Pearson Edexcel GCSE Maths · Topic G
Angles, constructions, transformations, mensuration, trigonometry and vectors. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme. Then practise with the original printable pack.
Tier
Total recorded: 0 of 65 marks
19 original questions written in the style of the real papers. Attempt one on paper, then reveal the mark scheme and award yourself the marks. Marks you award are saved in this browser against that question.
1
This question tests standard geometric vocabulary.
(a)
(1)
(b)
(1)
(c)
(1)
(Total for Question 1 is 3 marks)
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2
A garden sprinkler is fixed at a point P in the middle of a large flat lawn. The sprinkler waters every part of the lawn that is within 4 m of P.
(a)
(1)
(b)
(1)
(Total for Question 2 is 2 marks)
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3
This question is about the properties of quadrilaterals.
(a)
(1)
(b)
(1)
(c)
(1)
(Total for Question 3 is 3 marks)
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4
In triangle ABC, AB = 7 cm, BC = 5 cm and angle ABC = 40°. In triangle PQR, PQ = 7 cm, QR = 5 cm and angle PQR = 40°.
(a)
(1)
(b)
(1)
(Total for Question 4 is 2 marks)
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5
Triangle T has vertices A(1, 1), B(1, 4) and C(3, 1). Triangle T is enlarged by scale factor 2, with centre the origin (0, 0), to give triangle T′.
(a)
(2)
(b)
(2)
(Total for Question 5 is 4 marks)
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6
A shape is reflected in the x-axis and then the image is reflected in the y-axis.
(a)
(2)
(b)
(1)
(Total for Question 6 is 3 marks)
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7
This question is about the properties of 3D solids.
(a)
(1)
(b)
(1)
(c)
(1)
(Total for Question 7 is 3 marks)
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8
This question is about converting between standard units.
(a)
(1)
(b)
(1)
(c)
(1)
(Total for Question 8 is 3 marks)
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9
The point A(3, −2) is translated to the point B(−1, 5).
(a)
(1)
(b)
(2)
(Total for Question 9 is 3 marks)
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10
A trapezium has parallel sides of length 8 cm and 12 cm, and a perpendicular height of 5 cm.
(a)
(2)
(b)
(2)
(Total for Question 10 is 4 marks)
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11
The sizes of the three angles of a triangle, in degrees, are x + 20, 2x and 3x − 20.
(a)
(3)
(b)
(1)
(c)
(1)
(Total for Question 11 is 5 marks)
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12
A right-angled triangle has its two shorter sides of length 6 cm and 8 cm, with the right angle between them.
(a)
(2)
(b)
(1)
(c)
(2)
(Total for Question 12 is 5 marks)
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13
A circle has radius 7 cm. Use the π button on your calculator.
(a)
(2)
(b)
(2)
(Total for Question 13 is 4 marks)
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14
A sector of a circle has radius 9 cm and a sector angle of 80°. Use the π button on your calculator.
(a)
(2)
(b)
(2)
(Total for Question 14 is 4 marks)
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15
This question is about the areas of 2D shapes.
(a)
(2)
(b)
(1)
(Total for Question 15 is 3 marks)
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16
This question is about angle facts.
(a)
(2)
(b)
(2)
(Total for Question 16 is 4 marks)
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17
A ship sails from a port P.
(a)
(2)
(b)
(1)
(Total for Question 17 is 3 marks)
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18
This question uses Pythagoras' theorem. Give your answers in the units stated.
(a)
(2)
(b)
(2)
(Total for Question 18 is 4 marks)
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19
Triangle ABC is isosceles with AB = AC. The angle at the apex, angle BAC, is 40°.
(a)
(2)
(b)
(1)
(Total for Question 19 is 3 marks)
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517 questions shown as printed, 33 reference-only · newest series first
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 · Q7 · 3 marks
3 marks
Examiner report
In part (a) of this question learners were asked to measure the length of the side of the given triangle giving their answers in centimetres. The majority of learners were able to do so successfully. Where incorrect answers were seen some appeared to be due to errors in reading the scale on the ruler leading to answers around 10.5 cm or answers around 95 where millimetres were incorrectly used. Only a very small proportion of learners seemed to have worked without a ruler or could not use the ruler they had. Some candidates were unable to measure a non-integer value, or assumed they should round and either 9 or 10 were common incorrect answers. In part (b) learners needed to identify the angle type that would describe the angle marked x on the diagram. The majority gave the correct answer of ‘acute’ although a variety of spelling errors (which were condoned) were seen. Common incorrect answers included a range of the other angle types with ‘obtuse’ or ‘right’ being the most popular of these. Part (c) asked learners to measure the size of the angle marked x. Most learners were able to do this successfully. Where incorrect answers were seen these appeared to be the result of incorrect reading of the scale on the protractor, for example an answer of 45°, or errors in lining up the protractor in order to measure the angle, with many incorrect answers only being a 1 or 2 degrees outside of the allowed range.
For this text adaptation of the accurately drawn triangle, use a centimetre scale and draw P=(-4, 0), Q=(5.5, 0) and R=(0.4, 3.2), then join the points. The angle marked x is the interior angle PQR at Q.
Edexcel 1MA1 · November 2024 · Paper 1F · Q10 · 3 marks
3 marks
Examiner report
In part (a) learners were asked to reflect a shape in a given mirror line. Around half of learners were able to give a fully correct answer. There were also a significant proportion of learners who were able to reflect the shape, but did not use the given mirror line correctly which commonly led to their answer being translated vertically from the correct position. In a small number of cases learners reflected in a vertical mirror line. In the minority of responses where the answer was not a reflection this was commonly due to the learner performing a different transformation, most commonly a rotation around the origin, or an attempt at a reflection but with errors in the drawing of the reflected shape. In part (b) learners were asked to write down the equation of the mirror line. This was answered correctly by less than half of the learners. The correct answer was = 1. Common incorrect answers were = 1, 1 and = .
On a coordinate grid, a quadrilateral has vertices (1,2), (1,3), (3,4), and (4,2). The dashed mirror line passes through (-5,1) and (5,1).
Edexcel 1MA1 · November 2024 · Paper 1F · Q14 · 5 marks
5 marks
Examiner report
In part (a) learners were asked to work out the height of the given cuboid having been given its volume, length and width. This proved slightly more challenging than the earlier questions on the paper and was answered correctly by around half of the learners. Where incorrect answers were seen there was often evidence of a start to the method to find the height, most often finding the area of the base rectangle (10 × 4 = 40) in this way the majority of learners were able to gain the first mark. In part (b) learners were asked to work out the surface area of a cuboid. Learners found this challenging and the majority were not able to gain marks. The most common error was to work out the volume of the cuboid rather than the surface area. A minority of learners were able to find the area of one face or the total surface area for three or more faces. Some learners worked with the area of six faces but incorrectly assumed that there were 4 faces that were 18 (6 by 3) having struggled to extract the information from the three-dimensional diagram of the cuboid. A number of learners multiplied each dimension by 4 and then added the answers; some learners worked out the perimeter of each face.
A cuboid has length 10 cm, width 4 cm and volume 200 cm³. A different cuboid has dimensions 6 cm by 5 cm by 3 cm.
Edexcel 1MA1 · November 2024 · Paper 1F · Q21 · 3 marks
3 marks
Examiner report
A significant majority of learners struggled with this question which used the angle properties of parallel lines and it was very rare to see a fully correct answer. The majority of learners did not correctly identify any of the angle results that could be obtained by using angles in parallel lines. Where marks were awarded, this was often for identifying angle ACD as being e and gaining the first mark. Some learners identified ABC as 3e or referenced the sum of angles in a triangle being 180° but these alone did not gain marks. It was common to see angle DAC being incorrectly identified as e or multiple angles labelled with e, 3e or similar. Very few learners were able to give an appropriate reason even when they had found a correct angle.
ABCD is a parallelogram with diagonal AC. Angle BAC is e° and angle ADC is 3e°.
Edexcel 1MA1 · November 2024 · Paper 1F · Q23 · 4 marks
4 marks
Examiner report
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.
A pentagon has interior angles a, b, c, d and e. Angle a = angle c, b = 155°, d = 3c and e = 2c.
Edexcel 1MA1 · November 2024 · Paper 1H · Q4 · 3 marks
3 marks
Examiner report
This question was not answered as well as might have been expected for an early question on the paper. It is a little different to the usual type of question assessing angles on parallel lines because the sizes of the angles are given with letters rather than numbers. Those learners that earned the method mark usually did so for indicating that angle ACD = e, which was often marked on the diagram. Stating that angle ADC + angle BAD = 180 was seen less often. Relatively few learners managed to find an expression, in terms of e, for the size of angle CAD. Some got as far as 4e + x = 180 but gave this as their final answer or rearranged it incorrectly. The C mark was awarded for an appropriate reason relating to parallel lines and most commonly this was ‘alternate angles’ to explain why angle ACD = e. For this angle, ‘corresponding angles’ was a common incorrect reason given. A large number of learners failed to gain any marks at all. Some marked angle ABC as 3e on the diagram but made no correct progress using angles on parallel lines. Many seemed to think that they needed to find a value for e.
ABCD is a parallelogram with diagonal AC. Angle BAC is e° and angle ADC is 3e°.
Edexcel 1MA1 · November 2024 · Paper 1H · Q6 · 4 marks
4 marks
Examiner report
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.
A pentagon has angles a,b,c,d,e where a=c, b=155°, d=3c and e=2c.
Edexcel 1MA1 · November 2024 · Paper 1H · Q12 · 4 marks
4 marks
Examiner report
It was pleasing to see some fully correct solutions but, overall, this question was answered very poorly with relatively few learners showing that they understood the relationships between lengths, areas and volumes in similar solids. Starting the process to find the ratio of the radius of A to the radius of B by finding 3 64 or 3 125 gained the first mark and some learners reached the ratio 4 : 5. The second mark was awarded for using the information about the radius of sphere C and working out 5 ÷ 2 or for starting to work with area and working out 42. A process to find the ratio of the area of A to the area of C, such as 42 : 2.52, was needed for the third mark but few were able to get this far. Some of those that did gave 16 : 6.25 as the final answer and were not awarded the accuracy mark because the question asked for an answer in the form a : b where a and b are integers. A successful route used by a few learners was to combine the ratios 4 : 5 and 2 : 1, giving the ratio of the radii as 8 : 10 : 5 from which they could easily see that the required ratio is 82 : 52. Many learners could not make a correct start. It was common to see 125, the volume of sphere B, divided by 2 and then 64 : 62.5 given as the final answer. Since the question involved three spheres some learners attempted to use the formulae for the surface area or volume of a sphere.
Solid spheres A and B have volumes 64 cm³ and 125 cm³. Sphere C's radius is 50% of sphere B's radius.
Edexcel 1MA1 · November 2024 · Paper 1H · Q14 · 3 marks
3 marks
Examiner report
Some learners gained the first mark for stating the value of sin30 or the value of sin45. Those who did not know these values were still able to gain the second mark for a correct sine rule statement such as 3 2 sin 45 sin30 AB . More than half of those who gained the first two marks did not get a correct final answer, some because they had an incorrect value for sin30 or sin45, some because they could not rearrange their sine rule statement correctly and some because they could not combine the surds correctly. Many learners showed a lack of understanding of the mathematical requirements of the question with some using right-angled trigonometric calculations.
Triangle ABC has angle A=30°, angle C=45°, and BC=3√2 cm.
Edexcel 1MA1 · November 2024 · Paper 1H · Q15 · 4 marks
4 marks
Examiner report
Many of the higher attaining learners gained either partial or full credit for their answers to this question and the vector algebra seen was generally clearly presented. Using the information given to write a correct expression for a vector earned the first mark and often this was awarded for a correct expression for OM or ON . A good proportion of those that gained the first mark were able to make further progress and earn the second mark which was most commonly awarded for a correct expression for MN . Many of those that earned the third mark for correct expressions for both MN and XN did not reach the ratio 4 : 1. Sometimes this was because of errors in the simplification of the expressions found. However, some simplified correctly but gave 4 b a : 1 b a as the final answer. Learners could be encouraged to write expressions such as 1 4 a , 1 3b and a b on the diagram, with arrows to show their direction, as this may aid their understanding of the direction when combining vectors.
OXYZ is a parallelogram with OY vector a and OZ vector b. M divides OY with OM:MY=1:3. N divides OZ with ON:NZ=1:2.
Edexcel 1MA1 · November 2024 · Paper 1H · Q17 · 4 marks
4 marks
Examiner report
Most of the learners who earned the first method mark for finding that angle CAB = 40º did so by using the alternate segment theorem. A few found angle COB = 80º and then used the angle at the centre is twice the angle at the circumference to find the size of angle CAB. Many of those who found angle CAB = 40º went on to earn the second method mark for using it to find either angle OAC = 10º or angle OAB = 30º. The third method mark could be awarded for angle OCD = 90º which was often marked on the diagram or for angle OCB = 50º and for some learners this was the only mark they got. Complete methods to find angle ACD were quite rare. Some that showed a correct method leading to angle ACD = 100º and earned the three method marks then failed to correctly state the one circle theorem that was needed for the final mark. Some gave no reason, others gave a description which did not include the required key words. Many learners had no idea how to proceed in this question and misconceptions such as angle OBC = 40º and angle ACD = 90º were very common. Learners should ensure that they clearly identify which angles they are working with by marking any found angles on the diagram and by using the correct three letter angle notation.
A, B and C lie on a circle with centre O. Chords AB, BC and AC and radii OA and OB are drawn. CD is tangent at C, with ray CB lying inside angle ACD (the ray order at C is CD, CB, CA). Angle BCD = 40° and angle OAB = 3 × angle OAC.
Edexcel 1MA1 · November 2024 · Paper 2F · Q4 · 1 mark
1 mark
Examiner report
Naming the shape in this question appeared to cause very few issues for the majority. Incorrect spellings were often seen but learners were rewarded with the mark if the intention was clear.
The polygon is shown by this outline: ______ / \ / \ \ / \______/
Edexcel 1MA1 · November 2024 · Paper 2F · Q10 · 3 marks
3 marks
Examiner report
This problem-solving question presented a challenge to learners, with many not able to reach the correct answer and only being able to gain one of the three marks available by making a valid start to solving the problem. The most common error was to add up the perimeters of all five of the squares seen in the diagram, taking no account of the fact that this included lengths on the interior of the shape. These attempts led to answers of 200 and scored only 1 mark. Where learners began by working out that the side length was 10, often shown on the diagram, they tended to have more success. Sometimes they had difficulty counting up how many 10s were needed to go round the outside, so answers of 110 and 130 were often seen.
A square has perimeter 40 cm. Five identical squares form a plus shape: one central square with one attached on each of its four sides.
Edexcel 1MA1 · November 2024 · Paper 2F · Q13 · 4 marks
4 marks
Examiner report
Performance in this angle reasoning question was varied, with the full range of marks between 0 and 4 being awarded. Whilst the majority of learners were confident to attempt this question, with many gaining 2 or 3 of the marks available, it was very rare to see a fully correct response worthy of 4 marks. A good start of working correctly with angles at a point or angles on a straight line to find missing interior angles of the triangle was common to see. This gained the two method marks and many also gained a communication mark for stating a valid angle fact if this linked clearly to their method used. It is surprising to note the number of learners who misunderstand the concept of angles on a straight line, believing that angles at two different points on the line will also sum to 180°rather than angles at a single point on the line. However, the quality of the reasons stated for the stages of their working was very mixed and some learners wrote no reasons at all. Others attempted reasons but were not able to do so clearly enough to gain credit due to not including the underlined words on the mark scheme which indicate the minimum requirements. To gain the final mark, learners needed a statement that two angles were equal or similar, correct figures, and fully correct reasons for each stage of their method, with the statement confirming the triangle was isosceles or clearly identifying the size of angle CBA being the most commonly omitted elements.
