of showing that this is equal to the right-hand side. Many attempted to do this by expanding 4(m + n)(m – n + 1) and arriving at the same simplified expression. Some attempted to do it by factorising 4m2 + 4m – 4n2 + 4n to get 4(m + n)(m – n + 1) but convincing attempts at factorising were rarely seen. Very few students were able to give a correct and complete answer in part (b) and this part was often not attempted. Many students made the wrong decision and said that Sophia is not correct. Some of those who made the correct decision tried to justify the decision using numbers and consequently did not gain the mark. Students needed to explain that 2m + 1 and 2n – 1 are odd numbers and the right-hand side is a multiple of 4.

  Text version

  Complete all parts of this question.

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• Edexcel 1MA1 · November 2022 · Paper 1H · Q19 · 5 marks

  5 marks

  19

  
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  Some students correctly wrote the two fractions with a common denominator and gained the first mark but many were then unable to carry the algebraic solution any further. Those that did reduce the equation to a 3-term quadratic sometimes made errors when rearranging and did not get the second mark. Substitution into the quadratic equation formula was generally done well. Some students who did not get a correct quadratic equation were still able to gain the third mark for dealing correctly with their 3-term quadratic equation. The first of the two accuracy marks was awarded for two correct values of x, e.g. 4 − . Some students gave this as the final answer and others made errors when attempting to write it in the required form. Only a few students could demonstrate the necessary skills of algebraic manipulation to solve the equation and give the answer in the required form.

  Text version

  Solve 1/x − 1/(x + 1) = 4.

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• Edexcel 1MA1 · November 2022 · Paper 1H · Q21 · 2 marks

  2 marks

  21

  
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  This question proved to be a good discriminator between the most able students. When only one of the two values was correct it tended to be the y coordinate and (90, – 1) was a common response that gained 1 mark. There were some students who gave the answer as (–1, 180) instead of (180, –1) and they gained 1 mark.

  Text version

  A sketch shows y = cos x° for 0 ≤ x ≤ 360. P is the unique minimum point on this part of the curve. Write down the coordinates of P.

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• Edexcel 1MA1 · November 2022 · Paper 1H · Q23 · 5 marks

  5 marks

  23

  
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  For a question towards the end of the paper part (a) was answered reasonably well. Some students started by using consecutive terms to find the common ratio. Having identified the common ratio as 2 5 many worked out the next term as 4000 but 400 5 2 5  = was a common error. Instead of finding the common ratio some students used alternating terms of the sequence and the calculation 200 × (200 ÷ 10) gave the next term without them having to deal with surds. Part (b) proved to be a challenging question with very few students knowing how to work out the first term of the sequence. Some students gained the first mark for finding the second term of the sequence and others gained it by finding the ratio of the 4th and 6th terms. Those who were able to show a complete process to find the first term usually went on to gain full marks. The most common misconception was to create an arithmetic sequence of the denominators of the fractions leading to the final answer being given as 5 2 − .

  Text version

  The first five terms of a geometric sequence are √5, 10, 20√5, 200, 400√5.

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• Edexcel 1MA1 · November 2022 · Paper 1H · Q24 · 6 marks

  6 marks

  24

  
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  In part (a) many of the students who gained the first mark for equating the two volumes were not able to go on and find the correct ratio. Those that got as far as writing 4r = h often gave the ratio r : h as 4 : 1 instead of 1 : 4. The majority of students made no attempt at part (b) and very few managed to gain more than one mark. The first mark was awarded for equating the surface area of the sphere with the surface area of the cone but a common error here was to omit the area of the base of the cone and write 4πr² = πrl. Students making this error could still be awarded the next two process marks if they went on to substitute for l and isolate the terms in r2. Summary Based on their performance on this paper, students should: ● practise their arithmetic skills, particularly division and operations with decimals ● take care when carrying out arithmetic operations and check their working to avoid careless errors ● read each question carefully and ensure that their final answer matches the question asked ● practise subtracting one algebraic expression from another, especially expressions with negative terms, and use brackets more efficiently ● practise finding angles using the circle theorems and stating the circle theorems used

  Text version

  A solid sphere of radius r and a solid cone of radius r and height h have equal volumes. Use volume of sphere = (4/3)πr³ and volume of cone = (1/3)πr²h.

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• Edexcel 1MA1 · November 2022 · Paper 2F · Q3 · 1 mark

  1 mark

  3

  
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  Students found this question slightly more challenging. Even so, the majority gave the correct answer.

  Text version

  Write down the 7th odd number.

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• Edexcel 1MA1 · November 2022 · Paper 2F · Q8 · 2 marks

  2 marks

  8

  
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  This question was very well answered with only a very few students failing to full marks. The main errors seen were arithmetic.

  Text version

  Safiya wants to hire a van. She uses this rule to work out the cost of hiring a van for a number of days. Cost = £45 × number of days Safiya is going to hire the van for 7 days. Work out the cost. £

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• Edexcel 1MA1 · November 2022 · Paper 2F · Q14 · 7 marks

  7 marks

  14

  
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  In part (a) this simplifying question was answered very well, with the majority of students being awarded both marks. The method mark was often awarded for correctly collecting like terms for one variable but often lost the second mark due to an error when working with 7d and −d. Commonly seen errors included an extra step to combine 7c and 6d to get a final answer of 13cd or, when working with the d variable, either misinterpreting the negative sign to get 7c – 6d or forgetting that this represented 1d, resulting in 7c + 7 being stated as the final answer. A less frequent error was to use powers when collecting like terms. In part (b), students often struggled to expand the bracket correctly by only multiplying 5 by 2m to give 10m but failing to multiply 5 by 6 as well. Students that did correctly expand the bracket generally solved the equation successfully, with almost half of students scoring full marks. A significant number of students attempted to solve the equation using trial and error rather than algebraic methods. This approach scores either full marks or zero marks and should not be encouraged. In part (c), responses were encouraging. Students demonstrated a good understanding of how to write the expression 3x + 2y correctly, with again almost half of the cohort scoring full marks. Where candidates failed to gain full marks, some gained one for having one of the terms correct but most failed to score at all. Often this was due to using powers or a subtraction sign or even missing the sign altogether.

  Text version

  Complete all parts of this question.

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• Edexcel 1MA1 · November 2022 · Paper 2F · Q20 · 4 marks

  4 marks

  20

  
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  Examiner report

  For part (a) students had mixed success finding the nth term, with almost equal numbers gaining either 0 or 2 marks. Errors made were finding the common difference but not recognising this as 6n and stated a response of n + 6 or 6 – n. Those students who identified 6n sometimes paired it with –1 instead of +1 or put 6n + 7 or left as 6n. A common incorrect answer was 7 – 6n which gained no credit. In part (b) some students were successful in listing the terms of the sequence. But a common listing error was to start with 6 and subtract 8, rather than the other way around. There were very few attempts to set up and solve an equation. Those that set up the equation 8 – 6n = – 58 were then unable to successfully show that n was 11 as many arithmetic errors occurred. Centres should ensure that students practising creating equations and solving them, especially involving negative numbers.

  Text version

  Here are the first five terms of an arithmetic sequence. 7            13            19            25            31

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• Edexcel 1MA1 · November 2022 · Paper 2F · Q24 · 4 marks

  4 marks

  24

  
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  This question was completed with varying degrees of accuracy. It was rare to give full marks for all parts of the question but many students scored 1 or 2 marks within different parts of the question. Part (a) was probably answered most, although it was clear that some students did not understand what was being asked as there were a large number of incorrect responses. Some students understood how to identify the y intercept and wrote their answer expressed as a co- ordinate, which was accepted. In part (b) some blank scripts were seen and it was disappointing to see a significant number the coordinates written in reverse i.e. (– 5, 3) suggesting that the term turning point was understood but the order of the coordinates was the stumbling block. For part (c) there were a large number of blank responses. When the right number was there it was often in the form of a coordinate. Only a handful of responses gained 1 mark for marking the roots on the graph. It is clear that many did not understand the meaning of ‘root’ of an equation. Centres would benefit from reminding students to mark all key points on the diagram. If mistakes are made method marks may still be awarded.

  Text version

  The graph of y = x² − 6x + 4 is drawn for −1 ≤ x ≤ 7 on axes with unit divisions.

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• Edexcel 1MA1 · November 2022 · Paper 2H · Q3 · 4 marks

  4 marks

  3

  
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  Linear sequences were being tested in this question, and again was a familiar question allowing many to score some good marks. Part (a) was asking for the nth term rule of the given sequence. Most candidates scored at least 1 for a sequence in the form 6n + c with a good number gaining both marks for a fully correct expression. Common incorrect answers were n + 6. 6n – 1 and 7n + 6. Part (b) required students to show if a particular number was in the sequence. The negative aspects of this sequence caused some problems, with many students generating a sequence by subtracting 8 rather than 6. A significant number of students on the higher paper used algebra to form and solve an equation, and most who did gained both marks. Incorrect approaches included using n = 58 in the formula.

  Text version

  Here are the first five terms of an arithmetic sequence. 7            13            19            25            31

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• Edexcel 1MA1 · November 2022 · Paper 2H · Q7 · 4 marks

  4 marks

  7

  
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  This 3-part Quadratic Graph question was answered generally well on the higher paper. Almost all candidates scored full marks in part (a) and part (b) for reading off the y-intercept and turning point correctly. Many were then unsure as to what the roots were and didn't gain marks in part (c). Some did gain 1 mark for identifying the roots on the graph, but being unable to read off the values to gain both marks. Some candidates tried to solve algebraically despite being told to use the graph - this approach gained no credit.

  Text version

  The graph of y = x² − 6x + 4 is drawn for −1 ≤ x ≤ 7 on axes with unit divisions.

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• Edexcel 1MA1 · November 2022 · Paper 2H · Q9 · 2 marks

  2 marks

  9

  
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  This question required candidates to rearrange one of the two equations, so that both were in the same form (y = mx + c or by = ax + c) so that the gradient, or coefficients, can be compared. Most candidates attempted to manipulate one or the other of the equations, and mostly they were successful. Of those who successfully manipulated, most were able to draw a correct conclusion. A small number of candidates gained both marks by giving a full statement such as 3 divided by 6 is equal to ½ so both lines have the same gradient. Comparing the equations as they were given lead to incorrect conclusions that the gradients were different.

  Text version

  Here are the equations of two straight lines. y = 1 2 x − 6                6y = 3x + 7 Oscar says that these lines are parallel. Is Oscar correct? You must give a reason for your answer.

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• Edexcel 1MA1 · November 2022 · Paper 2H · Q13 · 2 marks

  2 marks

  13

  
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  This is a topic that has not been assessed regularly, and it showed in the quality of candidates' responses. Very few candidates realised this required the use of Pythagoras. The most common incorrect response seen was to find the gradient of the line AB. It is evident that centres have not spent much time on the topic and further work is needed.

  Text version

  A and B are points on a centimetre grid. A is the point with coordinates (−7, 6) B is the point with coordinates (8, −5) Work out the length of AB. Give your answer correct to 1 decimal place. cm

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• Edexcel 1MA1 · November 2022 · Paper 2H · Q16 · 4 marks

  4 marks

  16

  
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  In this inequalities question candidates had to first find the equations of the 4 lines. There was 1 mark for either of the horizontal or vertical lines, and then 1 mark each for the two diagonal lines. For these marks we could accept use of an equals sign, or any inequality symbol. The final mark was for then applying the correct inequalities to all four correct equations. Most candidates gained some credit, even if this was just a single mark for y = 6 or x = ⧿3. Candidates did however sometimes mix these two up, giving answers of x = 6 or y = ⧿3. The line that caused the most problems was y = ⧿x2 + 1 with most being unable to deduce this equation correctly, especially the negative gradient. It was very rare to see a fully correct response, with most candidates using the same symbol for all equations. Candidates who had no access to this question simply gave the points of intersection.

  Text version

  A shaded quadrilateral is bounded by the lines x = −3, y = 6, y = 3x + 6 and y = −x/2 + 1. Its interior contains the point (−2, 3). Find the four inequalities that define the shaded region.

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• Edexcel 1MA1 · November 2022 · Paper 2H · Q19 · 3 marks

  3 marks

  19

  
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  Again, a high proportion of learners trying to use the formula would suggest that the additional information provided was well used. Those using the formula had very mixed results. It was evident that many candidates were not confident in their substitution, and missing signs, short root notation and short fraction notation led not only to incorrect answers but often to the withholding of marks. Those who attempted factorisation typically struggled, probably down to the coefficient of x2 being 6. All that said, a significant proportion of candidates scored 2 or 3 marks, for either one or two correct values given. Again, it is important that candidates remember when rounding values such as 23 one decimal place is not enough, and as such an answer given of 0.6 (without first seeing 2/3 or 0.6̇) did not get credit.

  Text version

  Solve 6x² + 5x − 6 = 0

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• Edexcel 1MA1 · November 2022 · Paper 2H · Q21 · 4 marks

  4 marks

  21

  
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  Well into the Grade 8 questions on the paper and it is not surprising that fewer candidates are now scoring marks. For part (a), failure to draw a tangent at t = 17.5 meant that no marks could be scored at all. It was very common to just see 10.5 17.5 which came from reading the volume at t = 17.5, and this scored zero. Those who drew a suitable tangent often then went on to score full marks for an answer in range. However, some were unable to use the scale correctly and then lost marks for their method to find the gradient. Part (b) showed improvements on previous series, and there were many suitable descriptions of what the gradient represented although students still relate this answer to a correlation type response. ‘Acceleration’ was a commonly seen incorrect answer.

  Text version

  A smooth graph gives the volume V, in litres, of water in a container t seconds after filling starts. Approximate plotted readings are (0,0), (2.5,0.8), (5,1.8), (7.5,3.0), (10,4.5), (12.5,6.3), (15,8.2), (17.5,10.6), (20,13.5), (22.5,17.3) and (25,23.0).

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• Edexcel 1MA1 · November 2022 · Paper 2H · Q22 · 3 marks

  3 marks

  22

  
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  This functions question required candidates to first find a composite function, and then to find the inverse function. A pleasing number of candidates were able to find the composite function. Those who were unsuccessful typically gave √2𝑥 + 3 as their incorrect composite. However, the second mark here was not dependent upon the first, so providing a composite function was found (correct or not), the second mark could be awarded if the first step to find the inverse was correct. Again, a good number of candidates were able to get this mark.

  Text version

  f(x) = ∛x, g(x) = 2x + 3 and h(x) = f(g(x)).

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• Edexcel 1MA1 · November 2022 · Paper 2H · Q24 · 4 marks

  4 marks

  24

  
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  As with some of the other higher grade questions, a pleasing number of candidates gained some credit on Question 24. The first mark here was to find the gradient of the radius, and was seen both as 43 and as 2.82.1. The second mark was for a method to find the gradient of the tangent using -1m where the m came from the correct method. The final method mark was for substitution of the given coordinate into a suitable equation. In these first steps, many who gained the first went on to gain the second, and often the third mark. The most common errors were to forget the minus sign when working with the gradient of the tangent, or for using the wrong gradient when substituting in. Very few candidates were then able to gain the accuracy mark as they were unable to rearrange the equation into the stated form, normally with a, b and c being left as fractions or decimals. It is unclear if this was due to not reading the question correctly or for being unable to complete the manipulation. Summary Based on their performance on this paper, students/ centres should: • Make use of calculators where applicable rather than relying on mental method • Understand the difference between standard and reverse percentage and the method for compound percentages • Not prematurely round • Encourage pupils to mark significant points and lines on diagrams and graphs, in order to gain method marks that are available • Further practice finding length and gradient between two given points, emphasising the difference between the two things being found • Learn to use and apply basic geometrical formulae such as the area of a triangle = ½ × base × height. This will be particularly important once the use of the formulae sheet is no longer available for students • Interpreting values such as the median and quartiles from cumulative frequency graphs • Understand that to estimate the gradient of a curve, a tangent must be drawn at the point in question to gain any credit

  Text version

  A circle has equation x² + y² = 12.25 The point P lies on the circle. The coordinates of P are (2.1, 2.8) The line L is the tangent to the circle at point P. Find an equation of L. Give your answer in the form ax + by = c, where a, b and c are integers.

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• Edexcel 1MA1 · November 2022 · Paper 3F · Q9 · 3 marks

  3 marks

  9

  
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  Candidates were well practiced with coordinates, and only a few failed to gain the marks in both part (a) and part (b), usually by reversing the coordinates. In contrast, it was rare to see a correct answer in part (c). Frequently the y axis (and sometimes the x-axis) was often just marked with a cross. Drawing x = −3 was a common incorrect answer, although many candidates felt it needed to be a diagonal line, often going through point B. Another common error was to draw a line through A and B.

  Text version

  On a coordinate grid, point A is at (−1,2).

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• Edexcel 1MA1 · November 2022 · Paper 3F · Q10 · 2 marks

  2 marks

  10

  
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  In part (i), most students were able to find one of the two solutions given in the mark scheme, 11, 10 being seen most frequently. Errors occurred when candidates assumed it was a linear sequence and only found the first difference and applied that throughout, or where they had difficulty processing a decreasing sequence and added onto the starting number and found the terms that came before the start of the sequence instead of terms 4 and 5. Occasionally arithmetic errors lost the mark. In part (ii) candidates had difficulty in explaining the decreasing difference, often referring to the amount or sequence going down by one rather than the difference. The most common successful strategy for gaining the mark was for writing the differences, (-4, -3, -2, -1), sometimes on the sequence itself. Those that did not use numbers had less success here as many of the worded answers without examples were too vague or incoherent. Candidates are advised to practice talking about differences in terms of what operation they are performing and its value. There was also some confusion as to whether it was an increasing or decreasing sequence.

  Text version

  Here are the first three terms of a sequence. 20            16            13

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• Edexcel 1MA1 · November 2022 · Paper 3F · Q14 · 3 marks

  3 marks

  14

  
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  In part (a), the answer of 30 was most often correct. But in part (b) many candidates were unprepared for the requirements for reading from the graph and scaling their reading. There were very few who chose the most appropriate value of 20 to scale to 80, and many who took inaccurate readings from the axis. This question highlighted that candidates need more practice with this topic of the curriculum.

  Text version

  A straight conversion graph for ounces and grams passes through (10 ounces, 300 grams) and (35 ounces, 1000 grams). Use the graph for the conversions below.

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• Edexcel 1MA1 · November 2022 · Paper 3F · Q19 · 4 marks

  4 marks

  19

  
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  The axes scaling proved challenging for some candidates in this question and thus they were unable to correctly read off values from the graph. Though the majority of candidates gave the correct answer of 15 for part (a), a significant number gave the incorrect answer of 20 instead. In part (b) the majority of candidates gave a value within the acceptable range. The most common answer was 4.5 and the most common incorrect answer was 7. Candidates who drew guidelines on the graph had greater success. In part (c) candidates were less successful and a multitude of errors were seen stopping many candidates from scoring all or some of the marks. These included not knowing or being able to substitute correctly into the speed/distance/time formula, being unable to convert minutes into hours or even reading the correct values from the graph. It was common to see answers of 5 and 0.2 from 20 ÷ 4 (no marks) and 4 ÷ 20 (1 mark). It was rare to see a correct answer of 12, but when it was seen it was often due to a scaled approach to this question rather than 20 minutes being converted to a fraction of an hour or attempting to convert 0.2km per minute.

  Text version

  Carly cycles to her friend’s house, stays there, then cycles home. Her travel graph joins the points (08:00, 0 km), (08:20, 4 km), (08:30, 7 km), (08:45, 7 km) and (09:00, 0 km) with straight line segments.

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• Edexcel 1MA1 · November 2022 · Paper 3F · Q21 · 2 marks

  2 marks

  21

  
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  Poorly attempted with many not completing a correct first step. A = 3p - 9 was a very common answer. Some were able to score 1 mark for the first step and showing the result of adding 9 to both sides getting p + 9 = 3a. Very few attempted dividing by 3 as a first step.

  Text version

  Make a the subject of the formula p = 3a − 9

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• Edexcel 1MA1 · November 2022 · Paper 3H · Q1 · 2 marks

  2 marks

  1

  
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  Most candidates found this question to be a good introduction to the paper and scored both of the marks available. Successful candidates usually added 9 to both sides of the formula as a first step rather than the alternative of dividing throughout the formula by 3. Nearly all candidates gained some credit for their answers. Where there were errors, a common response was a = p + 9 ÷ 3 which earned one mark for candidates who carried out the correct inverse operations but did not put brackets round the p + 9 or write it as a fraction.

  Text version

  Make a the subject of the formula p = 3a − 9

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• Edexcel 1MA1 · November 2022 · Paper 3H · Q10 · 1 mark

  1 mark

  10

  
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  This question was quite well answered. It was expected that candidates would refer in some way to the change in signs needed for the last two terms in the expansion. The question asked “what is wrong with Peter’s working”. Candidates who simply gave a correct final answer had not responded to the question and so could not be awarded the mark. There were also many answers which were too vague. In particular, examiners regarded the response “he should expand the brackets first” as not giving sufficient clarity or detail to be deserving of the mark. Some candidates were concerned that the order of the two pairs of brackets should be the other way round.

  Text version

  Peter has to subtract (x² − 2x − 4) from (x² + 3x + 5) Here is his working (x² + 3x + 5) − (x² − 2x − 4) = x² + 3x + 5 − x² − 2x − 4 = x + 1 Explain what is wrong with Peter’s working.

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• Edexcel 1MA1 · November 2022 · Paper 3H · Q11 · 2 marks

  2 marks

  11

  
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  There were many fully correct answers to this question. However, a significant minority of candidates could not take into account all the information given and were awarded just one of the two marks available for listing a correct set of integers for x (or y). A small number of candidates tried to use inequality signs, therefore ignoring or not understanding the statement that “x and y are integers”.

  Text version

  x and y are integers such that 3 < x < 8 4 < y < 10 and x + y = 14 Find all the possible values of x.

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• Edexcel 1MA1 · November 2022 · Paper 3H · Q14 · 5 marks

  5 marks

  14

  
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  A minority of candidates scored full marks for their responses to this question. However, most candidates were able to gain some marks and the question was a good discriminator between candidates who could manipulate algebraic expressions with accuracy and those who made errors. A fully simplified expression was asked for in answering part (a) of this question. It was disappointing to see so few candidates being completely successful and it was much more usual for candidates to gain just one of the two marks available e.g. for 3x20y24. A common error was for candidates to add the indices 20 and 24 and give their answer in the form cxy44. Candidates usually scored well in part (b) which required the expansion of a product of three linear expressions to give a fully simplified cubic expression. Though there were some low attaining candidates who attempted to expand the product in one stage, working was usually shown in an organized way or a grid was used to show terms in the products. Errors were usually restricted to incorrect terms or difficulties in dealing with the signs when collecting terms together, rather than a flawed strategy. There were some candidates who omitted terms from their expansion. For candidates who did not give a fully correct answer, it was commonplace for them to earn 2 of the available marks in part (b).

  Text version

  Complete all parts of this question.

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• Edexcel 1MA1 · November 2022 · Paper 3H · Q21 · 3 marks

  3 marks

  21

  
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  Some candidates gave a very clear, correct and concise answer to this question by first completing the square in the form 4((x – 7)2 – 49) before giving an explanation how this leads to the given result. Examiners expected candidates to show the connection between 4((x – 7)2 – 49) and –196 as well as make a conclusion to gain the last mark. Only a very small proportion of candidates used the alternative form (2x – 14)2 – 196. Many candidates showed a lack of confidence in the process of completing the square and as a result were unable to gain marks because their algebra was full of errors.

  Text version

  The equation of a curve is y = 4x² − 56x The curve has one turning point. By completing the square, show that the coordinates of the turning point are (7, −196) You must show all your working.

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• Edexcel 1MA1 · November 2022 · Paper 3H · Q22 · 3 marks

  3 marks

  22

  
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  Fully correct answers to this question were rarely seen. However, a good number of candidates were able to score at least one mark for using a correct common denominator. Only the higher attaining candidates were able to work accurately and/or gain more than one mark for their responses. In particular mistakes were often made with signs when −3(x2 − 25) or equivalent was written as −3x2 − 75. Much of the work seen showed little understanding of the method required to combine the 3 fractions concerned. For example, a significant number of candidates produced work involving expressions which were not fractions and could not be awarded any credit. (2x + 3)(x + 5) + (x − 4)(x − 5) is an example of such an expression.

  Text version

  (2x + 3)/(x − 5) + (x − 4)/(x + 5) − 3 can be written in the form (ax + b)/(x² − 25), where a and b are integers.

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• Edexcel 1MA1 · November 2022 · Paper 3H · Q23 · 2 marks

  2 marks

  23

  
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  For part (a) of this question, while there were attempts by a majority of candidates, few were correct. There were many more candidates who sketched the graph of y = f(x − 2), that is, translated the graph to the right instead of to the left. Some candidates tried to sketch y = f(x) + 2. Those candidates who did identify the correct translation invariably sketched a graph meeting the criteria in the mark scheme and so scored the mark available. Part (b) also attracted an incorrect response from the majority of candidates who attempted the question. The incorrect response, y = g(−x) was seen more frequently than the correct answer y = −g(x).

  Text version

  The graph of y = f(x) crosses the x-axis at (−1,0), (1,0) and (2,0), has a local maximum at (0,2), and has a local minimum at approximately (1.5,−0.6).

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• Edexcel 1MA1 · June 2022 · Paper 1F · Q2 · 1 mark

  1 mark

  2

  
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  The usual error in collecting these four algebraic expressions was to give an answer of e4 instead of 4e, carelessly reading the operation as ‘×’ instead of ‘+’.

  Text version

  Simplify e + e + e + e.

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• Edexcel 1MA1 · June 2022 · Paper 1F · Q9 · 2 marks

  2 marks

  9

  
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  Both parts of this question were answered well with very few mistakes seen.

  Text version

  Patterns are L-shapes of unit square tiles: pattern n has n tiles across the top and n tiles down the left, sharing the corner tile.

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• Edexcel 1MA1 · June 2022 · Paper 1F · Q14 · 2 marks

  2 marks

  14

  
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  Basic arithmetical errors prevented many students gaining full marks here. 24 – 5 and/or 6 × 4 were often seen calculated wrongly. The most common error was through incorrect substitution, usually resulting in 64 – 5 = 59 or 6 + 4 −5. Some students attempted to balance and solve the equation.

  Text version

  y=6x−5.

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• Edexcel 1MA1 · June 2022 · Paper 1F · Q23 · 2 marks

  2 marks

  23

  
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  It was common to see the critical value of 5 identified by many students, usually as the solution of a linear equation, but less were able to actually solve the inequality. Attempts to add 27 to both sides as a first step were seen but were poorly executed in many cases. 7x < 35 was often seen alone with no subsequent conclusion offered. Some students employed trial and improvement methods, quite often ending up with an answer of 4 (the greatest possible integer value of x)

  Text version

  Solve 7x−27<8.

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• Edexcel 1MA1 · June 2022 · Paper 1F · Q28 · 6 marks

  6 marks

  28

  
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  Accurately completing the table of values in part (a) was rare with the usual mistake occurring when substituting the negative value of x. Some ignored the quadratic expression given and simply tried to complete an arithmetic sequence. If a mark had been awarded in part (a) for at least two correct values, then at least one mark was usually earned in part (b) for correctly plotting their values. The drawing of a graph following a fully correct table in part (a) was sometimes spoiled by a ‘flat bottom’ to the quadratic curve through (1, -1) and (2, -1) or by joining the points with line segments. Many students did not understand what was required in finding solutions to the quadratic equation in part (c). Those that did, often wrote their solutions in a coordinate form; one mark was still available for this.

  Text version

  For y=x²−3x+1, use x=−1,0,1,2,3,4.

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• Edexcel 1MA1 · June 2022 · Paper 1H · Q1 · 2 marks

  2 marks

  1

  
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  This question was answered very well with most students using an algebraic approach to solve the inequality. It was quite common though to see 7x < 35 followed by an answer of 5 or x = 5 or even x > 5 which resulted in the loss of the accuracy mark. Answers of x < 35/7 and x < 35 were also seen quite often. Students who attempted to solve the inequality by substituting values of x gained the method mark if they identified 5 as the critical value.

  Text version

  Solve 7x−27<8.

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• Edexcel 1MA1 · June 2022 · Paper 1H · Q6 · 6 marks

  6 marks

  6

  
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  This question was well answered by the majority of students. The table in part (a) was often completed correctly. Most errors occurred with the substitution of x = –1 into the equation and y = 3 and y = –1 were common incorrect values of y. In part (b) the plotting of the points was usually accurate. Most students realised that a curve was needed to join the points and it was pleasing to see freehand curves drawn with an appropriate turning point. Some students, however, drew a graph with a flat bottom which resulted in the loss of a mark. Few graphs consisting solely of line segments were seen. These can get at most one mark. Curves that were sloppy and missed a point also gained at most one mark. Students should be encouraged to make sure that their curve passes through all of the points and doesn’t consist of more than one curve between any two points. Part (c) was answered quite well although some students did not know how to use their graph to find estimates for the solutions of the equation. A common error was giving the solutions as coordinates rather than as values. Some students gave the coordinates of the turning point and others gave the values when y = 1. Centres should encourage students to mark the intercepts with the x-axis to show where they are attempting to read off values, few students did this. Some students attempted to answer this part of the question by trying to factorise the quadratic rather than reading from the graph as the question asked. Inevitably their factorising was incorrect and they received no marks.

  Text version

  For y=x²−3x+1, use x=−1,0,1,2,3,4.

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• Edexcel 1MA1 · June 2022 · Paper 1H · Q14 · 4 marks

  4 marks

  14

  
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  Where tangents were drawn they were generally accurate and students often went on to gain all three marks for working out an estimate of the gradient that was within the acceptable range. Some answers were given in the form a/b and the final mark was lost when a and b were not integers. Errors were often the result of reading the scales incorrectly although some students divided the change in x by the change in y. A significant number of students drew a tangent, thus gaining the first method mark, but could not then complete the method to find the gradient. Many students used the coordinates of the point on the curve at t = 2 and simply divided 2.8 by 2. Some of these students had already drawn a tangent to the curve at t = 2. In part (b) many students knew that the area under the graph represents the distance travelled. Incorrect answers often referred to acceleration or to time. Some students gave the answer as velocity or speed.

  Text version

  A smooth speed-time curve rises from (0,0) through about (2,2.8), peaks near (4.5,4), and falls to (9,0).

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• Edexcel 1MA1 · June 2022 · Paper 1H · Q16 · 4 marks

  4 marks

  16

  
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  Overall, this question was answered very poorly with a large number of students unable to find an appropriate strategy to work out the probability of passing the practical test. Probability tree diagrams were very common but these were of little use to many students as the probability of passing only one of the two parts, 0.36, was often placed incorrectly on branches of the tree diagrams. Some students wrote down a relevant product, often this was 0.75 × x, and gained the first mark but many were not able to go on and make any further progress. Statements such as 0.75 × x = 0.36 were common because many students did not recognise that there are two ways of passing only one of the two parts. Some of the students who did consider the two ways wrote 0.75 × x + 0.25 × x = 0.36 and gained no more marks. It was also common to see 0.75 × x + 0.25 × y = 0.36 which gained the second mark but this did not necessarily get translated into an equation in one variable and so no further progress was made. Many of those that did show a correct equation such as 0.75(1 – x) + 0.25x = 0.36 were able to complete the process and give a correct final answer. A small number of students who got as far as forming and solving a correct equation in one variable were confused about what their variable represented and lost the accuracy mark.

  Text version

  A first-aid test has independent theory and practical parts. P(pass theory)=0.75 and P(pass exactly one part)=0.36.

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• Edexcel 1MA1 · June 2022 · Paper 1H · Q19 · 4 marks

  4 marks

  19

  
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  Relatively few students could demonstrate the necessary skills of algebraic manipulation to solve the equation and give the answer in the required form. Many students were able to correctly write the two fractions with a common denominator and gain the first mark but a significant number were then unable to carry the algebraic solution any further. Those that did reduce the equation to a 3 term quadratic often made errors when rearranging and did not get the second mark. Substitution into the quadratic equation formula was generally done well but some students attempted to use completing the square and this was done less well due to not dealing with the coefficient of 2 correctly. Some students who did not get a correct quadratic equation were still able to gain the third mark for dealing correctly with their 3 term quadratic equation. The final step to write 10  in the required form proved difficult with  a common incorrect answer.

  Text version

  Solve 1/(2x−1) + 3/(x−1) = 1. Give answers in form (p±√q)/2.

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• Edexcel 1MA1 · June 2022 · Paper 1H · Q20 · 4 marks

  4 marks

  20

  
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  It was pleasing to see many students attempting to draw diagrams to help them structure their response. Those who did often then started by finding the gradient of the line from the centre of the circle to the point A and gained the first mark. A good number were then able to find the gradient of the tangent and use y = mx + c to find an equation of the tangent. Those who got this far and gained the first three marks often failed to gain the accuracy mark. Sometimes this was due to an error when finding the value of c but more commonly it was because the final equation was not given in the form ax + by + c = 0 that was required by the question. A common error at the first stage was for students to attempt to work out a gradient without using both (–1, 3) and (6, 8) and give gradients such 6 or 3 . These students gained no marks as did students who changed the circle centre to the origin. A minority of students found an incorrect gradient from using change in x divided by change in y but they could still get two marks if their final equation was 5x + 7y – 86 = 0.

  Text version

  Circle centre (−1,3) passes through A=(6,8).

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• Edexcel 1MA1 · June 2022 · Paper 2F · Q8 · 4 marks

  4 marks

  8

  
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  As to be expected part (a) was well answered with most students able to give the correct coordinates for the point shown. The second part of this question was also very well answered with the majority of students gaining the mark for demonstrating the ability to plot the pair of coordinates accurately. The most common error seen was to transpose the values and plot the point as (3, −4) or use the incorrect positive or negative version of one or both coordinates. Students usually marked the point clearly with a cross but many failed to label the point. Students are reminded to read the instructions given in questions carefully. In part (c) a wide variety of answers were seen with varying levels of success; some students did not attempt to draw a circle. A good number of students did not have a pair of compasses and drew a freehand circle. Many students had the centre of the circle as (-1,1) and this gained one mark providing the radius was still 4 cm. Several students drew circles at the correct centre but with an incorrect radius, this gained one of the two marks available. Centres could significantly improve the success rate of this question by ensuring students are both correctly equipped and have had opportunity to practice using all necessary equipment as part of their exam preparation.

  Text version

  A coordinate grid is shown with one-unit spacing. Point A is plotted three squares to the right of the origin and two squares above the origin.

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• Edexcel 1MA1 · June 2022 · Paper 2F · Q12 · 3 marks

  3 marks

  12

  
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  Students were able to use the number machine in part (a) and the vast majority of students gave the correct answer and scored the mark available. Part (b) was generally well answered with most students achieving full marks. An answer of 6 was often given with no further working shown. Of the successful answers with working, by far the most common method seen was the use of the inverse operation working backwards from 154. Those unable to give the final answer often gained a mark for partially correct working out, usually 154÷11 =14. Some students did try to work with an unknown and 88 or 154 – 88 was often seen but not many using this method seem to complete the process and give an answer of 6. Those students who got as far as 154 – 88 or stating 66 were able to gain one mark. The most common incorrect answer seen was 13.27 from subtracting 8 from 154 first and then dividing by 11.

  Text version

  Number machines.

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• Edexcel 1MA1 · June 2022 · Paper 2F · Q17 · 3 marks

  3 marks

  17

  
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  This is a commonly asked question but on this occasion the topic was not well answered. The lack of a table to fill in seemed to confuse some students. Centres should encourage students to set up their own table of values if one is not given. Those students that did this were the most successful in scoring full marks. Even though this graph had a negative gradient many students tried to draw a line with a positive gradient going up in ones instead of down. It was also disappointing to see many students had plotted the correct points but failed to join them up with a line, hence losing the accuracy mark. Another common error seen was to have the correct graph for the positive values of x but the incorrect values for the negative values of x. It would help students to recognise the format for the equation of a straight line and know that their graph should be both straight and have all points joined.

  Text version

  Draw y=4−x for −2≤x≤4.