B, C and D lie on a straight line. At A the reflex angle outside triangle ABC is 310°. Exterior angle ACD is 115°.
Edexcel 1MA1 · November 2024 · Paper 2F · Q18 · 2 marks
2 marks
Examiner report
This familiar and standard question required learners to describe a single transformation using a given diagram. When marks were awarded, a partially complete response was given, and it was common for the centre of enlargement to be omitted. Many learners were only able to identify the transformation as an enlargement and give the scale factor of 2 which was described in a variety of ways but with the intention needed to be clear for credit to be given. No marks were awarded if learners described more than one transformation, but this was seen less frequently compared to previous series which was pleasing and when seen, it was invariably a translation that was described, sometimes using informal language such as “then move right and up”. When marks were not awarded, it was often due to either only one element of the three being provided or for simplistic statements such as “its doubled in size” or is “twice as big” which were unacceptable descriptions on their own.
On a coordinate grid, triangle A has vertices (2,3),(3,3),(3,1); triangle B has vertices (4,6),(6,6),(6,2).
Edexcel 1MA1 · November 2024 · Paper 2F · Q26 · 5 marks
5 marks
Examiner report
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.
Isosceles triangle ABC has AC=BC. AC=2x+11, BC=4x-4, AB=2x+5 cm; perimeter 72 cm.
Edexcel 1MA1 · November 2024 · Paper 2F · Q28 · 3 marks
3 marks
Examiner report
Although it was the last question on the paper, many learners attempted it and very few blank responses were seen. The majority were not able to calculate the volume of the cylinder correctly, with a wide variety of wrong attempts, mainly calculating the area or circumference of the circle instead, seen. However, when marks were awarded, it was often a single mark for realising the need to divide the mass by the volume, and they were able to gain the second mark for showing this even when the first mark was not awarded. Work on finding volumes of cylinders and prisms would be helpful for future exam practice. Summary Based on their performance on this paper, learners should: ensure they are well-practised in efficient use of their calculator, especially unfamiliar functions such as standard form read questions carefully, including checking whether the magnitude of an answer is sensible, units are appropriate, and the level of accuracy required is shown show every stage of working, particularly if the question specifically requests this practice questions involving percentage change practice questions requiring learners to form algebraic expressions use formal methods when working with percentages remember to cross out incorrect work or positively identify their chosen method when using a range of approaches to avoid losing marks due to choice.
A steel cylinder has radius 3 cm, height 10 cm and mass 2250 g.
Edexcel 1MA1 · November 2024 · Paper 2H · Q7 · 5 marks
5 marks
Examiner report
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.
Isosceles triangle has sides 2x+11,4x-4,2x+5; equal first two; perimeter 72.
Edexcel 1MA1 · November 2024 · Paper 2H · Q9 · 3 marks
3 marks
Examiner report
This question required learners to show that a shape is not a square using geometrical reasoning. The first mark was for finding the interior or exterior angle of the regular nonagon. It was pleasing to see so many learners being able to do this. The second mark was then for a method to find the angle at ABC, or to find the angle sum around the point B assuming the shape was a square. The final mark required a correct method and conclusion. This was typically of the form of showing angle ABC = 80 and that it is not 90. Those who were less successful often assumed there was another nonagon above the diagram and adjoining the side AD. This was incorrect and normally resulted in 0 marks.
Two congruent regular 9-sided polygons meet only at vertex B. AB is a side of the left polygon and BC is a side of the right polygon. The interior region of each nonagon lies outside quadrilateral ABCD at B. D lies outside both polygons and is joined to A and C, forming quadrilateral A–B–C–D.
Edexcel 1MA1 · November 2024 · Paper 2H · Q11 · 2 marks
2 marks
Examiner report
This enlargement question was a challenging one, and as a result many did not gain full marks. Those who achieved 1 mark typically did so for stating ‘enlargement’ along with a correct centre, using ‘ray lines’. The scale factor proved a real challenge to many, and if stated, was often given as 1.5. It is important that learners read the question carefully and realise which shape is the ‘object’ and which is the ‘image’. They must not assume that an enlargement is always increasing in size. The other error seen with the scale factor was to give an answer of −1.5. We did see some cases of learners using ‘ray lines’ inaccurately and giving a centre that had decimal coordinates. This is an error that has not been evident much in the past.
On a coordinate grid, triangle A has vertices (2, 7), (5, 7) and (5, 1). Triangle B has vertices (0, 4), (2, 4) and (2, 0).
Edexcel 1MA1 · November 2024 · Paper 2H · Q12 · 5 marks
5 marks
Examiner report
This shape problem had 4 stages, and each one was available to candidates, even with prior errors. The first step was to find the volume of the cylinder, which many struggled to do. There were various errors from finding the surface area to multiplying the radius by the height. The next mark was for using the volume and given density to find the mass. This value was then to be used to find the volume of the cube using a given density. The final step was to cube root the volume to find the length of the cube. Although many struggled with the initial volume, their working was clear enough to establish that the value they used for the next step was what they believed to be the volume. Given this was the case the second mark could be gained, and often was. As was the case with the next mark for the second use of the density formula. The final process mark for cube rooting, a bit like the first, was less often awarded. It was quite common to see candidates dividing by 3 or even 6 as a last step or square rooting.
Steel cylinder radius 3 cm, height 8 cm, density 7.86 g/cm³. Brass cube side h, density 8.5 g/cm³. Masses equal.
Edexcel 1MA1 · November 2024 · Paper 2H · Q18 · 3 marks
3 marks
Examiner report
There were two main methods to approach this proof, and we saw a few fully successful attempts. The first method was to prove congruence of triangles AOB and COB, using SSS, and therefore prove that angles CBO and ABO are equal. Or they could extend BO to a point on the circumference, D, forming a diameter. This allowed proof on congruence of triangles ABD and CBD using RHS. In either case we rarely saw more than one mark being awarded for stating a single pair of equal sides. The second approach was longer, and even less successful. This method used a series of isosceles triangles and matching base angles. First recognising ABC is an isosceles triangle and so BAC and ACB are equal. The next step was to recognise AOC as isosceles due to the two radii and therefore OCA and OAC are equal. Next, learners had to use these two found angles to show that OCB and OAB are equal. The final step being to recognise that AOB and COB are isosceles and as OCA and OAC are equal then so are OBC and OBA. Much like the first method, it was rare to see more than a single mark awarded.
A,B,C lie on circle centre O and BA=BC.
Edexcel 1MA1 · November 2024 · Paper 2H · Q20 · 3 marks
3 marks
Examiner report
Although there were not as many steps as we often see on a 3D trigonometry problem here, many candidates struggled to make a start to the question. The first step to find FC proved conceptually too difficult and many either didn’t attempt, or used a method that would lead to VC not FC. This could then lead to FC with a second step using Pythagoras, but this was normally missing. If learners did find FC, they then had to use the ratio to find MC along with Pythagoras to form an equation for the side length. Alternatively, they could use MC along with either the 60 or 30 angle and trigonometry. Neither were seen often, but those who got this far generally gained the accuracy mark.
Pyramid VABC has equilateral base ABC. M midpoint AB; F on MC with MF:FC=1:2; V vertically above F; VA=VB=VC; VF=8 cm; angle VCM=52°.
Edexcel 1MA1 · November 2024 · Paper 2H · Q21 · 5 marks
5 marks
Examiner report
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.
P=(-4,5), Q=(6,-6), R=(k,k+3); angle PRQ=90°.
Edexcel 1MA1 · November 2024 · Paper 3F · Q7 · 2 marks
2 marks
Examiner report
Part (a) saw mixed results, a good number of this cohort were able to give a correct answer of trapezium; common incorrect answers included parallelogram, rhombus and prism. Part (b) was answered well with many able to gain one mark for a correct triangle. Those that used the given gridlines to draw their right angle generally gained B1, those that didn’t often struggled, although angles of 88 − 92° were accepted for the right angle. Common incorrect answers included drawing isosceles or scalene triangles and failing to draw the hypotenuse of the triangle.
On a 9-by-5 square grid whose lower-left corner is (0,0), an unlabelled shaded quadrilateral has vertices (2,1), (3,4), (5,4) and (7,1), joined cyclically. A separate blank 9-by-6 square grid is provided for part (b).
Edexcel 1MA1 · November 2024 · Paper 3F · Q13 · 4 marks
4 marks
Examiner report
This question saw the full range of marks awarded, with a small number of learners gaining 4 marks. Of those that didn’t, a variety of errors were seen such as, no/incorrect conversions (but could still gain 1 or 2 marks for the process), correctly converting the dimensions for the crate or the box and dividing the corresponding lengths but then adding the results rather than multiplying finding the volumes for both the crate and the box but dividing box by crate instead of vice versa. Learners who tried to convert units of volume after finding the volume of the crate or a box often used the incorrect conversion.
A crate measuring 3 m by 1 m by 1.5 m is completely filled with boxes measuring 20 cm by 10 cm by 75 cm.
Edexcel 1MA1 · November 2024 · Paper 3F · Q19 · 2 marks
2 marks
Examiner report
It was rare to see a correct set of relevant arcs drawn as most learners did not know how to begin their answer. Some learners were able to produce a bisector within the allowed tolerance without the relevant arcs, gaining 1 mark.
For this text adaptation, draw B=(0, 0), A=(-4, 3) and C=(5, 1), then draw rays BA and BC to reproduce angle ABC.
Edexcel 1MA1 · November 2024 · Paper 3F · Q24 · 2 marks
2 marks
Examiner report
Some were able to work with trigonometry correctly, those that produced a correct trigonometric statement or equation generally went on to gain 2 marks; a small number had the correct method but prematurely rounded their answer to 14, without first seeing an answer in range this was awarded M1A0. Many attempts to use Pythagoras were seen.
ABC is right-angled at B. BC = 7.6 cm and angle ACB = 62°.
Edexcel 1MA1 · November 2024 · Paper 3H · Q3 · 2 marks
2 marks
Examiner report
A significant number of learners were not familiar with the method required to accurately construct an angle bisector using compasses. Those who knew how to draw the relevant arcs usually gained full marks for completing the bisector accurately. There were many responses with arcs in places that were not required. In some cases, constructions such as a perpendicular bisector of one of the lines was seen. There were responses where one mark was awarded for an angle bisector within the guidelines, presumably from measuring the angle and halving it. To gain full marks on a construction’s question all relevant construction lines must be seen.
For this text adaptation, draw B=(0, 0), A=(-4, 3) and C=(5, 1), then draw rays BA and BC to reproduce angle ABC.
Edexcel 1MA1 · November 2024 · Paper 3H · Q8 · 2 marks
2 marks
Examiner report
Of those who scored 0, they usually chose the cosine ratio instead of the tan ratio or used the tan ratio incorrectly, writing tan62 = . . A number of learners opted for the sine rule to successfully get their answer.
Right triangle: right angle B, BC=7.6 cm, angle C=62°.
Edexcel 1MA1 · November 2024 · Paper 3H · Q11 · 3 marks
3 marks
Examiner report
Responses on this question showed quite a lot of confusion between area and perimeter as it was not uncommon for learners to equate the area of the sector to 34.3 or even 24.9. Of the learners that were most successful in answering this question, most started with a complete equation.
Sector radius 4.7 m, perimeter 34.3 m.
Edexcel 1MA1 · November 2024 · Paper 3H · Q13 · 3 marks
3 marks
Examiner report
The majority of those who identified that the triangles were similar proceeded efficiently with answering this question and gained full marks. However, many incorrectly assumed that the lengths of AB and BC or the lengths of BC and BE were equal. Of those who didn’t initially recognise that the triangles were similar, some did successfully navigate to a correct answer using trigonometry. The majority of learners appeared confident in using the formula for area of a trapezium, with only a small number choosing to split it into a rectangle and triangle.
A, B and C are collinear in that order. A, E and D are collinear in that order. BE is parallel to CD, angle ABE = angle ACD = 90°, BE = 4.2 cm, CD = 6.3 cm and AC = 10.8 cm.
Edexcel 1MA1 · November 2024 · Paper 3H · Q16 · 5 marks
5 marks
Examiner report
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.
OBCD rectangle; D,C,E collinear; B=(2,-4), E=(12,-6.5).
Edexcel 1MA1 · November 2024 · Paper 3H · Q17 · 4 marks
4 marks
Examiner report
Of those who correctly recognised the need to use 0.5 sin , ab C some were unsure on how to calculate an angle. In some responses the learners simply substituted in all 3 given sides into the formula gaining 0 marks. Those who realised that they needed to use the cosine rule substituted correctly into the formula but often made errors in the rearrangement to find the angle. Those who were more successful at finding an angle had more often substituted straight into cos b c a A bc rather than substituting into cos a b c bc A and having to rearrange. Some benefitted from the use of square brackets being used in this question, gaining a mark for correctly substituting into 0.5 sin , ab C even if the angle had not been found correctly, provided it was clearly identified on the diagram or stated in the working.
Triangle sides 14.6,18.2,7.9 cm.
Edexcel 1MA1 · November 2024 · Paper 3H · Q19 · 4 marks
4 marks
Examiner report
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.
Cone radius r=3/4 height h; total surface area 54π cm²; slant l.
Edexcel 1MA1 · June 2024 · Paper 1F · Q3 · 1 mark
1 mark
Examiner report
Candidates found this question more challenging than many of the other early ones on the paper. Common incorrect answers included the other classifications of angle type – acute and obtuse.
Angle y is greater than 180° but less than 360°. Write down its mathematical name.
Edexcel 1MA1 · June 2024 · Paper 1F · Q8 · 3 marks
3 marks
Examiner report
In part (i), the majority of candidates were able to work accurately with the sum of angles around a point to find the missing angle of 50°. Where incorrect answers were seen this was only rarely due to errors in arithmetic. A minority of candidates had an incorrect method, either working with an incorrect angle fact, for example working with 380 or 180 rather than 360, or combining the angles in an incorrect calculation, for example 220 −90. A minority of candidates measured the angle as printed on the question paper rather than attempting a calculation. In part (ii) candidates were asked to give a reason for their answer. This was less well answered than part (i). Although there were a significant number of candidates who were able to indicate that the angles at a point add to 360° or that a full turn adds to 360°, there were many candidates who referenced angles in a circle or omitted the word “point” in their description, which was not an acceptable reason. Other candidates gave an angle fact not related to their working, for example angles on a straight line add to 180° or restated their calculation steps from (i). Candidates need to be aware of the difference between a mathematical reason and an explanation of their working.
OA, OB and OC are three straight lines.
Edexcel 1MA1 · June 2024 · Paper 1F · Q12 · 4 marks
4 marks
Examiner report
A minority of candidates were able to correctly work with the perimeter of the triangle, find a quarter of this and complete the process to find the missing side length of the rectangle. Some candidates had a correct complete process, but mistakenly gave the perimeter or the rectangle (20cm) as their answer. There were also a number of candidates who correctly reached the perimeter of the rectangle but did not know how to find the missing length in the rectangle often dividing by 4 (working with area) rather than subtracting the two 4cm sides and dividing by 2. Some found the perimeter of the triangle but could not progress further. Others subtracted 4cm and did not consider the other 4cm on the rectangle or divided 20cm by 4cm and gave an answer of 5cm. They then divided 16cm by 2 to get an answer of 8cm. Common incorrect approaches worked with area or tried to use the quarter as a scale factor dividing a single side length of the triangle by 4.
A triangle has side lengths 36 cm, 14 cm and 30 cm. A rectangle has one side 4 cm and unknown length L.
Edexcel 1MA1 · June 2024 · Paper 1F · Q19 · 2 marks
2 marks
Examiner report
This question on describing a transformation was not well answered. Only a minority of candidates were able to give either a fully correct or partially correct answer. Where candidates correctly identified that the transformation was a translation they were often not able to give the correct translation vector. Where a vector was attempted common errors were incorrect signs on one or both of the values, reversing the values, giving a fraction rather than a vector or giving a coordinate. Some candidates seemed to recognised that the shape had been translated but were not able to use the correct mathematical term to describe the type of transformation. Common incorrect examples included ‘translocation’, ‘transition’, ‘shift’, ‘moved’ or description of a movement in words. The most common incorrect transformation given was a reflection. It was uncommon to see more than one transformation given.
Triangle A has vertices (-3, 1), (-3, 3), (-2, 3). Triangle B has vertices (2, -3), (2, -1), (3, -1). Describe fully the single transformation that maps triangle A onto triangle B.
Edexcel 1MA1 · June 2024 · Paper 1F · Q22 · 5 marks
5 marks
Examiner report
This two-part question was attempted by the vast majority of candidates. It was well answered by slightly less than half of the candidates with many of these being able to give a fully correct answer to both parts. In part (a) candidates who were successful generally calculated the area of the floor by splitting it into two rectangles (5m by 10m and 3m by 6m or 8m by 6m and 4m and 5m). Some were able to go on to find the area that the paint would cover (75m2) or find the volume of paint required to cover the floor together with the volume of paint available and give a correct conclusion for full marks. Some candidates did not show full working to support their conclusion, for example, finding the area as 68 and the amount of paint as 7.5 litres but then not finding the comparable figures (6.8 or 75). A few candidates did not answer the question as yes/no, instead saying that Petra was correct. Where only partially correct answers were seen these often involved finding the total area of the floor without making further meaningful progress, finding the area of one shape, finding the area covered by the 3 tins available or finding the number of litres required for their area. Common incorrect approaches included working with the perimeter or multiplying together three or more of the side lengths. When considering the floor area that could be covered it was common to see candidates incorrectly working with 100 after misinterpreting the 10m2 coverage. Some candidates did not show their calculations and made arithmetic errors and therefore could not score the corresponding process marks: obtaining 70.5 instead of 75 was a common example of this. In part (b) candidates were told that the paint actually covered a greater area per litre than had been suggested and were asked if this affected their answer to (a). Where candidates had concluded that there was sufficient paint in (a) they were often able to correctly identify that there would be no effect as less paint would be required. Some candidates gave a correct reason – “she would still have enough” but paired incorrectly with the answer yes. Where candidates had incorrect answers to (a) they were often still able to identify the effect that the change would have made for their figures. Common errors included giving the incorrect decision on the effect for the figures that they had found in (a) and including incorrect figures in the statement which then prevented the award of the C mark.
A floor is an L-shape formed from a 10 m by 8 m rectangle with a 4 m by 3 m rectangular corner removed. Petra has 3 tins containing 2.5 litres each, and she thinks each litre covers 10 m^2.
Edexcel 1MA1 · June 2024 · Paper 1H · Q3 · 5 marks
5 marks
Examiner report
It was pleasing to see many fully correct solutions to the problem in part (a) and the majority of students gained at least one mark. Most students divided the floor plan into two rectangles and were usually able to work out that the area is 68 m2. A few students made simple arithmetic errors but still gained the mark for a complete process to find the total area. After finding a floor area, students needed to find out whether Petra has enough paint to cover the floor. The most common approach was to work out that the 3 tins will cover an area of 75 m2. Alternatively, some students divided their area by 10 to find the number of litres needed and multiplied 3 by 2.5 to find that Petra has 7.5 litres of paint. Some students used approximation successfully, stating the area and stating that 7 litres would be sufficient and this was less than the 7.5 litres Petra had available. Whichever approach they chose, students usually went on to give the correct decision that Petra has enough paint. The majority of students did specify a “yes/no” answer and very few lost the final mark if the rest of the work was correct. Incorrect approaches included working with the perimeter or finding the product of all the given lengths. Part (b) was also well answered with many students stating that it would not affect the answer to part (a) and giving a correct reason such as she will need less paint or she will have more paint left over. Some students made a comment about Petra needing less paint but could not be awarded the mark because they did not make a decision. If figures were included in the answer to part (b) they had to be correct for the area that had been calculated in part (a). Many students correctly found that Petra would now be able to cover 82.5 m2.
A floor is an L-shape formed from a 10 m by 8 m rectangle with a 4 m by 3 m rectangular corner removed. Petra has 3 tins containing 2.5 litres each, and she thinks each litre covers 10 m^2.
Edexcel 1MA1 · June 2024 · Paper 1H · Q11 · 3 marks
3 marks
Examiner report
Most students attempted to draw triangle B and triangle C on the grid and many gained the first mark for showing at least one of the triangles in the correct position or for rotating their incorrectly positioned triangle B by 90o clockwise about the point (1, 2). Often both triangles were correctly positioned. Students who drew triangle B and triangle C in the correct positions were often able to recognise that the single transformation that will map shape A to shape C is a rotation. Descriptions of the transformation were generally good although sometimes when the angle of rotation was given as 90o the direction of ‘clockwise’ was missing or given as ‘anticlockwise’. The centre of rotation was sometimes missing or incorrect. It was apparent that some students had used tracing paper when answering this question and this was generally a successful strategy for determining the centre of rotation. Many students gave more than one transformation, often using a rotation and a translation, and therefore got neither of the two marks available for the description as the question asked for a single transformation. Quite a few students simply restated the transformations given in the question.
Triangle A has vertices (-3, 2), (-2, 2), (-2, 4). It is translated by vector (6, -4) to give triangle B. Triangle B is then rotated 90° clockwise about (1, 2) to give triangle C. Describe fully the single transformation that maps triangle A onto triangle C.
Edexcel 1MA1 · June 2024 · Paper 1H · Q15 · 4 marks
4 marks
Examiner report
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. Many of those with some knowledge of sectors of circles went wrong at the first step, using πr2 or πr instead of 2πr and gained no marks. A good proportion of those who did make a correct start went on to get full marks. Marks were often lost because of arithmetic errors or poor algebraic manipulation. Students who cancelled down fractions before multiplying worked more accurately than those who left cancelling until the end. Those who started by forming the equation x = gained the first mark but rearranging the equation correctly to find the value of x was a problem for some. Students who worked with proportion and found that the arc length is 5 of the circumference of the circle gained the first two marks. They could then complete the process to find the area of the sector by working out 12 although many chose to first work out the size of angle AOB. After gaining the first three marks for a complete process to find the area of the sector a significant number of students failed to get an answer of 15 because they could not evaluate correctly. Those who used 12 made fewer arithmetic errors. Some students assumed that the sector was one third of a circle and worked with an angle of 120o and some measured the angle and used 130o in their calculations. These students gained no marks.
The diagram shows a sector AOB of a circle with centre O, radius 6 cm and angle AOB = 150°. Find the area of the sector, giving your answer in terms of pi.
Edexcel 1MA1 · June 2024 · Paper 1H · Q19 · 3 marks
3 marks
Examiner report
Overall, this question was answered poorly with the majority of students unable to find an appropriate strategy to work out the ratio AB : AD. Some students made a good start by using similar triangles to form an equation involving AB, e.g. AB AB = , or an equation involving the scale factor, e.g. 4 × scale factor = 25 ÷ scale factor. Many of those who formed a correct equation were able to complete the process to find the required ratio. Most students, however, were unable to identify the need to compare the sides of the two similar triangles, ABC and DAB. Many used the ratio 4 : 21 incorrectly and answers of 4 : 25 were common. Students are advised that drawing two separate triangles and marking in lengths on the sides may help them to use the similarity of the triangles. A small number of students used Pythagoras to find the perpendicular height of the triangles and then AB. Some students assumed that Pythagoras could be applied to triangle ACD to find AC but it is not a right-angled triangle.
ABC and DAB are similar isosceles triangles. AB = AC, AD = BD, and C lies on BD. Given BC : CD = 4 : 21, find AB : AD.
Edexcel 1MA1 · June 2024 · Paper 1H · Q21 · 5 marks
5 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2024 · Paper 1H · Q22 · 4 marks
4 marks
![Edexcel 1MA1 · June 2024 · Paper 1H · Q22 · 4 marks — (a) Sketch y = sin(x°) for 0 ≤ x ≤ 360. [2 marks]](/api/paper-crops/1ma1/2024-06/higher-paper-1/q22/question-r1-5ce1d8d4.webp)
Examiner report
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.
(a) Sketch y = sin(x°) for 0 ≤ x ≤ 360. [2 marks]
Edexcel 1MA1 · June 2024 · Paper 1H · Q23 · 5 marks
5 marks
Examiner report
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
C is a circle with centre (0, 0)
Edexcel 1MA1 · June 2024 · Paper 2F · Q7 · 3 marks
3 marks
Examiner report
Part (a) was well answered with many students being able to accurately measure the length of the line. An answer outside of the acceptable range suggests students place their rulers incorrectly either at the start of the ruler or at 1 cm rather than at 0. In part (b) only a small proportion of the cohort being was able to accurately measure the angle. Commonly seen incorrect answers were between 110° and 115° suggest that students were reading backwards from 180° and so should be encouraged to check they are using the correct scale. A number of blank responses were seen which suggests not having a protractor was an issue. In part (c) most students knew that a hexagon was a 6-sided shape and were able to draw it confidently, with the majority of attempts resembling regular hexagons. In some instances, students were very helpful by numbering the sides or vertices. Where a ruler was used there were clearer shapes drawn. Freehand drawings often resulted in the mark being lost because there was doubt as to whether there was an extra vertex along their lines. The most commonly drawn incorrect shapes were octagons and pentagons. It was however pleasing to note that where multiple attempts had been made, an indication of the final answer was given and this reduced the mark being lost due to a choice of answer.
Use the textualised ruler/protractor readings from the official measurement diagram.
Edexcel 1MA1 · June 2024 · Paper 2F · Q17 · 3 marks
3 marks
Examiner report
In line with similar questions requiring constructions to be performed, full marks were rarely awarded for this loci question. Many students were seemingly unsure how to start and many simply shaded an unbounded area that they thought made sense. When a mark was awarded, it was more often for being able to draw the 4cm arc than the perpendicular bisector. Arcs were often drawn freehand (which was condoned if within guidelines) and suggests many candidates lacked a pair of compasses despite these being listed in the required equipment. Bisectors were mainly drawn without construction lines (acceptable in this question) with students often citing "9.4cm" or "4.7cm", showing that they had just measured the line and attempted to draw a perpendicular line from the midpoint. When the arc and bisector were drawn within guidelines, the region was almost always identified correctly.
In triangle ABC, region R consists of points less than 4 cm from A and closer to C than to B.
Edexcel 1MA1 · June 2024 · Paper 2F · Q20 · 2 marks
2 marks
Examiner report
A great number of students were seemingly unable to successfully engage with this question despite Pythagoras’ theorem being on the formulae sheet provided, with many trying to find a solution using angles in a triangle or area formulae. Of those attempting to work with Pythagoras’ theorem, many incorrectly thought the unknown side was the hypotenuse, finding 192 + 102 instead of 192 – 102 or not squaring the lengths before subtracting. Some attempted to use trigonometry rather than Pythagoras and this was often unsuccessful.
ABC is right-angled at C, with hypotenuse AB = 19 cm and AC = 10 cm.
Edexcel 1MA1 · June 2024 · Paper 2F · Q26 · 4 marks
4 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2024 · Paper 2F · Q27 · 2 marks
2 marks
Examiner report
This penultimate question required students to use knowledge of similarity and, of the students who knew how to calculate the scale factor, most then went on to calculate the correct length and got the answer correct so it was pleasing to see that many of the cohort were successful. The method mark was often awarded for making an attempt to find the relationship between the corresponding sides but often showed the scale factor as the final answer. A few students demonstrated the common misconception of adding or subtracting to find the missing sides, rather than realising a multiplicative relationship was needed, indicating they didn’t understand the relationship between the triangles. Another common error was to compare 6 to 8 as adding 2, so adding 2 to the 1.5 to get 3.5, or subtracting 1.5 from 6 to get a scale factor of 4.5 instead of 6/1.5 to get 4. A surprisingly larger than expected number of students seemed to measure the side of the smaller triangle with a ruler, despite the question saying similar shapes, leading to answers around 2.5.
ABC and DEF are similar isosceles triangles: AB = AC = 8 cm, BC = 6 cm, DE = DF and EF = 1.5 cm.
Edexcel 1MA1 · June 2024 · Paper 2H · Q1 · 2 marks
2 marks
Examiner report
This question was a straightforward Pythagoras theorem application and was a positive start for most students, with them gaining the two marks available. A small proportion of students, as is expected, mixed up the formula and worked as if they were finding the hypotenuse rather than one of the shorter sides. This typically resulted in no marks. Some knew to square and subtract but forgot to take the square root. These students were able to gain the first mark. Some attempted to use trigonometry but tended to get no further than trying to find a missing angle.
ABC is right-angled at C, with hypotenuse AB = 19 cm and AC = 10 cm.
Edexcel 1MA1 · June 2024 · Paper 2H · Q7 · 4 marks
4 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2024 · Paper 2H · Q8 · 3 marks
3 marks
Examiner report
There were two processes in this question, a use of scale and a use of the formula for the area of a triangle. The more successful students used the scale first on RQ to get a drawing length of 8 cm. This was then used correctly with the area of the scale drawing to find PQ, which was then scaled back up to get the real length. The other method was to scale up the area first, and then use the real length of RQ to find the real length of PQ. This method was the better of the two as it seems to have less steps. However, most who attempted this method only multiplied the area by the length scale factor of 5, not the area scale factor of 52, and as a result could only gain the one mark. It was common to see a correct scaling of 8 for the length together with an incorrect scaling of 140 for the area, and this was considered to be a choice of methods.
A playground is a right-angled triangle PQR. Dan makes a scale drawing using 1 cm to represent 5 m. The area of the triangle on the drawing is 28 cm^2, and the real length QR is 40 m. Work out the real length PQ.
Edexcel 1MA1 · June 2024 · Paper 2H · Q10 · 3 marks
3 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2024 · Paper 2H · Q17 · 4 marks
4 marks
Examiner report
This question was attempted by most, but only the stronger students were able to score. The first mark was for working with the volume scale factors of L and M to find the length scale factors, and the second similarly to work with the area scale factors of M and P to find the length scale factors. Both of these processes required students to take some form of root, and it is perhaps not surprising that students were better when the square root was involved. Many of the students who were able to find the 2 separate length scale factors or ratios, were then unsure as to how to combine them and scored no further credit. Those who did know, typically scored both the 3rd and 4th marks. The A mark was ‘or equivalent’ and we saw numerous acceptable possibilities, including 9.6 : 12 : 4 which had non-integer values.