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• Edexcel 1MA1 · June 2022 · Paper 2F · Q21 · 5 marks

  5 marks

  21

  
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  Part (a) of this question was not well answered with students displaying many misconceptions concerning powers. Many students added the powers instead of multiplying them or simply wrote 15x as the answer. Other common incorrect responses were 15, x -2 and 243. Part (b) was well answered by students, with the vast majority scoring at least one mark and many scoring both the available marks. Those who were unable to reach the correct simplified answer of 40 ̶ 10x were usually able to multiply the first bracket to obtain 4x + 12. The order of the terms in the second bracket seemed to confuse some students and the negative 14x caused confusion with students not confident with negative numbers failing to simplify this to ̶ 10x. Students scored better when they fully expanded both brackets and then collected like terms together. An incorrect answer of 40 + 10x was common to see. Students would benefit from more exposure to questions with negative terms and questions where the x term is in different positions within the brackets given. Students found the third part of this question challenging. With many displaying little understanding of extracting a common factor and just giving an answer with just one term. Some students did gain 1 mark for a partial factorisation often extracting 3 or 3x. A common error seen was to assume that two brackets were needed and this approach obviously led to incorrect answers. Students would benefit from understanding the meaning of ‘factorise fully’ as opposed to just extracting a single factor and also from being taught to look at their factorised expression and ensure it had no further common factors and that it multiples out to the original expression.

  Text version

  Algebra.

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• Edexcel 1MA1 · June 2022 · Paper 2F · Q28 · 3 marks

  3 marks

  28

  
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  Student’s responses to this last question were very mixed, some not attempting it at all. However, a good number of students gained full marks and others two marks. The last mark was often lost where the students failed to solve an equation correctly giving the value of y as 2.75. It was pleasing to note that many students used the correct method but were, unfortunately, let down by careless arithmetic. Summary Based on their performance on this paper, students should: • read carefully questions • practice questions involving negative numbers and using their calculator to work with these • give succinct explanations when a written answer is required • use a calculator rather than relying on paper methods, particularly when working with percentages

  Text version

  Solve 5x+2y=27 and 6x+4y=28.

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• Edexcel 1MA1 · June 2022 · Paper 2H · Q1 · 5 marks

  5 marks

  1

  
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  The first question on the paper assessed some basic algebra skills. Part (a) assessed the third law of indices and was generally answered correctly. The most common wrong answer being to add the index numbers rather than multiply. In part (b) students had to expand a pair of single brackets and simplify. Most students were able to gain at least one mark, for expanding one bracket correctly. Errors tended to come in the collection of terms, and in particular the x terms, often leading to 18x. Part (c) was a factorisation question looking for the student to bring two terms out of the expression. A good proportion of the students scored at least one mark for a suitable partial factorisation of the expression, commonly taking out the factor of 3x rather than 3x2 . There was however, a pleasing proportion of students that gained both marks for the correct factorisation.

  Text version

  Algebra.

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• Edexcel 1MA1 · June 2022 · Paper 2H · Q12 · 2 marks

  2 marks

  12

  
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  This question was answered better than it has been on previous papers, with many students understanding the need to rearrange one of the equations, and in most cases being successful to some extent. The first mark could be gained by correctly moving one of the variables in either equation, or by correctly scaling one of the equations by multiplying or dividing by 5. To gain both marks the equations had to be manipulated enough for a comparison of coefficients to be possible, and for a comparison to be drawn – this was generally done well with the most common incorrect comparison being to state that the gradient was 10 when both equations were in the form 5y = 10x ...

  Text version

  L1: y=2x+3. L2: 5y−10x+4=0.

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• Edexcel 1MA1 · June 2022 · Paper 2H · Q15 · 4 marks

  4 marks

  15

  
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  Many students struggled to start this problem. There were 2 main methods to solving this problem. The first method was to use the 2 ratios to form an equation, which could then be solved. This is the method that most tried to attempt but struggled to combine the different parts. The second more efficient method was to compare 12 : 10 with 13 : 11 and realise this means the £1.50 is worth one part of this ratio. Students looking to earn the higher grades need to practice more with using equating ratios to form equations.

  Text version

  Marta:Khalid hourly pay=6:5. Both rise £1.50; new ratio=13:11.

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• Edexcel 1MA1 · June 2022 · Paper 2H · Q17 · 5 marks

  5 marks

  17

  
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  This volume problem required students to work with a hemisphere and a cone. The first process mark was for a correct method to find the volume of the hemisphere or for a correct expression for the volume of the cone. Many students were unable to get past this point for 2 reasons. Firstly, they didn’t halve the volume of the sphere to find the volume of the hemisphere, or they used ‘y’ as the height of the cone, which was unacceptable as ‘y’ was the height of the overall solid. The second mark was for linking the 3 elements of the problem correctly. This was sometimes seen as an equation, but most commonly seen in multiple steps, normally subtracting the volume of the hemisphere from the total volume, and then equating to the volume. The third mark was for correct manipulation to find the height of the cone and was typically seen again in multiple steps of working. Those who correctly found the height of the cone scored 3 marks. It was quite common to see students use 120 instead of 120π, but this was dealt with as a misread as per the general marking guidance. Common incorrect methods worth no marks were to use 7 or 4.5 (without showing where 4.5 had come from) as the radius. Part (b) required students to understand the inverse proportional relationship between radius and height when the volume is constant. Many were able to do so, and thus gained this mark.

  Text version

  Solid T is a cone joined base-to-base to a hemisphere. Both diameter 7 cm. Total volume 120π cm³; total height is y cm.

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• Edexcel 1MA1 · June 2022 · Paper 2H · Q19 · 4 marks

  4 marks

  19

  
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  Part (a) was a simple substitution mark which many were able to score. Some students were put off by the function notation and didn’t attempt the question. In part (b) students were required to understand function notation clearly and realise in this case they needed to find the inverse of g(x) before then substituting that into h(x). This question posed a number of challenges. For many the notation alone was the stumbling block and they tried to put g(x) into h(x) and then find the inverse. Others struggled to find the required inverse function. However, these students were able to gain the second mark for correctly substituting their inverse function into h(x). A few of those who gained method marks, were then unable to gain the accuracy mark as they left it as a fraction within a fraction, rather than simplifying fully.

  Text version

  g(x)=(2x−5)/3 and h(x)=1/x.

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• Edexcel 1MA1 · June 2022 · Paper 2H · Q21 · 3 marks

  3 marks

  21

  
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  Part (a) of this question required students to understand the effect of a reflection on a function and then reflect the graph in the correct axis. As is expected by this stage of the paper, this was very challenging for a great number of students. Some clearly did understand a reflection was required but didn’t complete it with sufficient accuracy to gain the mark. A free-hand curve was acceptable, but students need to ensure key points are plotted correctly. Part (b) was more in-depth as it required a thorough understanding of the graph of tanx and its key intercepts. It also required students to understand the effects of a translation on a function. Those who identified a horizontal movement of 270 were able to score one mark, as were those who applied the correct vertical translation to the function. Very few students were able to score both marks.

  Text version

  Graph f is a smooth curve through (−5,4),(−3,0),(−1,1),(0,0),(2,−4). A tan x° graph has point Q=(360,0), which moves to R=(90,−5) under a translation.

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• Edexcel 1MA1 · June 2022 · Paper 2H · Q22 · 5 marks

  5 marks

  22

  
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  This demanding question required students to solve two quadratic inequalities and then select the correct critical values to satisfy both. The mark scheme has 2 sets of 2 marks, a method and an accuracy, for the solving of each quadratic inequality. Students could gain the accuracy mark for both by drawing a suitable diagram, either inequalities on a number line, or a sketch of the quadratic with the correct regions drawn. However, to score the final mark students needed to state both linear inequalities that satisfy both quadratics inequalities. A large number of students were able to gain one mark for a method to solve the first quadratic, normally by factorising, but typically didn’t correctly state the solutions. The second quadratic proved much more challenging. Many were unable to factorise, and thus chose to use the formula. The problem with this was their substitution was often inaccurate, either with incorrect values or with missing brackets. Summary Based on their performance on this paper, students: • should improve their use of calculators in general, but in particular when completing complex calculations involving powers, or fractions. • learn correct geometrical terms and use these when giving reasons in geometry questions. • work on more efficient methods when working with percentages when a calculator is available, and refrain from attempting to use ‘non-calculator’ methods. • ensure they do not round prematurely in multistep calculations, as it often leads to answers outside of the acceptable range for the accuracy mark. • check answers make sense in the context of the question.

  Text version

  Find algebraically x satisfying both x²−49>0 and 5x²−31x−72>0.

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• Edexcel 1MA1 · June 2022 · Paper 3F · Q4 · 1 mark

  1 mark

  4

  
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  This question was well answered with most giving the correct answer of 6m. Common incorrect answers were 5m or 2m3.

  Text version

  Simplify 2m×3.

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• Edexcel 1MA1 · June 2022 · Paper 3F · Q8 · 2 marks

  2 marks

  8

  
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  Most found the two terms 28 and 33 in part (a). In part (b) most students could work out the next two terms but justifying why 50 wasn't in the sequence was not done as well; a significant number of students left this section blank. The most common and efficient answer was to state that all the terms end in a 3 or an 8. Students who chose another method were not as consistent with achieving the mark. A number of students would state the 48 or 53, but not the two together and therefore not fully proving why 50 was not in the sequence. Some stated the nth term as 5n − 2 but this needed to be supplemented by a further statement linked to the question. Students should be encouraged to always write out the sequence if it is possible; those who did this secured a sound answer for both parts on this occasion.

  Text version

  Sequence: 3,8,13,18,23.

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• Edexcel 1MA1 · June 2022 · Paper 3F · Q17 · 4 marks

  4 marks

  17

  
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  Part (a) was generally well answered. Some students only partially processed the answer leaving it as 12 + − 6x . Others went on from the correct 12 – 6x to write 6x on the answer line, this scored no marks since it is the answer on the answer line which must be accepted. In part (b) the vast majority of students made some attempt. For those students who were successful their answers mainly came from an informal method of inverse operations, usually showing 12 × 4 ÷ 3= 16 with a few using the formal approach of 3y = 48. Another alternative method was to see 0.75y = 12 as the first step. Students scoring one mark mainly got this for 3y = 48 or an embedded answer of 16, usually shown as 3 × 16 ÷ 4=12; unfortunately, many then wrote the 12 on the answer line rather than the 16. Students who knew to multiply by 4 as the first step often failed to apply this to both sides of the equation and only stated the 48, which resulted in no marks since they wrote 4 × 12 = 48 rather than 3y = 48. Students scoring zero were usually unsure of how to approach the first step and would try to add or subtract 4 rather than multiplying. In part (c) students with understanding of factorisation in algebra did well. The most popular incorrect answer was 10p where students simply added 4p and 6 together without appreciating that these were unlike terms.

  Text version

  Algebra.

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• Edexcel 1MA1 · June 2022 · Paper 3F · Q23 · 4 marks

  4 marks

  23

  
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  In part (a) most students failed to understand the algebraic meaning of 4m2. Fortunately, they were usually able to gain a mark for substituting −3 in the expression. Common errors for dealing with 4× (−32) included: 4 × −3 × −3 = −36 −11 with the answer of −47 being most popular. Also common was 4 × 3 = 122 = 144 which could not be awarded any marks. A successful answer of 25 was usually as a result of using brackets around −3. Part (b) was poorly answered by most students. Many chose to try to simplify the expression, ending up with 7p. Some felt they should simply swap the d with the p to get p = 3d + 4. Of those who realised it required inverse operations to rearrange, most were unable to perform these in the correct order. d p − = and d p = − were common, or an answer written ambiguously such as p = d – 4 ÷ 3. Most frequently students obtained no marks on this question due to the error in their first step. Function machines were used by a small number of students but even when these were used correctly, the students could not then write the equation with p as its subject.

  Text version

  T=4m²−11 and d=3p+4.

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• Edexcel 1MA1 · June 2022 · Paper 3F · Q24 · 5 marks

  5 marks

  24

  
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  In this question many students were able to find the number of counters each person would get. There were two main routes to these values: algebraic and numeric. Weaker students started by dividing 54 by 3 but were then unsure of where to go from there. Students who adopted a clear algebraic approach often went on to solve the question completely. However, students occasionally ended up with 4x – 6 = 54 rather than 5x – 6 = 54, due to not understanding that Tony had six less counters than Selma rather than Rick. Most students adopted a numeric approach, and most were successful in arriving at the three numbers 12, 24 and 18 scoring three marks. Students were less successful in converting the counters each had into a ratio between Rick and Tony. Where students managed this, they usually then went on to gain the full marks for p = 1.5 but those who did not recognise this unitary ratio expression were sometimes put off from writing their ratio of Rick to Tony.

  Text version

  Rick has r counters, Selma twice as many, Tony 6 fewer than Selma; total 54. Rick:Tony=1:p.

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• Edexcel 1MA1 · June 2022 · Paper 3F · Q28 · 3 marks

  3 marks

  28

  
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  Most students left this question blank. Almost all of the successful students split QRVU into a rectangle and a triangle but complete methods were rare. A few were able to gain a mark for the area of the rectangle or triangle. Students showed a lack of understanding in cases where attempts were made by trying to find the area of the trapezium by using numerical values for the sides of the rectangle, trying to write an expression by adding the perimeter or parts of the perimeter or trying to rearrange A= 2x2 +20x. The majority of the marks that were given were for 4x × 5 = 20x.

  Text version

  Rectangle STUV has height TS=4x and width SV=3x+5. On top and bottom, TQ=2x and SR=3x, so QUVR is a trapezium.

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• Edexcel 1MA1 · June 2022 · Paper 3H · Q2 · 4 marks

  4 marks

  2

  
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  There was a disappointing proportion of fully correct answers to part (a) of this question. It has to be said that there was a great deal of sloppiness in students’ notation when making the substitution and far too many students failed to write the −3 in brackets. They very often wrote 4 × −3² − 11, with most of these students then going on to use their calculators with little understanding of how to deal with the square of –3. This resulted in many answers of −47 instead of the correct 25. Students are advised to put brackets round any negative numbers when working with algebraic expressions involving substitution both when writing down their working and when inputting any expression into their calculator. Part (b) of the question was done quite well. Students usually changed the subject by subtracting 4 from both sides first and then dividing by 3. Only a small minority of students divided by 3 as their first step. Students who did do this often forgot to divide the 4 by 3. Students are reminded that they should give a complete answer on the answer line, including the “p =”. A number of students just wrote “ 𝑑−4 3 ” on the answer line.

  Text version

  T=4m²−11 and d=3p+4.

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• Edexcel 1MA1 · June 2022 · Paper 3H · Q3 · 5 marks

  5 marks

  3

  
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  A majority of students scored full marks on this question. Many students in this higher tier paper used an algebraic approach, formulating and solving an equation, then using the solution to find a ratio and the value of p. Students usually worked accurately, but there were a number of students who wrote down a correct equation but collected terms in x to get 4x instead of the correct 5x. Other students, having got the correct equation, 5x − 6 = 54, then made an error in the first step to solve it, writing 5x = 48. A significant proportion of students adopted a purely numerical approach, finding, by trial and improvement, three numbers which satisfied the criteria given in the question, then using them to form a ratio. Some students found the number of counters for Rick and for Tony but did not form a ratio. There were also some other students who did form a ratio but did not write it in the ratio 1 : p.

  Text version

  Rick has r counters, Selma twice as many, Tony 6 fewer than Selma; total 54. Rick:Tony=1:p.

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• Edexcel 1MA1 · June 2022 · Paper 3H · Q7 · 3 marks

  3 marks

  7

  
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  This question was a good discrminator. There were a number of routes which could be followed in order to show the given result. The most popular method was to use the formula for the area of the trapezium QUVR. Once a student had found the length of QU (= x + 5), they could make substitutions into the formula then simplify it to show A = 2𝑥2 + 20x. Many students found this to be quite straightforward. A second commonly seen and generally successful method was to use Area of trapezium QUVR = Area of rectangle TUVS − Area of trapezium TQRS. Perhaps the most straightforward way was to find the area of the rectangle with length and width 4x and 5 then add the area of a triangle of base x and height 4x. It was surprising how few students took this approach. The presentation of students answers was often not very clear with examiners having to search the working space for relevant working. However, many students did use algebra in a confident way to show the given result. The absence of brackets to make working clear sometimes led to a student losing marks but examiners were sympathetic where intentions were clear.

  Text version

  Rectangle STUV has height TS=4x and width SV=3x+5. On top and bottom, TQ=2x and SR=3x, so QUVR is a trapezium.

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• Edexcel 1MA1 · June 2022 · Paper 3H · Q8 · 3 marks

  3 marks

  8

  
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  Many students found this question to be straightforward. They often used a triangle where readings were easy to take and gave a value for the gradient of the graph within the accepted range, more often than not 0.14. For some other students who drew a small triangle on the line getting an accurate answer was less likely. However, if the value given for the gradient was outside the range 0.135 to 0.145, examiners awarded both marks in part (a) provided a correct method was clearly shown. A small number of students used “increase in x ÷ increase in y” or counted squares rather than using the scales on the axes. These students could not be credited with any marks in part (a). Part (b) was answered less well and many incorrect or ambiguous statements were seen. An answer equivalent to “cost per unit of electricity” was expected but many students merely described what the graph, not the gradient, showed and statements such as “cost of how many units used” were often seen.

  Text version

  A straight-line electricity cost graph passes through (0 units, £0) and (100 units, £14).

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• Edexcel 1MA1 · June 2022 · Paper 3H · Q14 · 5 marks

  5 marks

  14

  
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  A minority of students scored full marks for their responses to this question. However, most students were able to gain some marks and the question was a good discriminator between students who could manipulate algebraic expressions with accuracy and those who made, sometimes unnecessary, errors. A fully factorised expression was asked for in answering part (a) of this question. It was disappointing to see so few students factorise the expression fully and it was much more usual for students to gain just one of the two marks available for getting as far as, for example, 4(p2 − 9) or (2p − 6)(2p + 6). Students usually scored well in part (b) which required the expansion of a product of three linear expressions to give a fully simplified cubic expression. Students generally showed their working in an organized way or used a grid to show terms in their products. Errors were usually restricted to incorrect terms or difficulties in dealing with the signs when collecting terms together, rather than a flawed strategy although some students omitted terms from their expansion. For students who did not give a fully correct answer, it was commonplace for them to earn 2 or 3 of the available marks in part (b).

  Text version

  Algebra.

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• Edexcel 1MA1 · June 2022 · Paper 3H · Q19 · 4 marks

  4 marks

  19

  
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  Fully correct answers to this question were seen infrequently. However, a significant proportion of students taking this paper were able to score at least one mark for using a correct common denominator and writing at least one of the three terms with that common denominator, usually the first fraction as 3𝑥(𝑥−2) 𝑥2−4 . Only the best students were able to work accurately in order to gain more credit for their responses. The question discriminated well between high ability students aiming to get one of the two top grades. For many students, this question proved to be the most challenging question on the paper. In particular mistakes were often made with signs when the fraction (𝑥+2)(2𝑥+1) 𝑥2−4 was subtracted from the fraction 3𝑥(𝑥−2) 𝑥2−4 or in writing −(x2 – 4) as −x2 − 4.

  Text version

  Show 3x/(x+2) − (2x+1)/(x−2) − 1 has form (ax+b)/(x²−4).

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• Edexcel 1MA1 · June 2022 · Paper 3H · Q20 · 4 marks

  4 marks

  20

  
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  Though a significant number of students did not attempt this question, many of those students who did attempt it scored full marks for a complete and correct solution. The most common error seen was for students to write down a correct equation, 29 600 = 24 000a + 800 but then make an error in solving the equation. It was not unusual for students to divide 29600 by 24800 as a first step, then subtract 800. Some students who found a correct value for “a” went on to use the iterative relationship correctly but only went as far as finding the profit made by the shop in the year 2020.

  Text version

  Pn+1=aPn+800. Profit was £24,000 in 2018 and £29,600 in 2019.

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• Edexcel 1MA1 · June 2022 · Paper 3H · Q22 · 5 marks

  5 marks

  22

  
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  This question was accessible to and discriminated well between the most high attaining students sitting this paper. These students often gave a complete and fully correct solution and scored full marks. They usually used the approach of eliminating y to get an equation in x. This was more straightforward to deal with than the corresponding equation in y. The most common errors made by students included expanding (2x − 5)2 incorrectly, often as 4x2 + 25, and making errors when re-arranging their quadratic equation to get all terms on one side of the equation. Some students who did get the correct quadratic equation to solve, made errors when trying to factorise and so restricted themselves to the award of 2 marks. Summary Based on their performance on this paper, students are offered the following advice: • make sure you put negative numbers in brackets when substituting them into algebraic expressions or when using a calculator. • check all your working for careless errors and to make sure you have not miscopied numbers or algebraic expressions either given in the question or in a previous line of your working. • practise the addition and subtraction of algebraic fractions. • check your working when expanding brackets and collecting terms, especially when finding the product of three linear expressions or when removing brackets after a subtraction of one expression from another. • ensure you have a good understanding of how to use multipliers in problems concerning reverse percentages. • practise drawing histograms and using them to solve problems.

  Text version

  Line L: y=2x−5. Curve C: y²=6x²−25x−8.

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• Edexcel 1MA1 · November 2021 · Paper 1F · Q13 · 3 marks

  3 marks

  13

  
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  In part (a), the great majority of students recognised and were able to explain why 7 could be the next term of the given sequence. The usual incorrect response was from those who insisted that Emma must be wrong since the pattern was one of doubling resulting in 8 as the next term. Many students did not explain the rule well or gave an ambiguous explanation. Those that annotated the original sequence made their explanation clearer. Another common incorrect response was 1+2+4=7, using the sum of all previous terms. In part (b) most students usually gained at least one mark for a 5th term of 15. Many correctly completed subsequent terms of the sequence but then chose 45 as the 8th term, applying eight differences by mistake.

  Text version

  A number sequence starts 1, 2, 4. Emma says that the next term is 7. The first four triangle numbers are 1, 3, 6, 10.

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• Edexcel 1MA1 · November 2021 · Paper 1F · Q15 · 4 marks

  4 marks

  15

  
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  The first two parts of this question were not answered well. 2ad or 2a + d were common incorrect answers in part (a), whilst in part (b) very few demonstrated any understanding of factorisation. y and y2 were often incorrect answers offered here. Students had greater success with part (c), gaining full marks for an answer of 11. Many students showed the value of x as 11 in their working but then gave an answer of 44 on the answer line. Very few students used algebra explicitly in the question and many took the wrong first step by subtracting instead of adding 7.

  Text version

  Complete three algebra tasks.

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• Edexcel 1MA1 · November 2021 · Paper 1F · Q16 · 4 marks

  4 marks

  16

  
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  In part (a) many argued that Jasper was not correct because x could not be a decimal. The few who did substitute x = 0.5 into the given expression usually got the right idea and correctly explained the error. Only a very small minority gained any credit in part (b). Use of algebra was very nearly non-existent with students instead preferring to attempt numerical methods which were usually doomed to failure. The most common incorrect approach was to assume all 4 sides of the kite were equal and divide 64 by 4. A few students gained a mark for writing 3(4x – 2) or 12x – 6, some then went on to incorrectly equate that to 64. Some thought they were dealing with angles and introduced “360 degrees” into their working. Even when students correctly calculated AD, they were then unable to link this to an equation or expression for perimeter. Lots of incorrect responses featured students simply setting 4x – 2 = 64 and then attempting to solve from there; this gained no credit.

  Text version

  ABCD is a kite. AB = (4x - 2) cm. Jasper says that x could be 0.5. Also AD = 3AB and the kite has perimeter 64 cm.

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• Edexcel 1MA1 · November 2021 · Paper 1F · Q18 · 3 marks

  3 marks

  18

  
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  Tables of values were seen but very often with very many errors in them, especially for the calculations involving negative values of x. Often correct values shown in a table were contradicted by incorrect points plotted on the grid, thus negating any award. It was still not uncommon to see the correct points plotted, but not joined with a straight line. However, there were some indications that stronger students managed this by using the gradient and y- intercept rather than by drawing a table of values.

  Text version

  For -2 <= x <= 3, consider y = 2x - 2 on a coordinate grid.

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• Edexcel 1MA1 · November 2021 · Paper 1F · Q26 · 4 marks

  4 marks

  26

  
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  This question was not well answered. Some students gained a mark, usually for substituting 3 into the given formula for the surface area of the sphere. Some gave a numerical value for pi and attempted to find the surface area of the sphere, with varied success.

  Text version

  A cube has edges of length x cm. A sphere has radius 3 cm. The surface area of the cube equals the surface area of the sphere. You are given that the surface area of a sphere is 4πr^2.

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• Edexcel 1MA1 · November 2021 · Paper 1F · Q28 · 2 marks

  2 marks

  28

  
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  In this cohort, very few students showed any understanding of y = mx + c Closest attempts were 3 or 4 in part (a) and (3, 4) or (3, -4) in part (b). Summary Based on their performance on this paper, students should:  show clear, ordered working for all questions  show full working if using a build up method for finding a percentage of an amount  practice long multiplication and long division of decimal values  ensure that they read the question carefully and so provide the answer required by the question

  Text version

  The equation of a straight line L is y = 3 - 4x.

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• Edexcel 1MA1 · November 2021 · Paper 1H · Q7 · 4 marks

  4 marks

  7

  
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  Examiner report

  This question was not answered particularly well. Many students gained one mark for using the given formula to write a correct expression for the surface area of the sphere, 4 × π × 32 for example, but not many of these students attempted to find the surface area of the cube. It was very common to see 4 × π × 32 followed by x  = . Those students who did find an expression for the surface area of the cube generally went on to form a suitable equation and complete the method to gain full marks. Some students found an expression for the area of one face of the cube rather than for the surface area, often writing x  = , and gained the first two marks only.

  Text version

  A cube has edge x cm. A sphere has radius 3 cm. Their surface areas are equal. Sphere surface area = 4πr^2.

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• Edexcel 1MA1 · November 2021 · Paper 1H · Q8 · 3 marks

  3 marks

  8

  
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  Students who rearranged the given equation to get x2 – 5x – 24 = 0 often went on to give a fully correct answer. Some, however, lost marks because they were unable to factorise correctly or they factorised correctly but then wrote (x – 8)(x + 3) as the final answer. A surprising number of students did not realise that a sensible first step would be to rearrange the equation and attempted to factorise straight away. Those who gave a factorisation of the form (x ± 8)(x ± 3) gained the second method mark.

  Text version

  Solve x^2 = 5x + 24.

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• Edexcel 1MA1 · November 2021 · Paper 1H · Q11 · 4 marks

  4 marks

  11

  
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  This question was not well answer with relatively few students managing to gain more than the first two marks. Writing an expression for the area of each rectangle and forming an equation ought to have been a straightforward process for the first mark, particularly as missing brackets were condoned. Some students wrote two expressions and failed to form an equation but many students were unable to write two correct expressions. A significant number of students worked with perimeter rather than with area. Having gained the first mark, many students were not able to make any further correct progress. Those without brackets in their equation usually failed to expand correctly. Some with brackets made errors expanding and some expanded correctly but then simplified incorrectly. Many students did not know how to deal with the fact that y appeared on both sides of the equation. Those who isolated the terms in y often went on to factorise correctly and were able to find a correct expression for y in terms of w.

  Text version

  Rectangle A has sides 6 and (2w+y) cm. Rectangle B has sides 7w and (3y+6) cm. Their areas are equal.

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• Edexcel 1MA1 · November 2021 · Paper 1H · Q14 · 3 marks

  3 marks

  14

  
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  It was pleasing to see many fully correct answers with clear accurate algebra but there were also very many responses where algebraic weaknesses were apparent. Most of the students who indicated that they were going to find the area of the L-shape by splitting it up into two rectangles were able to score at least one mark for a correct expression for the area of one rectangle. A complete expression for the total area was needed for the second mark; most commonly this was given as 4(x + 1) + (x + 7) (2x + 6), with (x + 1)(x + 11) + (x + 7)(x + 5) and (x + 11)(2x + 6) – 4(x + 5) used less often. Incorrect expressions for the total area were often the result of working out missing lengths in the shape incorrectly. Many students could not handle the algebra accurately enough to show a complete chain of reasoning leading to A = 2x2 + 24x + 46. Omitting necessary brackets and expanding brackets incorrectly were common errors.

  Text version

  An L-shape is an outer rectangle of width (2x+6) cm and height (x+11) cm with a rectangular corner cutout of width (x+5) cm and height 4 cm. Equivalently, the remaining left leg is (x+1) cm wide.

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• Edexcel 1MA1 · November 2021 · Paper 1H · Q15 · 3 marks

  3 marks

  15

  
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  Most of the students who used 10x as a common denominator went on to score full marks. Exceptions to this were those who made errors in expanding brackets, for example 5(4x + 3) = 20x + 3, and those who wrote 5 as 10𝑥 . Some students wrote only one of the fractions with a suitable denominator, often this was ( 20𝑥+15 10𝑥) and gained the first mark only. Many students did not know how to tackle this question.

  Text version

  Show that (4x+3)/(2x) + 3/5 can be written as (ax+b)/(cx), where a,b,c are integers.

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• Edexcel 1MA1 · November 2021 · Paper 1H · Q17 · 3 marks

  3 marks

  17

  
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  Many students gained no marks because they were not able to draw two of the three lines correctly. When only two of the three lines were drawn correctly it was most commonly the lines x = 3 and y = 6 – 3x. The graph of 2y + 4 = x caused the most problems. A common error was to draw this line with the correct gradient but the wrong y-axis intercept. Some students confused the line x = 3 with the line y = 3. Even after gaining the first two marks for drawing all three lines correctly many students failed to give a fully correct solution because they identified the wrong region – often this was the region enclosed by the three lines. Students should check whether points in the region they choose satisfy all three inequalities.

  Text version

  On coordinate axes, show the region satisfying 2y+4<x, x<3 and y<6-3x.

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• Edexcel 1MA1 · November 2021 · Paper 1H · Q20 · 3 marks

  3 marks

  20

  
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  This question was poorly answered with relatively few students achieving any marks. For those who drew the graph of y – 2x = 1 it was common to see at least two marks awarded if not all three. When one of the accuracy marks was lost this was usually due to the values not being given as pairs or an error being made when reading from the graph. Not all attempts at drawing the graph of y – 2x = 1 were successful. Some students attempted an algebraic approach despite the question directing them to use the graph but these attempts were doomed to failure on this non-calculator paper and gained no marks.

  Text version

  A graph shows the circle x^2+y^2=30.25 on axes from -7 to 7.

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• Edexcel 1MA1 · November 2021 · Paper 1H · Q21 · 6 marks

  6 marks

  21

  
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  In part (a), the most popular approach to a correct answer was to first find f(1) = 4 and then work out g(4) although some students first found gf(x) and then substituted x = 1. A surprising number of students could not evaluate 3×12 + 1 correctly and lost the accuracy mark. Misunderstanding of composite functions was evident in many answers and a common incorrect method was to find both f(1) and g(1) and then multiply the two results or add one to the other. Part (b) was not well answered. Students with some knowledge of functions often gained the first mark for finding fg(x) but most then made little or no progress in finding the inverse function. Errors in algebraic manipulation frequently spoilt the attempts of those who did carry on and only a small number of students gave a fully correct answer.

  Text version

  For x>0, f(x)=3x^2+1 and g(x)=4/x^2. A function h is defined by h=(fg)^-1.

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• Edexcel 1MA1 · November 2021 · Paper 1H · Q22 · 4 marks

  4 marks

  22

  
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  Very few students showed a full understanding of what was required to find the coordinates of the turning point on the curve. Some of those who realised that they needed to use completing the square tried to do this as the first step and invariably failed. Some students divided 9 + 18x – 3x2 by –3 and worked with x2 – 6x – 3, often gaining one mark for starting the process to complete the square. Those who started by factorising often went on to give fully correct answers but sign errors and missing brackets prevented some students from achieving full marks. In Summary Based on their performance on this paper, students should: ● practise their arithmetic skills, particularly division and operations with mixed numbers ● consider whether or not an answer to a calculation is of a sensible size ● practise working out estimates by rounding numbers and develop an understanding of the purpose of rounding so that they can choose appropriate rounded values ● show their method when working out a percentage of an amount ● know and be able to use area formulae for triangles and trapezia ● practise answering questions involving algebraic fractions

  Text version

  Find the turning point of y=9+18x-3x^2. Show all working.

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• Edexcel 1MA1 · November 2021 · Paper 2F · Q4 · 2 marks

  2 marks

  4

  
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  In part (a) there was generally one of two answers given here, either 4m or m4. The incorrect answer was popular and only about two thirds of the cohort gave the correct answer. Part (b) was answered more accurately with over 80% of the cohort giving an acceptable answer.

  Text version

  Simplify two algebraic expressions.

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• Edexcel 1MA1 · November 2021 · Paper 2F · Q14 · 3 marks

  3 marks

  14

  
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  The vast majority of students answered both parts of this question correctly, demonstrating a good understanding of the skills being assessed. In part (a) many did gain the first mark by stating an answer within the range of 14 to 16. A common error seen was to read off the value of 75 for 5 hours but not make the adjustment to a payment for one hour or not being able to interpret the divisions on the scale correctly as £2 per square, with £10.60 being seen frequently. A very small number of students added together readings for 1, 2, 3, 4 and 5 hours. The most common and successful approach to part (b) was to use their hourly rate found in part (a) and multiply by 36. A build up method using one or more readings from the graph was less common and less successful, particularly as many using this method only found the pay for 35 hours by using the reading at 5 hours and not adding an extra hour of pay to show a complete build-up.

  Text version

  A straight-line pay graph passes through (0 hours, £0), (1, £15), (2, £30), (3, £45), (4, £60) and (5, £75).

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• Edexcel 1MA1 · November 2021 · Paper 2F · Q20 · 2 marks

  2 marks

  20

  
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  Examiner report

  Both parts of this question were similarly accessible to students. Approximately half of the cohort got each part correct. In part(b) the most common incorrect answer was d7

  Text version

  Simplify two index expressions.

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• Edexcel 1MA1 · November 2021 · Paper 2F · Q21 · 3 marks

  3 marks

  21

  
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  The first part of this question was beyond most students and many failed to use any reasonable inequality notation. In part (b) The most commonly awarded score was 1 mark for either a line of the correct length and position, a closed circle at −3 or an open circle at 4, which was seen less frequently. Some knowledge of inequalities was demonstrated but the accuracy mark was often lost due to not knowing how to translate the inequality symbols into open and closed circles correctly. Whilst it was common to see a circle shown at −3, sometimes correctly closed, the majority of incorrect answers terminated the line with an arrowhead at the number 4.

  Text version

  A number line has an open circle at -1 and is shaded to the right.

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• Edexcel 1MA1 · November 2021 · Paper 2F · Q23 · 5 marks

  5 marks

  23

  
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  Part (a) of this question was not often correctly answered. Many students were able to correctly identify that they needed to divide distance by time to arrive at the answer but very few went on to convert minutes into hours. Common incorrect responses were often 1.3 or 20. Those students that obtained the correct answer of 80 often did so by scaling ie multiplying 20 by 4. It would be beneficial for centres to ensure students are confident answering questions involving compound measures. Part (b) was often awarded part marks with only a few students scoring all the marks for a fully correct graph. Those who scored one mark often did so by drawing the correct horizontal line for the stop in Sam’s journey. A few students scored two marks for drawing a line representing the correct distance travelled in the last 20 minutes but forgot to show the stopping time on the graph. Rarely was 25 seen as part of calculations and a common error seen was for students to continue their line to (1050,50).

  Text version

  A travel graph shows Sam travelling 20 km from 10:00 to 10:15. At 10:15 he stops for 10 minutes, then drives for 20 minutes at 75 km/h.

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• Edexcel 1MA1 · November 2021 · Paper 2F · Q24 · 6 marks

  6 marks

  24

  
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  This question was not well answered. Many students continue to find working with negative numbers difficult and hence 5 was rarely seen in the correct place in the table, 3 was the most popular incorrect answer. Students were usually able to plot their points but far too many did not join them with any form of a curve. Because few quadratic graphs were drawn the answer to part (c) was often blank. This question is at the higher levels required on the foundation paper and so its lack of success is not surprising. The correct answers seen were from drawing a line at y = 4 but even with this line most students then just read off the positive answer and so still only gained one mark.

  Text version

  For y = x^2 - 2x + 2, use x values -2, -1, 0, 1, 2, 3 and 4.

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• Edexcel 1MA1 · November 2021 · Paper 2F · Q27 · 3 marks

  3 marks

  27

  
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  Only a few students recognised that factorising was the most suitable method to solve the quadratic equation. Of the students who did manage to factorise some were unable go on to solve the equation and identify the two correct solutions. Some students found the solution of 9 by a substitution method, so gaining one mark for 𝑥𝑥= 9 Many tried to ‘solve’ the equation by rearranging to isolate x by adding 18 or 7𝑥𝑥 to both sides, clearly this led to no marks being awarded.

  Text version

  Solve x^2 - 7x - 18 = 0.