L, M and P are three similar solid cylinders made from the same material. L has mass 64 g. M has mass 125 g and total surface area 144 cm^2. P has total surface area 16 cm^2. Work out height of L : height of M : height of P.
Edexcel 1MA1 · June 2024 · Paper 2H · Q19 · 5 marks
5 marks
Examiner report
The first mark was for finding 𝐶𝐵⃗⃗⃗⃗⃗ or 𝐵𝐶⃗⃗⃗⃗⃗ in terms of a and b. the second then required them to apply the ratio of 4 : 5 correctly to this. These marks were awarded relatively often, although some did apply the ratio incorrectly, normally when working from N to either C or B. The last two marks were for an un-simplified and then a simplified expression for 𝑀𝑁⃗⃗⃗⃗⃗⃗⃗ . Again, a good number gained the third mark, and of these a significant number were able to simplify correctly. This is a place where using a calculator to simplify coefficients might have been a good idea. This step normally involves working with fractions and negatives, and most students attempted it manually. In part (b) students had to recognise that 𝑀𝑁⃗⃗⃗⃗⃗⃗⃗ was not a multiple of 𝑂𝐵⃗⃗⃗⃗⃗ . It was clear that many did actually understand that point, but lots struggled to articulate themselves accurately and so lost the communication mark.
The diagram shows quadrilateral OACB.
Edexcel 1MA1 · June 2024 · Paper 2H · Q22 · 4 marks
4 marks
Examiner report
The final question required students to prove that two tangents from a fixed point are equal in length. Most attempted this using congruency, with varying degrees of success. As it was a proof, it required reasons to be given at each stage, and although a good number knew the facts, they did not give correct, or complete reasons and struggled to gain any credit. It is important again that students practise stating the geometric properties and reasons. There was an alternative method seen using isosceles triangles, and this too was able to gain credit when complete reasons were given. When this method was used it was common for the marks to be lost for writing only isosceles rather than isosceles triangle. Alternate segment theorem was also a possible approach and was an efficient method. Other errors include claiming that OP bisected the angle with no justification for this, describing the relationship without actually stating that, e.g. OA = OB. Summary Based on the performance on this paper, students should: • Ensure they know how and when to use a calculator efficiently, but not to forget manual methods. Use of brackets and edit function will help with substitution on a calculator. • Practise articulating their mathematics in verbal discussions to aid them in written responses. • Use tables and diagrams to extract information from a question so they can better navigate through the problem. • Know the relationships between volume, area and length scale factors. • Provide unambiguous explanation whenever required by the question. • Cross out any work they do not wish the examiners to mark, leaving one clear solution. • Whenever possible use accurate figures without prematurely rounding or truncating in presenting work.
A and B are points on a circle, centre O.
Edexcel 1MA1 · June 2024 · Paper 3F · Q10 · 2 marks
2 marks
Examiner report
For part (a), most of the cohort provided a correct drawing of the radius. Of those that didn’t, a whole variety of incorrect answers were seen, including sector, chord, diameter and tangent. It should be noted that if a student drew two radii, this was awarded 0 marks as it was unclear whether the student was drawing two radii or a sector, unless the radius had been clearly labelled as such. Some students had not properly crossed or rubbed out an incorrect response; students should make it clear to the examiner which line they wish to be considered as their answer. Part (b) was poorly answered with few students giving the correct answer. A whole variety of incorrect answers were seen such as diameter, segment, tangent, area, arc, circumference, section, showing that this cohort need to learn the names of the parts of the circle.
Two circle questions.
Edexcel 1MA1 · June 2024 · Paper 3F · Q18 · 3 marks
3 marks
Examiner report
For part (a), a minority of the cohort were able to produce a correct rotation. It was common to see the shape rotated by 180° with an incorrect centre; this was awarded 1 mark. A similar number of students were unable to gain any marks, common incorrect answers included reflections and translations. Part (b) was answered less well with a small number of students gaining 1 mark for a correct explanation. For those who did gain the mark, explaining that he has reflected in y = 3 was the most common correct answer seen. A whole variety of incorrect answers were seen, the most common of which are highlighted in the mark scheme. Some students drew the correct reflection, but this was only awarded the mark if accompanied by an explanation such as ‘it should be here’. It was clear that many students were unable to describe lines with the correct equation.
A quadrilateral has vertices (4,2), (2,2), (1,4) and (5,4).
Edexcel 1MA1 · June 2024 · Paper 3F · Q20 · 5 marks
5 marks
Examiner report
This question was not well attempted by students. Many homed in on the statement ‘Angle BCD = 4 × angle ABC’ and multiplied 81 by 4 and went from there, gaining 0 marks. It is advisable for future cohorts to approach angles questions by taking in all the information given, including the diagram, and then working through systematically using the geometrical rules they know. For the other students, 1, 2, 3, 4 and 5 marks were all awarded, with 1 and 5 being most common; those that scored 1 mark achieved ABC = 18 but were then unable to interpret the other information correctly and made no further progress. A few students thought that angle ABD was a right angle and hence labelled angle CBD as 72 from 90 – 18. Correct 3-letter notation was rarely seen and a significant number of students lost marks as a result. Some students misinterpreted the isosceles triangle and thought that angle BAC = angle ABC. Marks were lost by students writing the angles in the incorrect positions on the diagram thus creating a contradiction where no 3-letter notation were used for the angles in the working, with the most common being 72 being placed as angle CBD on the diagram.
ABC and BCD are isosceles triangles.
Edexcel 1MA1 · June 2024 · Paper 3F · Q24 · 3 marks
3 marks
Examiner report
Part (a) was answered poorly with most students gaining 0 marks. Many were not able to get across that the height of the rectangle was wrong, nor could they understand that it is the perpendicular height of the prism that Rana should be using as the height of the rectangle. Many did not understand that a side elevation will make a 2D shape and a sizable number of students mistook the side elevation for the plan view, e.g. saying the dimensions of the rectangle should be 7 cm by 6 cm. Many incorrect answers were focusing on the shape of the side elevation, saying it should be a parallelogram rather than a rectangle. Part (b) also saw very few students gain 2 marks. Some were able to gain 1 mark, usually for a 7 cm by 6 cm rectangle without a dividing line or drawing an unacceptable ‘dashed’ line. Most of the students scored 0 marks and a whole variety of incorrect answers were seen. Many drew a 3D shape or an incorrect rectangle or drew a rectangle which was part of a net.
A triangular prism is 7 cm long. Its isosceles triangular cross-section has equal sides 4 cm and base 6 cm.
Edexcel 1MA1 · June 2024 · Paper 3F · Q28 · 4 marks
4 marks
Examiner report
A significant number of this cohort gained 0 marks on this question. It is clear that many did not know the formula for the volume of a cylinder. Of those that did not gain 0 marks, gaining 1 mark for [volume] ÷ 250 was often seen, with the most common incorrect volume used being 3000. Success beyond 1 mark was rarely seen and a minority of students were able to go on and gain 4 marks for a correct answer in the range 62 to 63.
A cylindrical paddling pool has radius 100 cm and depth 30 cm. It is filled from empty at 250 cm^3 per second. Work out the number of minutes needed to fill it, correct to the nearest minute.
Edexcel 1MA1 · June 2024 · Paper 3H · Q3 · 3 marks
3 marks
Examiner report
Students gained the mark in part (a) for recognising that the height was incorrect, with some correctly stating that the height should be sqrt7 or 2.6. However, some seemed to think that the front elevation had been drawn as they were under the wrong impression that the arrow always points to the front and so the side elevation would be a triangle, whereas others believed that the elevation should be a trapezium. Some students’ explanations were too ambiguous, for example stating ‘it should be smaller’ with no clear indication as to what ‘it’ referred to, did not gain the mark. Similar statements mentioning the slant were only given credit if it was clear that the slant was associated with height as otherwise it could have meant that the student thought that the rectangle should have been a parallelogram. Most students drew a rectangle in part (b) although sometimes this was drawn as part of a 3-D shape or as part of a net and in these cases no marks could be awarded. A frequent mistake was to draw one of the dimensions incorrectly, usually the 6 cm, which meant that a maximum of only one mark was awarded. Those students who drew a correct 7 cm by 6 cm rectangle for the plan often forgot to draw in a solid line in the middle and so only gained the first mark. There were a significant number of students who appeared to be unsure of what was meant by a plan and drew a mixture of elevations labelled as front, top or birds eye view. Without the correct label a mixture of drawings was treated as choice and gained no marks. Students should be advised to draw just the one diagram if they are unsure as a mixture almost never scored a mark even if the correct diagram was seen (as it often was).
A triangular prism is 7 cm long. Its isosceles triangular cross-section has equal sides 4 cm and base 6 cm.
Edexcel 1MA1 · June 2024 · Paper 3H · Q7 · 4 marks
4 marks
Examiner report
This multi-step question required the use of a variety of skills being combined to solve a problem and appeared to be challenging to many students. Many students multiplied the radius by the height or finding the surface area or squaring pi instead of the radius or dividing the volume by 3 and, hence spoiling this process. The second process mark was often the only mark being awarded, usually for [volume] ÷ 250. On this occasion [volume] could be anything the student believed to be the volume provided it was not a value that was not allowed as stipulated in the mark scheme. It was unfortunate and relatively rare when students divided something labelled other than volume by 250 and hence not achieving this mark. Again, it is important that students show their method and process so that they have greater access to these marks. A number of students who had a fully correct process to find the time in seconds often stopped short of finding the time in minutes and hence only gained 2 of the 4 marks. With this question, full marks were given for an answer within the range from 62 to 63 inclusive but as the questions stipulated that ‘you must show all your working’ a correct answer with no supportive working scored no marks. It is important that students heed this instruction given on specific questions to ensure that they do not fall foul of this.
A cylindrical paddling pool has radius 100 cm and depth 30 cm. It is filled from empty at 250 cm^3 per second. Work out the number of minutes needed to fill it, correct to the nearest minute.
Edexcel 1MA1 · June 2024 · Paper 3H · Q8 · 3 marks
3 marks
Examiner report
Many students were able to score at least one mark, and often two, in this question showing some understanding of vector arithmetic. Some students wrote out 2 2 − + but were unable to proceed any further and pick up any marks. Some students tried to square the vector a rather than multiply it by 2 giving 9 − + . Students that gained 2 marks often did so for writing the vector 5 although there were differing levels of success at drawing it. Unconventional notation was condoned for the method marks with many students missing out the brackets or even writing their answer as a fraction. The coordinate (5, 8) achieved M1 whereas just plotting the point (5, 8) without identification was M0. Most often the last mark was deducted for missing the arrow on the vector. When drawing vectors it is essential to indicate direction. Some students attempted to find the vector 2a + b by drawing but this method was often less successful. A number of these students had marks deducted as they only drew the vectors 2a and b without drawing the diagonal line of 2a + b, ie from start to finish, unfortunately without this diagonal line only one mark was awarded.
a = (3, 2) and b = (-1, 4). Draw and label the vector 2a + b.
Edexcel 1MA1 · June 2024 · Paper 3H · Q9 · 3 marks
3 marks
Examiner report
The vast majority of students gained at least one mark for this question usually for the 6 cubed. The majority of students also gained the second process mark for the process to form an equation or for a process to find h with 0< k ≤ 1, it was only the minority who were able to use the correct value of 1/3 for k. Students were, on the whole, very successful at rearranging their formula to find an answer for h. It is important to note that in future series students will be required to have learnt and to be able to apply the formula for the volume of a pyramid in questions such as these. For the accuracy mark students had to have an answer of 10.125 or an equivalent mixed number eg 10 8 . A final answer of 81 8 did not gain A1 if the 10.125 or equivalent mixed number were not seen. In questions such as these it is important that students give a suitable answer, ie one which allows an easy appreciation of its magnitude without carrying out a calculation. Similarly, an answer of 10.1 without the 10.125 or equivalent mixed number seen did not gain A1 – students are expected to give the exact answer to a question if it is a terminating decimal.
A cube has side length 6 cm. A square-based pyramid has base 8 cm by 8 cm and perpendicular height h cm.
Edexcel 1MA1 · June 2024 · Paper 3H · Q12 · 2 marks
2 marks
Examiner report
Some students were able to pick up one of the two marks for having a triangle of the correct size and orientation but in the wrong position or for 2 out of 3 vertices correctly placed. However, there were a significant number of students who either misinterpreted the negative for a fractional enlargement or ignored the negative altogether and completed an enlargement of scale factor 2 or –1 or ½ or –½. Others misinterpreted the centre of enlargement and drew their triangle so that it touched the origin and others seemed to draw a triangle that was not similar in a random position on the grid. Many students attempted drawing the projection lines first and trying to fit a triangle within these lines – they knew the orientation and quadrant the triangle should end up in but often did not place the vertices correctly or check that it was the correct size. A few students wrote calculations at the side to work out the directions or new coordinates, this was often successful. Many students used little jumps drawn on the squares to help them count, which was often successful and helped them place the triangle in the correct position.
Triangle T has vertices (1,2), (1,3) and (3,2).
Edexcel 1MA1 · June 2024 · Paper 3H · Q18 · 5 marks
5 marks
Examiner report
It was encouraging to see that so many students were able to gain the first mark for the correct application of 0.5absinC, and many could take their value for AC and gain further marks for a correct application and rearrangement of the sine rule. Errors in the first step to find the length of AC included treating the triangle ACB as right angled or as an isosceles triangle, with angle ABC assumed to be 35°. Those who were unable to make a correct start on this question could pick up marks later for correct use of the sine rule and the formula for the area of a triangle due to the use of square brackets being used throughout the scheme. Access to these later marks after an incorrect start was dependent on clear working being shown and sides being clearly identified by labelling or on the diagram with no contradiction. As with previous questions it is important that students show their full working to have an increased access to these marks. Of those who gained just four of the five marks, this was usually due to premature rounding and hence having a final answer out of range. It is important that students working at this level do not round their values throughout their processes and only round their final answer.
In triangle ABC, AB = 17 cm, angle BAC = 35° and the area of triangle ABC is 54 cm^2. Point D is such that angle ACD = 48° and angle ADC = 57°. Calculate the area of triangle ACD.
Edexcel 1MA1 · June 2024 · Paper 3H · Q21 · 4 marks
4 marks
Examiner report
A challenging question, as intended, at the end of the paper. When students were awarded one mark, it was usually for identifying that a side of the smaller hexagon was equal to the radius of the circle or for Perimeter of PQRSTU = 6r. Some students arrived at this result by using trigonometry whereas others used their knowledge of equilateral triangles. Students found the work required to gain the second mark much more challenging, with most trying to go down the route of finding the length of half of a side of the larger hexagon with various levels of success. A few students attempted to work the question backwards to appear as though they were finding the correct perimeter for the larger hexagon. This was a question where clear layout was essential. Those students that took care to present their work in a logical way were often successful but where working was muddled, students often lost their way and had marks deducted they might otherwise have gained. Many students had marked deducted because their diagrams were not properly labelled or it was unclear what length they were trying to find an expression for. There was also a tendency to calculate areas instead of perimeters. The students that were able to gain the first two marks, often went on to gain all the marks; scores of 2 and 3 were few and far between. The main reason for a student to score 3 rather than the 4 marks was because they had made a slip within their working, usually dropping an r or for giving a final answer of 3 : π : 2 3 instead of 3 < π < 2 3 . It is worth reminding students that in questions in which the answer is given to them in the form of a 'show that' then examiners are expecting to see sufficient working to prove that the result has been derived accurately (and not just stated after only a line or two of either correct or incorrect working). Summary Based on their performance on this paper, students are offered the following advice: • Write down all steps in your working, ensuring work is structured and easy to follow to have access to as many marks as possible. This can partly be done by annotating given diagrams eg labelling sides and angles. • Check all your working for any 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. Along with this check that you have answered the question fully eg giving the answer in the required form. • Practice factorising quadratics, particularly those where a ≠ 1 or b = 0 and check this by expanding the brackets. • Take care with algebraic notation and ensure that brackets are used when necessary. • Learn all the necessary formulae eg the formula for the volume of a pyramid, the Sine Rule and the formula for the area of a triangle. • Only round your answer at the final stage to avoid losing out on the final accuracy mark. • Be concise when answering questions that require an explanation and don’t add in unnecessary calculations that risk the addition of an incorrect statement and the mark(s) being deducted.