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• Edexcel 1MA1 · November 2021 · Paper 2H · Q1 · 3 marks

  3 marks

  1

  
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  Examiner report

  Part (a) was answered very well. Almost every candidate was able to extract the correct critical value, and most provided the correct inequality symbol too. It was not uncommon to see −1 < x as an acceptable alternative answer. Those who did drop marks were fairly equally spread between choosing < over > or for including an inclusive inequality. Part (b) was again very well answered. Almost all students scored at least one mark, either for a line of correct length, or for a closed circle at −3, with a large proportion gaining the second mark also. Those who failed to score typically had their circles the wrong way round, or the line stopping at 3 rather than 4.

  Text version

  A number line has an open circle at -1 and is shaded right.

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• Edexcel 1MA1 · November 2021 · Paper 2H · Q3 · 5 marks

  5 marks

  3

  
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  In both parts of this question it was evident that many students did not have a secure understanding of time and calculations involving time. In part (a) students typically scored either two marks as they understood the question, or no marks as they incorrectly worked in minutes rather than hours. Part (b) allowed students who struggled with time to score at least one mark for the horizontal line segment to represent the stationary part of the journey. Those who calculated the time correctly and gained a method mark then normally gained the communication mark for the correct sloping section to the graph. A note to students is that 3 is not equal to 0.3

  Text version

  A travel graph shows Sam travelling 20 km from 10:00 to 10:15. He then stops 10 minutes and drives 20 minutes at 75 km/h.

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• Edexcel 1MA1 · November 2021 · Paper 2H · Q4 · 6 marks

  6 marks

  4

  
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  This question was one where the performance was below that of a typical higher tier cohort. Quite a number of students were unable to complete the table, and therefore scored no marks at all. Of those who were able to gain a single mark in (a), most were unable to identify their error (usually the point with x co-ordinate -1) when plotting, and hence showed a real lack of understanding of the shape and symmetry of a parabola. It is important that students learn how to use their calculators when calculating with negative numbers. Part (c) was answered poorly. Most students had no method at all for using a quadratic graph to solve a quadratic equation, and hence didn’t attempt the question.

  Text version

  For y=x^2-2x+2 use x=-2,-1,0,1,2,3,4.

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• Edexcel 1MA1 · November 2021 · Paper 2H · Q11 · 5 marks

  5 marks

  11

  
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  Examiner report

  Part (a) was not well answered, with most students not understanding the ratio and how to apply it to the coordinates. It was rare to award both marks, with some students able to find one coordinate (typically the x value) or for getting as far as 4.5 when working with the y-coordinate. Part (b) was a more common type of question and it showed that students have completed work on gradients and, compared to previous questions, a good number showed knowledge of gradients of perpendicular lines. The first mark was awarded in equal numbers for a correct process to find the gradient of L or for an equation in the form y = mx + 3. Those who correctly (or incorrectly) found a gradient for L, were still able to gain the second mark for a correct process to find the gradient of the perpendicular line, and this mark was awarded nearly as frequently as the first. Few gained full marks, either due to an error in finding the first gradient or for an incorrect y intercept, despite this being given in the question.

  Text version

  Line L passes through P(-6,0) and Q(0,3). R is on the same straight line beyond Q and PQ:QR=2:3.

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• Edexcel 1MA1 · November 2021 · Paper 2H · Q12 · 3 marks

  3 marks

  12

  
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  This question was answered well, with the majority of students scoring at least one mark for expanding two brackets, and many gaining two marks for the expansion of all three brackets. It was possible for both of these marks to be awarded with mistakes in the working, and this was usually the reason why full marks were not awarded for the final answer given. However, it was clear students understood the processes required to complete this question. It is interesting to see that the use of grid method to expand two brackets, once a common method, was very rarely seen. This may go some way to explaining why it was quite common to see 3 instead of 4 terms for the first expansion, or 7 instead of 8 terms, for the full expansion.

  Text version

  Expand and simplify (x-2)(3x+2)(2x+3).

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• Edexcel 1MA1 · November 2021 · Paper 2H · Q16 · 4 marks

  4 marks

  16

  
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  The majority of students did not know how to work with general iterative processes such as this. It was common to see students substituting in a succession of integers starting with 0 or 1. This typically brought up one correct value (1.817…) but didn’t gain credit as it was not a correct process to find x1. Of those who did correctly find the first value, most were able to go on to gain all three values. Some however went too far, and then put the wrong values as their answers (such as x2, x3 and x4) and therefore lost the accuracy mark. Very few students had any idea how to complete part (b) with many either leaving it blank or just guessing random values.

  Text version

  Use x_(n+1)=cuberoot(10-2x_n), starting x_0=2.

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• Edexcel 1MA1 · November 2021 · Paper 2H · Q18 · 4 marks

  4 marks

  18

  
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  At this stage of the paper many students had no knowledge to help in accessing the marks. If marks were awarded in part (a) it was usually 1 mark, for 2 correct values, rather than both marks. This is because even those who did know to read values off from the graph, then either didn’t read the range for solutions, or more commonly, didn’t understand how to find the values beyond 360°. In part (b) it was again quite rare to award the mark for a correct equation as many didn’t understand the effect on the function caused by a reflection. Of those who did gain credit, y = −sin x was the most common response. More students realised what was needed in part (c), that a translation was required, but to the left or down were more common than the correct direction of right.

  Text version

  A sine graph is supplied for 0°<=x<=360°. A second graph y=f(x) is the polyline through (-4,1),(-2,-1),(-1,-1),(2,1).

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• Edexcel 1MA1 · November 2021 · Paper 2H · Q21 · 4 marks

  4 marks

  21

  
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  Examiner report

  This grade 9 problem was a struggle for most, however there was scope for a good number to gain a single mark for one correct bound. This was normally a bound for T rather than l. The bounds for l most commonly seen were 51.5 and 52.5. These were outside the range that was acceptable for the substitution mark, so those who made this error were unable to gain the third mark even if they chose the correct pair. The second mark was for rearrangement, either before or after substitution, and proved to be unattainable for the majority of students due to the complex nature of the stated formula. To gain the third mark students had to select the correct pairs, eg LB of l and UB of T, or vice versa, and substitute these into a formula. In Summary Based on their performance on this paper, students should: • ensure they are familiar with sufficient higher tier content to be able to access the paper • work on understanding standard topics such as trigonometry, graphs, and box plots • spend more time working with ratio and proportional reasoning to allow them to work confidently with problems relating to these areas. • be confident in the different applications of trigonometry, depending on the type of triangle presented • practise solving problems which require conversion between units, in particular here for time in minutes and hours • make use of the units given in questions to determine formulae eg density units of g/cm3 lead to density = mass/volume • use brackets around negative numbers when entering these into a calculator

  Text version

  Pendulum period T seconds and length l cm satisfy T=2πsqrt(l/g). Katie measures l=52.0 and T=1.45, both 3 s.f.; use π=3.142.

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• Edexcel 1MA1 · November 2021 · Paper 3F · Q6 · 3 marks

  3 marks

  6

  
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  In part (a) the minority of students who answered incorrectly did so since they either only partly simplified, leaving in a × sign, or wrote part of the expression in index form, usually either ab4 or 4ab. In part (b) many students got 3x + 8, but some tried to simplify further to 11x.

  Text version

  Simplify two algebraic expressions.

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• Edexcel 1MA1 · November 2021 · Paper 3F · Q11 · 3 marks

  3 marks

  11

  
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  In part (a) most students used a non-calculator method and obtained the correct answer. It was obvious when calculators were used where an answer of 196 ÷ 603 = 3 h 26 min was seen. Many gained 1 mark for getting as far as 3 hours. In part (b) it was very rare to see any correct answers. Some wrote down the formula triangle for speed distance but could not relate this to an algebraic expression. The closest many students got was 2x, the most common incorrect algebraic answer.

  Text version

  Complete a time conversion and an average-speed expression.

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• Edexcel 1MA1 · November 2021 · Paper 3F · Q17 · 7 marks

  7 marks

  17

  
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  Examiner report

  In part (a) many students correctly expanded the first part of the bracket to get y2 but did not then multiply y by 5 to obtain 5y. Part (b) was less well answered. An incorrect factor of 4 was commonly seen outside the bracket. In part (c) more students found success. It was pleasing to see the layout of the process to solve the equation clearly structured. Some students were more confident at expanding brackets than at solving equation algebraically, though some who expanded incorrectly were able to access the mark for rearranging their equation ready for solution. Unsuccessful efforts included those who tried a trial and improvement approach of substituting a variety of numbers into the equation. Responses to part (d) were varied. Some secured full marks, but rarely. Most commonly students processed the numerical term but then multiplied the indices. Some failed to recognise that the lone e or f had an index of 1. Those who correctly remembered the rules of indices frequently wrote 9e3f4

  Text version

  Complete four algebra tasks.

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• Edexcel 1MA1 · November 2021 · Paper 3F · Q25 · 3 marks

  3 marks

  25

  
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  Examiner report

  Many students left this question blank or gave poor responses. Some students managed to gain 1 mark for finding the next term of the sequence, or for taking a step to use their (incorrect) fifth term in working towards a value for a. These included a significant minority who thought the fifth term was 7a and then gained a mark for 228 ÷ (1+2+3+5+7). Some students used a trial and improvement method but these were unsuccessful.

  Text version

  The first four terms of a Fibonacci sequence are a, 2a, 3a, 5a. The sum of the first five terms is 228.

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• Edexcel 1MA1 · November 2021 · Paper 3F · Q28 · 2 marks

  2 marks

  28

  
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  This question was not attempted by the majority of students. Of those who did attempt it, the most popular response was to sketch the line y = x or a quadratic graph. In some instances, students attempted to plot the graph using a table of values but failed to draw the graph due to the lack of numbers on the axis; in a few cases an attempt was made to draw the graph in just one quadrant. Summary Based on their performance on this paper, students should  show all working out and ensure that their written work is legible  transcribe figures taken from the question correctly  avoid rounding or truncating answers to calculations and use the most accurate values where appropriate  practice working with time and conversion of all types of metric units, including units of area  practice questions assessing algebraic manipulation and derivation, application of ratios, scaling and rates

  Text version

  Sketch the graph of y = 1/x.

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• Edexcel 1MA1 · November 2021 · Paper 3H · Q5 · 3 marks

  3 marks

  5

  
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  Examiner report

  This question rewarded students for finding the next term of the Fibonacci sequence but it also rewarded those students who formulated an equation to solve, even if they used an incorrect fifth term for the sequence. The most common mistake for the fifth term of the sequence was to give the term as “7a” instead of the correct “8a”. Unfortunately, many students used only the four terms given to form an equation instead of the five terms demanded in the question. Again, a more careful reading of the question might have led to more students obtaining some marks for their answers.

  Text version

  The first four terms of a Fibonacci sequence are a, 2a, 3a and 5a. The sum of the first five terms is 228.

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• Edexcel 1MA1 · November 2021 · Paper 3H · Q12 · 2 marks

  2 marks

  12

  
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  This question was not well answered and only about a quarter of all students scored one mark or more. Students who scored one mark usually scored this for giving the correct value for n. In general, little working was seen in the working space so it was rare for examiners to be able to award the process mark where both answers were incorrect.

  Text version

  (a x^6)^(1/n) = 7x^3.

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• Edexcel 1MA1 · November 2021 · Paper 3H · Q13 · 5 marks

  5 marks

  13

  
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  This question was generally not well answered. Many students did not attempt the question and there were many attempts where students started by calculating the gradient of the line joining point A to point B. Where students were more successful, there were some well presented and clear solutions showing the derivation of two equations together with their solution, usually to find the length and width of each rectangle. These students usually completed the question and scored the full 5 marks.

  Text version

  Four identical rectangles, with sides parallel to the axes, form the published two-step pattern. A=(3,4) is the lower-left corner and B=(11,20) is the top-right corner. If each rectangle has length L and width W, the arrangement gives horizontal displacement A to B of 2L−W, vertical displacement 2L+W, and C lies L units right and W units above A.

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• Edexcel 1MA1 · November 2021 · Paper 3H · Q15 · 3 marks

  3 marks

  15

  
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  Higher attaining students often scored full marks for their answers to this question. Where tangents were drawn they were generally accurate and allowed them to be used to get a good estimate of the gradient. However there was a significant number of students who used the coordinates of the point on the curve at t = 12 and who simply divided the y coordinate by the x coordinate (17 ÷ 12) to get the gradient. Some of these students had already drawn a tangent to the curve at t = 12.

  Text version

  A smooth decreasing graph gives water volume V litres against time t seconds. Approximate readings from the official curve are (0,40), (5,29), (10,22), (15,16), (20,12), (30,7), (40,4) and (50,2.5).

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• Edexcel 1MA1 · November 2021 · Paper 3H · Q16 · 4 marks

  4 marks

  16

  
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  Many students gave a concise, clear and accurate algebraic argument to score full marks for their answers to this question. Lower attaining students often made errors in the manipulation of the equations involved. It was common for examiners to award 3 marks for students completing the algebra correctly but not the fourth communication mark because the student had not made a statement to conclude their findings.

  Text version

  Curve C has equation y=x^2+3x−3. Line L has equation y−5x+4=0.

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• Edexcel 1MA1 · November 2021 · Paper 3H · Q21 · 4 marks

  4 marks

  21

  
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  A good number of students demonstrated how to estimate the area under the curve by using 3 trapeziums though some students used only one or two trapeziums. Unfortunately, a large number of students were unable to access full marks for their answers because they used incorrect y values for t = 3 and/or for t = 4. Most students did gain some credit for a correct expression for at least one of the areas, usually for the trapezium between t = 1 and t = 2. It was encouraging to see students using the formula for the area of a trapezium rather than splitting the trapezium into a rectangle and triangle. These students were usually more successful in terms of the number of marks they got for their answers. Answers to part (b) were often correct though there were a significant number of answers stating that the area represented either acceleration or speed. Units of measurement for the distance were not required but where these were given they were invariably correct.

  Text version

  A speed-time curve starts from rest. Readings from the official graph at t=1,2,3,4 seconds are approximately 4, 6, 7.2 and 7.8 m/s respectively.

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• Edexcel 1MA1 · November 2021 · Paper 3H · Q22 · 3 marks

  3 marks

  22

  
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  There were some excellent clear and concise answers to this question leading to the correct answer 𝑥(𝑥+4). Many students were able to show that to divide the algebraic fractions, inverting the second fraction and multiplying it by the first fraction was needed. Fewer students realised the need to factorise the denominator of the first fraction and much fruitless algebra was seen with students frequently multiplying out expressions.

  Text version

  Consider [6x^3/(9x^2−144)] ÷ [2x^4/(3(x−4))].

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• Edexcel 1MA1 · November 2020 · Paper 1F · Q8 · 3 marks

  3 marks

  8

  
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  Text version

  On a coordinate grid, point A is at (2, 3) and point B is at (0, -1).

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• Edexcel 1MA1 · November 2020 · Paper 1F · Q12 · 3 marks

  3 marks

  12

  
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  A number machine multiplies the input by 2, then subtracts 3.

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• Edexcel 1MA1 · November 2020 · Paper 1F · Q19 · 4 marks

  4 marks

  19

  
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  Text version

  Expand x(x - 4)

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• Edexcel 1MA1 · November 2020 · Paper 1F · Q20 · 2 marks

  2 marks

  20

  
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  Text version

  The first five terms of an arithmetic sequence are 1 4 7 10 13 Write down an expression, in terms of n, for the nth term of this sequence.

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• Edexcel 1MA1 · November 2020 · Paper 1F · Q22 · 2 marks

  2 marks

  22

  
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  Four graphs are described as follows. Graph A is a reciprocal curve with branches in quadrants I and III. Graph B is a decreasing cubic through the origin. Graph C is an increasing cubic through the origin. Graph D is an upward-opening parabola with vertex at the origin.

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• Edexcel 1MA1 · November 2020 · Paper 1F · Q30 · 3 marks

  3 marks

  30

  
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  Text version

  Make q the subject of p = 6q + 7

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• Edexcel 1MA1 · November 2020 · Paper 1H · Q1 · 2 marks

  2 marks

  1

  
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  The first five terms of an arithmetic sequence are 1, 4, 7, 10, 13. Write an expression, in terms of n, for the nth term.

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• Edexcel 1MA1 · November 2020 · Paper 1H · Q3 · 2 marks

  2 marks

  3

  
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  Four graphs are described as follows. Graph A is a reciprocal curve with branches in quadrants I and III. Graph B is a decreasing cubic through the origin. Graph C is an increasing cubic through the origin. Graph D is an upward-opening parabola with vertex at the origin.

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• Edexcel 1MA1 · November 2020 · Paper 1H · Q11 · 6 marks

  6 marks

  11

  
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  Text version

  Evaluate each expression.

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• Edexcel 1MA1 · November 2020 · Paper 1H · Q15 · 3 marks

  3 marks

  15

  
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  Text version

  Line L1 is y = 3x - 4. Line L2 is perpendicular to L1 and passes through (9, 5). Find an equation of L2.

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• Edexcel 1MA1 · November 2020 · Paper 1H · Q17 · 4 marks

  4 marks

  17

  
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  Make f the subject of d = 3(1 - f)/(f - 4).

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• Edexcel 1MA1 · November 2020 · Paper 1H · Q19 · 5 marks

  5 marks

  19

  
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  Text version

  Functions f and g are defined by f(x) = 12/sqrt(x) and g(x) = 3(2x + 1).

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• Edexcel 1MA1 · November 2020 · Paper 1H · Q22 · 5 marks

  5 marks

  22

  
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  The diagram shows two shaded shapes, A and B. Shape A is formed by removing a sector of a circle with radius (3x - 1) cm from a sector of the circle with radius (5x - 1) cm. Shape B is a circle of diameter (2 - 2x) cm. 45 degrees A B 2x cm (3x - 1) cm (2 - 2x) cm The area of shape A is equal to the area of shape B. Find the value of x. You must show all your working.

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• Edexcel 1MA1 · November 2020 · Paper 2F · Q3 · 1 mark

  1 mark

  3

  
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  Text version

  Simplify 3e - e + 4e

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• Edexcel 1MA1 · November 2020 · Paper 2F · Q10 · 2 marks

  2 marks

  10

  
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  Solve 3m = 36 m = .......................................................

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• Edexcel 1MA1 · November 2020 · Paper 2F · Q15 · 4 marks

  4 marks

  15

  
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  Text version

  T = 3x + 4y

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• Edexcel 1MA1 · November 2020 · Paper 2F · Q24 · 4 marks

  4 marks

  24

  
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  Text version

  For y = 5 - x^3, use x-values -2, -1, 0, 1 and 2.

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• Edexcel 1MA1 · November 2020 · Paper 2F · Q29 · 1 mark

  1 mark

  29

  
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  Write down the gradient of the line with equation y = 2x + 3

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q4 · 4 marks

  4 marks

  4

  
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  For y = 5 - x^3, use x-values -2, -1, 0, 1 and 2.

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q10 · 4 marks

  4 marks

  10

  
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  Text version

  Simplify each expression.

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q12 · 3 marks

  3 marks

  12

  
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  Text version

  A straight-line graph of petrol volume v litres against distance d kilometres passes through (d, v) = (0, 27) and (300, 0).

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q14 · 4 marks

  4 marks

  14

  
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  Here are two squares, A and B. A B The length of each side of square B is 4 cm greater than the length of each side of square A. The area of square B is 70 cm2 greater than the area of square A. Find the area of square B. Give your answer correct to 3 significant figures. You must show all your working. ....................................................... cm2

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q16 · 3 marks

  3 marks

  16

  
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  Text version

  Here are the first five terms of a quadratic sequence. 10 21 38 61 90 Find an expression, in terms of n, for the nth term of this sequence.

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q17 · 1 mark

  1 mark

  17

  
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  Write down the coordinates of the turning point on the graph of y=(x+12)^2-7.

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q19 · 4 marks

  4 marks

  19

  
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  A hot air balloon is descending. The height of the balloon n minutes after it starts to descend is hn metres. The height of the balloon (n +1) minutes after it starts to descend, hn + 1 metres, is given by hn + 1 = K x hn + 20 where K is a constant. The balloon starts to descend from a height of 1200 metres at 09 15 At 09 16 the height of the balloon is 1040 metres. Work out the height of the balloon at 09 18 ....................................................... m

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q20 · 8 marks

  8 marks

  20

  
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  A bag contains n red sweets and 8 yellow sweets. Sajid takes one sweet at random and eats it.

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q21 · 4 marks

  4 marks

  21

  
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  The graph y = f(x), for -4 <= x <= 0, is a smooth curve through (-4, 2), a local minimum at (-3, 0), (-2, 1), a local maximum at (-1, 2), and (0, 0). The curve C has equation y = 5 + 2x - x^2.

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• Edexcel 1MA1 · November 2020 · Paper 2H · Q22 · 5 marks

  5 marks

  22

  
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  C is a circle with centre the origin. A tangent to C passes through the points (-20, 0) and (0, 10) Work out an equation of C. You must show all your working.

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• Edexcel 1MA1 · November 2020 · Paper 3F · Q10 · 3 marks

  3 marks

  10

  
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  A graph has time in minutes on the horizontal axis from 0 to 12 and temperature in degrees C on the vertical axis from 0 to 110. A smooth decreasing cooling curve starts at (0,100), passes approximately through the neutral read-off points (1,89), (4,67), (5,61), (6,57), (7,53), (10,44) and (11,42), and continues smoothly to time 12 minutes.

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• Edexcel 1MA1 · November 2020 · Paper 3F · Q12 · 3 marks

  3 marks

  12

  
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  The first sequence is 45, 40, 35, 30, 25, ... . A different sequence has nth term 4n + 3.

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• Edexcel 1MA1 · November 2020 · Paper 3F · Q14 · 4 marks

  4 marks

  14

  
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  Simplify 3x + 5y + 2x - 4y

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• Edexcel 1MA1 · November 2020 · Paper 3F · Q18 · 3 marks

  3 marks

  18

  
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  A currency-conversion graph is a straight line through the origin and the point (GBP 50, USD 65). Rosie paid 345 US dollars.

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• Edexcel 1MA1 · November 2020 · Paper 3F · Q23 · 5 marks

  5 marks

  23

  
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  Simplify n3 x n5

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• Edexcel 1MA1 · November 2020 · Paper 3F · Q28 · 3 marks

  3 marks

  28

  
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  Two coordinate graphs are supplied. On the first graph, 5x - 9y = -46 is a straight line through (-8, 2/3), (1, 17/3) and (4, 22/3), while y = -2x is a straight line through (-4, 8), (0, 0) and (2, -4). On the second graph, the curve y = x^2 - 4x + 2 is drawn smoothly through (-1, 7), (0, 2), (1, -1), (2, -2), (3, -1), (4, 2) and (5, 7).

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• Edexcel 1MA1 · November 2020 · Paper 3H · Q1 · 5 marks

  5 marks

  1

  
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  Text version

  Simplify n3 x n5

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• Edexcel 1MA1 · November 2020 · Paper 3H · Q6 · 3 marks

  3 marks

  6

  
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  Two coordinate graphs are supplied. On the first graph, 5x-9y=-46 is a straight line through (-8,2/3), (1,17/3) and (4,22/3), while y=-2x is a straight line through (-4,8), (0,0) and (2,-4). On the second graph, the curve y=x^2-4x+2 is drawn smoothly through (-1,7), (0,2), (1,-1), (2,-2), (3,-1), (4,2) and (5,7).

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• Edexcel 1MA1 · November 2020 · Paper 3H · Q12 · 6 marks

  6 marks

  12

  
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  Algebraic fractions and expansion.

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• Edexcel 1MA1 · November 2020 · Paper 3H · Q13 · 5 marks

  5 marks

  13

  
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  Part (a) uses the inequalities x > 0, x <= 2, y <= x + 3 and 2x + 3y >= 6. Part (b) uses y < 4x, y > 0.5x and x + y <= 6.

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• Edexcel 1MA1 · November 2020 · Paper 3H · Q16 · 4 marks

  4 marks

  16

  
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  A car's speed-time graph is a smooth concave-down curve from (0, 0) to a maximum of about (75, 13.5), then back to (150, 0). During the first 30 seconds the graph gives approximate speeds 0, 3.4, 6.4 and 8.8 m/s at 0, 10, 20 and 30 seconds.

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• Edexcel 1MA1 · November 2020 · Paper 3H · Q21 · 5 marks

  5 marks

  21

  
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  Given 5c + d = c + 4d, and separately 6x^2 = 7xy + 20y^2 with x>0 and y>0.

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• Edexcel 1MA1 · November 2019 · Paper 1F · Q10 · 2 marks

  2 marks

  10

  
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  There were many fully correct answers to this question but also many students who scored one mark for plotting the point correctly in part (a) of the question but in part (b) wrote down (0, −1) instead of the correct (−1, 0) for the coordinates of the midpoint of BC.

  Text version

  On a coordinate grid, B = (-3, 2) and C = (1, -2).

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• Edexcel 1MA1 · November 2019 · Paper 1F · Q13 · 3 marks

  3 marks

  13

  
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  Part (a) of this question was answered very well and it was unusual to see students not receiving any credit for their response to part (b). The most common incorrect response to part (a) was p7. In responses to part (b) there was a significant proportion of students who collected the terms in x correctly and who collected the terms in y correctly only to give a final answer of 8x – y. These students were awarded one mark. 8xy was also seen as a final answer quite frequently. Where this answer was preceded by “8x” or “y” as a standalone term in the working space, examiners could award one mark.

  Text version

  Simplify each expression.

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• Edexcel 1MA1 · November 2019 · Paper 1F · Q16 · 3 marks

  3 marks

  16

  
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  About three quarters of students gave a correct answer to part (a) of this question. A fully correct answer to part (b) was seen much more infrequently. However, a fair proportion of students could score at least one mark for either an attempt to find the gradient of the line representing Steve’s journey home or for using the speed, distance, time formula correctly. Students often scored a mark for ଶହ ଷ଴ but could then not convert this to a speed in km/h. Those students who did give the correct answer, 50, had usually argued that 25 km in half an hour was equivalent to an average speed of 50 km/h. Other common errors included multiplying 25 by 30 and dividing 25 by 2 instead of by ଵ ଶ.

  Text version

  A travel graph shows Steve 25 km from home from 12:45 until 13:30, then travelling home from 13:30 to 14:00.

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• Edexcel 1MA1 · November 2019 · Paper 1F · Q17 · 3 marks

  3 marks

  17

  
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  A high percentage of students correctly obtained the value of x from the equation given but only about a quarter of students were able to use the correct order of operations to find the correct value (18) of the expression 2x2 when x = 3. Instead most students worked out the value of (2x)2, giving 36 as their final answer. The working 2 × 3 = 6, 6 ×6 = 36 was seen on many scripts.

  Text version

  Given x - 1 = 2, work out 2x^2.

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• Edexcel 1MA1 · November 2019 · Paper 1F · Q19 · 2 marks

  2 marks

  19

  
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  This question was not well done with only a small proportion of students gaining any marks and few students gaining two marks. The great majority of responses were in a form connecting the two numbers –3 and 1 with an inequality sign (often incorrect), for example –3 ≤ 1. Such responses could not be given any marks. A number of students listed integer values only.

  Text version

  A number line has a closed endpoint at -3, an open endpoint at 1, and is shaded between them.

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• Edexcel 1MA1 · November 2019 · Paper 1F · Q26 · 3 marks

  3 marks

  26

  
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  About a half of students taking this paper answered part (a) correctly to score 1 mark. Examiners saw the incorrect response “p7” on many occasions. About the same proportion of students scored at least one mark for their response to part (b) for giving a final answer which included at least two out of three correct components of the expression 2x4y2.

  Text version

  Simplify the expressions (p^2)^5 and (12x^7y^3)/(6x^3y).

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• Edexcel 1MA1 · November 2019 · Paper 1F · Q28 · 3 marks

  3 marks

  28

  
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  This, the penultimate question on the paper, targeted the most able students sitting the examination. It was rare to see a fully correct solution and most responses could not be credited with any marks. A very common error made by students was to think that an obtuse angle was one greater than 180̊ or to mistakenly use one of the results that “angles in a triangle sum to 180̊” or “angles on a straight line add to 180̊”. Students using 180° but carrying out subsequent processes accurately to get an answer of 34 were rewarded with some credit. There were very few students who used algebra to formulate and solve an equation or inequality. Instead, most students who attempted the question used a numerical approach, sometimes by trial and improvement. Students who used such an approach and achieved an answer in the range 15 ≤ x <16 could be given some credit.

  Text version

  In triangle AOB, two internal rays from O meet side AB and partition angle AOB into three adjacent angles. From ray OA towards ray OB, these angles are (2x) degrees, (3x) degrees and 10 degrees. Angle AOB is not obtuse.

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• Edexcel 1MA1 · November 2019 · Paper 1H · Q7 · 3 marks

  3 marks

  7

  
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  In general, both parts of this question were answered well. In part (a), p7 was the modal incorrect answer offered. In part (b), a significant number of students used correct rules of indices to derive expressions of x4 and/or y2 but often these were not seen in a product or a fraction. It was also common to see 12 – 6 = 6 instead of 12 ÷ 6 = 2.

  Text version

  Simplify the expressions (p^2)^5 and (12x^7y^3)/(6x^3y).

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• Edexcel 1MA1 · November 2019 · Paper 1H · Q12 · 1 mark

  1 mark

  12

  
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  Although the question asked for identification of Spencer’s mistake in line one, credit was given for answers which correctly explained how q could be made the subject of the formula, so answers of “he should have subtracted the 5 first before multiplying by 2” were awarded the mark. Any ambiguity in answers was penalised.

  Text version

  T = q/2 + 5. Spencer incorrectly rearranges this as q = 2T - 5.

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• Edexcel 1MA1 · November 2019 · Paper 1H · Q13 · 3 marks

  3 marks

  13

  
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  Any marks gained in this question were usually in part (a), only a very small number of students gained the mark in part (b). In part (a), many students correctly found the common denominator of 3x(x + 1) or equivalent. Unfortunately, numerators of 5 and 2 were often retained gaining no credit. A numerator 5 × 3x + 2(x + 1) was often not or incorrectly simplified resulting expressions of 15x + 2x + 2(or 1). It was also disappointing to see so many students having found the correct simplified expression then try to simply further. In algebraic questions, such subsequent incorrect working can never be ignored, and the final accuracy mark is lost. In part (b), the modal approach was to expand the brackets rendering the resulting expression near impossible to factorise.

  Text version

  Algebraic simplification.

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• Edexcel 1MA1 · November 2019 · Paper 1H · Q14 · 4 marks

  4 marks

  14

  
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  Very few students were able to derive a correct quadratic equation from which a complete solution could be found. A common mistake was in finding the area of the triangle, (x – 2)(x + 4) and (x – 2) + (x + 4) were often seen, many times without brackets shown. One mark was available in a special case for a correct product expanded and equated to 27.5. Many students after seeing a right-angled triangle, often started their solutions by attempting to use Pythagoras’ theorem. When a trial and improvement method was employed, credit could only given if the final answer was correct. Those students who realised that the area of the rectangle of sides (x – 2) and (x + 4) was 55 were often successful in getting the correct answer.

  Text version

  A right-angled triangle has perpendicular legs of lengths (x-2) cm and (x+4) cm. Its area is 27.5 cm^2.

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• Edexcel 1MA1 · November 2019 · Paper 1H · Q18 · 6 marks

  6 marks

  18

  
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  In part (a), only a very few students correctly found an algebraic expression for the inverse of f(x). The most successful approach was in showing f(3) = 50 but a great number of students found f(50) instead which was clearly incorrect. Notation was often poor with the cube root symbol not encompassing the whole expression in some cases. Only a minority of students correctly quoted hg(x) = (x + 2)2, in part (b), and were able to make any progress. The most common mistake was in substituting x + 2 into hg(x). Those who managed to substitute correctly and form an equation often failed to simplify and solve the resulting quadratic equation.

  Text version

  The function f is given by f(x)=2x^3-4.

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• Edexcel 1MA1 · November 2019 · Paper 1H · Q20 · 3 marks

  3 marks

  20

  
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  Both parts of this question were poorly attempted. A translation of the given function was often seen in part (a), but rarely was this a correct one. A correct translation of either, -1 in y direction OR -3 in x direction obtained 1 mark. In part (b), it was rare to see a correct reflection of f(x) drawn, many students either guessing the coordinates of B or using a diagram found in part (a).

  Text version

  The graph y=f(x) is the polyline through (-4,3), (-3,3), (-2,1), (1,4), (3,4). Point A=(-2,1) lies on it.

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• Edexcel 1MA1 · November 2019 · Paper 1H · Q21 · 5 marks

  5 marks

  21

  
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  Again, only a very few students were able to make any headway in sketching a graph of the given quadratic. The y-intercept and the turning point were most often found by constructing tables of values. Some students did attempt to solve the quadratic, usually by completing the square, however this was often littered with mistakes. Many of the students who attempted this question drew a parabola which was symmetrical about the y-axis.

  Text version

  Sketch y = 2x^2 - 8x - 5, showing the turning point and exact axis intercepts.

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• Edexcel 1MA1 · November 2019 · Paper 2F · Q8 · 2 marks

  2 marks

  8

  
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  Part (a) was answered very well. A few students misread the question and worked out the input for an output of 6. Part (b) was also answered well. Instead of giving the answer as –20 or ÷3 some students omitted the operation and wrote just 20 or 3 in the number machine. They gained no mark.

  Text version

  A number machine multiplies the input by 8 and then subtracts 5. A different number machine adds 13 and then applies one further operation.

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• Edexcel 1MA1 · November 2019 · Paper 2F · Q11 · 3 marks

  3 marks

  11

  
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  In part (a) some students did not read the graph with sufficient care when changing 3 stones to kilograms. Answers such as 10.9 or 20 were quite common. The most common approach seen to change 80 kilograms to stones in part (b) was reading from the graph at a factor of 80, which was often 40, and scaling up to 80. A few students used two values that sum to 80. The scale on the horizontal axis was often not read correctly and statements such as 40 kg = 6.1 stones or 10 kg = 1.3 stones were common. An answer outside the range 12.4 to 12.8 was awarded full marks if it came from a complete method that included a correct reading from the graph. Some students, for example, read from the graph to find 1 stone = 6 kg and then divided 80 by 6 to complete the method. Too many students gave an answer with little or no working out. When the answer was not in the range 12.4 to 12.8 marks could only be awarded if a complete method was shown. Students should be encouraged to show a clear method on the graph as, without this, incorrect conversions can be given no marks. Attempts at answering the question by extending the graph were usually unsuccessful.

  Text version

  A straight conversion graph passes through (0 stones, 0 kilograms), (3 stones, 19 kilograms) and approximately (10 stones, 63.5 kilograms).

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• Edexcel 1MA1 · November 2019 · Paper 2F · Q18 · 4 marks

  4 marks

  18

  
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  In part (a) it was pleasing that many students did manage to write a formula for T in terms of n even if it wasn’t a fully correct formula that gained full marks. Students that started by writing down Chloe’s age as 2n or Dan’s age as n – 5 were awarded the first mark and tended to be the most successful. Common errors were expressing Chloe’s age as n2 or Dan’s age as 5 – n or –5n. Students who used both 2n and n – 5 often went on to give a fully correct formula although some gave the answer as n + 2n + n – 5 or as 4n – 5 and scored only two marks. Students with an incorrect expression for either Chloe’s age or Dan’s age were still able to get two marks for an answer such as T = n + n2 + n – 5 or T = n + 2n – 5n. Some students failed to include the original n for Ben’s age. Answers of T = 2n – 5 with no working were common and these gained the first mark only for T = a linear expression in n. Some students thought they had to work out a numerical answer. Many students were unable to select the identity in part (b). A popular incorrect answer was the inequality x + 7 ≤ 12.

  Text version

  Ben is n years old. Chloe is twice as old as Ben. Dan is five years younger than Ben. Their total age is T years.

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• Edexcel 1MA1 · November 2019 · Paper 2F · Q20 · 4 marks

  4 marks

  20

  
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  Most of the students that substituted f = 110 into the formula obtained 452 and were awarded the first two marks. Many students, however, were not able to complete the solution by showing that the difference between 452 and the real height is less than 5% of the real height. Some made no attempt to do so. Others did make an attempt but could not show a complete method. Some found the difference of 10 but did not work with percentages whereas others found 5% of 442 but not the difference of 10. A common mistake was to find 5% of 452 instead of 5% of the real height. A significant number of students did not use the formula correctly and gained no marks. Attempts at solving an equation such as 442 = 4f + 12 or 110 = 4f + 12 were quite common. Some students thought that they needed to round values because of the word ‘estimate’.

  Text version

  The height H metres of a building is estimated using H = 4f + 12, where f is its number of floors. A building is 110 floors high and its real height is 442 m.