A circle of radius r is inscribed between a regular hexagon whose sides are tangent to the circle and a regular hexagon whose sides are chords. By comparing the three perimeters, show that 3 < pi < 2sqrt(3).
Edexcel 1MA1 · November 2023 · Paper 1F · Q4 · 1 mark
1 mark
Examiner report
Measuring angles is an area of the specification that needs attention. Although some students did gain a correct answer in range, with 40 most commonly seen, there were a whole range of incorrect answers given such as 35, 50 and 140.
On a square coordinate grid, plot O=(0,0), A=(10,0) and B=(7.66,6.43). Draw rays OA and OB; x is the angle AOB. These coordinates reconstruct the official accurately drawn angle without stating its size.
Edexcel 1MA1 · November 2023 · Paper 1F · Q11 · 2 marks
2 marks
Examiner report
Less than half of students in this cohort were able to gain 2 marks for a correct enlargement. Of those that didn’t, the main error was in the height of the triangle and the positioning of the top vertex; most were able to gain B1 for a base of 6 cm. Some students misread the scale factor and enlarged the shape by a different scale factor – usually 2 – if done correctly B1 could still be gained.
On a square grid, a triangle has vertices (0,0), (2,0) and (1,3).
Edexcel 1MA1 · November 2023 · Paper 1F · Q22 · 2 marks
2 marks
Examiner report
It was rare to see a fully correct construction and 90° angle at P on this question. Almost all students on this paper did not know what arcs were required. Many students did however gain 1 mark for drawing a 90° angle at P without the correct construction lines.
Point P lies on straight line AB.
Edexcel 1MA1 · November 2023 · Paper 1F · Q23 · 4 marks
4 marks
Examiner report
Very few students were able to gain full marks on this question. A small number made a positive start and gained one of the first two P marks, the equations most commonly seen were x = 2y, 2x + y = 180 and y + w = 180. Division by 3, rather than 5, of the 180° in the triangle was commonly seen. Most students understood the correct angle facts for angles in a triangle / on a straight line. Many students ignored the fact the triangle was isosceles and calculated the angles as if it was an equilateral triangle, giving 60° angles which led to 180 − 60 = 120 as the incorrect answer. Many students attempted to assign values to x and y, these values were usually incorrect but if the correct values were seen this usually led to a correct value for w as well.
ABD is an isosceles triangle with BA = BD. ABC is a straight line. The angle at A is x degrees, the interior angle at B is y degrees and x:y = 2:1. The exterior angle at B on line BC is w degrees.
Edexcel 1MA1 · November 2023 · Paper 1H · Q6 · 4 marks
4 marks
Examiner report
Many students gained one mark for using angle facts to write an equation such as y + w = 180 or x + x + y = 180. Further progress then depended on being able to use the ratio x : y = 2 : 1 correctly. Those that could deal with the ratio usually went on to give fully correct solutions. A significant number of students, though, could not use the ratio correctly and incorrect statements such as x = 120, y = 60 or x = 90, y = 45 were very common. It was disappointing that a large proportion of the students failed to make any meaningful progress with this question.
An isosceles triangle ABD is drawn on the straight line ABC, with D above the line and C beyond B. BA = BD. Angle DAB = x°, angle DBA = y° and angle DBC = w°. Also x : y = 2 : 1.
Edexcel 1MA1 · November 2023 · Paper 1H · Q10 · 3 marks
3 marks
Examiner report
This question was answered quite poorly. Given that a statement such as πd = 10 or 2πr = 10 was sufficient for the first mark to be awarded it was disappointing that more than half of the students scored no marks. Many of those who did score the first mark were unable to make any further correct progress. It was not uncommon to see 2πr = 10 followed by r = 5π. Students were awarded the second mark for substituting a correct expression for the radius in πr2 to give an expression such as for the area. Often this expression was not simplified correctly to show a single value of π and the final mark could not be awarded. Marks were also lost because necessary brackets were omitted.
The circumference of a circle is 10 m.
Edexcel 1MA1 · November 2023 · Paper 1H · Q13 · 3 marks
3 marks
Examiner report
This question was very poorly answered. Relatively few students showed that they understood the relationships between lengths, areas and volumes in similar figures. Those who did recognise that they should use the ratio of the heights, 2 : 5, to find the ratio of the volumes or the volume scale factor were usually able to give a complete method to find the volume of solid B. Many students worked out the scale factor as 2.5 but the majority then proceeded to use this as their volume scale factor, giving an answer of 30.
Solid A and solid B are similar. The ratio of the height of solid A to the height of solid B is 2 : 5. The volume of solid A is 12 cm³
Edexcel 1MA1 · November 2023 · Paper 1H · Q20 · 4 marks
4 marks
Examiner report
This proved to be a challenging question and finding the size of angle ABC was beyond most students. The first mark was awarded for using the cosine rule for any angle but many students could not even get that far despite the cosine rule formula being given on the formula sheet. Many lost marks for not using brackets when squaring 5 7 and this often resulted in 35 as only the √7 was squared. Many of those that did correctly substitute into the cosine rule for angle ABC could not then rearrange the equation to make cosB the subject. Mistakes in the order of operations and sign errors were very common. Relatively few students got as far as cosB = –0.5 and many of those that did then failed to give the answer as 120o. Many students showed a lack of understanding of the requirements of the question and attempted to use the sine rule or right-angled trigonometry.
In triangle ABC, BA = 10 cm, BC = 5 cm and AC = 5√7 cm.
Edexcel 1MA1 · November 2023 · Paper 2F · Q7 · 2 marks
2 marks
Examiner report
The one mark available for stating that the 3-D shape was a cone was awarded to a great many students. Common incorrect answers often included a combination of a number of 3-D shapes and sometimes a mixture of 2-D and 3-D shapes but this was less common. Triangle cone, cylinder cone or cylinder pyramid were the most commonly seen answers that gained no credit. Question 7(b) This question was answered very well by students, with many able to show a clear intention to draw a triangular prism. Unfortunately, not all students were able to show ‘hidden’ lines but this did not detract from their understanding of what the shape looked like and credit was still given. The most common incorrect shape was the drawing of a triangular pyramid.
A diagram shows a solid with one circular base tapering to one vertex.
Edexcel 1MA1 · November 2023 · Paper 2F · Q13 · 2 marks
2 marks
Examiner report
Both parts (a) and (b) required students to identify errors made in calculations or answers to calculations and were generally well answered. The main loss of marks was where the students contradicted their otherwise good response with incorrect statements such as an incorrect figure or correctly identifying that Milo needed to multiply the length and width once only BUT then added that Milo had found the perimeter or volume. Similarly, in part (b) many students answered this question really well, referring to the need for square units. Some said ‘Anya should have squared’ which was ambiguous and scored no marks. Of the students not gaining the mark in part (b), it was often due to not referring to units at all and focussing on what the answer should be based on part (a). Centres should make students aware that a contradictory, or wrong, response on an explain question loses the marks.
A rectangle is 5 cm by 4 cm. Milo claims its area is 5 x 4 x 5 x 4 = 400. Anya gives an area as 86 cm.
Edexcel 1MA1 · November 2023 · Paper 2F · Q19 · 4 marks
4 marks
Examiner report
Although many students were able to make a start on this geometry question utilising angle facts, fully correct responses were rarely seen. Students could gain the method marks either from labelling the diagram or from calculations in the working space, and many were able to identify an angle in the equilateral triangle for the first mark. Completing the method as far as 78 degrees was more challenging, however, usually because they were unable to identify the alternate angle in the quadrilateral as being 60 degrees or subtracting from 180° rather than from 360°. The question asked for a reason for each stage of the working that was relevant to their chosen method, and the attempts at these were generally poor or completely absent. Some students managed to gain one of the C marks, usually for reasons involving angles in a triangle or a quadrilateral, or for identifying the equilateral tringle. Acceptable reasons relating to the parallel lines were seldom seen, without which students were unable to gain full marks. Centres should give students guidance and practice in how to express their reasons for angle work. The minimum wording accepted when awarding marks is indicated by the underlining in the mark scheme. Students should also be encouraged to only list angle facts used as part of their working.
ABDE is a quadrilateral and BCD is an equilateral triangle. CB is parallel to DE. Angle ABD is 90 degrees and angle AED is 132 degrees. Let x = angle EAB.
Edexcel 1MA1 · November 2023 · Paper 2F · Q21 · 2 marks
2 marks
Examiner report
Describing a single transformation is still found to be challenging for a large number of students. A high proportion of the students thought that it was a rotation of 180 degrees, not a reflection. Of the students that gave the single transformation as a reflection, very few scored the second mark for stating a correct equation of the line such as y = −x. Incorrect terminology was also used in place of reflection such as mirroring or flipping the object. It was rare to see any fully correct answers with the majority of students including a centre of rotation, angle of rotation or stated a translation vector which provided a contradiction by implying a second transformation, thus leading to no marks being awarded.
Triangle A has vertices (1,2), (1,5), (3,5). Triangle B has vertices (-2,-1), (-5,-1), (-5,-3).
Edexcel 1MA1 · November 2023 · Paper 2F · Q27 · 2 marks
2 marks
Examiner report
Very few students were able to demonstrate an ability to calculate with bearings in this question. A common incorrect answer was 360 – 147 = 213 and suggests a lack of understanding of this type of bearings question. Very few students attempted to draw a diagram, which would have been helpful in calculating their answer and occasionally gained 1 mark. Students should be encouraged to draw a diagram as a starting point for this type of question.
The bearing of B from A is 147 degrees.
Edexcel 1MA1 · November 2023 · Paper 2F · Q28 · 4 marks
4 marks
Examiner report
This multi-step question required the use of a variety of skills being combined to solve a problem and appeared to be very challenging to a large number of students. Whilst many students knew they needed to find the volume, only a very few students were able to correctly calculate this, with many simply multiplying the radius by the height or the circumference by the height. Converting cm3 to litres or 0.47 litres to cm3 was generally only seen in fully correct responses. The final skill of working with a flow rate was often the only mark being awarded, namely for [volume] ÷ 0.47 to find the time taken. However, a large number multiplied by the flow rate rather than divided.
A cylindrical tank has radius 15 cm and height 43 cm. Liquid flows in at 0.47 litres per minute.
Edexcel 1MA1 · November 2023 · Paper 2F · Q29 · 2 marks
2 marks
Examiner report
This standard question involving trigonometry was not answered well. Of the students able to link the question to trigonometry, many used the formula sheet to effectively rewrite SOHCAHTOA next to the question but many were unable to identify the correct trigonometric ratio to use. When tangent was used, the answer was often calculated correctly and given to the appropriate degree of accuracy but some often incorrectly substituted the angle in place of a side in the formula. Quite a few students tried using Pythagoras’ Theorem to calculate the lengths or tried to use angle properties of a triangle to try to find the unknown angle.
ABC is right-angled at B. BC = 28 cm and angle ACB = 49 degrees.
Edexcel 1MA1 · November 2023 · Paper 2H · Q6 · 2 marks
2 marks
Examiner report
Very few students were able to access this question. Despite it being only grade 5 this is a topic area students seem to really struggle with. There was credit given for a suitable diagram, but very few were able to provide one, showing very limited understanding of bearings in general. In most cases, they drew the two points the wrong way round, or were unable to correctly mark on the reverse bearing.
The bearing of port B from port A is 147°
Edexcel 1MA1 · November 2023 · Paper 2H · Q7 · 4 marks
4 marks
Examiner report
Most were able to gain some credit on this volume and flow problem. This was either for correctly working out the volume of the cylinder, or for correctly using the rate of flow. Unfortunately, very few were able to put all the steps together. Many didn’t work with the conversion at all which could have been used with either the rate of flow or the volume, and this meant in most cases the maximum they could score was 2 marks out of the 4.
An empty tank is in the shape of a cylinder. The cylinder has radius 15 cm and height 43 cm. Water flows into the tank at a rate of 0.47 litres per minute.
Edexcel 1MA1 · November 2023 · Paper 2H · Q12 · 3 marks
3 marks
Examiner report
This problem was unfamiliar to most and there was very little success. A small number were able to find one or both midpoints of the given coordinates, but for almost all, that was all they could do. There was little evidence of students knowing to use Pythagoras to find the distance between two points, and so it was rare to see the award of more than a single mark.
A is the point with coordinates (7, 13). B is the point with coordinates (−3, 21). C is the point with coordinates (15, 23). M is the midpoint of AB. N is the midpoint of BC.
Edexcel 1MA1 · November 2023 · Paper 2H · Q16 · 4 marks
4 marks
Examiner report
This problem required students to use the Sine rule followed by the area of a triangle using sin 2 ab C . In comparison to many other questions at similarly high grade, this was attempted by many, and there was some good evidence of success. Those who recognised the Sine rule normally went on to score the first 2 marks for using it correctly, which was really encouraging. However, many then either didn’t use the correct formula, or used it incorrectly when finding the area of the triangle. Often 64 was used as the angle or candidates worked with 90 degrees rather than recognising the alternate angle. It seems that many didn’t use the available formula sheet, and this is something that can be prepared for in Summer when the formula sheet will be available again.
ABCD is a trapezium with AD parallel to BC. AB = 7.9 cm, BC = 15.3 cm, angle BAD = 112° and angle ADB = 42° (the diagonal BD is drawn).
Edexcel 1MA1 · November 2023 · Paper 2H · Q18 · 4 marks
4 marks
Examiner report
The level of vector algebra was low and in was common to see blank responses for this question, with a majority having no idea how to start the proof. We did see a small number of candidates gaining some marks, normally one or two, typically for a correct expression for or EH BC , but very few were able to complete the proof.
ABCD is a quadrilateral. E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Vector AH = a, vector AE = b, vector DG = c.
Edexcel 1MA1 · November 2023 · Paper 3F · Q10 · 3 marks
3 marks
Examiner report
In part (a) most candidates were able to correctly identify a pentagon. Common incorrect answers were polygon and hexagon. In part (b) very few candidates actually used the word 'edges' in their working box. The most common error was using 10 vertices, shown on the diagram. Many candidates gained 1 mark for marking the edges as they counted them and then multiplying the wrong total of edges by 7.5. Most candidates who scored 0 marks did so because they made no attempt to count the lines (either with ticks or overwriting). This meant that whilst they carried out a multiplication with 7.5, they did not gain marks as their answer was incorrect or ambiguous and there was no indication of a correct method.
A diagram shows a five-sided polygon. A second diagram shows a prism with two congruent five-sided ends and five joining edges; every edge is 7.5 cm.