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• Edexcel 1MA1 · November 2019 · Paper 2F · Q25 · 2 marks

  2 marks

  25

  
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  Both parts of this question were poorly answered. Relatively few students were able to identify the graph with equation y = x3 in part (a). It was perhaps to be expected that many students would choose graph B instead of graph F but it was surprising that both graph A and graph E were very popular incorrect choices. In part (b) graph F was a popular incorrect choice for the graph with equation y = 1/x.

  Text version

  Six graph options are labelled A to F. A is an upward parabola touching the origin; B is a cubic with two turning points; C is a reciprocal-type graph in quadrants II and IV; D is a reciprocal-type graph in quadrants I and III; E is a straight line; F is an increasing cubic through the origin with a horizontal tangent there.

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• Edexcel 1MA1 · November 2019 · Paper 2F · Q26 · 3 marks

  3 marks

  26

  
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  This question was not answered well. It was disappointing that many students could not find at least two terms by substituting values of n. Many of those that did attempt to find terms of each sequence did not generate sufficient terms to satisfy the demand of the question. Some students found that the number 31 is in both sequences but did not show that that there is only one number that is in both sequences by generating the first five terms of the first sequence and the first six terms of the second sequence. Errors in generating terms using 2n2 – 1 were often the result of using an incorrect order of operations, most often finding 2n and then squaring.

  Text version

  The nth term of one sequence is 2n^2 - 1. The nth term of a different sequence is 40 - n^2.

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• Edexcel 1MA1 · November 2019 · Paper 2F · Q28 · 3 marks

  3 marks

  28

  
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  This question was generally answered quite well and it was pleasing that many students were able to gain the first mark for finding the number of workers used each day by Ali’s company or by Hayley’s company. Most went on to give a fully correct solution. Having found the two correct values (18 and 24) a few students did not complete the process by subtracting 18 from 24. Some students could not use the formula correctly and a common error was to work out 720 × 40 and 720 × 30.

  Text version

  The number of days d needed to build a house is d = 720/n, where n is the number of workers used each day. Ali's company takes 40 days and Hayley's company takes 30 days.

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• Edexcel 1MA1 · November 2019 · Paper 2H · Q5 · 2 marks

  2 marks

  5

  
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  This was not well answered. B was a common incorrect answer for part (a).

  Text version

  Six graph options A-F are shown: A upward parabola touching origin; B cubic with two turns; C reciprocal in quadrants II/IV; D reciprocal in I/III; E straight line; F increasing cubic through origin with horizontal tangent.

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• Edexcel 1MA1 · November 2019 · Paper 2H · Q6 · 3 marks

  3 marks

  6

  
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  A significant minority of student left this question blank, seemingly unable to work out how to proceed with the question. Of those who started working out terms, many were able to gain two marks for generating at least 3 terms of each sequence, but often failing to go as low as the 4 on the second sequence. A common error was seen by many students who attempted to equate the two expressions and solve the equation, which led them nowhere.

  Text version

  The nth terms of two sequences are 2n^2 - 1 and 40 - n^2.

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• Edexcel 1MA1 · November 2019 · Paper 2H · Q10 · 2 marks

  2 marks

  10

  
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  There were some very confused attempts at doing this. Many tried to move items around without squaring first and got nowhere. Many made a good start getting to y2 = 2m-k then about half of these correctly rearranged to get full marks. Many were unprepared for manipulating the square root, often showing 2m2 as well as k.

  Text version

  Make k the subject of y = sqrt(2m - k).

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• Edexcel 1MA1 · November 2019 · Paper 2H · Q15 · 3 marks

  3 marks

  15

  
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  Many students could successfully find 3 out of 4 values of a correct expansion (more successfully if they used the positive brackets). A large number tried to do the multiplication in one step, which always led to an incorrect response as they could not keep track of all the multiplications necessary. Students who tackled the question in two steps often gained additional credit or the fully correct answer. Common misconceptions were errors in signs once the negative bracket was used, or when collecting the like terms such as 3x × 2x = 6x.

  Text version

  Expand and simplify (3x + 2)(2x + 1)(x - 5).

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• Edexcel 1MA1 · November 2019 · Paper 2H · Q21 · 5 marks

  5 marks

  21

  
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  There was a lot of misunderstanding of what was being asked for in this question, particularly confusion between speed and acceleration. In part (a) some found the half way mark on the time axis and gave the answer as 50 or used distance/speed/time. Of those that realised they needed to work with the area under the graph a large number made an error with the vertical scale using 12.5 or 14 for the speed. Some made errors using the formula for the area of a triangle. However, many still gained the first make for starting to find an area. In part (b) few students realised the first part was acceleration and the last part deceleration, though they did notice that the last part was greater than the first part because the gradient was steeper. Some responses gave no comparison.

  Text version

  A train speed-time graph has vertices (0,0), (30,12), (80,12), (100,0), with straight segments; total journey time is 100 s.

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• Edexcel 1MA1 · November 2019 · Paper 2H · Q23 · 6 marks

  6 marks

  23

  
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  In part (a) very few responses gained any credit. Nearly all students failed to recognise the sine rule for area was needed and instead just multiplied the two sides of the parallelogram. Just expanding the two brackets was the most common strategy adopted or students did the work for part (b) and just tried to solve the quadratic. Part (b) was more successfully attempted as students could either factorise, or more often use the quadratic formula to solve the equation. Once the two answers were found, all too often they were expressed as x = or with an incorrect inequality, especially with 2.5. Students who gave the correct inequality often did so by sketching a curve and using it to assist them in interpreting their answer. Some students just used the brackets from the parallelogram, evaluating them to zero.

  Text version

  A parallelogram has adjacent sides (2x - 1) cm and (10 - x) cm with included angle 150°. Its area is greater than 15 cm^2.

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• Edexcel 1MA1 · November 2019 · Paper 2H · Q25 · 5 marks

  5 marks

  25

  
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  It was clear that many students knew what to do but lacked the skills to do what they wanted. As a result this question was rarely attempted at all. Where the question was attempted and credit was achieved it was usually for finding the gradient of line L, although several forgot to divide the whole expression by 2. Once a gradient had been established, it was common for students to then use -1/m to find the gradient of the perpendicular, although -2/3 was seen. It was common to see the correct equation here, but then students did not know how to progress further, although some then sketched the equation identifying the y coordinate of B. Even in the rare cases when the coordinate of C was established forming the correct triangle and then successfully calculating the area was infrequent. Summary Based on their performance on this paper, students should:  ensure that figures taken from the question, and from their own work, are transcribed accurately.  avoid rounding or truncating answers part way through a solution and use the most accurate values where possible.  ensure that calculators are used correctly.  remember to include working out to support their answers.  spend time on topics such as algebraic manipulation and proof, and application of ratios, bounds, iteration and histograms when preparing for future examinations.  spend time on practicing response type questions where a written explanation is required.

  Text version

  Line L is 3x + 2y = 17. A=(0,2). Line M is perpendicular to L through A. L crosses the y-axis at B and intersects M at C.

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• Edexcel 1MA1 · November 2019 · Paper 3F · Q7 · 2 marks

  2 marks

  7

  
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  A good number of students gained both marks, but where that wasn’t the case, many gained one mark for the first step. These students then typically completed the calculation in the wrong order, often arriving at 44. Of those who scored zero this was often a result of not including the “×” when substituting and simply writing 48 rather than 4 × 8, typically leading to an answer of 51.

  Text version

  w = 4u + 3.

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• Edexcel 1MA1 · November 2019 · Paper 3F · Q8 · 2 marks

  2 marks

  8

  
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  It is clear that many of the students were familiar with this sequence (triangle numbers) and were simply able to write down the next two terms. Many others were able to decipher the correct pattern and generated the next two terms by addition of 6 then 7. Some of these students only gained one of the two marks, for, despite knowing what to add, they made arithmetic errors along the way.

  Text version

  Sequence: 1, 3, 6, 10, 15.

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• Edexcel 1MA1 · November 2019 · Paper 3F · Q14 · 2 marks

  2 marks

  14

  
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  Both these response questions were well attempted by students and, in general, answered well. Part (a) was looking for students to realise that the area had been found rather than the perimeter, and many did. There were some cases where the response included mathematically incorrect statements, such as P = 7 + 3 = 10, and where this occurred the mark was not awarded. In part (b) students were expected to realise that a triangle cannot have a negative length. This part was answered less well than the previous one, with many students either being unclear in their response, or making incorrect statements. It was not uncommon to see arguments relating to 180 (degrees) or students trying to find the value of x themselves, without sufficient information to do so.

  Text version

  Coby finds the perimeter of a 7 cm by 3 cm rectangle by writing 7×3. A triangle has side lengths x, x+7 and x+8 cm; Iram obtains x=-2.

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• Edexcel 1MA1 · November 2019 · Paper 3F · Q17 · 4 marks

  4 marks

  17

  
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  A really well answered question. There were relatively few cases of incorrect tables, and where there were incorrect values they tended to be when x was negative. It was rare to award no marks in part (a). Part (b) was answered very well, and in many cases better than part (a). It was not uncommon for students to have incorrect values in their table and to then draw a fully correct graph. The method mark in part (b) was able to be awarded as a follow through mark from their table in (a) provided one mark had been scored in part (a). The most common cause for a lost mark in this part was to not complete the graph by joining the points plotted with a line.

  Text version

  Complete the table of values for y = 4x - 6. The x-values are -1, 0, 1, 2, 3, 4. The table already gives y = -2 when x = 1 and y = 10 when x = 4.

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• Edexcel 1MA1 · November 2019 · Paper 3F · Q19 · 2 marks

  2 marks

  19

  
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  The majority of students attempted the expansion as their first step, and in many cases, were successful. Of those who were not successful this was for a variety of reasons. Common errors were to exchange the minus sign for a plus; to fail to multiply the second term when expanding, or to expand but not as part of an equation. The second mark was for the correct answer. As under general guidance, markers were able to award the first mark for an embedded “17” in working provided it wasn’t contradicted by the answer line.

  Text version

  Solve 4(x-6)=44.

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• Edexcel 1MA1 · November 2019 · Paper 3F · Q22 · 4 marks

  4 marks

  22

  
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  Most students were able to gain at least one mark in part (a), normally for 3 of their 4 terms correct. It was very common to see 2x in place of x2 or to have the constant term incorrect. Many students who did gain one mark were unable to gain the full marks as they often over simplified their expression by combining non-like terms. Part (b) was testing students ability to factorise into a single bracket. One mark could be awarded for a correct partial factorisation, removing either the factor of 3 or x correctly from the expression. The first mark could also be awarded if one of the correct factors, 3x or (3x + 2), was found. Another frequently seen incorrect response was an attempt to factorise into two brackets due to the 9x2. A small proportion of students correctly factorised the expression and checked that the expansion would lead them to the original expression but then wrote the original expression on the answer line rather than the factorisation.

  Text version

  Algebraic expressions.

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• Edexcel 1MA1 · November 2019 · Paper 3H · Q1 · 4 marks

  4 marks

  1

  
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  Algebraic expressions.

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• Edexcel 1MA1 · November 2019 · Paper 3H · Q10 · 4 marks

  4 marks

  10

  
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  Equation and algebraic fraction.

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• Edexcel 1MA1 · November 2019 · Paper 3H · Q15 · 3 marks

  3 marks

  15

  
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  Consecutive even numbers.

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• Edexcel 1MA1 · November 2019 · Paper 3H · Q17 · 4 marks

  4 marks

  17

  
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  The graph y=x^2-3 is supplied on axes with -4≤x≤4 and -4≤y≤10. The smooth parabola passes through the neutral reconstruction points (-3,6), (-2,1), (-1,-2), (0,-3), (1,-2), (2,1) and (3,6).

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• Edexcel 1MA1 · November 2019 · Paper 3H · Q19 · 3 marks

  3 marks

  19

  
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  A distance-time graph for part of a cyclist's journey has time in seconds on the horizontal axis (0 to 9) and distance travelled in metres on the vertical axis (0 to 20). The graph is a smooth increasing curve, concave upwards, starting at (0,0) and passing approximately through (1,0.2), (2,1.0), (3,2.4), (4,4.4), (5,6.9), (6,9.8), (7,13.8), (8,18.2), reaching 20 m at about 8.4 s.

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• Edexcel 1MA1 · November 2019 · Paper 3H · Q20 · 2 marks

  2 marks

  20

  
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  Text version

  Sequence first terms: -1, 0, 3, 8, 15.

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• Edexcel 1MA1 · November 2019 · Paper 3H · Q22 · 4 marks

  4 marks

  22

  
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  Text version

  Algebraic fraction: [ (7x-14)/(x^2+4x-12) ] ÷ [ (x-6)/(x^3-36x) ].

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• Edexcel 1MA1 · June 2019 · Paper 1F · Q10 · 4 marks

  4 marks

  10

  
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  In general, all three parts of this question were well answered. The answers to part (a) and part (b) were often just written down. Working was often seen in part (c), either the traditional algebraic method or a trial and improvement method. No marks are awarded for a trial and improvement approach to solving an equation unless the answer is correct so it is not the best approach to use. Many students gave the correct answer but for those that did not and who used an algebraic approach many often wrote -2 beneath each side but then failed to do this correctly so did not score as they did not complete a correct first step. Some subtracted 2 from 6w and 2 and 20, others subtracted only from the 2.

  Text version

  Three equations.

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• Edexcel 1MA1 · June 2019 · Paper 1F · Q13 · 1 mark

  1 mark

  13

  
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  This question was very poorly answered. Only a relatively small proportion of students were able to express multiply x by 10 in an acceptable format. Some mixed up multiply with divide, e.g. x/10 was seen, others knew the connection was a factor of 10 but were unable to write an expression, e.g x cm = 10 mm was given and another incorrect response was x = 10x. There were many instances of units being included within the expression they offered and this often made the expression invalid. Sometimes m was also given in the expression. It is fair to say that the majority of foundation students found this question challenging.

  Text version

  A line has length x cm.

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• Edexcel 1MA1 · June 2019 · Paper 1F · Q16 · 2 marks

  2 marks

  16

  
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  Overall, students performed well with the expansion in part (a). However, a small number of students tried to simplify their correct answer of 10m – 15, commonly giving 15m and therefore losing the mark. Another common mistake was to only multiply the first term by 5, giving 10m – 3 or 5×2m – 3 but full processing was required. Equally, some students thought the expression was an equation which they tried to solve. Part(b) was generally answered quite well with many students gaining the mark. In general, most students either got the correct answer or did not attempt the question. Common mistakes included 3(n + 12), 3n + 4 and n + 4.

  Text version

  Algebra.

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• Edexcel 1MA1 · June 2019 · Paper 1F · Q28 · 4 marks

  4 marks

  28

  
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  Many students were able to score marks on this question. The starting point of calculating PQ as 4.5 cm was often seen. Those students that understood they then needed to use the perimeter of ABCD were usually able to find the required answer, with a small number making arithmetical errors. Some could not calculate 17 ÷ 2 or 26 − 9 accurately. Some tried to start with rectangle ABCD using trial and improvement to find the lengths of sides. Some, just guessed the sides of the rectangle, usually 10, 3, 10, 3 whilst others divided 26 by 4 to give a final answer of 6.5. A method of trial and improvement cannot score marks unless the fully correct answer is seen. For those who did not get the correct answer some struggled to understand the connection between the area given of PQRS and the side length. Many used the side length of 10 on the rectangle ABCD to give an answer of 3.

  Text version

  Rectangles ABCD and PQRS: QR=10 cm; BC=PQ; perimeter ABCD=26 cm; area PQRS=45 cm^2.

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• Edexcel 1MA1 · June 2019 · Paper 1F · Q29 · 3 marks

  3 marks

  29

  
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  This question was not well answered. There were some blank responses seen. It is not clear if the terms ‘turning point’ or ‘roots’ were understood by some students. Part (a) was usually correct or blank. Many who gave answers to part (b) stated all the intercepts with both axes or tried to give coordinate solutions. A few students tried to solve the quadratic algebraically, usually unsuccessfully. Summary Based on their performance on this paper, students should:  take care when carrying out arithmetic operations, checking answers to avoid careless errors  set out working clearly for all questions, particularly those with three or more marks  practice answering ‘explain’ and ‘give a reason for your answer’ type of questions  practice working with questions involving the speed-distance-time formula and working in fractions of an hour and decimal equivalents of time

  Text version

  The graph of y=x^2-2x-3 is supplied on axes with -2.5≤x≤4.5 and -5≤y≤7. The smooth parabola passes through the neutral reconstruction points (-2,5), (0,-3), (2,-3) and (4,5).

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• Edexcel 1MA1 · June 2019 · Paper 1H · Q7 · 4 marks

  4 marks

  7

  
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  It was pleasing that many students gained full marks for finding the length of AB, often making use of the diagram as part of their working. Most students gained at least one mark with relatively few students unable to use the information that the area of PQRS is 45 cm2 and QR = 10 cm to work out that PQ = 4.5 cm. Having found PQ = 4.5 cm the next step for many students was 26 – 9. Some formed an equation such as 2x + 9 = 26 but algebraic approaches were quite rare. Whichever approach was used most students were able to show a complete process to find the length of AB although a few got as far as 26 – 9 = 17 and stopped. Arithmetic errors when subtracting 9 from 26 (26 – 9 = 15 was common) or when dividing 17 by 2 meant that the accuracy mark was sometimes lost. A small number of students thought that BC is equal to QR or tried to work with the area of ABCD rather than with the perimeter and gained no marks.

  Text version

  Rectangles ABCD and PQRS satisfy QR=10 cm, BC=PQ, perimeter ABCD=26 cm and area PQRS=45 cm^2.

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• Edexcel 1MA1 · June 2019 · Paper 1H · Q10 · 2 marks

  2 marks

  10

  
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  Fewer students than expected realised that the estimates of the solutions of the simultaneous equations could be found from the point of intersection of the two straight lines. Many attempts at solving the equations algebraically were seen even though the question instructed students to use the graphs. Those students that did focus on the point of intersection of the two straight lines did not always score both marks. Errors were frequently made in reading the required values from the graph (x = 2.5 was a common error) and some students transposed the x and y values or omitted the minus sign from the y value. Surprisingly few students marked the graph to help them estimate the values. A common mistake was to use the points of intersection of the line 3y + 2x = 1/2 with the x-axis and y-axis instead of the point of intersection of the two straight lines. Some students gave a correct value for x using the point of intersection but then used the y-intercept as the value for y.

  Text version

  The graphs 3y+2x=1/2 and 2y-3x=-113/12 are drawn on a grid with -1≤x≤3 and -3≤y≤1. For reconstruction, the first line runs from (-1,5/6) to (3,-11/6), while the second runs from (41/36,-3) to (3,-5/24).

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• Edexcel 1MA1 · June 2019 · Paper 1H · Q13 · 2 marks

  2 marks

  13

  
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  This question was answered surprisingly poorly. Many students did not appreciate what is required in a proof and simply substituted different values of n into n2 – n and stated that the result was never odd. Such responses were very common and gained no marks. Students who gained one mark for factorising n2 – n to n(n – 1) were often unable to complete the proof by explaining that n(n – 1) must be even because either n or n – 1 must be even. Fully correct proofs were often based on correct reasoning for n even, eg even × even = even and even – even = even with similar reasoning for n odd. A number of students gave reasoning for n odd or for n even but not for both. The use of general expressions for even and odd numbers such as 2n and 2n + 1 was quite common. For many students this was not a wise choice because this approach is unnecessarily time consuming and error prone. For n odd, for example, it is necessary to simplify (2n + 1)2 – (2n + 1) to 4n2 + 2n and then show that 4n2 + 2n is always even, which is usually done by factorising. Mistakes in simplifying and factorising were common, often arising from a failure to include brackets around (2n + 1). Reasoning for n even often went wrong at the first step with students writing 2n2 – 2n rather than (2n)2 – 2n. Some students completed an argument for n odd but didn’t realise that a similar argument is needed for n even. Attempts at an algebraic proof using n and (n + 1) were quite common and some students simply stated the algebraic statement n2 – n in words.

  Text version

  n is any integer with n≥1.

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• Edexcel 1MA1 · June 2019 · Paper 1H · Q17 · 4 marks

  4 marks

  17

  
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  Many of the students who used the given statement, x2 : (3x + 5) = 1 : 2, to form a correct equation were able to give fully correct solutions. Marks were sometimes lost through factorising 2x2 – 3x – 5 = 0 incorrectly. Some students gained only the first mark because they could not write their equation (which was often 2x2 = 3x +5) in a suitable form ready for solution. Many students scored no marks at although some of these students had attempted to form an equation. Common incorrect equations were x2 = 2(3x + 5) and x2 + 3x + 5 = 3 and some students used x2 = 1 and 3x + 5 = 2. Many students, though, simply substituted different values of x into the given statement. This approach sometimes resulted in an answer of x = –1 but gained no marks because two values of x are needed.

  Text version

  Given x^2:(3x+5)=1:2, find the possible values of x.

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• Edexcel 1MA1 · June 2019 · Paper 1H · Q19 · 3 marks

  3 marks

  19

  
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  A large number of students did not recognise this as a completing the square question or understand how to complete the square. Students who did tackle part (i) by completing the square were generally the most successful and many earned the method mark for getting as far as (x – 3)2. A common mistake made by students who completed the square correctly to get (x – 3)2 – 8 was writing a = – 3 and b = –8 on the answer lines. These students lost the accuracy mark. Those who started by expanding (x – a)2 – b were less successful. Errors were frequently made in the expansion but many of the students who did get to x2 – 6x + 1 = x2 – 2ax + a2 – b went on to rearrange the equation and did not identify a = 3. Students who had found (x – 3)2 – 8 in part (i) were often able to identify the coordinates of the turning point as (3, –8) in part (ii). It was not unusual to see students who had used an incorrect method in part (i) complete the square in this part to find the coordinates of the turning point. Any working in part (ii) could not be credited in part (i). An answer of (–3, 8) was a common incorrect answer following a correct answer in part (i).

  Text version

  For all x, x^2-6x+1=(x-a)^2-b.

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• Edexcel 1MA1 · June 2019 · Paper 1H · Q21 · 7 marks

  7 marks

  21

  
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  Students who knew about inverse functions usually answered part (a) by rearranging y = 3x – 1 to make x the subject or rearranging x = 3y – 1 to make y the subject. Some mistakes were made in the first step with y = 3x – 1 being followed by y – 1 = 3x for example. Most of the students who made a correct first step went on to give a correct answer. Those who completed the rearrangement correctly sometimes gave the answer in terms of y and lost the accuracy mark. Some students found the inverse function by using a flow diagram but this approach appeared to be less successful. A common incorrect answer was 1/(3x – 1) because some students interpreted f−1 (x) as 1/f(x), confusing the notation for inverse functions with the notation for a negative power. Part (b) was very well answered considering that it is one of the more challenging questions on the paper. It was pleasing that many of the students who demonstrated knowledge of composite functions were able to give fully correct answers. Working was generally set out well when students understood what to do. A few otherwise correct solutions were spoilt by a final line that was not 15x2 – 12x – 1 = 0 and these gained 4 of the 5 marks. Most errors occurred in the simplification of gf(x) or 2gf(x) and in working that was poorly set out. Incorrect terms in the expansion and simplification of (3x – 1)2 + 4 were quite common. Sometimes only (3x – 1)2 was multiplied by 2. If the composite functions were not labelled as fg(x) and gf(x) then this could be implied by the setting up of the equation fg(x) = 2gf(x) or by multiplying the correct composite function by 2. If labelling was not seen or implied then the answer scored no marks. Students with little or no idea of composite functions interpreted fg(x) as f(x)×g(x) or started with 15x2 – 12x – 1 = 0 and attempted to solve it.

  Text version

  f(x)=3x-1 and g(x)=x^2+4.

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• Edexcel 1MA1 · June 2019 · Paper 1H · Q22 · 5 marks

  5 marks

  22

  
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  This proved to be a challenging question. Many students failed to find a successful strategy and gained no marks. Working out was often messy and difficult for examiners to follow. Students attempting an algebraic approach often started by forming two equations such as g/(r + g) = 3/7 and (g + 3)/(r + g + 5) = 6/13. To solve these equations students needed to deal with the fractions and eliminate one of the variables and a successful outcome proved to be beyond many. Students who formed the equations g/r = 3/4 and (g + 3)/(r + 2) = 6/7 had a slightly easier route to a solution. Some students formed only one equation, using either 3/7 or 6/13, and scored one mark only. A different approach used by some students was to form an equation such as (3x + 3)/(7x + 5) = 6/13. These students were generally more successful. Some of the students that attempted an algebraic approach were not able to derive the necessary equation(s). Answers of 12 red counters and 9 green counters did not always come from an algebraic approach. Some students realised that the number of counters in the bag at the start must be a multiple of 7 and the number of counters in the bag at the end must be a multiple of 13. Since 5 counters had been added they knew that they were looking for a multiple of 13 that is 5 more than a multiple of 7 and quickly came up with 26 and 21. These students usually went on to score full marks. Many of the unsuccessful attempts were based on using fractions with a denominator of 91 (from 7 × 13). Since 3/7 = 39/91 and 4/7 = 52/91 a common incorrect answer was 52 red counters and 39 green counters. There were many attempts at representing the information in tree diagrams. These were generally unhelpful. It seemed that most students wanted to jump straight to probability tree diagrams and had no other strategies to use. Summary Based on their performance on this paper, students should:  practise working out estimates by rounding numbers and develop an understanding of the purpose of rounding so that they can choose appropriate rounded values  have a greater awareness of the graphical solution of simultaneous equations  practise writing clear concise explanations when using the median and measures of spread to compare two sets of data  learn what is required in a proof as opposed to giving specific numerical examples  memorise the necessary exact trig values or learn how to work them out

  Text version

  A bag initially has r red counters and g green counters only. P(green)=3/7. The counter is replaced, then 2 red and 3 green counters are added; now P(green)=6/13.

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• Edexcel 1MA1 · June 2019 · Paper 2F · Q7 · 1 mark

  1 mark

  7

  
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  This question was not well answered; y = 7 and 7 + y were the most common incorrect answers seen.

  Text version

  There are y boats on a lake and 7 people in each boat.

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• Edexcel 1MA1 · June 2019 · Paper 2F · Q8 · 4 marks

  4 marks

  8

  
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  Parts (a) and (b) were generally well answered. In part (a) a common error was to leave a multiplication sign in the answer, for example, 7 × ab and in part (b), incorrect answers of 3y and y3 were often seen. Part (c) was poorly answered although many students were awarded one mark usually for either a correct numerator or a correct denominator. The most common mistakes made were either to simply count the number of each variable, for example, 3 2 2 ef e f or to sum the powers, e f e f   both of which received no credit. An answer of 𝑒ଵ𝑓ି ଵ was given full marks although writing terms with an index of one is not to be encouraged. Cancelling of terms in the original expression was a correct method but this had to be carried out completely to gain any credit. The partially simplified answer of e f e f was very common and gained one mark. Another common incorrect answer was e f , where students had clearly just counted the number of each variable.

  Text version

  Simplify algebraic products and a quotient.

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• Edexcel 1MA1 · June 2019 · Paper 2F · Q11 · 2 marks

  2 marks

  11

  
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  Although a correct answer of 23 was often seen, many students are clearly unable to evaluate expressions by substitution, particularly when negative values are included. 3 × – 4 was often seen written as 3 – 4 leading to an incorrect answer of 34 (35 – 1). Many students ignored the negative sign and simply worked out 7×5 + 3×4, 47 being a very common incorrect answer. A number of students were unable to deal with the addition of a negative number and incomplete answers of 35 + – 12 were not uncommon; which is a concern especially given that this is a calculator paper. Only a very few students wrote an incorrect substitution of 75 – 34 or 75 + 34. Some weaker students wrote 7 + 5 + 3 – 4 = 11, scoring no marks.

  Text version

  P=7r+3q.

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• Edexcel 1MA1 · June 2019 · Paper 2F · Q20 · 5 marks

  5 marks

  20

  
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  Both parts of this question were poorly answered. In part (a), correct manipulation of the inequality was rare although some did show an intention to subtract 11n from both sides. However, many chose wrongly to subtract 6 from both sides. Algebraic methods did sometimes lead to n = 2 for the award of one mark. Sometimes this value was found by trial and improvement methods but this method often led to an incorrect answer of n = 3. In part (b), very few students solved the two inequalities and values of – 5 and 1 were rarely seen. The most common attempt was a line drawn between – 2 and 4 sometimes with the correct circle notation, gaining one mark. There were several attempts to select integer values satisfying the inequalities but this rarely resulted in any credit being earned. Some students put arrows or crosses on the number line showing a complete lack of understanding of the correct convention.

  Text version

  Solve inequalities.

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• Edexcel 1MA1 · June 2019 · Paper 2F · Q21 · 3 marks

  3 marks

  21

  
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  It was pleasing to see so many correct graphs drawn, although it is still a surprise that so many students plot the correct points and then fail to join them up, denying themselves of an easy additional mark. Incorrect points plotted were usually the inability of some students to work with negative numbers. Disappointing also, to see that those with one or two incorrect plots failed to realise that a straight line was the correct answer. A few students drew only part of the line, failing to go completely from x = – 2 to x = 4. Some students used the – 2 and 4 to mean co-ordinates and plotted them. A number of responses saw the students confused due to the difference in scales on the two axes.

  Text version

  For -2≤x≤4, draw y=2x-3 on the supplied coordinate grid.

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• Edexcel 1MA1 · June 2019 · Paper 2F · Q28 · 3 marks

  3 marks

  28

  
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  Both parts of this question were poorly answered. In part (a), 21 (15 + 6) or 18 (15 + 3) were common 6th terms. In part (b), many chose incorrect patterns to follow; 2a, 3a, 3a, 4a, 4a, etc. and 3a, 4a, 5a, 6a, etc. and 4a, 6a, 8a, 10a, etc. were typical incorrect patterns seen. Students who got the correct pattern, perhaps with no understanding of a Fibonacci sequence, usually succeeded in both parts.

  Text version

  A Fibonacci sequence begins 3,3,6,9,15. Another begins a,a,2a.

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• Edexcel 1MA1 · June 2019 · Paper 2H · Q1 · 5 marks

  5 marks

  1

  
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  Part (a) was answered relatively well; many students were able to solve the inequality. However, a significant proportion of students either made a mistake in the isolation of n, or used the wrong symbol in their answer. There were also a surprising number of students who got to 3n > 6 but then had a final answer of n > 3. Part (b) was generally poorly answered. A majority of students did not deal with the inequality first to get to x alone in the middle. These students typically plotted the inequality −2 < x ≤ 4 to score one mark only. Most of those students who correctly worked with the inequality to rearrange it first to −5 < x ≤ 1 went on to score full marks. Occasionally students used the wrong type of circles at the ends and scored two marks for the line connecting −5 and 1. A small number thought integers were needed and used −4 as a lower limit rather than −5 (losing 2 marks as they didn't show −5 < x).

  Text version

  Solve inequalities.

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• Edexcel 1MA1 · June 2019 · Paper 2H · Q2 · 3 marks

  3 marks

  2

  
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  This linear graph question was incredibly well answered, with a great number of students scoring all three marks for the correct graph. Those who dropped marks did so for a variety of reasons, the most frustrating of which was when the correct line was plotted but not all the way from x = −2 to x = 4 as required by the question and thus scored two marks only. It was also relatively common to see students not correctly interpreting the scale on the x axis and, as a result, draw a graph with the wrong gradient but correct intercept. The mark scheme allowed such students to score two of the three marks. A few non-linear graphs were seen, and very few students used the space at the top to calculate values; those who did often got the values for y when substituting negative values of x wrong. Students need to learn how to use calculators to work with negative numbers.

  Text version

  For -2≤x≤4, draw y=2x-3 on the supplied coordinate grid.

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• Edexcel 1MA1 · June 2019 · Paper 2H · Q13 · 4 marks

  4 marks

  13

  
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  This algebraic fraction problem was answered quite well, with the majority of students attempting it, and most gaining at least some credit. The question had three steps of method and each was awarded a method mark. Students needed to factorise the given quadratic expression. This could be done at any stage. The second step was the need for a method to divide fractions, for example by multiplying by the reciprocal of the second fraction. The final step was to be able to add a fraction to an integer by using a common denominator. The first two steps were very commonly seen and for most, at least one mark was awarded. The division mark was often lost by failing to deal with the reciprocal of ( 1) ( 5) "( 5)( 2)" x x x x      having cancelled the (x + 5). The third mark was the hardest to achieve, with students regularly losing the addition sign and then multiplying in the last step. It was disappointing to see a number of those who, having gained the first three marks, then lose the final one by either failing to correctly simplify the numerator, or for adding denominators.

  Text version

  Show that 6 + [(x+5) ÷ ((x^2+3x-10)/(x-1))] simplifies to (ax-b)/(cx-d), where a,b,c,d are integers.

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• Edexcel 1MA1 · June 2019 · Paper 2H · Q14 · 7 marks

  7 marks

  14

  
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  This new topic was a challenge to many students. A large number failed to draw a tangent, often simply dividing the speed at t = 15 by 15, and so scored zero marks. Those who did draw a tangent then often scored all three marks. Some students took little care over the reading of the scales and therefore had an incorrect method for their gradient. It is important to realise that these questions need to have a tangent drawn. Several students gave an answer within the acceptable range, but they scored no marks as their answer clearly came from an incorrect method. Part (ii) proved more challenging than the earlier one. This is a new topic and many students do not fully understand that gradients represent the instantaneous rate of change. In this context that being acceleration. Part (b) brought in another new topic and students appeared better prepared than in previous series. However, the method for finding the area under the curve was often crude. Many students attempted to use rectangles, and whilst this does give an estimate, it is not a very good one. These students were able to score two marks typically for the processes, but their answer was usually outside the range given in the mark scheme. Students should be encouraged to use trapezia to find a better estimate. Many responses showed attempts to estimate distance using distance = speed × time. Many forgot to divide the triangle product for the first strip by two.

  Text version

  A car moves from rest. Its speed-time graph has time t in seconds on the horizontal axis (0 to 40) and speed v in m/s on the vertical axis (0 to 25). The source graph is a smooth curve passing approximately through (0,0), (5,4), (10,12), (15,18), (20,20), (22,20), (25,19.8), (30,18), (35,14.5), (40,12.7).

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• Edexcel 1MA1 · June 2019 · Paper 2H · Q15 · 3 marks

  3 marks

  15

  
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  At this stage in the paper, the algebra proved a challenge to many. That said, almost all students attempted this question, and a good number gained the first mark at least. Those who didn’t get past the first mark, and there were a good number, struggled because they did not use a bracket when multiplying by “m – 1”. This meant that from that point on their terms were incorrect as they had “−1” rather than “−f”. As is often seen with these harder rearranging questions, many students failed to factorise once terms had been isolated and therefore lost the final mark.

  Text version

  f=(3m+4)/(m-1).

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• Edexcel 1MA1 · June 2019 · Paper 2H · Q16 · 3 marks

  3 marks

  16

  
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  It is evident that most students have some knowledge of y = mx + c and the properties of the gradients of perpendicular lines, but many got into trouble because of the form in which line L was given. A good number of students failed to rearrange the given equation to find the gradient, often working with 4. These students were able to gain the second mark, provided they had clearly stated the gradient of L as being 4. Failing to do that meant they could gain no credit for their knowledge of the properties of the gradients of perpendicular lines. The students who knew the relevant property often scored the second mark, however, quite a number forgot either the negative or the reciprocal and only used one of these aspects, meaning the second mark was not awarded. There was no prerequisite in the question regarding the form of the final equation, and as such, any equivalent equation was accepted for the accuracy mark.

  Text version

  Line L has equation 3y=4x+7. Point A is (3,-5).

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• Edexcel 1MA1 · June 2019 · Paper 3F · Q9 · 2 marks

  2 marks

  9

  
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  This was well answered for an algebra question, with many gaining the full two marks. If only one mark was awarded this was usually for 11e. The most common incorrect answers were 11e – 5f or 11e5f.

  Text version

  Simplify 4e + 6f + 7e - f.

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• Edexcel 1MA1 · June 2019 · Paper 3F · Q13 · 2 marks

  2 marks

  13

  
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  The most common answer in part (a) was 40, though some gave 80. A significant minority lost the mark by giving the full sequence of numbers, but not highlighting 40 as their chosen answer. Part (b) was also well attempted but at times was poorly explained by students who would just say they counted up. The most common explanations included the numbers 80 and 88, or focussed on the fact that the sequence was all even numbers, factors of 8, or that 85 was an odd number. Students who deduced the algebraic rule and did a division to show it was not an integer could also gain the mark, as long as their working was correct.

  Text version

  A sequence starts at 24 and has term-to-term rule 'add 8'. Josie says, 'No number in this sequence is in the 5 times table.'