Edexcel 1MA1 · November 2023 · Paper 3F · Q18 · 5 marks
5 marks
Examiner report
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.
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.
Edexcel 1MA1 · November 2023 · Paper 3F · Q19 · 3 marks
3 marks
Examiner report
It was very rare for a candidate to gain full marks for a rectangle of 7 × 8. Writing the dimensions of the given elevation on the diagram would have been an easy first mark, but many did not do this. Drawing a rectangle of length 7 or 8 was the most common way that candidates gained 2 marks, but whether they understood the term ‘plan’ is unclear. The majority of candidates did not appear to know how to find the missing length or didn’t realise that they needed to.
A cuboid has a front elevation 8 cm wide and 4 cm high. Its volume is 224 cm^3.
Edexcel 1MA1 · November 2023 · Paper 3F · Q27 · 3 marks
3 marks
Examiner report
Some candidates scored 1 mark for correctly substituting the values correctly into the equation or were able to gain the mark for 144. Very few scored marks beyond this point. The most common error was to rearrange the formula into a multiplication instead of a division. Most did not substitute the values into the equation as a first step and hence missed out on the method marks. Most of the candidates showed a lack of understanding of pressure and where an object presses onto another. Once they found the area of one face, they often made no further progress as they failed to understand this was one face of the cube and not the whole cube. Several candidates divided 900 by 144 (area of one face) = 6.25 but did not then show the 6 to compare it to, failing to realise they had demonstrated that 900 cm2 was more than enough to cover 6 faces. Sometimes 3.5 was substituted as 3.52 as candidates sometimes confuse the square on the units and apply it to the numbers also. The candidates that answered successfully for full marks understood that a cube has 6 faces. 864 was the most common correct answer.
A cube exerts force 504 N on a table and pressure 3.5 N/cm^2. Pressure = force/area.
Edexcel 1MA1 · November 2023 · Paper 3F · Q29 · 4 marks
4 marks
Examiner report
This question was poorly attempted with only a small minority recognising the need to use Pythagoras to calculate the diagonal of the square. A few students earned M1 for 2 × 3.52 but were unable to progress further. A lot of candidates demonstrated a good understanding of calculating the circumference but were unable to be awarded marks without identifying the radius or diameter correctly. There was lots of use of π but with the wrong values for radius or diameter. Summary Based on their performance on this paper, centres should note that: • Written work needs to be legible for examiners to consider awarding marks. Figures taken from the question, and taken from candidate’s own work, need to be transcribed accurately. • Candidates need to be trained to avoid rounding or truncating answers to calculations, and to use the most accurate values where possible. • At this level of Foundation, this year candidates showed significant weaknesses in handling issues to do with time, and proportion. • There is a continued need for emphasis to be given to algebraic manipulation and derivation, and application of ratios, scaling and rates in preparing for future examinations. • The inclusion of working out to support answers continues to need emphasis.
A square of side 3.5 cm is inscribed in a circle, with all four vertices on the circle.
Edexcel 1MA1 · November 2023 · Paper 3H · Q12 · 2 marks
2 marks
Examiner report
About one third of students scored both marks on this question. Students who found the angle for the minor sector were awarded 1 mark. Some students used the formula for the circumference of a circle rather than the area. There were also some students who combined different measures, for example working out the area of the circle then subtracting it from 360(°).
A sector POQ of a circle with centre O and radius 6.2 cm is shaded. The shaded sector is the major sector, and the angle of the shaded sector at O is x. The area of the shaded sector is 82.6 cm^2.
Edexcel 1MA1 · November 2023 · Paper 3H · Q15 · 2 marks
2 marks
Examiner report
The hint in the question given by the ordering of the vertices was lost on most students and this question was very poorly done. Most students were unable to identify the need to compare the pairing of AD and AC with the pairing of AE and AB or equivalent. Many students stopped after finding a scale factor between AD and AB (eg, 54 ÷ 32) and the scale factor between AE and AC (eg 80 ÷ 21.6). As they got different values, they stopped. There were many erroneous attempts involving “Pythagoras’ rule” or areas of triangles. A small number of students who did get two figures to compare did not get the final mark because they did not write a conclusion. Students are advised that drawing two separate triangles and marking in the lengths of the sides may help them to identify and use the similarity of the triangles.
In triangle ABC, the point D lies on side AB and the point E lies on side AC, and DE is joined to form triangle AED. BD = 22 cm, DA = 32 cm, AE = 21.6 cm and EC = 58.4 cm.
Edexcel 1MA1 · November 2023 · Paper 3H · Q19 · 4 marks
4 marks
Examiner report
Students who were able to use the given formulae for the volume of cones and spheres were able to gain some credit in this question. Many students took advantage of this by showing a correct process to find the volume of the larger cone or of the hemisphere. However, some students found the volume of the sphere but did not halve it to find the volume of the hemisphere. The majority of students realised that they needed to find the radius of the smaller cone in order to find the volume of the frustum, but a relatively low proportion of them were able to use similarity to do this. Instead, students often appeared to halve the radius of the larger cone to find the radius of the smaller cone as 6 cm or to use the volume formula for the cone with the frustum height (5 cm) as the value of h to find the volume of the frustum. Students may find it helps to recognise that the problem involves similarity if they sketch the two cones and annotate them with known dimensions in such cases as this. A significant number of high attaining students scored 4 marks for a fully correct answer.
F is the frustum of a cone. The frustum is made by removing a cone with height 10 cm from a solid cone with height 15 cm and base diameter 24 cm; the frustum has depth 5 cm. The solid S is made by removing F from a solid hemisphere of diameter 24 cm (the frustum sits inside the hemisphere with their 24 cm circular faces together). You are given: volume of sphere = (4/3)πr^3; volume of cone = (1/3)πr^2 h.
Edexcel 1MA1 · November 2023 · Paper 3H · Q21 · 4 marks
4 marks
Examiner report
Faced with quite a complex diagram, nearly a half of students achieved some credit for their answers to this question by getting at least one, and often two of the missing angles correct. Angles ACE and ACB were often found first and examiners were also encouraged to see that a good proportion of students could use the “Angles in the same segment are equal” theorem to state or show on the diagram that angle CBD was 35°. Only a small number of these students could accurately quote the theorem to justify their reasoning. A significant number of students stated “alternate segment theorem” but did not state what the theorem is. An error seen commonly in responses from lower attaining students was that of assuming that triangle CDE was isosceles, whereas it was in fact right-angled. Students are advised not to jump to conclusions without evidence from the question or diagram to support their assumptions.
A, B, C and D are points on the circumference of a circle with centre O. AC is a diameter of the circle. Chords AD, BD, BC, CD and AB are drawn (D and A are on the same side, with B on the opposite side of the diameter from D). ADE and BCE are straight lines, meeting at the point E outside the circle. Angle DAC = 35° and angle AEB = 28°.
Edexcel 1MA1 · November 2023 · Paper 3H · Q24 · 4 marks
4 marks
Examiner report
This question discriminated well between students working towards the higher grades. Many students were able to use the given ratio in order to work out the length of AP and so gain the first mark. Only a handful of students awarded one mark out of four were awarded this mark for identifying the angle between TP and the base of the prism on the diagram. Significantly fewer students than those who gained one mark showed a correct method to find either the length MP or the length TP so that they could use trigonometry to find angle TPM. Those students who did get that far usually completed the question successfully. Many students mistakenly worked towards finding the size of angle TPA and so usually restricted the marks they could be given to 1 mark. Summary Based on their performance on this paper, students are offered the following advice: • practice working with time conversions, particularly converting a number of hours expressed in decimal form to hours and minutes • improve your skills in dealing with the conversion of units in the context of area problems • ensure you take into consideration whether the gradient of a straight line is negative or positive when you are finding its equation • practice using the quadratic formula to solve a quadratic equation, particularly when you are asked to give your solution to a certain degree of accuracy. • remember that when you multiply through an equation or a formula by a constant you must multiply every term by that constant
A triangular prism has a horizontal rectangular base ABCD, with AB = DC = 14.7 cm and DA = BC = 3.8 cm. M is the midpoint of AD. The vertex T of the prism is vertically above M, with MT = 2.3 cm. P is the point on AB such that AP : PB = 5 : 2.
Edexcel 1MA1 · June 2023 · Paper 1F · Q6 · 1 mark
1 mark
Examiner report
When the diameter was identified as the label which was wrong, a correct explanation usually followed although some found difficulty in describing a diameter unambiguously. Many students thought that the circumference was wrong and some even questioned the nature of the centre. A few candidates drew additional line segments on the diagram to support their answer. Some students said the diameter was wrong because it was not drawn at the correct angle, perhaps indicating the need to stress that radius and diameter can be at any angle, not just a horizontal or vertical line.
A circle diagram has three labels. 'Circumference' points to the boundary and 'Centre' points to the centre, but the arrow labelled 'Diameter' runs only from the centre to the boundary.
Edexcel 1MA1 · June 2023 · Paper 1F · Q8 · 3 marks
3 marks
Examiner report
Part (a) was usually answered well, although 130 (180 – 50) and 40 (90-50) were very common errors. In part (b), those students explaining that 50o is an acute angle were awarded the mark provided they did not contradict themselves with incorrect statements. Many realised that 50o was not an obtuse angle but often failed to give a satisfactory reason why. “An obtuse angle is over 180” and “an obtuse angle is 90o or above” were often seen. Students should think about their statements carefully to ensure they are correct, ‘obtuse angles are more than 90 degrees’ is a correct statement, whereas ‘all angles more than 90 degrees are obtuse’ is not. It was clear that most students knew what an obtuse angle was but did not use the correct vocabulary or terminology to gain marks here. Students were more likely to score if they defined the 50o angle as acute, rather than trying to explain why the angle isn’t obtuse.
At a point, one angle is 50 degrees and the remaining reflex angle is x.
Edexcel 1MA1 · June 2023 · Paper 1F · Q22 · 4 marks
4 marks
Examiner report
Many students found this question far too demanding, not realizing that the total surface area of a cube was 6 times the area of one face. 150 divided by 4 was a common error. Even when the area of one face was correctly found many were able to go no further, often giving the volume as 25 × 25 × 25 or attempting to divide 25 by 4 (rather than finding the square root). For the students who did gain credit on this question the vast majority only scored 1 mark for dividing 150 by 6.
A cube has total surface area 150 cm^2.
Edexcel 1MA1 · June 2023 · Paper 1F · Q27 · 3 marks
3 marks
Examiner report
Inability to find the area of the cross-section of the cylinder prevented any chance of success in this question, rarely was1200 divided by 40 correctly, 300 being a common incorrect result but that did lead to the possibility of picking up another mark by attempting to divide it into 90. Some tried to use the formula for the volume of a cylinder in an attempt to find the radius of the circular base and then its area.
A solid cylinder has volume 1200 cm^3 and height 40 cm. It stands on its circular base and exerts a force of 90 N on the floor. Pressure = force / area.
Edexcel 1MA1 · June 2023 · Paper 1F · Q30 · 1 mark
1 mark
Examiner report
Only the most able students were able to quote 0.5 oe as their answer. Some wrote √1 2 and we even saw √4 4 but these gained no credit unless simplified to give an ‘exact value’ as required.
Write down the exact value of cos 60 degrees.
Edexcel 1MA1 · June 2023 · Paper 1H · Q3 · 4 marks
4 marks
Examiner report
Many students gained the first mark for dividing 150 by 6 to find the area of one face of the cube and very often they went on to score full marks for working out the volume of the cube. After finding the area of one face some students could make no further correct progress. It was common to see 25 divided by 2 or by 4 in an attempt to find the side length and some thought that the volume was given by 25 × 25 × 25. A significant number of students could not show a correct first step and gained no marks. Some attempted to draw the cube and gave a first step of 150 ÷ 3 (presumably the 3 visible faces). Some misinterpreted the question and assumed that 150 was the surface area of one face and wrote 150 or divided 150 by 4.
A cube has a total surface area of 150 cm².
Edexcel 1MA1 · June 2023 · Paper 1H · Q8 · 3 marks
3 marks
Examiner report
A good proportion of the students who gained the first mark for dividing 1200 by 40 then went on to show a complete process. Many gained all three marks but some students made arithmetic errors and lost the accuracy mark. Errors such as 1200 ÷ 40 = 300 and 90 ÷ 30 = 30 were quite common. After dividing 1200 by 40 some students did not realise that they had found the value they needed to use in the pressure formula and it was common to see equated πr2 equated to 30 in an attempt to find the value of r. Some carried out further incorrect work in an attempt to find the ‘area’ and some made no further progress. A common error was attempting to find the surface area of the cylinder and using this in the pressure formula. Some attempts failed at the first hurdle when students did not know how to use the volume of 1200 cm3 and the height of 40 cm.
A solid cylinder rests on a horizontal floor. The cylinder has a volume of 1200 cm³ and a height of 40 cm. The cylinder exerts a force of 90 newtons on the floor. You are given that pressure = force ÷ area.
Edexcel 1MA1 · June 2023 · Paper 1H · Q10 · 4 marks
4 marks
Examiner report
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.
ABCDE is a pentagon in which angle EAB = 120°, angle BCD = 110° and angle CDE = 135°. Angle AED = 4 × angle ABC.
Edexcel 1MA1 · June 2023 · Paper 1H · Q16 · 4 marks
4 marks
Examiner report
Most students started with a process to find the surface area of the sphere. Some made no further progress but many continued with 4πr2 = 200π or 4r2 = 200 and went on to find the radius and gained the first three marks. Inconsistent use of π was a problem for some. It was not uncommon to see a surface area of 200π followed by 4πr2 = 200. After showing a correct process to find the radius a significant number of students then failed to get the accuracy mark for finding the diameter of the sphere in the required form. Many could not write as 5 2 with 50 = 10 5 a common mistake and the radius was often given as the final answer . Instead of starting with a process to find the surface area of the sphere some students started by forming the equation r = .
The surface area of a sphere of radius r is 4πr². 3/8 of the surface area of a sphere is 75π cm².
Edexcel 1MA1 · June 2023 · Paper 1H · Q22 · 2 marks
2 marks
Examiner report
Many students drew the lines FC and AC on the diagram and identified ACF as the angle they were attempting to find. Those who drew only the line FC often failed to identify angle ACF. After identifying the correct angle some students made no further progress and attempts at using Pythagoras were quite common. About half of those who got the first mark, often for a statement such as 6.8 sin 13.6 x = or 6.8 sin 13.6 x − = , were then able to give a correct answer. Some students evaluated 6.8 ÷ 13.6 incorrectly, obtaining 2 for example, but many did not attempt to simplify 6.8 13.6 . Had students recognised that 6.8 is half of 13.6 more might have gained the second mark. Some that did get to sin x = then gave an answer such as 45 or 60. Students who gained the first mark for a correct sine rule statement were often unable to rearrange this correctly to find the value of x. Some students presented the exact trig values in tabular form but were unable to apply them to the question.
ABCDEFGH is a cuboid in which the vertex F is directly above the vertex A, so that AF is a vertical edge and ABCD is a horizontal face. AF = 6.8 cm and FC = 13.6 cm.
Edexcel 1MA1 · June 2023 · Paper 2F · Q7 · 3 marks
3 marks
Examiner report
Part (a) was well answered with many students being able to accurately measure the length of the line. However, performance in part (b) was not as good, with far fewer being able to accurately measure the angle. A number of blank responses were seen which suggests not having a protractor was an issue. In part (c) the wrong name was often stated for the type of triangle, with equilateral being the most common incorrect answer. Symmetrical was also often seen but as it is only a property of an isosceles triangle and not a mathematical name, credit could not be given.
An accurately drawn triangle ABC is measured. On a ruler, A aligns with 0.4 cm and C with 9.7 cm. At B, the two rays align with 18 degrees and 124 degrees on the same protractor scale. A different triangle PQR has QP = QR.
Edexcel 1MA1 · June 2023 · Paper 2F · Q12 · 2 marks
2 marks
Examiner report
Students seemed to find drawing the mirror line in part (a) challenging and accuracy was not always acceptable even when they had ‘the right idea’. Where students had drawn lines connecting the matching vertices to aid them, often no reflection line was drawn possibly indication a lack of understanding of what was needed. Many answers showed little or no use of a ruler with lots of free hand lines or dashed lines. Frequent incorrect responses included simply to draw a line in the diagonal of the grid, drawing a vertical line, or a vertical and a horizontal line, or a diagonal that was clearly nearer one shape than the other. Many drew the line slightly incorrectly, having not realised the need to draw across the diagonals of the grid, often due to trying to link their mirror line to a corner of the squared grid. It was very pleasing to note the success of the first explanation response question, particularly so early in this paper. Most students correctly identified the mistake made and were able to explain this clearly. However, a correct description was occasionally contradicted by giving a combination of different transformations. Other common responses that didn’t gain credit included ambiguous or incomplete descriptions of where the shape should appear such as it should go down not across, or it is facing the wrong way.
For part (a), triangle A has vertices (2,4), (4,2), (5,4) and its reflected image B has vertices (7,9), (9,7), (7,6). For part (b), P has vertices (1,2), (1,4), (4,4), while Alex draws vertices (-1,2), (-1,4), (-4,4) when asked to reflect P in the x-axis.
Edexcel 1MA1 · June 2023 · Paper 2F · Q14 · 4 marks
4 marks
Examiner report
This multi-step problem proved challenging for many students. Whilst many were able to start this question sensibly, very few were able to complete the processes required to find the correct answer. Partial marks were often awarded with many students successfully gaining marks for correctly calculating the volume of one or both shapes. This was usually done before any unit conversion. Fewer students worked accurately with one dimension between the shapes to find the number of packets that could fit along one edge of the box. Students who found the number of packets along each edge (9, 8 and 12), often misunderstood what to do with these values, with addition of 9, 8 and 12 being a commonly seen next step rather than multiplying. Students choosing to work this way often marked their values on the diagram and this approach is to be encouraged. Students who gained the unconditional accuracy mark for conversion between units usually converted units before working out the volume. Some students attempted to convert after calculating the volume and were often confused with volume conversion from mm3 to cm3 and vice versa and so were unable to go on and achieve full marks. Quite a few students calculated the surface area rather than the volume or simply added all the three lengths together for each shape.
Each sweet packet is a cuboid 80 mm by 60 mm by 20 mm. Each box is 72 cm by 48 cm by 24 cm.
Edexcel 1MA1 · June 2023 · Paper 2F · Q24 · 4 marks
4 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2023 · Paper 2H · Q5 · 4 marks
4 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2023 · Paper 2H · Q12 · 4 marks
4 marks
Examiner report
This problem-solving question was answered really well, with a good proportion getting 2 or even all 4 marks. Many spotted the need for Pythagoras and typically gained the 2 marks for a full process. There were a number then that used the formula for the circumference incorrectly and used the wrong value of radius or diameter and gained no further credit. Some candidates also calculated area instead of circumference, particularly disappointing given the formula are on the Exam Aid given to candidates for this series. Those who did not spot the need for Pythagoras unfortunately scored no marks unless they applied trigonometry correctly.
The points A, B, C and D lie on a circle, centre O. ABCD is a rectangle. AB = 8 cm and BC = 10 cm.
Edexcel 1MA1 · June 2023 · Paper 2H · Q13 · 4 marks
4 marks
Examiner report
This trigonometry question proved to be a challenge for many as they either attempted to use the cosine rule to find the length BC, but substituted incorrectly, or they attempted the sine rule but incorrectly linked the side AB with the angle BAC rather than ACB. In all these cases candidates typically scored zero marks. We did see candidates though who were able to correctly use the sine rule to find ACB and those that did normally then gained the last 2 marks for a process to find BAC and a correct answer. Greater care in labelling the sides and angles of the triangle prior to substitution into one of the formulae would have helped many to not make the mistakes they did. Some assumed the triangle was right angled and tried Pythagoras' theorem to find the missing side. Most candidates who used the correct method gained full marks - in these cases the working out tended to be neater and more logical than those who did not gain full marks.
ABC is a triangle with AB = 15 cm, AC = 18 cm and angle ABC = 70°.
Edexcel 1MA1 · June 2023 · Paper 2H · Q19 · 3 marks
3 marks
Examiner report
Many candidates, including some of the weaker ones, were able to gain a mark here for a suitable bound, it was good to see that they were applying their knowledge of error intervals into a more complex context. However, many then struggled to select the correct bound in their expression for sin(x) and so gained no further credit. Candidates who did select the correct bounds and achieved 2 marks, typically then did gain the final mark, although a number forgot to find the inverse sine and were left with an answer of 0.731... rather than 46.989… Some candidates attempted the question in the incorrect order and worked with sin(x) and then attempted to find bounds of this answer. This method gained no marks.
ABC is a right-angled triangle with the right angle at B. Angle ACB is marked x. AB = 9.3 cm correct to the nearest mm. AC = 12.6 cm correct to the nearest mm.
Edexcel 1MA1 · June 2023 · Paper 2H · Q20 · 4 marks
4 marks
Examiner report
A slightly more straightforward vector problem allowed a good number of candidates to gain 1 or even 2 marks, for working with or and then applying the ratio to find or . Unfortunately, many selected the incorrect vectors to add to gain the third mark, and it may be worth candidates writing out a clear vector expression such as before working in terms of a and b. The other areas that caused candidates to lose marks were not clearly assigning their vectors, and much like in Question 5, not working correctly with brackets. At this grade the mark scheme is less lenient to incorrect algebra compared to earlier in the paper.
ORT is a triangle. The vector OT = a and the vector RT = b. M is the point on OR such that OM : MR = 2 : 3.
Edexcel 1MA1 · June 2023 · Paper 2H · Q23 · 3 marks
3 marks
Examiner report
Many candidates were caught out but the relatively familiar figures of 1, 8 and 27, and thought they had to work with cubing numbers. However, candidates had to find the length scale factor of the 3 triangles from the given ratio as 1:3:6 to gain the first mark, whereas many worked with the 1:2:3. Once they had the correct ratio of lengths they could then work out the correct ratio of areas for the 3 triangles to get 1:9:36. From here they had to work out the ratio of the areas of the 3 shapes, 2 of which were trapezia not triangles. There were a number of excellent responses, some working as above and others working nicely with algebra. There were also a number of candidates that gave lengths to the parallel sides and perpendicular height and worked correctly with areas.
Here are three similar triangles, ABG, ACF and ADE. ABCD and AGFE are straight lines, with GB, FC and ED parallel. AB : BC : CD = 1 : 2 : 3.
Edexcel 1MA1 · June 2023 · Paper 3F · Q6 · 2 marks
2 marks
Examiner report
In part (a) very few incorrect answers were seen. A common incorrect answer was 288 where they worked out the length of each side and multiplied these all together, usually showing 6 × 2 × 4 × 1 × 2 × 3 = 288 In part (b) the perimeter caused more issues when candidates missed some sides or double counted some sides. Some candidates confused area and perimeter by giving the answers the wrong way around, for which no marks could be awarded.
An L-shaped polygon on a centimetre grid has vertices (0,0), (3,0), (3,2), (2,2), (2,6), (0,6), joined in that order and back to the start.
Edexcel 1MA1 · June 2023 · Paper 3F · Q8 · 2 marks
2 marks
Examiner report
In part (a) there were many incorrect answers. Some recognised the shape was a 2-D quadrilateral but were unable to state the description in full; equilateral, quadrilateral and rectangle were common incorrect answers. Other candidates gave the name of a 3D shape, with both cube and cuboid common. In part (b) there were a mixture of responses. Many candidates gave a 2D shape name instead of a 3D solid. Those who recognised it was a 3D shape usually gave a correct response. The most common were rectangular prism and cuboid.
Identify a quadrilateral and a solid shape.
Edexcel 1MA1 · June 2023 · Paper 3F · Q16 · 2 marks
2 marks
Examiner report
Centres are encouraged to reinforce learning in this type of question at this level as it tends to be less successfully answered. Candidates found it difficult to isolate the single transformation of "rotation" and this stopped them from securing any marks. Some candidates did not use the correct mathematical language and chose to use the word "turn", others spoiling their answers by combining two or more transformations. The fully correct response, to gain two marks, required candidates to state rotation with an angle of 90 (direction only required for an angle of 270) and a centre of rotation. Some spoilt their description by stating a “Rotation clockwise of 90º” which of course was a rotation from B to A, rather than from A to B.
Shape A has vertices (2,-2), (5,-2), (2,-4). Shape B has vertices (2,2), (2,5), (4,2).
Edexcel 1MA1 · June 2023 · Paper 3F · Q17 · 2 marks
2 marks
Examiner report
It was disappointing to find that this question was answered so poorly with many candidates leaving this blank. Whether this was due to the absence of a ruler and protractor is unknown. The correct distance was the most successful part of their answers although indicating where exactly their point was sometimes led to confusion. The bearing was very poorly done with candidates often using the N line, which could be due to a failure to use a protractor correctly; certainly, there was evidence of some candidates using the wrong “63” on their protractor.
Town R is 55 km from town T on a bearing of 065 degrees. A map uses 1 cm to 10 km.
Edexcel 1MA1 · June 2023 · Paper 3F · Q22 · 5 marks
5 marks
Examiner report
There were three marks available for calculating each of the three angles in triangle ACD. There was an easy one step calculation using "angles on a straight line" to calculate angle ADC = 70. Some candidates then spoiled their work by then using the stated fact (isosceles triangle) to state the other two angles as 70 and 40. In order to complete this problem, candidates had to use a parallel line fact to work out another angle around point C, which they did by either directly calculating angle ACD or by finding another angle at the point that allowed another simple angle fact to be used to state angle ACD = 55. The final angle required the sum of angles in a triangle to deduce angle CAD = 55. The communication marks were awarded for stating a parallel line fact and/or another simple angle fact that was linked to their method. Both were required to secure the final two marks of this five-mark problem. On this occasion it was not necessary to state that the triangle ACD was isosceles or what this meant. Very few students were able to state a parallel line reason. "Parallel lines are the same" was often seen as an incorrect answer. The use of correct angle notation was rare. It should be noted that while most candidate’s work was linked and sequential, some work was spoiled by not naming angles or naming angles ambiguously. Candidates should be encouraged to write their angles on the diagram, which is not ambiguous unless contradicted by alternative working.
ACF and ADG are straight lines meeting at A. BCD and EFG are parallel straight lines, with C on AF, D on AG, and F and G below C and D. Angle EFC = 125 degrees and angle CDG = 110 degrees.
Edexcel 1MA1 · June 2023 · Paper 3H · Q3 · 5 marks
5 marks
Examiner report
This question proved to be a good discriminator of attainment at the lower grades awarded in this paper. It provided the opportunity for students to choose one of several routes to find the size of the angles in triangle ADC and hence show the triangle was isosceles. There were many clear and complete answers comprising of accurate working, the marking of angles on the diagram and well expressed reasons. Some students were less successful in giving a reason linked to parallel lines. The majority of students used a chain of reasoning which included finding angles CFG, ACD, and ADC before using “Angles in a triangle add up to 180” to show angle CAD was equal to angle ACD and hence the triangle was isosceles. Sometimes a single letter was used to describe an angle. Students should avoid this as it can be ambiguous. Similarly, reasons were not always linked clearly to working seen or to the size of angle found. Common errors made included showing that triangle AFG (not ACD) was isosceles, assuming that angles ACD and ADC must be equal and assuming that the triangle ACD was isosceles in order to find the size of angle CAD.
ACF and ADG are straight lines, meeting at A. BCD and EFG are parallel lines (BCD crosses AF at C and AG at D; EFG passes through F and G, below BCD). Angle CDG = 110° and angle EFC = 125°. Show that triangle ACD is isosceles. Give a reason for each stage of your working.
Edexcel 1MA1 · June 2023 · Paper 3H · Q10 · 3 marks
3 marks
Examiner report
A good discriminator, this question tested trigonometry within the context of right- angled triangles. Higher attaining students found the question to be routine but it also gave most students the opportunity to apply their knowledge and skills to show a process or processes to find the length of AC as an intermediate stage. A high proportion of students got this far. Many students were also able to progress to find a relationship involving DC, for example tan 55 = 4.46 𝐷𝐶 so scored the second mark. Fewer students could rearrange the relationship to get “DC =” so could not obtain the final accuracy mark. A common sense check that the length of DC, often calculated as 6.4, should be less than the length of AC may have alerted some students to the error made at this stage. A small minority of students used the sine rule as an alternative approach, sometimes, finding the length of AD as a first step. This was, of course acceptable for the award of the first mark. These students sometimes then progressed to get a correct answer by using right-angled triangle trigonometry and/or Pythagoras’ rule. A significant number of students lost the final accuracy mark because they rounded too much earlier in their working.
ABD is a triangle. C is a point on BD, and AC is perpendicular to BD (angle ACB and angle ACD are both right angles). AB = 6.8 cm, angle ABC = 41° and angle ADC = 55°. Work out the length of DC. Give your answer correct to 1 decimal place.
Edexcel 1MA1 · June 2023 · Paper 3H · Q14 · 4 marks
4 marks
Examiner report
Just under a half of students scored full marks for their responses to this question. However, many other students were able to gain some marks and the question was a good discriminator between students who could show a good understanding of the cosine rule and those who had little understanding of the application of trigonometry to non right-angled triangles. There were a significant proportion of students who could correctly write down an expression for 𝑂𝐵2 by using the cosine rule but then did not observe the correct order of operations and calculated 9 cos35° instead of 117 − 108cos 35°. Students who made a good start to the problem were often able to complete it successfully, but some students used OB2 rather than OB as the radius when finding the area of the sector. Other errors seen included using 4 rather than 360 when finding the area of the sector and assuming that the radius of the circle was 6 cm.
OAB is a triangle. OBC is a sector of a circle, centre O, with OB and OC as radii (A, O and C lie on a straight line, with the sector on the opposite side of O from A). BA = 9 cm, AO = 6 cm, angle BAO = 35° and angle BOC = 80°. Calculate the area of sector OBC. Give your answer correct to 3 significant figures.
Edexcel 1MA1 · June 2023 · Paper 3H · Q18 · 5 marks
5 marks
Examiner report
The successful completion of this question depended upon students being able to use direct proportion to find the radius of the top of the frustum. Many students realised this but there was also a significant number of students who merely halved the radius of the base of the frustum and used 1.5 in their calculations. This severely limited the number of marks examiners could award for these students’ responses. Students working below the attainment level for which this question was designed often scored the mark available for working out the area of the circular base of the solid or two marks for finding the radius of the top of the frustum and the curved surface area of the frustum. Higher attaining students often gave a complete and correct solution to this question to obtain full marks. Some students subtracted either the curved surface area or the total surface area of the cone with base radius 1.8 cm from the total surface area of the cone with base radius 3 cm and did not properly take into account the area of the top of the frustum.