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• Edexcel 1MA1 · June 2019 · Paper 3F · Q15 · 4 marks

  4 marks

  15

  
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  This was a high scoring question, in both parts. A significant minority of students lost marks through mis-reading the figures in the question, typically using 5 instead of 50. The most common error was adding 70 and 50 first before multiplying by 4. In part (b) the most common method was 680 – 50 then to divide by 70. Some students did 9 × 4 × 70 + 50 which was probably the result of trialling rather than making use of inverse operations. A common error was to double the answer to part (a), concluding that if 4 weeks cost £330 then 8 weeks must cost £880.

  Text version

  The total hire charge, in pounds, for a 3D printer is given by: total charge = 70 x number of weeks + 50.

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• Edexcel 1MA1 · June 2019 · Paper 3F · Q19 · 2 marks

  2 marks

  19

  
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  This was very challenging for many Foundation students. It was common to see the correct steps to rearrange written down but they were either not carried out or not completed correctly. A few stopped at 2x = y – 4 In most cases the first rearrangement was −4 followed by division of 2, but many spoilt this by writing 4 – y instead of y – 4. This remains an area of content that is weak.

  Text version

  Make x the subject of the formula y = 2x + 4.

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• Edexcel 1MA1 · June 2019 · Paper 3F · Q30 · 3 marks

  3 marks

  30

  
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  There were many who failed to attempt this question. There was unfamiliarity about adjusting the equations to give the same coefficient of x or y, evidenced when students started to manipulate just one of the equations, or tried a trial and improvement approach. Of those who made a good start, it was not uncommon to find them making an arithmetic error or dropping a minus sign. If this was the only error they could then go on to the next step and potentially gain two of the three marks. For the second stage in their calculation, there were as many who started again with elimination as there were who substituted back in, though substitution was done with greater success. Summary Based on their performance on this paper, students should:  ensure they transcribe figures accurately both within their working and when copying figures from a calculator  avoid truncating or rounding values until the final answer  ensure that they know how to use their calculator  ensure that figures are written clearly, for example, avoid any possible confusion when writing 1s and 7s, when writing 4s and 9s, and even when writing 0s and 6s  include working out to support answers  practise questions on topics such as algebraic manipulation and derivation, and application of ratios, scaling and rates  practise answering response type questions where a written explanation is required

  Text version

  Solve the simultaneous equations 3x - 4y = 11 and 9x + 2y = 5.

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• Edexcel 1MA1 · June 2019 · Paper 3H · Q5 · 4 marks

  4 marks

  5

  
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  A good discriminator, this question attracted many concise and fully correct solutions. Despite there being a number of possible routes that students could take, there were also many incorrect or incomplete solutions. The most common approach was to use the total of the angles in a hexagon together with the relationship between angles BCE and CDE to work out the size of angle AFE. The most commonly seen alternative route was to use the symmetry of the hexagon to form two pentagons and work from there. A minority of students joined the points F and C and found the size of angle BCE. These students did not usually make significant further progress beyond labelling angles BCD and CDE as 2x and x respectively. A surprising number of students marked angles BCD and CDE as x and 2x respectively. In most cases, however, labelling of the diagram proved to be a help to students and is to be encouraged. About two in every five of students obtained full marks for their solution to this question.

  Text version

  ABCDEF is a hexagon with one line of symmetry. The symmetry pairs A with B, F with C, and E with D. FA is parallel to BC, and EF is parallel to CD. Angle ABC = 117 degrees. Angle BCD is twice angle CDE.

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• Edexcel 1MA1 · June 2019 · Paper 3H · Q15 · 2 marks

  2 marks

  15

  
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  Some students appeared to have little knowledge about the content covered in this question and either did not attempt the question or revealed little understanding through their responses to the question. In general, little working was seen and students may have benefited from writing down the transformations represented by the equations y = f(−x) and y = f(x) – 3. Very few students did this and so some denied themselves the award of a mark for showing an understanding of the types of transformation involved in the question. However, examiners were able to award a good number of students at least one mark for getting an x coordinate as −7 or a y coordinate as −1. Students who gave a final answer of (−10, −5) following on from an intermediate stage of (−7, −2) could not be awarded any marks because they had not discriminated between which coordinates were changed by which transformation. Some students attempted to substitute values for x and y, often 7 and 2 respectively. The best students did show a good understanding of the notation and transformations involved and gave a fully correct answer.

  Text version

  The point (7, 2) lies on the curve y = f(x).

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• Edexcel 1MA1 · June 2019 · Paper 3H · Q16 · 3 marks

  3 marks

  16

  
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  This question discriminated well with many students being awarded each of the 0, 1, 2 or 3 marks available. Most students followed the strategy of finding second differences to help them find the term in n² and realised that the common differences of 4 meant they should write an expression for the nth term which included the term 2n². Many students went on to give a fully correct expression but a significant minority of students gave 2n2 − 3n as their answer. Students would be wise to check their answers to questions of this type by substitution.

  Text version

  The first six terms of a quadratic sequence are -1, 5, 15, 29, 47, 69.

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• Edexcel 1MA1 · June 2019 · Paper 3H · Q17 · 2 marks

  2 marks

  17

  
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  Matching the functions to the graphs in this question produced a good spread of marks with about a third of all students scoring each of the marks available. The graph of a trigonometrical function was identified correctly more frequently than the other graphs and the graphs showing one form of proportionality or another were the least well known.

  Text version

  Four graphs are labelled A to D. Graph A crosses the x-axis at a negative x-value, falls to a minimum in quadrant III, passes through the origin, rises to a maximum in quadrant I, and returns to the positive x-axis. Graph B has two branches, one in quadrant I and one in quadrant III, with the coordinate axes as asymptotes. Graph C starts at the origin and increases in quadrant I with decreasing gradient. Graph D is positive and increasing, crosses the y-axis above the origin, and approaches the x-axis for negative x-values.

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• Edexcel 1MA1 · June 2019 · Paper 3H · Q18 · 6 marks

  6 marks

  18

  
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  Students usually scored well on part (a) of this question requiring the expansion of a product of three linear expressions to give a fully simplified cubic expression. Errors were usually restricted to incorrect terms rather than a flawed strategy although some students omitted terms from their expansion. It was usual for students to earn two or three of the available marks. Less able students lacked a clear strategy and sometimes tried the multiply all three brackets together at once. Part (b) of the question proved to be more challenging than part (a) and it was relatively rare for examiners to award full marks. Credit awarded to students was often restricted to the mark gained for taking the square root of each side of the inequality or, in the case of students who started by expanding the left hand side, for getting a quadratic equation or inequality with all the terms on one side. Very few students used the expected method of taking the square root of both sides, taking into account both square roots of ଽ ଶହ. Students who expanded (1 – x)² were more likely to gain two or three marks for their responses but only about a third of all students earned one or more marks. The best answers demonstrated a very good understanding of the topic together with the good practice of including a graph sketch with the rest of their working.

  Text version

  Expand (2x + 1)(x + 3)(3x + 7), then solve an inequality.

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• Edexcel 1MA1 · June 2019 · Paper 3H · Q20 · 5 marks

  5 marks

  20

  
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  This question also proved to be a good discriminator between the most able students. Many of the more able students were able to make a good start, realizing the need to make a substitution, usually using y x   from the second equation, to eliminate x from the first equation. Fewer students could carry out the substitution and expand ቀ ଻ିସ௬ ଷቁ ଶ correctly to score the second method mark. Often the denominator of the fraction was ignored in this process. Students who rearranged 3x + 4y = 7 often stopped at 4y = 7 – 3x, then substituted this into x2 − 4y 2 = 9 to give the incorrect equation x2 − (7 – 3x)(7 – 3x) = 9. A further error seen quite commonly occurred when students multiplied through their equation by 9 but forgot to multiply all the terms by 9. Only the most able students could obtain a correct quadratic equation in the form ay² + by + c = 0 (or ax² + bx + c = 0 in the cases where y had been eliminated). Some students who did not get a correct quadratic equation were able to score a further mark which was given independently for solving a three term quadratic equation. Full marks were scored in this question by only a very small proportion of all students who were entered for the examination. Some students who arrived at the correct 4 values were not awarded the final accuracy mark as the values were not clearly paired.

  Text version

  Solve algebraically the simultaneous equations x^2 - 4y^2 = 9 and 3x + 4y = 7.

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• Edexcel 1MA1 · June 2019 · Paper 3H · Q22 · 4 marks

  4 marks

  22

  
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  Fully correct answers to this question were seen only rarely. It was found to be the most challenging question on the paper. Having said that, a significant proportion of students were able to score at least one mark for a successful start by identifying that angle OAB is a right angle or for a correct process to find the radius of the circle. Few students could see how to use the coordinates of P in order to derive an equation by either using the equation of the circle or by using Pythagoras’s Theorem. The evidence from students’ scripts showed that more students took the former approach, quoting x2 + y2 = r2 and then substituting x = 3p, y = p, r = 8 to get an equation in p. Unfortunately, the lack of of use of correct notation, that is a pair of brackets, meant that many students wrote 3p2 + p2 = 82 instead of (3p)² + p2 = 8². These students usually went on to simplify their equation to 4p2 = 82.

  Text version

  A circle has centre O = (0, 0). The point P = (3p, p) lies on the circle. The point B = (16, 0), AB is tangent to the circle at A, and angle OBA = 30 degrees.

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• Edexcel 1MA1 · November 2018 · Paper 1F · Q8 · 3 marks

  3 marks

  8

  
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  This question was generally well answered. The majority of students were able to write down the coordinates of point A and to mark point B on the grid. The most common incorrect response in part (a) was (−1, −2) and in part (b) some students plotted the point with coordinates (3, 2). Part (c) was less well done. Many students either drew the line y = −4 or another line which passed through (0, −4). A number of students plotted a point at (−4, 0) rather than the line required.

  Text version

  On a coordinate grid, point A is at (-2,-1).

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• Edexcel 1MA1 · November 2018 · Paper 1F · Q9 · 2 marks

  2 marks

  9

  
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  This question acted as a good discriminator. It was encouraging to see clear substitutions shown in the working space and this usually led to students scoring at least one mark for their response. The majority of students went on to obtain the correct value. However, some weak students added 29 and 34 or gave answers which were expressions in g and h.

  Text version

  g = 9 h = 4 Work out the value of 2g + 3h

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• Edexcel 1MA1 · November 2018 · Paper 1F · Q12 · 3 marks

  3 marks

  12

  
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  There were many fully successful answers to this question though a good proportion of students could not relate the idea of a “fixed charge” to the graph and there were some students who gave £40 as their answer to part (b) this being the delivery cost corresponding to a distance of 20 miles instead of the difference between two delivery costs. Some students interpreted the vertical scale on the graph incorrectly and used 2 mm to represent £1.

  Text version

  A delivery-cost graph is a straight line through (0 miles, GBP 10) and (60 miles, GBP 100).

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• Edexcel 1MA1 · November 2018 · Paper 1F · Q21 · 4 marks

  4 marks

  21

  
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  Only the most able students sitting this paper were able to score both marks in part (a) of this question. However, most students scored at least one mark for a correct substitution of the values into the formula. Students who started to work out the value of v2 by breaking the formula up into parts were less likely to score this mark than those who started by substituting values into the equation before attempting to break it up and make any evaluation. Common errors which prevented students from getting full marks included stating 122 = 24, working out 2 – 3 × 18 for 2as or multiplying both –3 and 18 by 2. Part (b) of the question was also well done by some students but many students were unable to make a correct first step and so could not be given any credit for their attempt. The most common errors seen included the subtraction of 2a rather than division by it and having u2 - v2 instead of v2 - u2 as part of their rearranged equation.

  Text version

  v^2 = u^2 + 2as, where u = 12, a = -3 and s = 18.

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• Edexcel 1MA1 · November 2018 · Paper 1F · Q25 · 3 marks

  3 marks

  25

  
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  This standard question on solving simultaneous equations was attempted by a large proportion of students. Unfortunately, it was only rarely completed without error. Students sometimes tried to add or subtract the original equations, without multiplying first. Those students who did make better progress were often confused about which operation to use once the coefficients of x or y were equal and there were many errors when dealing with signs. There were very few completely correct solutions. A very small number of students successfully used a method involving substitution. Trial and improvement methods were also seen but were largely unsuccessful.

  Text version

  Solve the simultaneous equations 5x + y = 21 and x - 3y = 9.

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• Edexcel 1MA1 · November 2018 · Paper 1H · Q2 · 4 marks

  4 marks

  2

  
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  Part (a) was not well answered as many students were unable to deal correctly with the negative result of multiplying -3 × 18 by 2; 144 + 108 = 252 was a common error. Some students who correctly evaluated 144 – 108 = 36 then failed to even consider the square root. However, the great majority of students gained one mark for a correct initial substitution. Even so, it was disappointing to see 2 × -3 × 18 written as 2 × -3 +/× 2 × 18 so many times. In part (b), most students were able to score at least one mark in this question for a correct first step. This was usually by correctly subtracting u2 from both sides of the equation. Those choosing to divide both sides of the equation by 2a, often forgot to divide the u2 term and so gained no credit. Some students wrote v2 + u2 = 2as as their first step, again gaining no credit.

  Text version

  v^2 = u^2 + 2as, where u = 12, a = -3 and s = 18.

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• Edexcel 1MA1 · November 2018 · Paper 1H · Q6 · 3 marks

  3 marks

  6

  
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  Only a few students were able to score full marks on this question. The vast majority clearly knew how to eliminate a variable but failed to score any credit as a result of too many arithmetic errors. Again, working with negative values and non-integer answers proved too much for many students; many stopped at, for example, 16, failing to realise this would cancel down to 2 or 4.5, then started the process again, only to be stopped by another fraction answer usually 16 . Some were able to pick up two marks, usually by just making one subtraction error, and then going on to correctly substitute their found value. A significant number of students started from scratch to find the value of the second variable, rather than using the substitution method.

  Text version

  Solve the simultaneous equations 5x + y = 21 and x - 3y = 9.

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• Edexcel 1MA1 · November 2018 · Paper 1H · Q10 · 3 marks

  3 marks

  10

  
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  The first step of many students, in part (a), was to expand the denominator. Only a minority cancelled (x – 1) and those that did often ignored the fraction giving 5(x – 1) or equivalent as their answer. In part (b), many students picked up one mark for 2(25 – y2) or other partial factorisations eg. (10 + 2y)(5 - y) but could go no further. A significant number thought that y2 – 25 was the same as 25 – y2. Only a few students were able to score full marks.

  Text version

  Simplify an algebraic fraction, then factorise a difference of two squares.

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• Edexcel 1MA1 · November 2018 · Paper 1H · Q17 · 4 marks

  4 marks

  17

  
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  Some students realised the need to substitute the given points into the equation in an attempt to find the values of a and b. Once found, it was rare to see an acceptable method leading to the coordinates of the turning point. Some students successfully used –b/2a to find the x co-ordinate.

  Text version

  The graph of y = x^2 + ax + b passes through (0, -5) and (5, 0).

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• Edexcel 1MA1 · November 2018 · Paper 1H · Q18 · 2 marks

  2 marks

  18

  
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  In part (a), it was pleasing to see a good number of students attempting to sketch the required graph in the right place. The most common mistakes were translating the graph 2 units in either the negative x-direction or the negative y-direction. Part (b) was a different story with only a very few gaining any credit, y = - g(x) was a common error.

  Text version

  The graph y = f(x) is an upward-opening parabola segment with endpoints (-1, 3) and (3, 3), roots (0, 0) and (2, 0), and minimum point (1, -1). Separately, graph A passes through approximately (-3, 3), (-1, 0) and (0, -0.5), while graph B passes through approximately (0, -0.5), (1, 0) and (3, 3). Graph A has equation y = g(x).

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• Edexcel 1MA1 · November 2018 · Paper 1H · Q19 · 4 marks

  4 marks

  19

  
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  Of those that attempted to answer part (a), the most common incorrect approach was to find either the product or sum of the two functions or to equate the two functions. Students showing some understanding of combining functions often gained full marks; however, some did attempt to find fg(x) by mistake. Part (b) was poorly answered with the most common error simply to substitute x = 7 into g(x). Some did get the inverse operations in the wrong order giving 𝑥+1 2 . Some interpreted g-1(x) as the reciprocal of g(x).

  Text version

  f(x) = (x + 1)^2 and g(x) = 2(x - 1).

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• Edexcel 1MA1 · November 2018 · Paper 1H · Q22 · 6 marks

  6 marks

  22

  
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  Again, very few students were able to seriously attempt to answer this question. Many found the “12” a distraction and insisted upon using it to develop numerical probabilities. Of the very few making a correct start it was rare to see a solution beyond the product of two correct probabilities. Summary Based on their performance on this paper, students should: • Read questions carefully and ensure that their answers address the requirement of the question. • Avoid introducing a value for pi when pi is part of the answer. • Practise working out estimates by rounding numbers and develop an understanding of the purpose of rounding so that they can choose appropriate rounded values. • Practise substituting negative values into formulae and work with calculations involving negative numbers. • When using trigonometry, students should be encouraged to quote trigonometric equations before any attempting a calculation.

  Text version

  There are x green pens and x + 3 blue pens, with more than 12 pens in total. Two pens are taken at random without replacement. The probability that they have the same colour is 27/55.

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• Edexcel 1MA1 · November 2018 · Paper 2F · Q7 · 2 marks

  2 marks

  7

  
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  In part (a) many students simplified the expression correctly. A few students gave only partially simplified expressions, e.g. 6m – 2m or m + 3m. Part (b) was also answered well. Some students did not simplify 2 × n × p × 4 fully, giving answers such as 2n × 4p or 8 × n × p, and did not gain the mark.

  Text version

  Simplify the following expressions.

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• Edexcel 1MA1 · November 2018 · Paper 2F · Q9 · 4 marks

  4 marks

  9

  
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  In part (a) many students were unable to explain why 21 is not a term of the sequence because they didn’t know how to use the nth term. Nevertheless, some very good full explanations were seen. The most common approach was to show that the 5th term is 19 and the 6th term is 22. Some students formed the equation 3n + 4 = 21 and solved it to show that n would not be an integer. Many students gave a partial explanation and scored one mark. Common partial explanations included finding just one term of the sequence, stating that 21 is a multiple of 3 and interpreting the nth term as multiplying by 3 and adding 4. Incorrect explanations were usually due to the nth term being interpreted wrongly. A common mistake was to state that the sequence goes up in 4s and some students wrote out the sequence 7, 11, 15, 19, 23. A wide range of approaches were seen to continue the sequence in part (b) and these included the use of both conventional and creative (or unconventional) term to term rules. Students who continued the sequence using a standard rule of doubling or increasing the difference by 1 were often successful at generating two correct terms and explaining the rule used. Students who used this approach often gave an ambiguous statement or a contradictory nth term rule and subsequently lost a mark. Students should check that their rule works for each of the given terms and their continuation, particularly if attempting to state an nth term rule. For the students who used a more creative approach of +1 then +2, the common loss of a mark was due to not showing that their rule repeated itself every two terms.

  Text version

  A sequence has nth term 3n + 4. A different sequence begins 1, 2, 4.

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• Edexcel 1MA1 · November 2018 · Paper 2F · Q10 · 3 marks

  3 marks

  10

  
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  Part (a) was answered very well. Some students misread the question and worked out the input when the output is 8. Part (b) was also answered very well. Some students found flow diagrams useful. The incorrect answer 28.4 was given by some students and scored no marks unless supported by working of 28 + 2 ÷ 5. Another common mistake was to subtract 2 (instead of adding 2) before dividing by 5. Some students used 28 as the input and worked out the output as 138.

  Text version

  Here is a number machine. input ×5 –2 output

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• Edexcel 1MA1 · November 2018 · Paper 2F · Q19 · 4 marks

  4 marks

  19

  
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  Part (a) was generally answered quite well and it was pleasing that many students used an algebraic approach to solve the equation. The first step was usually expanding the bracket with hardly any students choosing to divide both sides by 3. Most of those who expanded the bracket to get 3x – 12 = 12 went on to give the correct solution. A few students, though, went from 3x – 12 = 12 to 3x = 0. Attempts to expand the bracket were not always successful and common mistakes were 3x – 4 = 12 and 3x – 12 = 36. Students who demonstrated knowledge of factorisation in part (b) did not always factorise the expression correctly. Some students identified 3b as the common factor but made a mistake inside the bracket, e.g. writing 3b(3b –b), and scored 1 mark. Partial factorisations such as 3(3b – b2) and b(9 – 3b) were quite common and scored 1 mark but in many responses with 3 or b identified as a factor the terms inside the bracket were not correct. Many students did not understand what was required and attempted to ‘simplify’ the expression, giving answers such as 6b2.

  Text version

  (a) Solve 3(x – 4) = 12 x = (2) (b) Factorise fully 9b – 3b2 (2)

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• Edexcel 1MA1 · November 2018 · Paper 2F · Q25 · 3 marks

  3 marks

  25

  
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  Working out the value of d proved to be beyond most students on this Foundation tier paper and many did not even attempt to answer the question. Those that did make an attempt often started by finding that the difference in the y coordinates is 6. Most students then simply added 6 to the x coordinate and gave an answer of 11. A few students, though, used the 6 and the gradient to work out the difference in the x coordinates as 2 and they were usually able to complete the solution. Some students wrote down y = 3x + c as a first step and a few started with (15 – 9)/(d – 5) but in both scenarios fully correct solutions were rare. A significant number of students attempted to draw a grid and mark the given coordinates and line on it and in some cases they were able to find correct values or the final answer and could gain credit for this. Correct answers of d = 7 were sometimes given with no working shown

  Text version

  A is the point with coordinates (5, 9) B is the point with coordinates (d, 15) The gradient of the line AB is 3 Work out the value of d.

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• Edexcel 1MA1 · November 2018 · Paper 2F · Q26 · 4 marks

  4 marks

  26

  
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  Those students that appeared to know a method for expanding two sets of brackets often achieved at least one mark in part (a). Mistakes were sometimes made with the signs. It was quite common to see the brackets expanded as 10x – 15x + 4x – 6, which scored 1 mark for 3 correct terms with signs, or even as 10x – 15 + 4x – 6, which scored no marks. An expansion of 10x2 – 15x + 4x – 6 did not always lead to the correct answer as some students were unable to simplify – 15x + 4x to –11x. It was disappointing to see that some students attempted to expand by adding rather than by multiplying. Part (b) was not answered very well. Many of those that did attempt to factorise into two brackets did not find two numbers which both multiplied to give 3 and added to give 4. Answers such as (x + 3)(x + 4) and (x – 1)(x + 4) were quite common. The fact that the expression being factorised has only ‘+’ signs should have been of assistance to students. Many students did not put the expression into double brackets at all but used some form of single bracket factorisation such as x(x + 4) + 3.

  Text version

  (a) Expand and simplify (5x + 2)(2x – 3) (2) (b) Factorise x2 + 4x + 3 (2)

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• Edexcel 1MA1 · November 2018 · Paper 2H · Q6 · 3 marks

  3 marks

  6

  
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  Many gained the first mark by showing the equation as y = 3x + c but could not get any further. Answers of 11 were common as students did 15 – 9 = 6 and then added 5. Students who started the question trying to find the difference of y and the difference of x either misplaced signs or had the relevant fraction upside down. Some drew sketches but these were of little help, especially when drawn incorrectly.

  Text version

  A is the point with coordinates (5, 9) B is the point with coordinates (d, 15) The gradient of the line AB is 3 Work out the value of d.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2018 · Paper 2H · Q12 · 6 marks

  6 marks

  12

  
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  Algebraic manipulation was a weakness shown by nearly all students taking this paper. Inappropriate cancelling, mistakes in multiplying out terms and in simplifying hindered most students in making much progress. In part (a) students regularly failed to get beyond simple algebra into the need to factorise in order to simplify, showing little understanding that such a step was necessary. In part (b) there was some knowledge of what to do to find a common denominator by some, either taking 1 or 2 stages to do this, but multiplying out the numerators caused further problems. Negative signs were badly handled: many ignored the minus sign in front of the 4. The most able students succeeded by using all 3 fractions with the correct numerators and common denominator.

  Text version

  Simplify algebraic fractions.

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• Edexcel 1MA1 · November 2018 · Paper 2H · Q14 · 2 marks

  2 marks

  14

  
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  If this question was attempted at all the graphs presented were varied with not many students getting the correct shape, with many most commonly related to quadratic or cubic curves. Very few were able to find, or indicate the point of intersection with the y axis.

  Text version

  Sketch y = 2^x and give all intercepts.

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• Edexcel 1MA1 · November 2018 · Paper 2H · Q15 · 1 mark

  1 mark

  15

  
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  There were many who failed to attempt this question. Of those who did make an attempt it was common to see 42.25 used as a radius in πd or πr2.

  Text version

  The equation of a circle is x2 + y2 = 42.25 Find the radius of the circle.

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• Edexcel 1MA1 · November 2018 · Paper 2H · Q17 · 5 marks

  5 marks

  17

  
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  From this point on through to the end of the paper there were many students who made few attempts at the remaining questions. In this question students sometimes gained the first mark for associating algebraic representations with the correct ratio. Beyond this step few knew what to do with their expressions. More able students went on to form simultaneous equations to solve.

  Text version

  p and q are numbers with p > q. Subtracting 5 from each gives (p - 5):(q - 5) = 5:1. Adding 20 to each gives (p + 20):(q + 20) = 5:2.

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• Edexcel 1MA1 · November 2018 · Paper 2H · Q18 · 4 marks

  4 marks

  18

  
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  Students sometimes gained the first mark for either showing a method to find the gradient of L1, or for stating L2 as y = -3x. A few tried working with the general equation y = mx + c but most did know what to do with it.

  Text version

  Line L1 passes through (4,6) and (12,2). Line L2 has equation y = -3x. The lines intersect at P.

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• Edexcel 1MA1 · November 2018 · Paper 2H · Q19 · 5 marks

  5 marks

  19

  
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  Some students made a start to this question by showing some rearrangement, recognising that m2 was 81 or 121, and some even went on to identify a pair of critical values. Most often this was 9 and 11, since it was rare to find students who wrote down the negative, as well as the positive values of the square root of m. There were some attempts at trial and improvement which rarely gained any marks since students gravitated to the value of 10 rather than the critical values for the inequality.

  Text version

  Solve 22 < (m^2 + 7)/4 < 32. Show all working.

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• Edexcel 1MA1 · November 2018 · Paper 3F · Q17 · 5 marks

  5 marks

  17

  
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  Many students got confused in their working with this question, with many completely missing part (a) and simply going on to attempt part (b), often with working in part (a). Those who attempted part (a) had mixed results. A good number failed to address the two un-labelled sides resulting in an expression of 7x + 3 rather than 10x + 6. Of those who did deal with CD and DE most went on to gain all 3 marks, although some lost the third mark by failing to link the 10x + 6 with 18. Part (b) was answered very well. Many of those who didn’t get two marks were able to get the first mark with a correct first step to solve the equation.

  Text version

  A pentagon ABCDE has one line of symmetry. AE = 4x cm, AB = DE = 2x + 1 cm, and BC = CD = x + 2 cm. Its perimeter is 18 cm.

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• Edexcel 1MA1 · November 2018 · Paper 3F · Q19 · 7 marks

  7 marks

  19

  
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  Many students seemed very unfamiliar with showing inequalities on a number line. There were various errors seen. It was common to see no circle at all at 4, or a vertical line, or for the line to terminate at a second “circle” rather than continuing to the left. Some students also dealt with the inequality as “greater than”, rather than “less than”. Part (b) showed up lots of misconceptions. Common mistakes were to have the 7 missing or to include the 3. This meant that students who made both these mistakes gained no marks. However, it wasn’t uncommon to see values completely outside the range of the inequality. Part (c) showed a level of algebra that many at this tier struggled with. It was common to see unsuccessful trial and improvement methods as a method of solution. It was also common to see students add x rather than subtract or to ‘add’ the x and number terms together. Many of those who were able to find the critical value of 6, then failed to give their answer as the correct inequality.

  Text version

  Represent and solve inequalities.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2018 · Paper 3F · Q22 · 5 marks

  5 marks

  22

  
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  Typically in part (a) the positive x values were correctly evaluated but the negative one wasn’t. When it came to plotting, the scale of one square = 0.5 really challenged students so many who gained one mark in part (a) then failed to get one in part (b) because they made mistakes in their plotting. Almost no students understood what part (c) was actually asking and as such normally left the response blank. In general, it is apparent the foundation students need greater exposure to quadratic graphs and in using a quadratic graph to solve an equation.

  Text version

  For y = x^2 + x - 4, a value table has x-values -3, -2, -1, 0, 1, 2 and 3. The given y-values are -2 at x = -2, -4 at x = -1, and -2 at x = 1. The y-cells for x = -3, 0, 2 and 3 are blank.

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• Edexcel 1MA1 · November 2018 · Paper 3F · Q26 · 2 marks

  2 marks

  26

  
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  The final question on the paper is a standard arithmetic sequence question, and many students were able to access one or both marks. Common incorrect answers were 1n – 6 and n + 6. Summary Based on their performance on this paper, students: • use efficient methods when working with proportion, such as the unitary method, rather than complex, and often incorrect, build up methods. • come to exams fully equipped to be able to answer any question, including bringing a protractor, ruler and pair of compasses, along with their calculator. • spend time working with significant figures, to understand their meaning and how to work with them. • continue to work on the new Foundation content such as trigonometry to be able to access these marks. • check the suitability of answers given for example, in question 7, where common wrong answers of 2.44m or 7.88m are not reasonable for the height of David.

  Text version

  Here are the first four terms of an arithmetic sequence. 5 11 17 23 Write down an expression, in terms of n, for the nth term of the sequence.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2018 · Paper 3H · Q3 · 5 marks

  5 marks

  3

  
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  This question was well answered by the majority of students. The most common error made was with the substitution of the negative value into the equation, the incorrect value of y = -8 for x = -3 was often seen. In part b the plotting was usually accurate but occasional misreads of the scale were seen. It was pleasing to see freehand curves drawn with an appropriate turning point. A few graphs using line segments were seen, these will never score full marks. Part c was not as well answered as the previous parts. Some blank responses were seen and a popular incorrect answer was -4 from reading the graph at x = 0 instead of at y = 0. Many students who had correctly drawn the quadratic curve did not try to use their curve to identify the value of x when y=0. Centres should encourage their learners to use the curve rather than trying to solve a quadratic equation.

  Text version

  For y = x^2 + x - 4, a value table has x-values -3, -2, -1, 0, 1, 2 and 3. The given y-values are -2 at x = -2, -4 at x = -1, and -2 at x = 1. The y-cells for x = -3, 0, 2 and 3 are blank.

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• Edexcel 1MA1 · November 2018 · Paper 3H · Q9 · 8 marks

  8 marks

  9

  
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  The majority of students gained one mark in part (a) for the expansion of two brackets. Many went on to then try and expand the third bracket. The main error was the failure to obtain the correct number of terms for their second expansion; often students did not fully expand the third brackets and hence failed to score the second mark. Other students were more successful and did gain the second mark but inaccurate simplification cost them the final mark. Unfortunately, some students attempted to expand all three brackets in one go and did not achieve any correct terms. Part (b) was either totally correct or incorrect there was very little in the way of working shown by many students. When working was evident the main error seen was to multiply and divide the powers i.e. 4×n÷2 = -3 which when solved gave rise to the incorrect value of -1.5. The last part saw many students gain a mark for substitution into the formula, although a number of students had the incorrect formula. Some students made the error of using -4 for –b and this error was carried through which resulted in the values having the wrong signs at the end. If only one error was seen the method marks could still be awarded as long as all steps were clearly shown. The use of calculators led to a number of students leaving their answers in surd form rather than evaluating the solutions as requested to 3 significant figures.

  Text version

  Expand, use index laws, and solve a quadratic.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2018 · Paper 3H · Q10 · 4 marks

  4 marks

  10

  
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  In part (a) many correct answers were seen. However the main problem for students was around rounding. Many had the correct answer and then rounded incorrectly, some just rounding one digit, often the third decimal place. Others removed the decimal point from the number altogether. Centres are advised to revise rounding to a given number of significant figures with students prior to examination sessions. The success rate for part (b) was lower than in part (a), errors seen included attempting to solve 34 = 2x − 3 or finding sin(65) and then dividing by 4. The last part of this question was not well answered by students. The concept of a positive and negative square root was missed by many. Although a few students did gain credit for stating that this quadratic should have two solutions or for actually solving the given quadratic to show that the other solution is -9.

  Text version

  f(x) = 4 sin(x degrees), g(x) = 2x - 3, and h(x) = (x + 4)^2.

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• Edexcel 1MA1 · November 2018 · Paper 3H · Q11 · 2 marks

  2 marks

  11

  
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  Students found this question difficult. Only a few fully correct answers were seen. There was no single misconception both sine and cosine curves were seen. However, so were cubics and quadratics. Some students drew lines which could have been asymptotes, but these often went through the curves drawn.

  Text version

  Sketch y = tan(x degrees) for 0 <= x <= 360.

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• Edexcel 1MA1 · November 2018 · Paper 3H · Q15 · 3 marks

  3 marks

  15

  
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  Many students did not score any marks for this question. A small minority of students were able to begin by writing down a general term for an odd number but those that did this then often lost the second mark through a lack of use of brackets. Squaring and subtracting expressions was not done well and little justification of results attained was seen. Some students wrote (2n+1)2 – (2n+3)2 and (4n2 +4n +1) – (4n2 +12n +9). This gained first the two marks as it is appropriate when finding a difference. It is advisable for centres to remind students that a numerical approach only is insufficient for this type of question.

  Text version

  Prove that the difference between the squares of any two consecutive odd numbers is a multiple of 8.

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• Edexcel 1MA1 · November 2018 · Paper 3H · Q19 · 5 marks

  5 marks

  19

  
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  This question was only partially successful for a small number of students. Many did not make any meaningful attempt to answer it and of those responses seen the main error was again inaccurate algebraic manipulation. However, some students did gain a mark for the correct rearrangement and substitution into the quadratic equation. The expansion of the brackets was only sometimes successful particularly where the students substituted the value for y. For the few students who obtained the correct quadratic equation and solved it by using the formula the final mark was not always awarded as their values of x and y were not paired correctly.

  Text version

  Solve simultaneously 2x^2 - y^2 = 17 and x + 2y = 1.

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• Edexcel 1MA1 · June 2018 · Paper 1F · Q6 · 2 marks

  2 marks

  6

  
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  Both parts to this question were answered well although part (a) had a greater success rate than part (b). In part (a) 7t was the most common incorrect answer whilst in part (b), many students initially worked out 3a + 2a and then subtracted from 8a to give 3a as the most common error.

  Text version

  Simplify each expression.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 1F · Q16 · 6 marks

  6 marks

  16

  
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  In part (a), whilst the majority of students correctly substituted the given values into the formula, many made mistakes evaluating +3× ̶ 2 and incorrect answers of ±26 were common. Many students correctly substituted x = 5 but then wrote +3 ̶ 2, working this out as +1 to give an incorrect answer of 21. Even when students correctly calculated 20 and ̶ 6, many then added incorrectly, stating ̶ 14 for example as their answer. In part (b), 4e2 + 8 and 4e2 + 2 were the most common incorrect answers seen. Expansions of 5e + 8e were not uncommon. Each of these partially correct answers were awarded one mark. Some lost the final mark for trying to simplifying to one term. Many students used trial and improvement methods to solve the equation in part (c) sometimes, but not always, successfully. After a correct first step leading to 3m – 12 = 21, a common error was then to subtract 12 from 21 resulting in an answer of 3 (9 ÷ 3). Expanding the bracketed term again proved a challenge for many students and 3m – 4 = 21 was seen often as was trying to add 4 to both sides as a first step. Fewer students’ first step was to divide both sides of the equation by 3, but those that did usually gave the correct answer of 11. Students who attempted a flow diagram method to solve the equation were usually unsuccessful.

  Text version

  P=4x+3y, where x=5 and y=-2. Also expand 4e(e+2) and solve 3(m-4)=21.

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• Edexcel 1MA1 · June 2018 · Paper 1F · Q24 · 5 marks

  5 marks

  24

  
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  Only a very small minority of students made any sensible attempt at finding both coordinates of point C and a fully correct answer was rare. Many students were able to find the width (38 – 6) and height (36 – 7) of the diagram. Values of 32 and 29 were then halved to give an incorrect answer of (16, 14.5). Some students thought that C was the midpoint of AB and gave (22, 21.5) as their answer. This did gain 3 marks for the correct x-coordinate. Most students did not recognise that there were four identical squares and so never made any attempt to work out the length of each side. Other common errors were to add 6 on the x axis and 7 on the y axis, interpreting the first coordinates as the side lengths of the squares. A minority of students drew their own scales on the axes to find the values of x and y, usually with little success.

  Text version

  Four identical squares have sides parallel to the axes. A=(6,7) and B=(38,36). B is four square-side lengths to the right of A. C is two side lengths to the right of A and two side lengths below B.

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• Edexcel 1MA1 · June 2018 · Paper 1F · Q25 · 3 marks

  3 marks

  25

  
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  This question was poorly answered simply because very many students were unable to substitute and deal with negative numbers. The positive values of x were often correctly substituted leading to 3 and sometimes 4 correctly plotted points. These were often accompanied by incorrect points on the ‘negative’ side of the diagram. A line of best fit was then a common attempt at the final graph. The most common wrong answer was a line with a positive gradient, negative values of x giving negative values of y and similarly on the positives.

  Text version

  Draw y = 1 - 4x for -3 <= x <= 3.

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• Edexcel 1MA1 · June 2018 · Paper 1H · Q6 · 5 marks

  5 marks

  6

  
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  A variety of different methods were used to work out the coordinates of C. A common first step was to find the difference between the coordinates of A and B, 38 – 6 = 32 and 36 – 7 = 29, and this earned the first mark. At this stage some students chose to work with the gradient of the line AB but this was not a useful strategy since the line AB does not pass through C. Many students, however, decided to work out 32 ÷ 4 = 8 to find the side length of a square or 32 ÷ 2 = 16 to find half the width of the diagram. Students who arrived at 8 or at 16 were in a good position to go on to find the coordinates of C, using for example 38 – 16 = 22 and 36 – 16 = 20, but many did not manage to reach (22, 20). When just one of the coordinates was correct it was most often the x coordinate. A common mistake was to work out 29 ÷ 4 = 7.25 or 29 ÷ 2 = 14.5 which meant that (16, 14.5) and (22, 21.5) were common incorrect answers. In order to find the y coordinate of C some students worked out the vertical distance of C from the bottom of the second square or from the top of the second square, usually by considering the difference between 36 – 7 = 29 and 3 × 8 = 24, and these attempts often lead to the correct y coordinate. Some of those who annotated the diagram seemed to find this beneficial. A very common approach was for students to find the midpoint of AB, (6 + 38) ÷ 2 = 22 and (7 + 36) ÷ 2 = 21.5. Although this approach gave the correct x coordinate it did not help students to find the correct y coordinate.

  Text version

  Four identical squares have sides parallel to the axes. A=(6,7) and B=(38,36). B is four square-side lengths to the right of A. C is two side lengths to the right of A and two side lengths below B.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 1H · Q12 · 4 marks

  4 marks

  12

  
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  For those who realised that an algebraic proof was needed the main obstacle to a successful outcome was writing a general expression for an odd number to start the proof, with n + 1 a common incorrect expression. The most common correct expressions used were 2n + 1 and, to a much lesser extent, 2n – 1. Some errors were made when expanding (2n + 1)2. A few students expanded it to 4n2 + 1 and occasionally the squared term was given as 2n2 rather than as 4n2. Most of the students who arrived at 4n2 + 4n + 1 were able to conclude the proof by factorising it to 4(n2 + n) + 1 or 4n(n + 1) + 1 or by explaining that since both 4n2 and 4n are multiples of 4 then 4n2 + 4n + 1 must be 1 more than a multiple of 4. Some students concluded with a statement such as 4n2 + 4n + 1 = 4n + 1 or substituted values into 4n2 + 4n + 1 and did not gain the final mark. Students who did not appreciate that an algebraic proof was required in this question simply squared various odd numbers and explained that each result was 1 more than a multiple of 4. Such responses were very common and gained no marks.

  Text version

  Prove that the square of any odd integer is one more than a multiple of 4.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 1H · Q15 · 4 marks

  4 marks

  15

  
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  Students who recognised the expression in part (a) as the difference of two squares almost always gave the correct answer, (a – b)(a + b). A common incorrect answer was (a – b)2. Students who used the result from part (a) to simplify (x2 + 4)2 – (x2 – 2)2 in part (b) were in a small minority but these students were often successful. Most students started again and expanded both (x2 + 4)2 and (x2 – 2)2 with many able to expand one of the two brackets correctly. Mistakes such as x2 × x2 = x3 or x2 × x2 = 2x2 or (x2 + 4)2 = x4 + 16 were frequently made in the expansions. A significant number of those who attempted to subtract x4 – 4x + 4 from x4 + 8x + 16 were not able to write a correct expression without brackets. The use of brackets and negative signs was poor. Students who did progress from (x4 + 8x2 + 16) – (x4 – 4x2 + 4) to 12x2 + 12 sometimes showed the intermediate step of x4 + 8x + 16 – x4 + 4x – 4. When x4 + 8x + 16 – x4 + 4x – 4 was shown it was sometimes simplified to 12x2 – 12. Many students wrote x4 + 8x + 16 – x4 – 4x + 4. Although some did recover to get to 12x2 + 12 most did not and incorrect answers of 4x2 + 20 were common. Having got to 12x2 + 12 some students ‘simplified’ their answer further to x2 + 1 and lost the accuracy mark. This question proved to be a good test of algebraic techniques including the use of brackets, expansion of brackets and working with negative signs.

  Text version

  Use the difference of two squares.

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• Edexcel 1MA1 · June 2018 · Paper 1H · Q17 · 3 marks

  3 marks

  17

  
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  This was a straightforward question for those students who were well practised in factorising quadratic expressions and there were many students who gained full marks. The denominator was correctly factorised more often than the numerator although some students did try to factorise the denominator into two brackets. Those that factorised correctly generally went on to gain full marks. When the correct answer was followed by incorrect cancelling, to x + 1 or to 2 for example, the final accuracy mark was not awarded. A significant proportion of students incorrectly cancelled values and letters without any attempt at factorising. Some students attempted to multiply the numerator and denominator together.

  Text version

  Simplify fully (3x^2 - 8x - 3)/(2x^2 - 6x).

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 1H · Q18 · 2 marks

  2 marks

  18

  
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  Many good attempts at the translation were seen. Some students lost marks because their sketches were hastily drawn and did not pass through the required points. Students would be well advised to look for those points where the graph passes through integer coordinates and transform these points carefully. Some students gained one mark for drawing a correct graph through four of the five key points or for translating the graph in the y direction. Most commonly this translation in the y direction was a translation of +2 or a translation of two 2 mm squares. Some students confused y = sin x° – 2 with y = 2sin x° and drew the corresponding stretch. Another common error was to reflect the curve in the x-axis.

  Text version

  Sketch y = sin(x degrees) - 2 for -180 <= x <= 180.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 1H · Q19 · 5 marks

  5 marks

  19

  
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  It was pleasing that many fully correct and well presented solutions were seen. Many students recognised that the first step was to rearrange the equation of the straight line to make y the subject. Attempts at rearranging were not always successful. Once y had been made the subject of the equation most students were able to identify the coefficient of x as the gradient. Some students, though, got to y = x  but made no further progress with the question. There was generally quite a good understanding that the gradient of PQ is obtained by finding the negative reciprocal of the gradient from the rearranged equation. A common error was to write the gradient as 3x  , sometimes followed by the perpendicular gradient being x . The most common approach to finding an equation for the line through P and Q was to use y = mx + c or the equivalent form y − y1= m(x − x1). Some of those who used y = mx + c did not substitute x = 3 and y = 4 to find the value of c; instead they just wrote y = 3 x + 3.5. Having got to y = 3 x + 2 some students failed to make the final step of substituting x = a and y = b. A method used less frequently was to find the gradient of the line through P and Q in terms of a and b and equate this gradient to 3 . The main difficulty faced by students using this approach was to rearrange   a b = 3 to get an expression for b in terms of a. There were many errors in the use of signs.

  Text version

  P=(3,4), Q=(a,b), and PQ is perpendicular to 3x+2y=7.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 1H · Q20 · 5 marks

  5 marks

  20

  
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  Many students gained one mark for working with the linear inequality. This involved either a method to solve 3n + 2 14 or for 3 × 4 + 2 = 14. It was disappointing that so many of the attempts to rearrange   n n did not lead to n2 – 6n + 5 < 0. It was not uncommon to see 6n > n2 + 5 followed by 6 > n + 5 and 1 > n. When rearrangement did lead to n2 – 6n + 5 this quadratic was usually factorised correctly but the inequalities that followed were often incorrect. A common mistake was to write n > 1 and n > 5 or n < 1 and n < 5. Drawing a sketch of the curve helped some students but there were others who drew a correct sketch and were still unable to identify the relevant inequalities. Many students used a trial and improvement approach. The mark scheme makes it very clear that trials must be correctly evaluated for marks to be awarded. As with the algebraic method the accuracy mark is dependent on the four method marks having been awarded so students with 2, 3 and 4 on the answer line did not get full marks unless the answer was supported by fully correct working. Students using trial and improvement frequently failed to carry out all the necessary trials and did not therefore receive the full marks possible for this approach. Summary Based on their performance on this paper, students should:  Be reminded to provide the units in a question on area (or volume) when the units are not given on the answer line.  Practise their arithmetic skills, particularly division and operations with fractions and decimals.  Practise working out estimates by rounding numbers and develop an understanding of the purpose of rounding so that they can choose appropriate rounded values.  Practise subtracting one algebraic expression from another, especially expressions with negative terms, and use brackets more efficiently.  Give correctly worded reasons when presenting a geometric proof.

  Text version

  Integer n satisfies 3n+2 <= 14 and 6n/(n^2+5) > 1.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 2F · Q11 · 3 marks

  3 marks

  11

  
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  This was well answered and all three parts had similar success rates. Centres should remind students to put their final answer on the answer line. This is particularly useful when students decide to show the correct answer and a check. Examiners cannot be expected to extract the correct answer from a list of calculations. The value of the unknown must be clearly communicated.

  Text version

  Solve three equations.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 2F · Q20 · 5 marks

  5 marks

  20

  
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  Most foundation students were able to answer part (a) however, they found part (b) particularly difficult. The most common incorrect answers in part (b) were 5np9, 5np6, 15np9 and 15np6. Where students were awarded 1 mark it was usually for 125np9 Part (c) was also difficult for the majority of the foundation students. Many did not apply the index laws correctly and often divided the powers as well as the numbers instead of subtracting the powers. The value 8 was sometimes seen but no other term was usually correct. Some students even failed to simplify 32 ÷8 correctly, writing the answer as 28. The most common incorrect answer was 8q3r 4, whilst the most common responses scoring 1 mark included 8q6r4 and 28q6r3

  Text version

  Simplify each expression.

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• Edexcel 1MA1 · June 2018 · Paper 2F · Q22 · 3 marks

  3 marks

  22

  
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  This question was not answered well. Many students did not attempt the question or just wrote down 2 and –6, sometimes as a coordinate pair or they just ringed these figures in the diagram. This did not usually lead to any meaningful working but sometimes led to the incorrect answer y = 2x–6, just using the x and y intercepts. Some students drew an appropriate triangle on the diagram but did not know how to use it, others did use it to find the gradient. However, few then went onto use this value correctly in an equation. It was also relatively common to see y = mx + c written but working showed that the student had little understanding of what this meant or how to use it. If the gradient was correctly used a common wrong answer was y = 3x + 2 where the x intercept was used rather than the y intercept. A significant number of students did try to use L in the equation rather than y or with both x and y.

  Text version

  A line passes through (0,-6) and (2,0).

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• Edexcel 1MA1 · June 2018 · Paper 2F · Q24 · 6 marks

  6 marks

  24

  
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  Fully correct answers were rarely seen. Part (c) was often left blank. Many students were able to achieve at least one mark on part (a). This was usually for finding the correct y coordinates using the positive values of x. Correct values for the negative x values were much less frequent. Common incorrect answers were –8 and –6. The use of negative values is often a weakness particularly when used in in calculations using a calculator. Also, centres should remind students that if their calculated points do not fit onto the given scaled grid then they should check their calculations carefully. In part (b) students who had values in their table were often able to plot five of their points but the mark for this was dependent on one mark being awarded in part (a). Of those students who were able to plot all of the points correctly most did join them with a smooth curve. A few students used line segments to connect points; a smooth curve is required for the accuracy mark. In part (c) not many students drew y = –2 and many scripts were blank. Some students tried to calculate a solution to the equation but very few showed any meaningful working.

  Text version

  For y = x^2 - x - 6 use x = -3, -2, -1, 0, 1, 2, 3.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 2H · Q1 · 5 marks

  5 marks

  1

  
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  In part (a) all students scored full marks by using the first law of indices. When wrong answers were seen they were when students multiplied the indices rather than adding. Part (b) provided evidence that students struggle most with the 3rd law rather than the others. When mistakes were made it was typically with either the 125 or the n3 whereas most students got the p9 correct. Commonly a 15 was seen where students multiply the 5 and 3, or many students don’t really understand that n means n1 so the sight of n rather than n3 following the removal of the brackets was common. Thus, 5np9 was a common incorrect response. Part (c) was answered better than (b), and the use of a fraction in the question rather than a division sign may have helped students. The common errors occured when students subtracted 4 from 32, or tried to divide the index numbers. In both parts (b) and (c) many students scored at least one of the two marks for two correct terms in a single product.

  Text version

  Simplify each expression.

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• Edexcel 1MA1 · June 2018 · Paper 2H · Q3 · 3 marks

  3 marks

  3

  
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  This question differentiated well, with the full range of marks awarded. Those who scored zero normally failed to show any method for finding the gradient, although in some cases a triangle was drawn under the graph, but then the height and base were incorrectly read off. There were cases of one mark being awarded for either a correct method to find the gradient, or for correctly identifying the y-intercept. It is worth highlighting to centres at this point that many students worked out the gradient as ˗3, therefore showing they don’t truly understand what it means for a linear graph to have a negative gradient. Students scoring two marks generally gained these from substituting their gradient into y = mx + c, or quite commonly for incorrect notation, such as L=3x − 6 or just 3x − 6. Not surprisingly at this grade a good number of students scored the full 3 marks.

  Text version

  A line passes through (0,-6) and (2,0).

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 2H · Q5 · 6 marks

  6 marks

  5

  
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  Almost all students were able to score at least one mark for two correct figures (typically positive x values), and a good number scored two marks. It is clear however that a large proportion of students are unable square negative numbers on their calculator and centres should focus on this. Good advice with these questions is to encourage students to check they can get the values already given in the table. Some students showed knowledge of the symmetry of a quadratic equation but used this incorrectly by assuming that the symmetry must be in the middle of the table. Once students gained a mark in part (a), almost all then went on to score in part (b). It was pleasing to see that most students recognised that this was a quadratic function and at least knew that a parabola should be drawn. A number of students did lose marks however for producing graphs that included line segments, normally seen with a graph with a “flat bottom”. There are still students who try and plot their calculated values even if they don’t fit on the axes. We would encourage centres to stress to students that if a point doesn’t fit on the grid then they should try and recalculate their value. Most students who were successful in part (c) did so by reading off the intersections with their graph from part (b) and y = 2, although in many cases the latter was either not seen, or was only partially seen. This meant in many cases students only scored one mark for one solution rather than two. There were a significant number of students who attempted an algebraic response to part (c). This almost always resulted in no marks as their algebra was incorrect, and possibly wasted a considerable amount of time. It is important that students are able to choose the most efficient strategy to solve each question; in this case the instruction to use the graph was given in the question.

  Text version

  For y=x^2-x-6 use x=-3,-2,-1,0,1,2,3.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 2H · Q11 · 3 marks

  3 marks

  11

  
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  Most students knew how to substitute into the function correctly, but many dropped marks as they couldn’t evaluate it properly on their calculators, meaning that ˗0.08 was a very common incorrect answer. Students generally struggled with the composite nature of the functions in part (b). A good number scored a single mark for finding g(1) = 4, but were then unable to substitute into f(x). Incorrect order of operations saw 4 ൈ1ଷൌ4ଷൌ64 occur which shows a lack of understanding. The most common wrong approach where students only gained the first method mark was to find f(x) × g(x) or f(x) + g(x).

  Text version

  f(x)=2/x^2 and g(x)=4x^3.

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• Edexcel 1MA1 · June 2018 · Paper 2H · Q14 · 3 marks

  3 marks

  14

  
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  This is the first time in this specification that this topic has been assessed. Commonly students read off the distance at time = 5 (28) and then worked out 28 ÷ 5. This obviously gained zero marks. Those who failed to recognise the need to draw a tangent were unable to gain any credit. Typically, those who drew a tangent went on to gain all three marks. Some work needs to be done in centres to recognise when a tangent is needed and when it is appropriate to find the area under a curve.

  Text version

  A tangent to the official distance-time curve at t=5 seconds passes approximately through (3 seconds, 0 metres) and (7 seconds, 60 metres).

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• Edexcel 1MA1 · June 2018 · Paper 2H · Q16 · 5 marks

  5 marks

  16

  
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  Examiner report

  At this level a significant proportion of students struggled, and this was one of few questions where a good number of blank responses were seem. There were some attempts at linear graphs and also quadratic graphs. However, a number of students recognised this as being the equation of a circle and although they were unable to get the correct radius, they scored a single mark for drawing a circle with centre (0, 0). There were many freehand circles, suggesting that many students had entered the examination without a pair of compasses. As with Question 5, a number of students chose not to use the most efficient method, despite being given very limited working space and the use of the word ‘hence’, and attempted an algebraic approach. Of those who tried to use the graph it was common to see at least two marks awarded if not all three. Some students who only scored one mark in part (a) were able to score follow through marks provided their method in part (b) was correct and complete.

  Text version

  Draw x^2+y^2=12.25. Hence find its intersections with the line 2x+y=1.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3F · Q4 · 3 marks

  3 marks

  4

  
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  Examiner report

  Sequences questions are usually popular, but students need to ensure they are answering the questions as stated. In part (a)(i) the correct answer was usually given, although a few students correctly identified “add 7” either in words or on the diagram for (a)(ii) but failed to correctly add 7 onto 23 for part (i). In part (a)(ii) the most common response seen was to give a simple sentence of “+7” or similar in words. Some students chose to over complicate their response by giving a step-by-step guide to checking the difference through the terms; these types of responses still managed to score the mark, provided they eventually stated the increase required or expressed this within the sequence. Unfortunately, some students failed to score the mark by not quantifying the addition or referencing an amount that the sequence increased by. Those students that opted for a nth term rule, although not required, were generally unsuccessful in arriving at the correct formula, a statement of 7n − 5 was awarded the mark as long as there was no contradiction elsewhere. In part (b) the most common method was to find all the terms of the sequence up to the 10th term, even if the expression 7n − 5 was seen. It was obvious that not all had used calculators to help with this part.

  Text version

  The sequence begins 2, 9, 16, 23.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3F · Q13 · 2 marks

  2 marks

  13

  
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  Examiner report

  Only about a fifth of students gained two marks for all graphs identified correctly but most students managed to gain one mark for one or two of the correct graphs, usually correctly recognising y = 2, giving 'D' in the first box. The most common incorrect answer was DCA, confusing y = x with y = −x.

  Text version

  Six straight-line graphs are described. A passes through (0,2) and (2,0). B is the vertical line x=2. C passes through the origin with gradient -1. D is the horizontal line y=2. E passes through (-2,0) and (0,2). F passes through the origin with gradient 1.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3F · Q17 · 3 marks

  3 marks

  17

  
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  Examiner report

  Very few gained full marks on this question, the main error being a failure to use the correct formula for the area of a triangle, usually missing dividing by 2. A common wrong answer was 54. Of those minority of students who used an algebraic method many forgot to find the square root of 36 in the final step and so lost the final mark or incorrect simplification of equation showing 6x or 9x instead of 9x2. Students that turned the right-angled triangle into a square were often successful in using a numerical approach in finding a solution of 6. A common incorrect response was to try to involve Pythagoras. A lot used a trial and improvement approach which had to be completely correct and give the correct answer to be awarded the marks.

  Text version

  ABC is right-angled at B, with AB=BC=3x cm. Its area is 162 cm^2.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3F · Q20 · 2 marks

  2 marks

  20

  
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  Examiner report

  This question provided the challenge of multiplying the terms in a single bracket by a negative value; this was a step too far for many students. Most students were successful in expanding the first bracket to gain 1 mark but then spoiled further work by unsuccessfully multiplying through the second bracket and either missed the required bracket after multiplying by 2 or wrote the incorrect sign for the final term. Other students spoiled further work by failing to simplify terms correctly or went on to add all the terms together numerically.

  Text version

  Expand and simplify 5( p + 3) - 2(1 - 2p)

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3F · Q25 · 3 marks

  3 marks

  25

  
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  Examiner report

  This question was beyond the algebraic skills of many. Of those who did make a start, there was a frequent intention to multiply both sides by 2, but a common error was to then go on to state 10 – 2x = 2x −7 or 5 – x = 2x − 14. Those who did give the correct equation sometimes lost the next marks because of incorrect rearrangement, for example adding 5 to both sides or subtracting x from both sides, and some had difficulty in dealing with negative numbers, showing −14 − 5 = −9 not −19. The best students did arrive at the correct value of 3.8, but there were many who did not attempt the question.

  Text version

  Solve (5 - x)/2 = 2x - 7.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3F · Q26 · 5 marks

  5 marks

  26

  
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  Examiner report

  Again, there were many blank responses for this question and it was rare to see the correct answer of 140. The weakest students were confused as to which angle was needed and a few thought that BCD meant B+C+D. The majority of students did not know that the sum of the interior angles of a pentagon is 540 and had no way of working it out. Without this it was difficult for a student to gain marks. Some credit could be gained by subtracting the sum of the known angles from a value in excess of 400 as this was an independent process mark. Some failed to spot that there was an angle of 90 degrees. Where students did move on and find 210 there was a tendency to divide the angle by 2 instead of splitting it in the ratio of 2:1. A small minority managed to find the correct angle of 140 with sufficient working to justify allocating full marks.

  Text version

  A pentagon has angles A=125, D=90, E=115 degrees, and C=2B.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3F · Q28 · 3 marks

  3 marks

  28

  
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  Examiner report

  This question was an opportunity for students to demonstrate correct algebra. However, very few could make the correct first step of squaring both sides or multiplying by root 2. Only a few made any progress with this question. Sometimes a string of errors led coincidentally to the correct answer, but in this case no marks could be awarded. Summary Based on their performance on this paper, students should:  present working legibly and in an organised way on the page so that the order of the process of solution is clear and unambiguous  show all working out particularly in questions where this is explicitly stated  ensure that they are familiar with the correct use of their calculator  practise algebraic manipulation and derivation, the application of ratios, scaling and rates  spend more time reading the fine detail of the question and avoid giving answers that do not answer the question posed  use the correct figures given in the question

  Text version

  Make g the subject of T = sqrt((g + 6)/2).

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• Edexcel 1MA1 · June 2018 · Paper 3H · Q2 · 2 marks

  2 marks

  2

  
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  Examiner report

  This question was a good discriminator. Just over a half of students entered for the examination scored full marks for their responses to this question testing the expansion of brackets and collection of terms. More able students could deal accurately with the signs involved in expanding and subtracting the second set of brackets whereas weaker students usually scored one mark for the correct expansion of the first set of brackets. Less successful students who found the expansion of brackets problematic often presented work along the lines, “5(p + 3) −2(1 − 2p) = 5p + 15 −2 − 4p = p + 13”.

  Text version

  Expand and simplify 5( p + 3) - 2(1 - 2p)

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3H · Q7 · 3 marks

  3 marks

  7

  
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  Examiner report

  Though there were many correct solutions to the equation seen, this question was poorly done by a large proportion of students taking the paper. Examiners were surprised by the number of students who could not carry out a correct first step with accuracy when all that was needed was 5 − x = 2(2x − 7) in order to score a mark. Instead, statements such as 10 − 2x = 2x − 7, 10 − 2x = 4x − 14 and 5 − x = 4x − 7 were often seen. Of those students who did carry out a first step correctly, some students then made errors in trying to isolate terms in x, obtaining equations such as 3x = 19.

  Text version

  Solve (5-x)/2=2x-7.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3H · Q8 · 5 marks

  5 marks

  8

  
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  Examiner report

  A majority of students were able to score some marks for their responses to this question involving the angles in a non-regular pentagon together with some work on ratio. More able students knew, or could quickly find, the sum of the interior angles of a pentagon and complete their answer to score full marks. Less able students did not know how to find the sum of the interior angles of a polygon and often assumed it was 360°. These students could not be awarded any credit. Students who used a total sum of over 400° could access up to 2 marks for their working if they were able to work with the statement “Angle BCD = 2 × angle ABC”. Few students followed either an algebraic approach or a method involving the sum of the exterior angles of a polygon. Those students who did follow such a strategy were rarely successful. A fairly common mistake was to split the pentagon into other shapes by drawing lines then make assumptions in order to calculate the size of angles, for example, a line joining vertex A to vertex B followed by the assumption that this creates two angles of equal size (62.5°) at A or that AE = ED or that AB = BC. Students are reminded that they should not make assumptions about the size of angles or the lengths of sides where this information is not stated in the question.

  Text version

  A pentagon has angles 125,115,90,B,C degrees with C=2B.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3H · Q9 · 4 marks

  4 marks

  9

  
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  Examiner report

  This question acted as a good discriminator. The most able students were able to provide a correct and complete answer to both parts of the question. Most students were able to make a start on the evaluation of T for the given values of w and d. Students generally showed a proficiency in being able to use their calculators with numbers in standard form and found a correct value for T. Unfortunately a significant proportion of these students did not give their answer in standard form so scored one of the two marks for their answers. There were some students who did not cube their value of d. Examiners were able to award one mark to students who worked out the correct value of ට ௗ and gave their answer in standard form. Part (b) of the question, involving percentage increases in the values of w and d was found to be more challenging and many students attempted to give an explanation of why Lottie was wrong without sufficient justification. Examiners expected to see either a re-calculation of T with increased and correct values of w and d followed by a comparison with the answer the student gave in part (a) or a full explanation involving a comparison of the effect of the 10% increase in w and the 5% increase in d on the value of T. This may have been demonstrated through a calculation of the total scale factor for the change in value of T or by a comparison of the effect of the percentage increases of 10% and 5% on the values of w and d³. Most students who were successful in part (b) took the re-calculation route.

  Text version

  T=sqrt(w/d^3), w=5.6 x 10^-5, d=1.4 x 10^-4.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3H · Q12 · 4 marks

  4 marks

  12

  
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  Examiner report

  There were many good answers to this question. Most students used the scales correctly to find the gradient of the straight line instead of counting squares though some students gave the gradient as “1.5x” or divided the increase in x values by the increase in y values. Good explanations were often seen in responses to parts (b) and (c) of the question. Examiners expected to see explanations in context so statements such as “the gradient represents the slope of the line” and “the value of y when x = 0” were not acceptable for the award of marks. Instead, in part (b), it was expected that students would describe, in some way, that the gradient represents the rate of flow of liquid into the container. The statement “litres per second” alone was not accepted nor was “as time increases volume increases”. A good proportion of the unacceptable answers seen were equivalent to the second of these two statements. In part (c) statements equivalent to “there were 4 litres of liquid in the container at the start” were expected.

  Text version

  A straight-line volume-time graph passes through (t,L)=(0,4) and (8,16), where t is time in seconds and L is volume in litres.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3H · Q15 · 4 marks

  4 marks

  15

  
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  Examiner report

  Many students worked out an estimate for the distance the car travelled by splitting the area into 4 trapezia, or a triangle and 3 trapezia. Work seen by these students was generally accurate and acceptable answers in the range 488 to 507 were common. These answers scored the full three marks available for part (a). Another possible route was to use 4 rectangles (or a triangle and 3 rectangles) though this usually lead to a poorer estimate outside the range acceptable for the accuracy mark. A number of students merely multiplied 20 by 35 (= 700). These students were not awarded any marks. Some students who used more than 4 strips gained one mark for showing an appreciation that they needed to work out an estimate for the area under the curve. Answers to part (b) were often clear and well expressed. The mark was awarded for a clear answer linked to their method for estimating the area in part (a) provided the student had gained at least one mark in part (a). Vague answers such as “an underestimate because of the curve” could not be given the mark. Some students based their answers on the effects of rounding values of the speed rather than comparing the estimates calculated for areas under the curve with exact areas.

  Text version

  A curved speed-time graph has approximate readings (time in seconds, speed in m/s): (0,0), (5,22), (10,28), (15,32), (20,35). Use four equal strips of width 5 seconds.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3H · Q16 · 6 marks

  6 marks

  16

  
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  Examiner report

  Relatively few students wrote down two simultaneous equations to solve in the first part of this question. Those who did often proceeded to complete their solution successfully and determine the sixth term of the sequence. A good number of students wrote down at least one equation then got no further. A large number of students wrote down the terms −2 and 12 in the hope of establishing a pattern. Many of these students found the difference between −2 and 12 (14) and gave their sixth term as 26 (12 + 14). Part (b) was answered successfully by a much greater number of students. Many students followed the strategy of finding second differences apparently realising that second differences all equal to 2 were the key to finding an expression for the nth term of the sequence. Most of these students gave an expression which included the term “n²” but some of them included “2n²”. The former usually gained at least one mark for their final answer. A common incorrect answer seen was n2 + n. This scored one mark. Many students gave a fully correct expression and scored two marks.

  Text version

  First sequence has nth term an^2+bn, term 2=-2 and term 4=12. A second begins 0,2,6,12,20.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3H · Q18 · 6 marks

  6 marks

  18

  
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  Examiner report

  Students often scored well on this question, particularly in part (c) where the application of an iteration was required. There were many good attempts to answer part (a) of the question by substituting both 1 and 2 into one of the expressions x3 + x or x3 + x − 7 then comparing the results with 7 or 0 respectively to reach the required conclusion. Some students used inequality notation to do this whilst other students gave clear written statements. Either was acceptable to examiners. Part (b) was usually well answered with the critical stage “x3 = 7 − x” being clearly shown. In part (c) values of ݔଵ, ݔଶ and ݔଷ were usually worked out with accuracy and the overwhelming majority of students who attempted this part of the question scored all three marks. Students are advised that they should write down the value of each of ݔଵ, ݔଶ and ݔଷ from their calculator and not just the value of ݔଷ. Some students did not restrict their calculations to the values of ݔଵ, ݔଶ and ݔଷ and unnecessarily worked out values for ݔସ, ݔହ, etc.

  Text version

  The equation is x^3+x=7.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2018 · Paper 3H · Q19 · 5 marks

  5 marks

  19

  
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  Examiner report

  This question proved to be a good discriminator between more able students sitting this paper and there was a wide spread of marks awarded. Some students did not attempt the question but many students were able to write down tan e and tan f and equate them to score the first mark. Attempts to deal with the fractions in order to obtain a quadratic equation in x to solve varied in success. Some students did not ensure all the terms of the quadratic equation were taken to one side of the equation before attempting to solve it. Students were split between those who used factorisation and those who used the formula to solve their equation. A substantial number of responses included two answers on the answer line despite the hint (“find the value of x”) to rule out one of the values as a possible length of a side. The best students produced concise, clear and accurate solutions to this question.

  Text version

  Two right triangles have opposite/adjacent pairs x/(4x-1) and (6x+5)/(12x+31). Their acute base angles e,f satisfy tan e=tan f.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1F · Q3 · 1 mark

  1 mark

  3

  
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  Examiner report

  It was pleasing to see that most students knew that y = 10.5 × 4, unfortunately a significant number could not evaluate this multiplication correctly. The need for accurate basic arithmetic on a non-calculator paper is obvious but unfortunately not displayed by students often enough. This question was only worth 1 mark and so students with an incorrect calculation scored no marks.

  Text version

  Solve y/4 = 10.5.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1F · Q5 · 1 mark

  1 mark

  5

  
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  This was also a well answered question with many correct answers seen.

  Text version

  Here are the first four terms of a number sequence. 2 5 11 23 The rule to continue this sequence is multiply the previous term by 2 and then add 1 Work out the 5th term of this sequence.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1F · Q6 · 3 marks

  3 marks

  6

  
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  Examiner report

  Students were able to access this question with many identifying the correct methods to form an expression or write an equation for L in terms of a to gain at least some marks. Many understood the need to add all terms to form an expression but were unable to gain full marks due to poor algebraic manipulation skills. Some solutions showed incomplete simplification such as L = 5a–1+4 or gave an incorrect simplification such as a5 in place of 5a or a3 when beginning to write an expression. A large number of students were able to correctly form a simplified expression but then did not convert this to an equation. Another common error was due to incorrect arithmetic when adding 4 to –1 and stating L = 3a – 5 as a final answer.

  Text version

  Here are five straight rods. a - 1 a a a a + 4 All measurements are in centimetres. The total length of the five rods is L cm. Find a formula for L in terms of a. Write your formula as simply as possible.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1F · Q7 · 4 marks

  4 marks

  7

  
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  Examiner report

  Part (a) of this question was well answered with the majority of students giving the coordinates in the correct order. Most used the answer line appropriately some put both numbers before the comma given, this was allowed for the 1 mark. In part (b)(i) the point was usually correct and labelled. A lack of labelling was condoned if this was the only point plotted on the gird. Students should be encouraged to label answers as this helps clearly communicate their intention to the marker. Part (b)(ii) was less well answered with many students drawing incorrect lines and then saying yes. The most successful approach seen was to set up a table and give x values including 2 and show the appropriate y values. Although all these points were not needed this approach was both popular and successful. Another successful approach seen was to plot at (0,1) and show the gradient of 4 as 1 across and 4 up twice to arrive at the point (2,9). For part (c) many candidates drew a horizontal line through (0, −2) ie y = −2 or no line at all or multiple lines. Some also just plotted the point (−2, 0) and gave no line at all.

  Text version

  On a coordinate grid, point A is (6,-2). (a) Write down A's coordinates. (b)(i) Plot B=(2,9). (ii) Decide whether B lies on y=4x+1, showing working. (c) Draw x=-2.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1F · Q24 · 4 marks

  4 marks

  24

  
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  Examiner report

  Some students tried to use algebra to solve this problem but the majority of attempts seen were based on a trial and improvement approach, normally in a non-systematic way, trying to find 3 ages that fitted the relationship requirements in the question, and summed to 77. Those that worked systematically were more successful than those that did not. Some candidates were able to find sets of numbers eg 1, 8 and 16 that fitted the relationships, but did not sum to 77, this start was given 1 mark. Those who used algebra often failed to use a single variable to base the relationship on. Students who identified Jay as being 14 then usually correctly found the age of the other two people. A few did not express this as a ratio as was required by the question. Provided the order of the ratios was made clear, variations on the correct order of ages was allowed for full marks, as was an equivalent ratio.

  Text version

  Kiaria is 7 years older than Jay. Martha is twice as old as Kiaria. The sum of their three ages is 77 Find the ratio of Jay's age to Kiaria's age to Martha's age.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1F · Q28 · 4 marks

  4 marks

  28

  
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  Examiner report

  This question proved challenging to many students and was not attempted by some. It was very common to see responses that confused area and perimeter, so an equation involving x and y was set equal to the perimeter which was incorrectly stated to be 48. A small number of students used a structured approach, either involving the identification that the two expressions in x had to be equal to 48÷3 = 16, or that the two expressions in x were equal. Those who did so normally identified that x was equal to 5, although there were algebraic errors seen in attempts to solve the equation. Far more students used trial and improvement methods, and some then realised that x=5 gave the same answer for both expressions, and then used this to show that y=3, the requirement of the question. The most successful start seen was to assume y=3 and show 48÷3 = 16 or set the expressions of the two sides equal. However many students did not develop the solution further.

  Text version

  A rectangle has top side y cm, vertical sides (2x+6) cm and (5x-9) cm, and area 48 cm^2. Show that y=3.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1F · Q29 · 1 mark

  1 mark

  29

  
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  Examiner report

  This question was accessible to many students who showed an understanding that the error lay in the joining of the points with straight lines. They expressed this in a variety of ways but often used words such as curve or smooth line or commented on the incorrect use of a ruler. However, incorrect answers showed a lack of confidence with plotting quadratics. There were several who thought that the wrong points had been plotted, the graph should have passed through 0 or that it should have been a straight line. Other common wrong answers did not relate to the graph itself but on the lack of a title or a table of values.

  Text version

  Brogan draws y=x^2+1 by joining the plotted points (-3,10),(-2,5),(-1,2),(0,1),(1,2),(2,5),(3,10) with straight segments. State one thing wrong with the graph.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1H · Q2 · 4 marks

  4 marks

  2

  
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  Examiner report

  The most successful approach to this first problem question on the paper was to form an equation based on the ages of three friends, solve to find the ages, and then write the ages as a ratio. Most students attempted an algebraic approach, but often only managing a single mark for two correct expressions. Once expressions had been formed some students were then able to sum and equate to 77 to gain the second mark. Those who managed to form an equation, were normally able to gain the 3rd mark for isolating their terms, or more commonly for actually solving to get 14. Of those who got this far many found all three ages and formed a suitable ratio. However, there were quite a number who forgot the demand of the question and lost a mark as they did not present their final answer as a ratio.

  Text version

  Kiaria is 7 years older than Jay. Martha is twice as old as Kiaria. The sum of their three ages is 77 Find the ratio of Jay's age to Kiaria's age to Martha's age.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1H · Q6 · 4 marks

  4 marks

  6

  
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  Examiner report

  This problem put a slightly different twist on a familiar concept, but was one that most students were able to access in one way or another. The most common approach was to equate the two lengths and solve for x, before substituting into one length and using the area to find y. Of those who were unable to form a suitable equation, many gained a mark for finding the width by assuming y=3, eg 48 ÷ 3 = 16. There were a number of students who scored zero as they went down a route of finding the product of the two lengths rather than equating them, or worked with perimeter.

  Text version

  A rectangle has top side y cm, vertical sides (2x+6) cm and (5x-9) cm, and area 48 cm^2. Show that y=3.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1H · Q7 · 1 mark

  1 mark

  7

  
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  Examiner report

  Students responded well to this question and a large number realised that the use of line segments on a quadratic graph was wrong. There were though, many students who talked about mis-plotting of points, or who wrongly thought that the graph should go through the origin.

  Text version

  Brogan draws y=x^2+1 by joining plotted points with straight segments. State one thing wrong with the graph.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1H · Q11 · 4 marks

  4 marks

  11

  
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  Examiner report

  Forming and solving simultaneous equations proved to be where many students stopped gaining marks. Many students attempted to solve this problem through a trial and improvement method, normally with little or no success. Of those who gained a mark for forming 2 equations, many then had no strategy for solving them. Those who did have a strategy often made arithmetic errors leading to incorrect answers.

  Text version

  3 teas and 2 coffees have a total cost of GBP7.80 5 teas and 4 coffees have a total cost of GBP14.20 Work out the cost of one tea and the cost of one coffee. tea GBP coffee GBP

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1H · Q14 · 3 marks

  3 marks

  14

  
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  Examiner report

  Students typically tried to start with the solution here and work backwards and as a result scored no marks at all. Very few students started with a correct statement, eg = k. Those who did very often had sufficient skill in manipulation to gain full marks.

  Text version

  (y+x):(y-x) is equivalent to k:1. Show that y=x(k+1)/(k-1).

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1H · Q17 · 2 marks

  2 marks

  17

  
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  Examiner report

  Most students understood the need to expand the brackets, but with only 2 marks available, both had to be expanded correctly to gain the method mark. This proved a step too far for many with one or both being expanded incorrectly. Those who did expand correctly were very rarely able to simplify and factorise correctly to complete the proof.

  Text version

  n is an integer. Prove algebraically that the sum of 1 2 n(n + 1) and 1 2 (n + 1)(n + 2) is always a square number.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1H · Q19 · 4 marks

  4 marks

  19

  
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  Examiner report

  Almost no students were able to start this question at all. The few that gained any marks typically rearranged the given equation into the form y = mx + c to find its gradient, or for showing an understanding perpendicular gradients. The concepts appeared to be beyond the vast majority of the cohort.

  Text version

  Line L is x+2y=12. A,E,B lie on L with AE=EB; B is the x-intercept and E the y-intercept. ABCD is a rectangle and A,D lie on line M. Find an equation for M.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1H · Q20 · 4 marks

  4 marks

  20

  
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  Examiner report

  A slightly better performance was seen here and it was evident that some effort has been made in centres to get students to learn the values of the trigonometric ratios, or how to find them using unit triangles. This meant that a reasonable number gained one mark for a correct value of cos(x). Very few were able to use this knowledge, combined with the information in the table to form any equations in a and b. As such more than one mark was rarely awarded.

  Text version

  y=a cos(x degrees)+b. The table gives y=3 at x=0, 1+sqrt(3) at 30, 2 at 60, 1 at 90, 0 at 120, 1-sqrt(3) at 150, and -1 at 180. Find y at x=45.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 1H · Q23 · 5 marks

  5 marks

  23

  
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  This grade 9 question did allow for some marks to be awarded to a small number of students who were able to write expressions for both areas. However it was disappointing to see how many failed to gain this first mark because they failed to use the correct formula for the area of a triangle, often forgetting the half. Very few were able to form a suitable inequality from their expressions, of those who did and had strategies to solve (typically by factorisation) almost no one gained full marks as they failed to discard the the critical value of 1 . Summary Based on their performance on this paper, students should:  Focus attention on basic arithmetic skills such as multiplication and division, and multiplying by powers of 10 to ensure marks are not dropped unnecessarily.  Practise non-standard procedure (AO1) questions and be able to answer them confidently. For example, finding the mean from a grouped frequency table.  Spend more time working with ratio and proportion.  Practice solving problems where translating the given information into algebraic expressions and equations is an efficient method of solution.

  Text version

  A rectangle has sides (x-1) and (3x-2). A right triangle has base x and height 2x. The rectangle's area is greater than the triangle's. Find the possible values of x.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 2F · Q3 · 3 marks

  3 marks

  3

  
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  In part (a) only a minority failed to simplify and left a “×” sign in. In part (b) weaker students wrote their answer as t2 rather than t2. In part (c) it was encouraging to find most students were able to simplify this expression.

  Text version

  (a) Simplify 3f x 5g. (b) Simplify t x t. (c) Simplify (2n+6n)/2.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 2F · Q14 · 4 marks

  4 marks

  14

  
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  In part (a) students had to give an example by choosing two odd numbers in the given expression, and calculate an answer that was a multiple of 4. This was usually done well, sometimes using the same odd number, though errors were not uncommon. It was surprising to see even, rather than odd numbers being used, and merely substituted and not worked out. In part (b) students had to show some reasoning by explaining how the use of (any) odd number in the expression could give a multiple of 4. Marks in this part were rare. Nearly all students thought that they just had to give multiple examples as in part (a). Some gained the first mark (only) by reasoning that doubling an odd number always gave an even number.

  Text version

  a and b are odd numbers. (a) Give an example to show that the value of 2(a + b) is a multiple of 4 (b) Show that, when a and b are both odd numbers, the value of 2(a + b) will always be a multiple of 4

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• Edexcel 1MA1 · November 2017 · Paper 2F · Q16 · 3 marks

  3 marks

  16

  
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  This question differentiated well across the ability range. Both trial and improvement and the flow chart method rarely resulted in the correct answer. Many gained a mark for expanding the bracket, but most then were unable to perform a correct manipulation of terms to get the second mark. Some students stated the answer as 2 rather than 3 .

  Text version

  Solve (5x-6)/3 = x-1.

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• Edexcel 1MA1 · November 2017 · Paper 2F · Q21 · 4 marks

  4 marks

  21

  
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  Examiner report

  In part (a) addition was required; some multiplied and gave the answer as 3. In part (b) multiplication was required; some added and gave the answer as 8. Students very rarely scored marks in part (c). A clear lack of conceptual understanding of standard form (or indices) was evident with most students missing the powers of 10 link to gain the correct algebraic power. Many tried to solve the equation to gain a numerical value with the use of the 3 and 2 as coefficients of 100 and 1000 rather than showing their derivation.

  Text version

  (a) pi^3 x pi^x = pi^9; find x. (b) (7^2)^y=7^10; find y. (c) If 100^a x 1000^b=10^w, show w=2a+3b.

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• Edexcel 1MA1 · November 2017 · Paper 2F · Q24 · 5 marks

  5 marks

  24

  
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  Examiner report

  In part (a) sight of a complete answer (both 6 and −6) was rare. Some credit was given where an answer was embedded, which was not uncommon. The main mistake occurred when students divided by 2 twice instead of dividing by 2 and then finding the square root. In part (b) the majority were familiar with what was required but many failed to multiply the 3x by 3x correctly, often writing this as 6x, but gained 1 mark if they multiplied their other terms correctly. Using a table format was very popular and generally successful for those students. In part (c) very few were familiar with the requirements of factorising into two brackets so often tried to “factorise” using only one pair of brackets. Common wrong answers like x(x + 6) + 9 were frequently seen. Summary Based on their performance in this paper, students should: • Present their working legibly and in an organised way on the page, sufficient that the order of the process of solution is clear and unambiguous. • Include working out to support their answers • Bring all necessary equipment to the examination, including ruler and protractor, and need to be trained in the correct use of their calculator. • Carry out a common sense check on the answers to calculations; for example you should expect the number of £1 coins in question 6 to be a whole number • Make sure they learn and understand algebraic manipulation and derivation, percentage calculations and application of ratios and rates when preparing for future examinations • Spend more time ensuring they read the fine detail of the question to avoid giving answers that do not answer the question, and to ensure they use the correct figures as given in the question.

  Text version

  (a) Solve 2x^2=72. (b) Expand and simplify (2x+1)(3x-2). (c) Factorise x^2+6x+9.

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• Edexcel 1MA1 · November 2017 · Paper 2H · Q1 · 3 marks

  3 marks

  1

  
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  Examiner report

  There were many fully correct solutions seen. Where full marks were not achieved, one mark was often awarded for a correct expansion of the brackets. Unfortunately this was frequently followed by an incorrect attempt to isolate ether the terms in x or the constant terms, the constant terms presenting the greater difficulty. Some students wrongly expanded the brackets as 3x − 1 but were then able to get some credit for the next stage in the manipulation of their equation.

  Text version

  Solve (5x-6)/3=x-1.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 2H · Q14 · 3 marks

  3 marks

  14

  
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  Examiner report

  This question discriminated well between the more able students. Many of these students scored 1 or 2 marks for drawing 2 or 3 lines correctly. Where 2 of the three lines were drawn correctly it was often y = 2x which was incorrect or not attempted. Students who drew all 3 lines correctly more often than not opted for the closed region bounded by these lines rather than the open region satisfying the three inequalities given. Weaker students could often only draw the line y = 1 correctly. Some students did not attempt to answer the question.

  Text version

  Shade and label R satisfying y>1, x+y<5 and y>2x.

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• Edexcel 1MA1 · November 2017 · Paper 2H · Q16 · 2 marks

  2 marks

  16

  
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  Examiner report

  Few students obtained full marks. The direct approach of taking square roots of each side of the equation was rarely seen. A more common approach was for students to expand and simplify ሺx െ 2ሻ², then use the quadratic formula. Some students who expanded ሺx െ 2ሻ² seemed to run out of steam and did not attempt to solve their resulting equation. Another very common error seen was for students to expand ሺx െ 2ሻ² as x² െ 4 and find a value or values of x from the resulting two term quadratic equation. Some students who worked accurately gave only one correct value for x (usually 3.73).

  Text version

  Solve (x-2)^2=3, giving solutions to 3 significant figures.

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• Edexcel 1MA1 · November 2017 · Paper 2H · Q19 · 4 marks

  4 marks

  19

  
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  Examiner report

  There were a small number of excellent proofs seen usually using gradients to show that the lines were parallel. Students who attempted the question but could not provide a full solution often gained one mark for a correct method to calculate the coordinates of at least one of the points M or N. Students often drew a diagram but without further work, these could rarely be awarded any marks.

  Text version

  A triangle has vertices P=(-3,-6), Q=(1,4) and R=(5,-2). M is the midpoint of PQ and N is the midpoint of QR. Prove that MN is parallel to PR. Show each stage of your working.

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• Edexcel 1MA1 · November 2017 · Paper 2H · Q22 · 5 marks

  5 marks

  22

  
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  Examiner report

  Though there were some fully correct answers, these were rarely seen. Few students were able to set up the required simultaneous equations, though some were able to score marks for 3a ൅ b ൌ20, gሺ1ሻ ൌ a ൅ b or for f‐1ሺ33ሻ ൌ 6. Some students confused f-1(x), with ଵ ௙ሺ௫ሻ.

  Text version

  f(x)=5x+3 and g(x)=ax+b. Given g(3)=20 and f^-1(33)=g(1), find a and b.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 2H · Q23 · 5 marks

  5 marks

  23

  
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  Examiner report

  The vast majority of students could make no progress with this question designed to test top grade students. Some students confused the geometric sequence with an arithmetic sequence and involved addition of the terms (rather than multiplication). For part (b) there were again few attempts worth any credit with some students starting their working by using their calculator to write down the value of 7 + 5√2 as a decimal. The best students gave clear, concise and full solutions to this question. Summary Based on their performance in this paper, students should: • practise solving linear equations. • learn standard techniques such as working out values in problems on compound growth by using a multiplier method. • carry out a common sense check on the answers to calculations, so for example you should expect the number of hours each of 3 cleaners need to clean all the rooms in a hotel to be more than the number of hours that each of 5 cleaners need to clean the same number of rooms. • use tracing paper to help in questions involving rotations. • check any readings taken from graphs to make sure scales on the axes have been interpreted correctly.

  Text version

  A geometric sequence starts sqrt(x)-1, 1, sqrt(x)+1. (a) Find x. (b) Show its fifth term is 7+5sqrt(2).

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 3F · Q2 · 1 mark

  1 mark

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  Examiner report

  Most students were able to give an acceptable form of the correct answer. However, some did not fully simplify their answer, for example, a few students gave an answer of 4y – 2y or y + y; this gained no credit.

  Text version

  Simplify y + 3 y - 2 y

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• Edexcel 1MA1 · November 2017 · Paper 3F · Q13 · 5 marks

  5 marks

  13

  
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  Examiner report

  Accurate completion of the table of values in part (a) was common; the most popular error was in working out the value of y when x = 1, often + 0.5 was seen. Surprisingly a correct graph, in part (b), often did not follow a correct table. Many students just plotted their points thinking that this was all that was required. Some started again and failed to produce a correct graph. Students should realise that a straight line graph is required and that any variation is an indication of an error. A number of students reversed the x and y coordinates when plotting. In part (c), some students correctly found the value of x by simply solving the equation. This was acceptable but a fully correct answer of 2.6 was required for any credit. Many students, even some with a correct graph, clearly had no idea how to use their graph to answer this question.

  Text version

  For y=0.5x-1: (a) complete the table for x=-2,-1,0,1,2,3; (b) draw the graph; (c) find x when y=0.3.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 3F · Q17 · 3 marks

  3 marks

  17

  
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  Examiner report

  In part (a), the most common incorrect response was an answer of 16m. Part (b) was poorly answered with many students seemingly selecting their choice at random. Clearly the meaning of these words is an area that needs clarification for students.

  Text version

  (a) Factorise 4m+12. (b) Complete: 5y is a ___ in the ___ 3x+5y, choosing from expression, equation, formula, identity, inequality, term, factor, multiple.

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• Edexcel 1MA1 · November 2017 · Paper 3F · Q18 · 4 marks

  4 marks

  18

  
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  Examiner report

  The predictable incorrect answer of n + 3 was the most common error made, in part (a). 4n + 3 was another incorrect answer seen. However, it was pleasing to see so many students scoring at least 1 for giving 3n as a part of their nth term, and often 2 marks. Only a minority of students were able to use their nth term to solve part (b). It was more common for students to gain success by laboriously writing out the complete sequence up to 91. Many students incorrectly tried to substitute 90 into their nth term.

  Text version

  Here is a sequence of patterns made with counters. pattern number 2 pattern number 3 pattern number 1 (a) Find an expression, in terms of n, for the number of counters in pattern number n. Bayo has 90 counters. (b) Can Bayo make a pattern in this sequence using all 90 of his counters? You must show how you get your answer.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 3F · Q24 · 5 marks

  5 marks

  24

  
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  Examiner report

  Only a very small minority of students made any sensible algebraic attempt to solve this problem. Those that did, usually made errors when giving the algebraic lengths in the perimeter of the 8-sided shape. 70 divided by 8 was a common, incorrect starting point. There were a few trial and improvement successes but usually this approach failed.

  Text version

  Here is a rectangle. The length of the rectangle is 7 cm longer than the width of the rectangle. 4 of these rectangles are used to make this 8-sided shape. The perimeter of the 8-sided shape is 70 cm. Work out the area of the 8-sided shape.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 3F · Q27 · 3 marks

  3 marks

  27

  
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  Examiner report

  The ability to solve a pair of simultaneous equations was only demonstrated by the few more able students, although many did attempt it. Of those who did make a correct start by multiplying one equation with the intention of eliminating one of the unknowns, often the method was flawed by more than one error. A common error was to use the wrong operation when attempting to eliminate a variable. Trial and improvement methods were very rarely successful. Summary Based on their performance in this paper, students should:  Learn standard metric conversions E.g. 1 litre = 1000 millilitres  Ensure they read each question carefully.  Show working whenever a calculation is carried out, no matter how trivial.  Learn the meaning of words such as ‘expression’, ‘identity, ‘inequality’ etc.  Learn all the necessary formulae for this tier of entry, including that for density

  Text version

  Solve x+3y=12 and 5x-y=4.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 3H · Q6 · 5 marks

  5 marks

  6

  
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  Examiner report

  It was pleasing that some good attempts that used an algebraic approach were seen though these were relatively few in number. Some students wrote x and x + 7 on the diagram of the rectangle but made no further progress. Successful attempts generally started with students writing expressions for the lengths of the sides on the diagram of the 8-sided shape. The most common mistake at this stage was for the two lengths of 7 to be incorrect or missing. Students often went on to write down an expression for the perimeter of the shape and equate this to 70. Those that formed an equation were usually able to solve it correctly. The final mark was awarded if students used their value of x correctly to find the area of the 8-sided shape. A common mistake was to multiply the area of one rectangle by 8 (number of sides). A few students mistakenly used 7x instead of x + 7. Trial and improvement approaches were seen; they were usually unsuccessful and so gained no marks.

  Text version

  A rectangle has width w and length w+7. Four copies form an 8-sided shape of perimeter 70 cm. Find the total area of the four rectangles.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 3H · Q13 · 2 marks

  2 marks

  13

  
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  Examiner report

  Those students with some idea about completing the square were often able to score one mark for (x + 3)2 but errors were frequently made with the ‘– 16’.

  Text version

  Write x^2+6x-7 in the form (x+a)^2+b.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 3H · Q15 · 9 marks

  9 marks

  15

  
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  Examiner report

  In part (a) those students that gained the first mark by substituting two appropriate values into x3 + 7x – 5 often failed to make a deduction about the roots. Students seemed to think that getting one positive answer and one negative answer was sufficient. Many students had no idea how to show that the equation has a solution between x = 0 and x = 1. Attempts at using the quadratic equation formula were very common. Many students were able to gain one mark in part (b) by showing a correct first step in the rearrangement, most commonly this was x3 + 7x = 5. Many, though, were then unable to continue with the rearrangement by using factorisation and show a complete method. When answers were seen in part (c) it was evident that some students had a good appreciation of the process of iteration and they were able to gain the first method mark for substituting the starting value of 1 into the formula. When the results of the next two iterations were not accurate the second method mark could only be awarded if the substitutions were shown. Rounding or truncating the value of x2 resulted in some final answers that were not sufficiently accurate. Some students carried out more than three iterations. In these responses the accuracy mark was awarded for the value 0.6704 and any further iterations were ignored. Part (d) was poorly answered. Those students who did gain one mark for substituting their answer to part (c) into x3 + 7x – 5 rarely compared the result of the substitution with zero to determine the accuracy of their estimate. Even when the correct value was substituted the result of the substitution was often incorrect.

  Text version

  For x^3+7x-5=0: (a) show a root between 0 and 1; (b) rearrange to x=5/(x^2+7); (c) from x0=1 iterate x_(n+1)=5/(x_n^2+7) three times; (d) substitute x3 and comment on accuracy.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 3H · Q18 · 4 marks

  4 marks

  18

  
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  Examiner report

  In part (a) relatively few students appreciated that working out an estimate for the distance the train travelled required them to find the area under the curve. Those that did usually showed 4 strips of equal width on the graph and made an attempt at working out the area. Some very good answers were seen but attempts were often spoilt by values being read incorrectly from the graph or by the formula for finding the area of a trapezium being used incorrectly. Some students worked with rectangles and triangles, often successfully. Many students, though, simply used distance, speed, time formulas and finished with wrong answers of 320 or 360. Part (b) was only accessible to those students who had attempted to work out an area in part (a). Some students did state that their estimate was an overestimate and gave a reason linked to their method. However, many of the reasons given had nothing to do with the method used to work out the area.

  Text version

  A speed-time curve has readings (0,0),(5,2),(10,5),(15,10),(20,18). (a) Use four equal trapezia to estimate distance in 20 s. (b) State whether it over- or underestimates, with reason.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · November 2017 · Paper 3H · Q19 · 5 marks

  5 marks

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  Examiner report

  Some students rearranged the equation of the straight line to make either x or y the subject and those who realised that they needed to solve the equations simultaneously then substituted into the equation of the circle. Mistakes were often made when expanding the brackets and when simplifying the resulting quadratic equation. Students usually solved the quadratic equation to show that the line and the circle only intersect at one point although in a few responses the discriminant was used to show that there is only one solution. Some students solved the equation but made no concluding statement about how this proved that the straight line is a tangent to the circle and they were not awarded the final mark.

  Text version

  Prove line x-2y=10 is tangent to circle x^2+y^2=20.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 1F · Q3 · 2 marks

  2 marks

  3

  
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  Examiner report

  A sizeable group of students scored the mark for algebraic simplification in part (a) of this question. A significant number of students had major misconceptions about how to simplify algebraic expressions leading to common incorrect answers such as 7e + 8f or 7e × 8f. Other students were let down by the recall of their times tables resulting in answers such as 54ef or 57ef. Most students were unable to answer part (b) of this question. It is possible that an integer on the right hand side would have resulted in more success, but the majority of students didn’t appear to know how to deal with the x being divided by 5. The multiplying of a mixed number by an integer then proved a further obstacle. Solving equations which include division appears to be an area for improvement for foundation students.

  Text version

  (a) Simplify 7 x e x f x 8. (b) Solve x/5 - 1/2 = 2.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 1F · Q11 · 6 marks

  6 marks

  11

  
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  Examiner report

  Students found this question more difficult than the normal arithmetic number patterns. That said, a lot of students gained credit for starting the process for part (a), either through diagrams, starting to generate the sequence or for looking at the differences. A good proportion also went on to get the correct answer. A common incorrect method was to take the number of squares for pattern 3 and double it, giving an answer of 18. Part (b) was the best answered part of the question with a large proportion of students gaining both marks. Failure to gain both marks was either because they went too far or not far enough when adding on 4s or mixed up the two sequences and calculated 20 × 20. In part (c) students generally struggled to get both marks. However, those who made a statement about a specific term correctly were able to gain a mark, which was a common score. To gain both marks a general statement relating to all odd terms was needed; it was this generalisation that was often lacking.

  Text version

  A tile pattern uses n^2 square tiles and 4n circular tiles in pattern n (as shown by the first three official diagrams). (a) Find the square tiles for pattern 6. (b) Find the circular tiles for pattern 20. (c) Derek says an odd pattern number always needs an odd number of square tiles. Is he right? Give reasons.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 1F · Q16 · 2 marks

  2 marks

  16

  
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  Examiner report

  This question threw up a major misconception for many students. Many omitted the “×” when substituting into “at”, often writing 3 ଵ ଶ rather than 3 × ଵ ଶ. Those who did substitute correctly and gained the method mark, then struggled with the arithmetic either because of not following the correct order of operation, or from struggling to carry out arithmetic with fractions, or dealing with negatives, even when they had achieved 1 + −1.5. The final correct answer of − ଵ ଶ or –0.5 was rare.

  Text version

  v = u + at, where u = 1, a = -3 and t = 1/2. Work out v.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 1F · Q17 · 4 marks

  4 marks

  17

  
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  Examiner report

  Although a good number of fully correct solutions were seen, this question proved demanding for a number of students despite the easy first mark for a correct process to find the weight of one tin of soup. Unfortunately, those who understood the steps in the process to solve this problem often failed to gain all four marks as they were unable to cope with the necessary arithmetic. A common incorrect starting point was to subtract the two total weights given or to divide by an incorrect amount of tins.

  Text version

  5 tins of soup have a total weight of 1750 grams. 4 tins of soup and 3 packets of soup have a total weight of 1490 grams. Work out the total weight of 3 tins of soup and 2 packets of soup.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 1F · Q19 · 4 marks

  4 marks

  19

  
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  Examiner report

  A familiar question and many students were able to gain the first mark at least in part (a), often for correctly expanding the bracket. Weaker students were then unable to solve the simple linear equation often subtracting rather than adding, or being unable to divide. It is worth noting that the improper fraction ଷ଼ ସ is an acceptable answer and gained full marks, but that the answer written as a process, 38 ÷ 4, is not acceptable unless followed by an accurate answer. Most students were also able to score some marks in part (b), although one mark was much more common than two marks. The usual errors regarding the end values (either including −3 or excluding 2) occurred frequently and a number of students excluded 0, presumably because they do not think it is an integer.

  Text version

  (a) Solve 4(x - 5) = 18. (b) t is an integer and -3 < t <= 2. Write down all possible values of t.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 1F · Q24 · 3 marks

  3 marks

  24

  
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  Examiner report

  This question proved very difficult for many students, even with the scaffolding of the individual sections shown on the diagram. Generally, most students scored zero, as they tried to work backwards from the given solution but with no real idea what to do or substituted various values in to the given equation. Those who scored marks normally got a single mark for two correct areas on the diagram or just below. Very few then simplified their expression and equated it to 10. The few students who did typically went on to complete the solution. It is essential that all steps are shown in a solution where the final answer is given in the question.

  Text version

  Square ABCD has side length (x + 3) cm and area 10 cm^2. Show that x^2 + 6x = 1.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 1F · Q26 · 2 marks

  2 marks

  26

  
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  Examiner report

  Success in this question was very rare. The method mark was available to those who realised the need that, in order to compare gradients, the coefficients of x needed to be compared with both equations in the form y = mx + c. Those who recognised what needed to be done were generally successful but the majority of students could not cope with the demand of this question.

  Text version

  The equation of the line L1 is y = 3x - 2 The equation of the line L2 is 3y - 9x + 5 = 0 Show that these two lines are parallel.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 1H · Q4 · 3 marks

  3 marks

  4

  
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  Examiner report

  Many candidates started by attempting to multiply the 2 side lengths algebraically. Most candidates were successful after starting with this approach, setting the sum of their expansion equal to 10 and re-arranging. A significant number started by finding the areas of each of the 4 sections on the diagram, then forming an equation equal to 10 or concluding that the 3 algebraic areas would sum to 1, but some candidates failed to show sufficient working when using this latter approach. Of those unable to gain full marks, many secured the first mark for showing at least two correct area expressions or forming an initial algebraic expression for the area i.e. (x + 3)(x + 3). The most common incorrect approach was finding the side length as 3x. Candidates should be encouraged to write down what they can see in a ‘show that’ question, making clear where expressions have come from and annotating in a clear manner.

  Text version

  Square ABCD has side length (x + 3) cm and area 10 cm^2. Show that x^2 + 6x = 1.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 1H · Q6 · 2 marks

  2 marks

  6

  
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  Most candidates understood the need to manipulate to get the two equations into the same form. The majority preferred to rearrange to make y the subject. Many began with L2 and gave 3y = 9x – 5 thus gaining the first method mark for starting to manipulate an expression. They then divided by 3 which successfully gave them two equations in the same form and gained the second mark. It is interesting that many candidates wrote 5 as a decimal - seemingly more comfortable to work with decimals than fractions even for this non-calculator question. Even more usual was for the division by 3 not to be applied to the constant. Often the constant term was incorrect but this was condoned as it was irrelevant to proving the lines parallel. Another successful route was to multiply L1 by 3 and subtract 9x. This often resulted in the constant term having an incorrect sign, again condoned.

  Text version

  The equation of the line L1 is y = 3x - 2 The equation of the line L2 is 3y - 9x + 5 = 0 Show that these two lines are parallel.

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• Edexcel 1MA1 · June 2017 · Paper 1H · Q10 · 3 marks

  3 marks

  10

  
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  As this question only involved positive terms most candidates were able to successfully expand a pair of brackets usually the first two brackets (x + 1)(x + 2) although a few still made arithmetical errors with multiplying simple values like 1 × 2 and writing 3 as their answer. Once one set of brackets had been expanded candidates generally seemed to be able to then expand this over a third bracket and were more successful when systematically multiplying each term across the bracket. They usually also then went onto get the second method mark for at least half the terms written correctly. There were some candidates who tried to do all three brackets in one step, usually leading to few marks being awarded. Candidates needed to be careful in copying their own work, often losing a mark when re-writing their answer out incorrectly in the next stage of their working. For example, having given x3 in their second stage of working, ending up writing x2 + 6x2 + 11x + 6 as their final answer.

  Text version

  Show that (x+1)(x+2)(x+3) has the form ax^3+bx^2+cx+d, where a,b,c,d are positive integers.

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• Edexcel 1MA1 · June 2017 · Paper 1H · Q11 · 3 marks

  3 marks

  11

  
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  In part (a) the turning point was well understood, with nearly all candidates gaining this mark. In part (b) most candidates knew they had to read off the values at the intersection of the curve with the x axis, but in part (c) the use of function notation confused a significant minority, who failed to give an answer; those who understood usually went on to read off from 1.5 as intended. In both parts (b) and (c) it was the most basic of errors that lost candidates marks. This included those who mis-read the scale, those who failed to include negative sings when needed, and those who gave coordinates as the roots rather than the values of x.

  Text version

  The graph of y = f(x) is drawn on the grid. 6 5 4 3 2 1 -2 -1 1 2 3 4 x O -1 -2 -3 -4 (a) Write down the coordinates of the turning point of the graph. ( (b) Write down estimates for the roots of f(x) = 0 (c) Use the graph to find an estimate for f(1.5)

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• Edexcel 1MA1 · June 2017 · Paper 1H · Q15 · 4 marks

  4 marks

  15

  
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  Candidates had varying degrees of success with this question. Many did much of the work using the original values and introduced estimated values only after they had completed nearly all their manipulation. When substituting the values into the formula, many omitted to square the radius. A significant number of candidates failed to write out a full equation, many calculating individual parts as opposed to using the given formula. There was poor rearrangement of the equation sometimes ending with subtraction rather than division. π = 3.14 was common, but without estimating this as 3 they then had difficulty cancelling π with 1 ; many kept 98 rather than rounding to 100. Those who gained full marks usually gave an answer of “4” from a calculation of 100 ÷ 25, because the actual working here was very simple for those who recognised what the question was asking. In many cases working was often poorly set out and difficult to follow. Candidates need to consider how they present their solutions. Using a systematic and logical sequence of steps leading to a final answer is highly recommended. In part (b) only the most able candidates were able to explain concisely the result of their division, which required a real understanding of the effects on calculation of rounding in both the numerator and denominator. Reference to the numerator and denominator was rarely seen. The majority stated that if they rounded more values down than up then their answer would be less, or that John would be more accurate as he had used a calculator.

  Text version

  A cone has volume 98 cm^3 and radius 5.13 cm. Use V=(1/3)pi r^2 h. (a) Estimate h. (b) State whether the estimate is above or below the calculator value, with reasons.

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• Edexcel 1MA1 · June 2017 · Paper 1H · Q16 · 4 marks

  4 marks

  16

  
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  Numerical solutions were attempted, despite the instruction being given to prove algebraically, which could not be awarded any credit. Almost all algebraic attempts correctly expanded the brackets for one mark but then failed to arrive at the correct expression for the second mark as they incorrectly dealt with the negative in front of the bracket. The final mark was available and largely awarded to those candidates who concluded by justifying their final expression as even; it was definitely easier to justify that the expression was even if the student factorised by putting 2 outside of the bracket and concluding it was therefore always even. Those that did not factorise needed to conclude by explaining that since both are even numbers (or multiples of 2) the expression is always even.

  Text version

  n is an integer greater than 1. Prove algebraically that n^2 - 2 - (n - 2)^2 is always even.

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• Edexcel 1MA1 · June 2017 · Paper 1H · Q18 · 4 marks

  4 marks

  18

  
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  Many candidates picked up a first mark by realising that the diagonals on a rhombus intersect at right angles. This was either shown on the diagram by inserting right-angles, by a statement, or by a correct start at using m  . Common errors with the gradient of AC involved giving the gradient as  (negative of 1 ) or 2 (the reciprocal of 1 ). Many candidates missed an easy mark by not stating the gradient of DB as 1 ; they went on to give the gradient of AC as  or 2 so they must have used the 1 to get their gradient, but it was not stated. Some candidates thought that the intercept should be +6 as in the equation of DB, whilst other candidates thought it should be −6. Those candidates who got as far as y = −2x + c then made often mistakes with their substitution or subsequent manipulation. For example, having got to 11 = −10 + c, they then stated c = 1 (or sometimes c= 22).

  Text version

  ABCD is a rhombus. A=(5,11) and diagonal DB has equation y=(1/2)x+6. Find the equation of diagonal AC.

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• Edexcel 1MA1 · June 2017 · Paper 1H · Q20 · 5 marks

  5 marks

  20

  
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  A minority of candidates incorrectly took the square root of the first equation to give x + y = 5 or attempted to square the second equation. They were therefore unable to gain any marks. Most candidates correctly rearranged the second equation and substituted y = 13 + 3x into the first equation, thus gaining the first method mark. They then usually expanded this correctly for further credit although a significant number of candidates expanded (3x + 13)2 as 9x2 + 169. A minority of candidates failed to write the quadratic in a form that could be solved, but tried to solve their expression equal to 25. A quadratic equation was presented by many attracting the third method mark, frequently the correct quadratic equation, but the factorisation proved to be beyond many. Even those who managed to factorise found it difficult to proceed to the four values in the final solution. Many attempted to use the quadratic formula rather than factorise which resulted in the usual confusion over signs, though this was eased as all the terms of the quadratic were positive. Very few successful attempts at completing the square were seen. The final mark was rarely awarded as it involved finding all four correct values given as two associated pairs. For a complex question it was reassuring to see so many candidates have the confidence to make a real effort and often pick up most of the marks.

  Text version

  Solve x^2+y^2=25 and y-3x=13 algebraically.

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• Edexcel 1MA1 · June 2017 · Paper 2F · Q1 · 4 marks

  4 marks

  1

  
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  Part (a) of this question requiring students to collect terms involving only one variable was quite well done. Common incorrect answers seen by examiners include 2p and 4p. A less common error was to give an expression involving a term in p². The second part of the question attracted only a small minority of correct answers and was the least well answered part of this question. The most common answer given was m6. 6m was also commonly given as the answer. Most students scored at least one of the two marks available for responses to part (c) for a correct term in either c or d. Many students gave a fully correct answer though some students lost a mark because they did not fully simplify + −4c. Where answers were not fully correct, this was usually due to carelessness when collecting terms or to a confusion with signs. Occasionally students tried to combine the constant term with the 3c and wrote 13c or combined terms in c with terms in d, for example by writing 7c + d as 7cd.

  Text version

  (a) Simplify 5p - 3p + p (b) Simplify m3 + m3 (c) Simplify 10 + 3c + 5d - 7c + d

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• Edexcel 1MA1 · June 2017 · Paper 2F · Q9 · 3 marks

  3 marks

  9

  
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  Examiner report

  Students could approach this question by forming and solving an equation in x or by a more informal numerical approach. The more able students often chose the former approach which led to 5x = 270 and x = 54. Weaker students were often able to make a start by subtracting 90° from 360° but fewer students could then identify the need to divide 270 by 5. Instead a significant number of students assigned angles to the 3x° and 2x° such as 150° and 120° which add to give 270°. These students appeared not to realise that was inconsistent with angles with sizes in the ratio 3:2. These students had more than likely measured the angles. Students are reminded of the rubric stated on the front of the examination paper, that “Diagrams are NOT accurately drawn unless otherwise indicated”.

  Text version

  Angles around a point are 3x degrees, 2x degrees and 90 degrees. Find x.

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• Edexcel 1MA1 · June 2017 · Paper 2F · Q14 · 3 marks

  3 marks

  14

  
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  Part (a) of this question was not well done and for many students part (b) was more successfully answered. For part (a) it was common to see −5, 5m and 5(−2m) given as answers. The factorisation of 2a²b + 6ab² in part (b) was well answered by more able students taking this paper but many students lost marks through not checking by using the reverse process of multiplying out their answers. Doing this would have helped a significant number of students detect errors. There were many attempts to combine the two terms 2a²b and 6ab² in other ways and responses such as 12a³b³, 8ab², 8ab4 or 8a³b³ were seen many times by examiners.

  Text version

  (a) Factorise 5 - 10m (b) Factorise fully 2a2b + 6ab2

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• Edexcel 1MA1 · June 2017 · Paper 2F · Q24 · 3 marks

  3 marks

  24

  
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  The method of solving simple quadratic equations by factorisation, a standard process, is now part of the Foundation level specification and should be accessible to the more able students taking this paper. A fully correct solution was rarely seen and attempts to factorise x² + 5x − 24 were seldom seen. Instead, much fruitless and incorrect algebra involving the manipulation and often combination of one or more of the three terms of the quadratic expression was common. A trial and improvement approach was common and often led to students obtaining one of the solutions (3) but not the other (−8). Students obtaining only one of the two solutions were not awarded any marks unless they had made a creditable attempt to factorise the expression.

  Text version

  Solve x2 + 5x - 24 = 0

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• Edexcel 1MA1 · June 2017 · Paper 2F · Q25 · 3 marks

  3 marks

  25

  
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  Examiner report

  Finding an expression for the nth term of an arithmetic sequence is not new to Foundation level students and this question was answered well by a good number of students who scored both marks. Many other students obtained one mark out of two for identifying the common difference and including the term 5n in their final expression. For example, 5n + 2 was commonly seen. Part (b) of this question also often attracted a correct and clear explanation that the value of 3n² was 48 and not 144 together with a correct conclusion. There were inevitably many students who thought Nathan was right because 3n² = (3 × 4)². A small number of students approached the question by trying to solve 3n² = 144 and stating that this leads to a non- integer value which is not possible in this context. This was a valid approach but usually students were unable to see the argument through to its conclusion. Summary Based on their performance in this paper, students should: • Remember that diagrams are not accurately drawn unless otherwise stated • Learn standard techniques such as writing a number in standard form and solving a quadratic equation by factorisation • Practise solving problems which require a chain of steps to find their solution • Carry out a common sense check on the answers to calculations; for example you should expect the number of £1 coins in question 6 to be a whole number • Use a calculator where possible, for example to add or for checking your addition but only do this after you have written down your method • Make sure you know the difference between simple interest and compound interest

  Text version

  Here are the first six terms of an arithmetic sequence. 3 8 13 18 23 28 (a) Find an expression, in terms of n, for the nth term of this sequence. The nth term of a different sequence is 3n2 Nathan says that the 4th term of this sequence is 144 (b) Is Nathan right? Show how you get your answer.

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• Edexcel 1MA1 · June 2017 · Paper 2H · Q11 · 4 marks

  4 marks

  11

  
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  Only a few students demonstrated enough convincing algebraic manipulative skills to carry the solution of this equation to its correct conclusion. Many were able to correctly write at least two fractions with a common denominator, usually the LHS, but then failed when trying to remove the fractional elements. One major failing of many students was to multiply both numerator and denominator by a common multiple, eg. 12(1−𝑥) 12 ×6 instead of simply 2(1 – x). Failure to correctly deal with negative signs was very common.

  Text version

  Solve (3x-2)/4 - (2x+5)/3 = (1-x)/6.

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• Edexcel 1MA1 · June 2017 · Paper 2H · Q14 · 3 marks

  3 marks

  14

  
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  Only a minority of students gained full marks here. There was no particular pattern to the confusion that surrounded the majority of students. Many clearly simply guessed.

  Text version

  Match y=sin x, y=x^3+4x, y=2^x and y=4/x to graphs A-I. A is a positive increasing exponential; B is a reciprocal curve in quadrants II and IV; C is a sine wave through the origin with positive initial gradient; D is a cosine-like wave with a maximum on the y-axis; E is an upward parabola; F is an increasing cubic through the origin; G is a downward parabola tangent to the x-axis at the origin; H is a reciprocal curve in quadrants I and III; I is a positive decreasing exponential.

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• Edexcel 1MA1 · June 2017 · Paper 2H · Q19 · 4 marks

  4 marks

  19

  
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  Only a small minority of students were able to faultlessly follow a solution to a correct conclusion. Many gained 2 marks for satisfactorily dealing with the two algebraic fractions but then failed to recognise that the sum of these two fractions was to be subtracted from 2. In fact, many students failed to recognise that 2 was part of the question. Poor recognition of the effect of the negative signs was the most common error.

  Text version

  2 - (x+2)/(x-3) - (x-6)/(x+3) can be written as (ax+b)/(x^2-9), where a and b are integers. Work out a and b.

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• Edexcel 1MA1 · June 2017 · Paper 2H · Q20 · 5 marks

  5 marks

  20

  
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  In part (a), very few students were able to derive the equation y = x + 4 from the information given. Students were generally more familiar with drawing tangents and in part (b), many students gained credit for drawing a tangent and attempting to find its gradient, often giving an answer within the accepted range. A number of students drew inaccurate tangents believing that they had to pass through the origin. Some obtained the correct solution by finding the gradient of a chord and not a tangent, this gained no credit. Some students used calculus to good effect usually resulting in a correct gradient of 2. This is outside this specification but did gain credit.

  Text version

  The diagram shows part of the graph of y = x2 - 2x + 3 10 8 6 4 2 2 4 O -2 (a) By drawing a suitable straight line, use your graph to find estimates for the solutions of x2 - 3x - 1 = 0 P is the point on the graph of y = x2 - 2x + 3 where x = 2 (b) Calculate an estimate for the gradient of the graph at the point P.

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• Edexcel 1MA1 · June 2017 · Paper 2H · Q22 · 3 marks

  3 marks

  22

  
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  This question was very well answered by a great many students of all levels of ability, gaining at least one mark for realising that the sequence was a quadratic sequence. Having found the second differences of ‘4’, many gave 4n2 as the first part of their nth term. Having found a correct first term of 2n2, many students continued to employ a differencing approach. Others successfully arrived at a correct solution from solving simultaneous equations. Where students were not successful it was common to see 4n, 2n or n instead of 2n2; thus many 2 term linear expressions.

  Text version

  Here are the first five terms of a sequence. 4 11 22 37 56 Find an expression, in terms of n, for the nth term of this sequence.

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• Edexcel 1MA1 · June 2017 · Paper 2H · Q23 · 3 marks

  3 marks

  23

  
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  Examiner report

  Of those students who understood what this question was asking, the majority gained at least two marks for the gradient of each of the normal and the tangent. Errors were often then seen in applying y = mx + c to find the value of c. It is evident many pupils are not confident working with surds. A few students lost marks for forgetting to put “x” in their final answer, although credit was given at the first sight of a correct equation. Summary Based on their performance on this paper, students should:  set out their solutions in a logical manner, showing all necessary working  learn to differentiate between simple interest and compound interest  ensure that the read the question carefully and that their final answer does answer the question as set  learn the sine rule and cosine rule and how to use these correctly, relabeling the sides and angles of a triangle if necessary  practise working with histograms

  Text version

  Circle L has equation x^2+y^2=4. P=(3/2,sqrt(7)/2) lies on L. Find the equation of the tangent at P.

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• Edexcel 1MA1 · June 2017 · Paper 3F · Q2 · 2 marks

  2 marks

  2

  
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  It was pleasing that many students were able to write an expression, in terms of p and b, for the total number of cups. Many students gained one mark for writing 12p or 18b or p + b (which was very common). Students who used both 12p and 18b often gave a fully correct expression although answers such as 12p × 18b and 12p, 18b were quite common. There were many students who went on to incorrectly “simplify” their algebraic expression. Those that wrote 12p + 18b = 30pb or 12p + 18b = 30 scored only one mark. Some students used unconventional algebraic notation such as p12 and p × 12 and this was accepted.

  Text version

  Cups are sold in packs and in boxes. There are 12 cups in each pack. There are 18 cups in each box. Alison buys p packs of cups and b boxes of cups. Write down an expression, in terms of p and b, for the total number of cups Alison buys.

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• Edexcel 1MA1 · June 2017 · Paper 3F · Q11 · 4 marks

  4 marks

  11

  
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  Examiner report

  Many students were able to substitute the value of v into the formula in part (a) and get an answer of 11 with working usually shown. Some students, though, did not know how to substitute correctly and answers such as 42 + 3 = 45, 4 + 2 + 3 = 9 and 42 + 3 = 19 were quite common. Far fewer correct answers were seen in part (b). When the answer was written as v = T – 3 ÷ 4 only one mark could be awarded due to brackets being omitted. It was not uncommon for students to gain one mark for getting as far as 4v = T – 3 and then either give this as the final answer or follow it with an incorrect step. Some students indicated the intention to subtract 3 from both sides of the formula but were unable to do this correctly. It was very rare to see an attempt at division by 4 as a first step. Many students wrote an incorrect answer such as v = 4T + 3 or  T v or 4  T v on the answer line without showing any working and could be awarded no marks. A few students attempted to use flow charts, which on the whole did not lead to the correct answer.

  Text version

  T = 4v + 3. (a) Find T when v=2. (b) Make v the subject.

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• Edexcel 1MA1 · June 2017 · Paper 3F · Q13 · 5 marks

  5 marks

  13

  
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  Examiner report

  Correct answers were quite common and it was pleasing that some of these came from an algebraic approach. However, such algebraic attempts were rare. Many of the students using algebra got no further than writing down expressions for the three angles. Sometimes only two of the three expressions were correct, usually x and 4x. The third angle proved to be the most problematic, with incorrect expressions such as x – 27 and just – 27 seen. Relatively few students were able to use their three expressions to set up an equation. The majority of correct answers came from a trials approach rather than from an algebraic method. Students who used a trials approach that did not result in the correct answer gained no marks.

  Text version

  The size of the largest angle in a triangle is 4 times the size of the smallest angle. The other angle is 27 degrees less than the largest angle. Work out, in degrees, the size of each angle in the triangle. You must show your working.

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• Edexcel 1MA1 · June 2017 · Paper 3F · Q16 · 3 marks

  3 marks

  16

  
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  This proved to be a challenging question for many students and it was frequently not attempted. Some of the students that appeared familiar with simultaneous equations realised that x could be eliminated straight away and subtracted to get 5y = –10 although errors were often made when subtracting. Having got to 5y = –10 and y = –2 students substituted this y value into one of the equations. Those that got to 3x + –2 = –4 were often unable to find the correct value of x. Some students could not isolate x correctly and some got to 3x = –2 but then wrote x = –1.5. Some of those that did succeed gave a decimal answer rather than the fraction  and when this decimal was –0.6 the accuracy mark could not be awarded. Many students appeared not to realise that x could be eliminated straight away and proceeded to multiply one or both equations. Students should be encouraged to look for matching terms to begin with. Nevertheless these students were often able to achieve a pair of equations which they could add or subtract to eliminate one of the variables. Some used an incorrect operation in an attempt to eliminate one variable and arithmetic errors were very common. Students who found one value usually went on to substitute this value into an equation in order to find the other value.

  Text version

  Solve 3x + y = -4 and 3x - 4y = 6.

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• Edexcel 1MA1 · June 2017 · Paper 3F · Q22 · 4 marks

  4 marks

  22

  
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  The table in part (a) was generally completed well and students were often able to use their values to draw a fully correct curve in part (b). Some students plotted the points correctly but then didn’t join them or else joined them with straight line segments or joined them with a curve that missed one or more of the points, resulting in one mark only. When plotting the points from the table most mistakes were made plotting (4, 1.5) and (5, 1.2).

  Text version

  (a) Complete the values for y=6/x at x=0.5,1,1.5,2,3,4,5,6. (b) Draw y=6/x for 0.5 <= x <= 6.

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• Edexcel 1MA1 · June 2017 · Paper 3H · Q2 · 3 marks

  3 marks

  2

  
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  Examiner report

  Many students answered this question correctly with appropriate methods and obtained three marks. The most successful method was to eliminate one variable, usually x, and obtain a value for y. Those who did this usually went on to successfully substitute their answer into the first equation given. However, arithmetic errors were seen. Students who lost marks on this question did so through their inability to deal with the subtraction of negative values. Many students realised that the equations had to be subtracted but ended up with incorrect results such as; 5y = 10, −5y = −10 or 5 = −2. Several students decided to equate the y values by multiplying one equation by 4 but then the basic elimination of the y values let them down. A few students rounded  to just 0.6 rather than as a negative recurring decimal. This was not acceptable for the final accuracy mark.

  Text version

  Solve 3x + y = -4 and 3x - 4y = 6.

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• Edexcel 1MA1 · June 2017 · Paper 3H · Q13 · 4 marks

  4 marks

  13

  
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  Examiner report

  Most students scored some marks on this question. The award of two marks was relatively frequent and shows that these students could correctly identify two of the lines bordering the region. Too often x = −2 was seen instead of y = −2 Sometimes y = x and y = −2 were given and not y = 0.5x + 1. Many students could not give the inequality signs correctly. For those that failed to score at all, the most common incorrect answer seen was just a list of coordinates with a complete failure to engage with the concept of boundary lines. Centres are advised to teach students to both plot and state equations of lines.

  Text version

  A solid-boundary triangular region has vertices (-6,-2), (-2,-2), (2,2). Its sides lie on a horizontal line, the line through (-2,-2) and (2,2), and the line through (-6,-2) and (2,2). Write the three inequalities defining the shaded interior.

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• Edexcel 1MA1 · June 2017 · Paper 3H · Q14 · 6 marks

  6 marks

  14

  
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  Examiner report

  It was pleasing to see a good number of correct answers to part (a) and part (b) almost in the same proportion. Part (a) was generally well attempted with the vast majority of students gaining at least one mark, normally for the factorisation of the two squares component. The factorisation of the other quadratic caused a problem, with many failing to obtain the correct solution often with incorrect signs seen. Some errors were also seen in the cancellation of the terms in the fraction. Also some further incorrect simplifications were seen. A common misconception still being seen is for students to just cross out part terms Eg x2 in this fraction. Obviously this issue should continue to be addressed by centres as it is an incorrect step. In part (b) a larger proportion of students gained the first mark than did in part (a) of this question. This mark was usually for clearing the v or expanding the brackets. The subsequent rearrangement of the equation caused more problems for the students. Several were able to rearrange the equation to the form vw + 30v = 15t only then to state that 31v = 15t/w, showing a lack of understanding of simplifying expressions. Some students struggled to read their own writing and often missed off a digit or letter for no apparent reason.

  Text version

  (a) Simplify (x^2-16)/(2x^2-5x-12). (b) Make v the subject of w=15(t-2v)/v.

  Pearson does not publish a direct link to this question paper.

• Edexcel 1MA1 · June 2017 · Paper 3H · Q15 · 5 marks

  5 marks

  15

  
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  Examiner report

  The statistics show that students found this question challenging. The main concept required to start the question was the use of 1 sin , ab C no longer given to students. The question also combined areas of mathematics, a requirement of the assessment objectives and some students struggled with the algebra, surds and the geometric concept. Some students worked accurately through all stages, showing elegant correct solutions. Whilst others were able to set up the initial algebraic expression or equation, many students were unable to manipulate the algebra into a suitable quadratic equation. Common errors seen included missing the half from the initial formula, adding the expressions for a and b instead of multiplying or not using an appropriate value for sin 45 Unfortunately arriving at a correct quadratic did not always mean full marks as some students could not go on to solve for x at this stage. The solving of a quadratic equation should be practised by centres as it can form part of many different questions.

  Text version

  Triangle ABC has AB=(2x-1) m, BC=(x+3) m and included angle ABC=45 degrees. Its area is 6sqrt(2) m^2. Find x to 3 significant figures.

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• Edexcel 1MA1 · June 2017 · Paper 3H · Q16 · 5 marks

  5 marks

  16

  
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  Examiner report

  This question was not attempted by all candidates suggesting possibly that this topic had not been covered by all centres. When answers were seen it was evident for part (a) that most students had a good appreciation of the process of iteration and successfully secured the first mark for putting the starting value of -2.5 into the iteration formula. Sadly, wrong answers appeared to come from the incorrect squaring of negative numbers. Students could avoid this error by placing negative numbers in brackets on their calculator. By showing their substitutions students could gain 2 marks. This question exemplifies the need for students to show their working out especially when the calculator is used heavily to ensure that method marks can be awarded. Part (b) was not well answered. Some students did state that the equation had been re arranged to give the iterative formula. Others correctly stated that iteration provides convergence towards a root of the cubic equation. A common misconception in part (b) was that the 3 answers from iteration in part (a) provided the 3 solutions to the cubic equation.

  Text version

  Use x_(n+1)=-2-4/(x_n^2), with x_0=-2.5. (a) Find x_1,x_2,x_3. (b) Explain their relationship to x^3+2x^2+4=0.

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• Edexcel 1MA1 · June 2017 · Paper 3H · Q19 · 3 marks

  3 marks

  19

  
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  Examiner report

  This was the penultimate question on the paper and required solving a quadratic inequality even so almost half the students scored at least one mark on this question. The use of the quadratic formula or factorisation was correctly done by most who attempted either process. Many who applied either process then accurately identified the two critical values of −2 and 0.5. Only a small number of students were able to correctly state the two distinct regions. Those who sketched a graph and identified the two regions on the graph were much more successful in gaining the final accuracy mark. The incorrect use of inequalities or giving one continuous inequality was often seen on complete solutions. Of those that found the correct critical values only about one quarter of these students went onto give correct inequalities.

  Text version

  Solve 2x^2+3x-2 > 0.

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• Edexcel 1MA1 · June 2017 · Paper 3H · Q20 · 4 marks

  4 marks

  20

  
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  Examiner report

  Part (a) was correctly answered by a significant proportion of the students. Unfortunately of the incorrect answers seen some gave (1, 0) instead of (0,1). In part (b) a significant number of students were unable to attempt this question even though they were asked to sketch a circle so could have started with that. A disappointing number of students were unable to write the co-ordinates as (x, y) and mixed up the order of the values thus losing the last mark because of poor and incorrect labelling. A common error seen was to translate in the positive y direction, but one mark could still be awarded if the radius was shown to be 4. Another common error was using a radius of 2 or 16 The quality of sketches varied greatly with some being drawn free hand and others with the aid of a pair of compasses, either was acceptable; on sketches clear labelling is helpful. Centres are advised to check that students use clear and consistent labels on this type of question going forward. Summary Based on their performance on this paper, students should:  show working and not just use a calculator and then write down the answer only  check their arithmetic carefully, particularly when negative numbers are involved  learn all formulae appropriate to this tier of entry  draw a diagram when one is not provided for a geometry question to aid understanding of the situation  ensure that the conventional three label lettering is used to identify angles in geometry questions.

  Text version

  (a) The curve y=a^x meets the y-axis at A. State A. (b) Circle C: x^2+y^2=16 is translated by vector (3,0) to circle B. Sketch B and label its centre and x-axis intersections.

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• Edexcel 1MA1 · sample · Paper 1F · Q3 · 1 mark

  1 mark

  3

  
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  Text version

  Write down the 20th odd number.

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• Edexcel 1MA1 · sample · Paper 1F · Q12 · 3 marks

  3 marks

  12

  
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  Text version

  The first three patterns in a sequence are made from triangles and rectangles. Pattern 1 has 6 triangles and 2 rectangles; pattern 2 has 8 triangles and 3 rectangles; pattern 3 has 10 triangles and 4 rectangles.

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• Edexcel 1MA1 · sample · Paper 1F · Q19 · 2 marks

  2 marks

  19

  
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  Text version

  Solve 4x + 5 = x + 26

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• Edexcel 1MA1 · sample · Paper 1F · Q22 · 2 marks

  2 marks

  22

  
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  Text version

  Expand and simplify (m + 7)(m + 3)

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• Edexcel 1MA1 · sample · Paper 1F · Q28 · 5 marks

  5 marks

  28

  
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  Text version

  In a parallelogram, the angles (2x + 43)° and (5x − 20)° are opposite. The angle (4y − 5x)° is adjacent to both of them.

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• Edexcel 1MA1 · sample · Paper 1H · Q2 · 2 marks

  2 marks

  2

  
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  Text version

  Expand and simplify (m + 7)(m + 3)

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• Edexcel 1MA1 · sample · Paper 1H · Q8 · 5 marks

  5 marks

  8

  
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  Text version

  Here is a parallelogram.

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• Edexcel 1MA1 · sample · Paper 1H · Q16 · 2 marks

  2 marks

  16

  
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  Text version

  Factorise fully 20x² − 5.

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• Edexcel 1MA1 · sample · Paper 1H · Q17 · 3 marks

  3 marks

  17

  
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  Text version

  Make a the subject of a + 3 = (2a + 7)/r.

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• Edexcel 1MA1 · sample · Paper 1H · Q19 · 3 marks

  3 marks

  19

  
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  Text version

  Solve x² > 3x + 4.

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• Edexcel 1MA1 · sample · Paper 1H · Q21 · 5 marks

  5 marks

  21

  
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  Text version

  A car accelerates uniformly from 0 m/s at 0 s to V m/s at 20 s, travels at constant speed V from 20 s to 60 s, then decelerates uniformly to rest at 100 s. It travels 1.75 km in total.

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• Edexcel 1MA1 · sample · Paper 1H · Q24 · 4 marks

  4 marks

  24

  
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  Text version

  John has an empty box.

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• Edexcel 1MA1 · sample · Paper 1H · Q25 · 5 marks

  5 marks

  25

  
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  Text version

  A(−2, 1), B(6, 5) and C(4, k) are the vertices of a right-angled triangle ABC.

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• Edexcel 1MA1 · sample · Paper 2F · Q10 · 3 marks

  3 marks

  10

  
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  Text version

  (a) Rob buys p packets of plain crisps and c packets of cheese crisps.

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• Edexcel 1MA1 · sample · Paper 2F · Q12 · 5 marks

  5 marks

  12

  
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  Text version

  You can use the information in the table to convert between kilometres and miles. miles 0 5 20 40 kilometres 0 8 32 64

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• Edexcel 1MA1 · sample · Paper 2F · Q18 · 2 marks

  2 marks

  18

  
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  Text version

  f = 5x + 2y

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• Edexcel 1MA1 · sample · Paper 2F · Q20 · 3 marks

  3 marks

  20

  
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  Text version

  (a) Factorise 3f + 9 [1 marks]

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• Edexcel 1MA1 · sample · Paper 2F · Q21 · 2 marks

  2 marks

  21

  
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  Text version

  q = p/r + s.

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• Edexcel 1MA1 · sample · Paper 2F · Q25 · 4 marks

  4 marks

  25

  
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  Text version

  Here are the first four terms of an arithmetic sequence. 6 10 14 18

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• Edexcel 1MA1 · sample · Paper 2H · Q3 · 4 marks

  4 marks

  3

  
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  Text version

  Here are the first four terms of an arithmetic sequence. 6 10 14 18

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• Edexcel 1MA1 · sample · Paper 2H · Q7 · 3 marks

  3 marks

  7

  
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  Text version

  The graph of y = f(x) has turning point (1, 4), crosses y = 2 at x ≈ −0.4 and x ≈ 2.4, and has f(0.5) ≈ 3.75.

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• Edexcel 1MA1 · sample · Paper 2H · Q9 · 5 marks

  5 marks

  9

  
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  Text version

  ABCD is a rectangle and EFGH is a trapezium. The two shapes have equal perimeters.

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• Edexcel 1MA1 · sample · Paper 2H · Q14 · 3 marks

  3 marks

  14

  
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  Text version

  Prove algebraically that

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• Edexcel 1MA1 · sample · Paper 2H · Q16 · 3 marks

  3 marks

  16

  
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  Text version

  Simplify a quotient of algebraic fractions.

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• Edexcel 1MA1 · sample · Paper 2H · Q19 · 2 marks

  2 marks

  19

  
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  Text version

  The graph of y = f(x) is the parabola y = (x + 2)². It has vertex (−2, 0) and passes through (−4, 4) and (0, 4).

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• Edexcel 1MA1 · sample · Paper 2H · Q20 · 5 marks

  5 marks

  20

  
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  Text version

  Solve x² + y² = 25 and y − 2x = 5 algebraically.

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• Edexcel 1MA1 · sample · Paper 3F · Q2 · 2 marks

  2 marks

  2

  
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  Text version

  (a) Simplify 8x – 3x + 2x [1 marks]

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• Edexcel 1MA1 · sample · Paper 3F · Q4 · 3 marks

  3 marks

  4

  
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  Text version

  On a coordinate grid, A = (3, 5), B = (5, 2) and C = (7, 5). The line segments AB and BC are drawn.

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• Edexcel 1MA1 · sample · Paper 3F · Q7 · 5 marks

  5 marks

  7

  
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  Text version

  Here is a number machine. input × 3 – 4 output

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• Edexcel 1MA1 · sample · Paper 3F · Q11 · 2 marks

  2 marks

  11

  
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  Text version

  Here is a list of five numbers. 14 15 16 17 18 From the list,

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• Edexcel 1MA1 · sample · Paper 3F · Q14 · 4 marks

  4 marks

  14

  
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  Text version

  The total weight of 3 tins of beans and 4 jars of jam is 2080 g.

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• Edexcel 1MA1 · sample · Paper 3F · Q17 · 4 marks

  4 marks

  17

  
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  A rectangle of card measures 2x cm by y cm. Lily uses four identical rectangles to make a frame: two rectangles stand vertically on the left and right, and two lie horizontally between them at the top and bottom. The inner opening therefore has horizontal sides 2x and vertical sides 2x − 2y. Its perimeter is P cm. Magda says, ‘When x and y are whole numbers, P is always a multiple of 4.’

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• Edexcel 1MA1 · sample · Paper 3F · Q20 · 6 marks

  6 marks

  20

  
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  Text version

  Here are the first six terms of a Fibonacci sequence. 1 1 2 3 5 8 The rule to continue a Fibonacci sequence is, the next term in the sequence is the sum of the two previous terms.

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• Edexcel 1MA1 · sample · Paper 3H · Q3 · 6 marks

  6 marks

  3

  
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  Text version

  Here are the first six terms of a Fibonacci sequence. 1 1 2 3 5 8 The rule to continue a Fibonacci sequence is, the next term in the sequence is the sum of the two previous terms.

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• Edexcel 1MA1 · sample · Paper 3H · Q9 · 6 marks

  6 marks

  9

  
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  Text version

  Answer the following algebra questions.

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• Edexcel 1MA1 · sample · Paper 3H · Q10 · 4 marks

  4 marks

  10

  
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  Text version

  The function f is such that f(x) = 4x – 1

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• Edexcel 1MA1 · sample · Paper 3H · Q11 · 3 marks

  3 marks

  11

  
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  Text version

  Solve x² - 5x + 3 = 0.

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• Edexcel 1MA1 · sample · Paper 3H · Q14 · 6 marks

  6 marks

  14

  
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  Text version

  Consider the equation x³ + 4x = 1.

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• Edexcel 1MA1 · sample · Paper 3H · Q17 · 5 marks

  5 marks

  17

  
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  Text version

  Louis and Robert are investigating the growth in the population of a type of bacteria. They have two flasks A and B. At the start of day 1, there are 1000 bacteria in flask A. The population of bacteria grows exponentially at the rate of 50% per day.

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• Edexcel 1MA1 · specimen2 · Paper 1F · Q5 · 3 marks

  3 marks

  5

  
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  Text version

  On a coordinate grid, A = (−2, 3) and B = (4, 5).

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• Edexcel 1MA1 · specimen2 · Paper 1F · Q10 · 4 marks

  4 marks

  10

  
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  Text version

  Answer both algebra questions.

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• Edexcel 1MA1 · specimen2 · Paper 1F · Q18 · 3 marks

  3 marks

  18

  
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  Text version

  Dimitar and Pip each start with 20 sweets. Dimitar gives Pip x sweets. Dimitar eats 5 of his remaining sweets; Pip eats half of hers.

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• Edexcel 1MA1 · specimen2 · Paper 1F · Q19 · 3 marks

  3 marks

  19

  
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  Text version

  Simplify or factorise each expression.

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• Edexcel 1MA1 · specimen2 · Paper 1F · Q24 · 2 marks

  2 marks

  24

  
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  Text version

  The first five terms of a sequence are 2, 8, 18, 32, 50.

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• Edexcel 1MA1 · specimen2 · Paper 1H · Q1 · 3 marks

  3 marks

  1

  
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  Text version

  Simplify or factorise each expression.

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• Edexcel 1MA1 · specimen2 · Paper 1H · Q9 · 3 marks

  3 marks

  9

  
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  Text version

  A cuboid has side lengths x, 2x and 3x cm, where x is a positive integer. Its volume is less than 900 cm³.

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• Edexcel 1MA1 · specimen2 · Paper 1H · Q11 · 5 marks

  5 marks

  11

  
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  Text version

  A solid consists of a hemisphere of radius x on a cone of radius x and height 4x (total height 5x). A cylinder of radius 2x and height h has the same volume. Use Vcone=(1/3)πr²h and Vsphere=(4/3)πr³.

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• Edexcel 1MA1 · specimen2 · Paper 1H · Q17 · 3 marks

  3 marks

  17

  
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  Text version

  Solve x²−6x−8=0.

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• Edexcel 1MA1 · specimen2 · Paper 1H · Q20 · 4 marks

  4 marks

  20

  
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  Text version

  Show that (3x+6)/(x²−3x−10) ÷ (x+5)/(x³−25x) simplifies to ax.

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• Edexcel 1MA1 · specimen2 · Paper 1H · Q21 · 4 marks

  4 marks

  21

  
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  Text version

  Solve x²>3(x+6).

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• Edexcel 1MA1 · specimen2 · Paper 1H · Q22 · 5 marks

  5 marks

  22

  
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  Text version

  Circle x²+y²=40 has centre O. The tangent at A=(2,6) meets the x-axis at P.

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• Edexcel 1MA1 · specimen2 · Paper 2F · Q10 · 3 marks

  3 marks

  10

  
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  Text version

  A speed-time graph for Karol passes through speed 6.5 m/s at t=3 s, has maximum 9.8 m/s, and crosses 9 m/s at t=5 and t=9.

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• Edexcel 1MA1 · specimen2 · Paper 2F · Q11 · 3 marks

  3 marks

  11

  
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  Text version

  A number pattern begins 1, 2, 4.

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• Edexcel 1MA1 · specimen2 · Paper 2F · Q14 · 2 marks

  2 marks

  14

  
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  Text version

  Mobile cost graph has fixed rental £20 and rises to £35 at 500 minutes.

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• Edexcel 1MA1 · specimen2 · Paper 2F · Q17 · 5 marks

  5 marks

  17

  
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  Text version

  Answer all three algebra questions.

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• Edexcel 1MA1 · specimen2 · Paper 2F · Q24 · 4 marks

  4 marks

  24

  
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  Text version

  Answer both algebra questions.

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• Edexcel 1MA1 · specimen2 · Paper 2H · Q3 · 4 marks

  4 marks

  3

  
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  Text version

  Answer both algebra questions.

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• Edexcel 1MA1 · specimen2 · Paper 2H · Q8 · 2 marks

  2 marks

  8

  
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  Text version

  Steve expands 5(x+2)=47 as 5x+2=47 and gets x=9. Liz solves 3x²+8=83, obtains x²=25, then gives x=5 only.

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• Edexcel 1MA1 · specimen2 · Paper 2H · Q9 · 5 marks

  5 marks

  9

  
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  Text version

  The functions are f(x) = 3(x − 4) and g(x) = x/5 + 1.

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• Edexcel 1MA1 · specimen2 · Paper 2H · Q13 · 3 marks

  3 marks

  13

  
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  Text version

  A sequence is defined by p₀ = 100 and pₜ₊₁ = 1.06pₜ.

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• Edexcel 1MA1 · specimen2 · Paper 2H · Q15 · 6 marks

  6 marks

  15

  
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  Text version

  A smooth speed–time curve describes Karol's motion. With time in seconds and speed in m/s, it passes approximately through (0, 0), (2, 4.5), (4, 7.8), (6, 9.5), (7, 9.8), (8, 9.6) and (10, 8).

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• Edexcel 1MA1 · specimen2 · Paper 2H · Q18 · 3 marks

  3 marks

  18

  
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  Text version

  Simplify (√a+2√b)(√a−2√b).

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• Edexcel 1MA1 · specimen2 · Paper 2H · Q19 · 4 marks

  4 marks

  19

  
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  Text version

  For the transformations in parts (b) and (c), the graph of y = f(x) is the polyline through (−3, 1), (−2, 1), (0, 3), (2, 1) and (3, 1).

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• Edexcel 1MA1 · specimen2 · Paper 2H · Q21 · 5 marks

  5 marks

  21

  
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  Text version

  The ratio equation is (2x − 1):(x − 4) = (16x + 1):(2x − 1).

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• Edexcel 1MA1 · specimen2 · Paper 3F · Q14 · 4 marks

  4 marks

  14

  
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  Text version

  A hire company calculates the total cost by multiplying the number of days by 7.8 and then adding 12.

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• Edexcel 1MA1 · specimen2 · Paper 3F · Q20 · 4 marks

  4 marks

  20

  
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  Text version

  Answer both inequality questions.

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• Edexcel 1MA1 · specimen2 · Paper 3F · Q21 · 2 marks

  2 marks

  21

  
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  Text version

  The first five terms of an arithmetic sequence are −3, 1, 5, 9 and 13.

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• Edexcel 1MA1 · specimen2 · Paper 3F · Q24 · 2 marks

  2 marks

  24

  
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  Text version

  The formula is y = t/3 − 2a.

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• Edexcel 1MA1 · specimen2 · Paper 3F · Q27 · 4 marks

  4 marks

  27

  
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  At a depth of x metres, the temperature of the water in an ocean is T °C. At depths below 900 metres, T is inversely proportional to x and is given by T = 4500/x. Four candidate graphs are shown: A is an increasing straight line, B is a decreasing straight line, C is an increasing curve, and D is a decreasing reciprocal curve approaching both axes.

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• Edexcel 1MA1 · specimen2 · Paper 3F · Q28 · 3 marks

  3 marks

  28

  
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  Text version

  A right-angled triangle has perpendicular sides x and y. Four congruent copies are joined to enclose the square ABCD, inside an outer square of side x + y.

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• Edexcel 1MA1 · specimen2 · Paper 3F · Q30 · 4 marks

  4 marks

  30

  
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  Text version

  A quadrilateral has side lengths 2x, 2x − 1, x − 3 and 3x cm. A triangle has side lengths x + 3, x + 3 and 5 cm. The perimeter of the quadrilateral is twice the perimeter of the triangle.

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• Edexcel 1MA1 · specimen2 · Paper 3H · Q3 · 2 marks

  2 marks

  3

  
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  Text version

  The formula is y = t/3 − 2a.

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• Edexcel 1MA1 · specimen2 · Paper 3H · Q7 · 3 marks

  3 marks

  7

  
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  Text version

  A right-angled triangle has perpendicular sides x and y. Four congruent copies are joined to enclose the square ABCD, inside an outer square of side x + y.

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• Edexcel 1MA1 · specimen2 · Paper 3H · Q10 · 3 marks

  3 marks

  10

  
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  Text version

  The surface gravity of a planet is given by g = 6.67 × 10⁻¹¹m/r², where m kilograms is the mass of the planet and r metres is its radius. For Earth, m = 5.98 × 10²⁴ and r = 6.378 × 10⁶. For Jupiter, m = 1.90 × 10²⁷ and r = 7.149 × 10⁷.

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• Edexcel 1MA1 · specimen2 · Paper 3H · Q11 · 4 marks

  4 marks

  11

  
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  Text version

  The simultaneous equations are 2x − 4y = 19 and 3x + 5y = 1.

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• Edexcel 1MA1 · specimen2 · Paper 3H · Q15 · 2 marks

  2 marks

  15

  
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  Text version

  Show (a+1)/(b+1)−a/b can be written (b−a)/[b(b+1)].

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• Edexcel 1MA1 · specimen2 · Paper 3H · Q17 · 3 marks

  3 marks

  17

  
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  Text version

  The product of two consecutive positive integers is added to the larger of the two integers.

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• Edexcel 1MA1 · specimen2 · Paper 3H · Q18 · 4 marks

  4 marks

  18

  
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  Text version

  A speed–time graph for a car is sampled every 2 seconds. The graph gives approximate speeds of 0, 2, 5, 9, 15 and 24 m/s at times 0, 2, 4, 6, 8 and 10 seconds respectively.

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• Edexcel 1MA1 · specimen2 · Paper 3H · Q22 · 7 marks

  7 marks

  22

  
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  Text version

  There are y black socks and 5 white socks in a drawer. Joshua takes two socks at random without replacement. The probability that he takes one white sock and one black sock is 6/11.

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• Edexcel 1MA1 · specimen2 · Paper 3H · Q23 · 4 marks

  4 marks

  23

  
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  Text version

  Consider 2x² + 16x + 35.

  Pearson does not publish a direct link to this question paper.