A solid frustum is made by removing a small cone from a large cone. The slant height of the small cone is 6 cm. The slant height of the large cone is 10 cm. The radius of the base of the large cone is 3 cm. [Formula given: curved surface area of a cone = πrl, where r is the base radius and l is the slant height.] Calculate the total surface area of the frustum. Give your answer correct to 3 significant figures.
Edexcel 1MA1 · June 2023 · Paper 3H · Q22 · 3 marks
3 marks
Examiner report
Fully correct answers to this question were seen infrequently and there were many students who made no attempt to answer the question. However, a significant proportion of students were able to score at least one mark for writing down a correct vector equation and/or two simultaneous linear equations from the information given. The most success was achieved by students who wrote down and solved the equations 2a + 8b = 13 and 6a + 2b = 6 to get b = 2 and a = 2 then used this to write down a correct expression for b in terms of a. Some students formed simultaneous equations but then used only one equation to find a relationship between a and b. Of those students who did solve the equations to get the value of a and the value of b, some failed to complete the question to get b in terms of a. Approaches using a(2 6) + b(8 2) = k(13 6 ) where k was treated as a general constant or using 2𝑎+8𝑏 6𝑎+2𝑏 = 6 as a start were rarely seen.
Given that the vector a(2, 6) + b(8, 2) is parallel to the vector (13, 6), find an expression for b in terms of a. (Vectors are written here as column vectors in coordinate form: (2, 6) means the column vector with top component 2 and bottom component 6.)
Edexcel 1MA1 · November 2022 · Paper 1F · Q9 · 2 marks
2 marks
Examiner report
Part (a) was well answered with the vast majority subtracting 70 from 360 to get a correct answer of 290o. However, a great many were unable to give a correct reason for this answer in part (b). The most common incorrect explanation related to a circle “having 360o”. Students need to be made aware that such an explanation is not acceptable. Many simply explained how they did it, rather than why e.g. “I subtracted 70 from 360”. Students should be reminded that a calculation is not a reason. In this case “angles at a point sum to 360o” or words to that effect is the reason required. A common incorrect approach was to work with 180° rather than 360°.
Two angles around a point are 70° and y°.
Edexcel 1MA1 · November 2022 · Paper 1F · Q11 · 4 marks
4 marks
Examiner report
Whilst many students correctly wrote down the scale factor of 2 in (a)(i), only a very small minority were able to pinpoint the centre of the enlargement in (ii). Many tried to position their centre either in between the two shaded triangles, in the centre of one of the triangles or on one (or two) of the vertices. In some cases, students who were able to draw lines to intersect at the point of enlargement identified it correctly but missed the mark because they also placed ‘X’ somewhere else on the graph. In part (b)(i), accurate drawing of the mirror line was good. Whilst lines drawn with a straight edge are preferred, freehand, dotted or dashed lines were acceptable. In part (b)(ii), unfortunately, very few students were able to give the equation of their line. Common incorrect lines drawn were x = 4 or y = 4, which in some cases allowed a follow through mark if they stated this as their equation for (ii).
On a square grid measured from its lower-left corner, triangle A has vertices (3,4), (6,4) and (5,2). Triangle B has vertices (5,6), (11,6) and (9,2). Triangle B is an enlargement of triangle A.
Edexcel 1MA1 · November 2022 · Paper 1F · Q15 · 2 marks
2 marks
Examiner report
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.
On a coordinate grid, P = (−3, −2) and Q = (2, 4).
Edexcel 1MA1 · November 2022 · Paper 1H · Q11 · 2 marks
2 marks
Examiner report
Descriptions of the transformation were generally very poor and surprisingly few students gained full marks for giving all three aspects of the description. Stating an incorrect scale factor of 3 or –3 was a common error and the centre of enlargement was often incorrect or not given. Many of the students who stated that the transformation was an enlargement did not get a mark because they failed to give either a correct scale factor or a correct centre of enlargement. Two of the three aspects of the description were needed for 1 mark. Some students tried to describe a combination of transformations – it was common to see translations combined with enlargements – and this gained no credit.
On a coordinate grid, trapezium P has vertices (6,14), (15,14), (12,8) and (9,8). Trapezium Q has vertices (2,6), (5,6), (4,4) and (3,4). Describe fully the single transformation that maps P onto Q.
Edexcel 1MA1 · November 2022 · Paper 1H · Q15 · 3 marks
3 marks
Examiner report
This question was not answered as well as might have been expected and a large number of students failed to gain any marks at all with many not attempting the question. Many of those with some knowledge of sectors of circles went wrong at the first step, using πr2 or πr instead of 2πr and gained no marks. A good proportion of those who did make a correct start went on to get full marks.
The diagram shows a sector of a circle of radius 18 cm. 18 cm 18 cm x° The length of the arc is 4π cm. Work out the value of x. x =
Edexcel 1MA1 · November 2022 · Paper 1H · Q18 · 3 marks
3 marks
Examiner report
This question was answered very poorly. Some students scored the first mark for angle OBC = 90º which was often marked on the diagram but could then make no further meaningful progress. Many could not even get that far. Some students correctly identified an isosceles triangle, but used arbitrary numerical values, e.g. 45, instead of an algebraic approach. Those with some idea usually found expressions for angles in terms of x although other approaches were occasionally seen. Some students labelled angle OBA or angle ABC as y and used the relationship between x and y to show the required result and a few students marked a point on the circle and formed a triangle with AB as the base. Full marks were awarded surprisingly rarely. Some answers were let down by poor algebra but some students who did show correct working leading to the required result then failed to correctly state the one circle theorem that was needed for the final mark. Some gave no reason, others gave a description which did not include the required key words.
A and B lie on a circle with centre O. The straight line DBC is tangent to the circle at B, and angle AOB = x°.
Edexcel 1MA1 · November 2022 · Paper 1H · Q22 · 4 marks
4 marks
Examiner report
A small proportion of students gained at least one mark for giving the value of sin30 as 0.5 or for a correct sine rule statement with 6.5, 10.7 and 30 substituted. Students who did not know the correct value of sin30 were still able to gain two of the four marks if they showed a complete process to find sin ABC. Many of those who gained the first three marks by showing a complete process to find sin ABC using sin30 = 0.5 did not gain the accuracy mark because they did not give their answer in the form m n where m and n are integers. The final answer was often given as 3.25 10.7 or 6.5 21.4 . Many students showed a lack of understanding of the mathematical requirements of the question, resorting to right-angled trigonometric calculations and/or Pythagoras’ Theorem.
In triangle ABC, AB = 10.7 cm, AC = 6.5 cm and angle ACB = 30°.
Edexcel 1MA1 · November 2022 · Paper 1H · Q24 · 6 marks
6 marks
Examiner report
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
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.
Edexcel 1MA1 · November 2022 · Paper 2F · Q6 · 2 marks
2 marks
Examiner report
The majority of the cohort were able to accurately measure the line but less were able to measure the angle. Centres are advised to ensure candidates have the correct equipment and can use it effectively.
EXCLUDED FROM TEXT-ONLY STUDENT DELIVERY: this official item requires measurements from the printed to-scale quadrilateral ABCD, which cannot be represented faithfully in the batch schema.
Edexcel 1MA1 · November 2022 · Paper 2F · Q19 · 2 marks
2 marks
Examiner report
This question was not well answered and very few full marks were awarded. A proportion of students gained 1 mark for a correct 5 by 3 rectangle drawn in the correct orientation; attempts to draw the horizontal lines to indicate features on the shape were rare. Even rarer was the use of the dashed line to indicate the hidden feature. This is an area of the specification students seem to find challenging and centres are advised to look at plans and elevations with students, emphasising the need for the internal lines.
On a square grid, the front elevation has boundary vertices (relative to its lower-left corner) (0,0), (0,5), (4,2), (4,3), (7,1), (7,0). The plan is a 7-square-by-3-square rectangle with an internal line across its 3-square depth at 4 squares from the left. Draw the side elevation viewed from the right.
Edexcel 1MA1 · November 2022 · Paper 2F · Q21 · 5 marks
5 marks
Examiner report
The majority of students found this multistage problem involving area, proportion and cost challenging, with a wide range of approaches and very few fully correct responses seen. Whilst many students were able to correctly calculate the area of the rectangular sections of the composite shape, very few were able to use the appropriate formulae for the area of a triangle and a circle to find the total area. Of those that did recall the formulae correctly, some used an incorrect dimension for the triangle by using 7cm instead of 9cm and some forgot to divide the area of the circle by 4 to find the area of the quadrant. It was also common to see areas of at least one rectangle being combined with angles of 90°, 180° or 360° for the circular area. The use of proportion to work out the number of bags needed to cover the area calculated was used correctly by some students, with many gaining a mark for finding the number of bags required to cover at least one correct area. These students often went on correctly to round the number of bags needed to an integer before multiplying by the cost of a single bag to gain another mark. A small number of students didn’t round up and therefore gained no credit for finding the cost using part bags. Students who were the most successful often showed working in a systematic way by splitting the shape into 4 sections before continuing with their calculations. It is important for students to be able to fully recall correct formulae for the areas of a number of shapes and the use of systematic working helped students work through this multistep question.
Jason’s garden is the non-overlapping union of a rectangle 11 m by 7 m, a rectangle 9 m by 7 m, a right-angled triangle with perpendicular sides 11 m and 9 m, and a quarter-circle sector of radius 7 m. Each bag of grass seed covers 14 m² and costs £10.95. Work out the cost of all the bags he needs.
Edexcel 1MA1 · November 2022 · Paper 2F · Q22 · 2 marks
2 marks
Examiner report
Many students did not recognise the need to use trigonometry in this 2 mark question, with many trying to work with the area, finding missing angles or using Pythagoras’ theorem. Of those who did recognise that the use of trigonometry was needed, many selected to use the wrong ratio, with Sine being the most common incorrect ratio used. When the correct ratio was used, the majority of students found a correct answer within the acceptable range. Of the students gaining only the method mark, it was often due to early rounding of figures. The incorrect use of the Cosine ratio due to writing cos (53 × 14.5) was occasionally seen so students should be encouraged to practice the correct use of trigonometric ratios and have a better understanding of the order of operations applied.
A right-angled triangle has hypotenuse 14.5 cm and an adjacent leg x cm. The angle between this leg and the hypotenuse is 53°.
Edexcel 1MA1 · November 2022 · Paper 2H · Q2 · 2 marks
2 marks
Examiner report
The side elevation on this question was more challenging than many candidates have seen in some time, and very few were able to score full marks. This was normally down to the hidden edge not being indicated with a dashed line which was needed to make the diagram fully correct. That aside many struggled to identify the other edges on the front face and those who scored one it was typically for drawing a rectangle 5 high by 3 wide.
On a square grid, the front elevation has boundary vertices (relative to its lower-left corner) (0,0), (0,5), (4,2), (4,3), (7,1), (7,0). The plan is a 7-square-by-3-square rectangle with an internal line across its 3-square depth at 4 squares from the left. Draw the side elevation viewed from the right.
Edexcel 1MA1 · November 2022 · Paper 2H · Q4 · 5 marks
5 marks
Examiner report
This is the first AO3 question on the paper and required 3 main processes. The first of which was to find the total area, and there were two marks assigned to this, the first for correctly finding 3 areas and the second for finding all 4. The second process was to find the number of bags of grass seed that is needed. This could be gained for finding this for a single area or a combination. Finally, the candidates had to round the number of bags up to gain the number needed to buy, and multiply by £10.95 to find the total cost. Almost all candidates were able to gain some marks, even if just for correct areas. This was as often seen for 2 rectangles and the quadrant, as it was for 2 rectangles and the triangle. Most were then able to gain the mark for dividing by 14. The two major causes of errors were learners not having all four areas correct, as this was needed to gain the 3rd mark. The other common error was for candidates to not round the number of bags up to the nearest integer and so found a cost that was inaccurate.
Jason’s garden is the non-overlapping union of a rectangle 11 m by 7 m, a rectangle 9 m by 7 m, a right-angled triangle with perpendicular sides 11 m and 9 m, and a quarter-circle sector of radius 7 m. Each bag of grass seed covers 14 m² and costs £10.95. Work out the cost of all the bags he needs.
Edexcel 1MA1 · November 2022 · Paper 2H · Q5 · 2 marks
2 marks
Examiner report
A straightforward question requiring candidates to use the cosine ratio to find a missing side. Typically candidates scored 2 marks or none. Those scoring none, normally chose the sine ratio, or tried to work with Pythagoras. Of those candidates who identified cosine correctly most gained 1 mark for a correct statement, but a small number struggled to rearrange the expression to correctly find x.
A right-angled triangle has hypotenuse 14.5 cm and an adjacent leg x cm. The angle between this leg and the hypotenuse is 53°.
Edexcel 1MA1 · November 2022 · Paper 2H · Q13 · 2 marks
2 marks
Examiner report
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.
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
Edexcel 1MA1 · November 2022 · Paper 2H · Q17 · 4 marks
4 marks
Examiner report
A similar solids question, but with an added layer of difficulty. It was pleasing to see so many candidates get some credit. This was normally for either finding the volume scale factor (by dividing the two volumes) or for a process to find the height of B, or in a good number of cases, both scoring 2. However, very few candidates were able to take this further. Typically candidates tried to use a scale factor of 27 when working with areas and lengths and gained no further credit e.g. 10.8 ÷ 27. There were also occasions where more marks could have been scored had candidates been able to deal with the half in the area of a triangle. A shame that such a low level skill was denying candidates high grade marks.
The diagram shows two similar solid triangular prisms, A and B. h cm 43.74 cm² 8.1 cm Prism A Prism B The volume of prism A is 58.806 cm³ The volume of prism B is 1587.762 cm³ The cross section of each prism is a right-angled triangle. For prism B the length of the base of the triangle is 8.1 cm the area of the triangle is 43.74 cm² The height of the triangle for prism A is h cm. Work out the value of h. h =
Edexcel 1MA1 · November 2022 · Paper 2H · Q18 · 2 marks
2 marks
Examiner report
Having access to a formula sheet has definitely helped students with this question, which was answered very well. Most candidates were able to use the given formula to find the area, and typically scored both marks. For those that only scored 1 it was normally due to incorrect rounding. Candidates should remember to write out unrounded answers first as this will normally gain the accuracy mark. It will be interesting to see if success on this topic drops when the formula sheet is removed. Common incorrect methods including treated the triangle as right-angled and doing ½ × base × height and using the Cosine Rule to find the missing side.
A triangle has two sides of length 11.2 cm and 4.3 cm, with included angle 118°.
Edexcel 1MA1 · November 2022 · Paper 2H · Q20 · 4 marks
4 marks
Examiner report
Three dimensional Pythagoras and Trigonometry is always a challenging subject for candidates, and so it proved again. However, it was good to see many make a start and gain some credit. A decent number of candidates were able to complete a process to find AF using Pythagoras. It is worth noting, that at this grade a full process was required for a single mark, whereas at a lower level we may see 2 marks for Pythagoras. The step that troubled many was knowing to, and how to, find FH. Many took the base to be square and incorrectly used Pythagoras rather than Trigonometry. Of those who recognised the need to use trigonometry, far too many chose the wrong ratio, and were unable to gain further credit. The final step was for a correct trigonometry statement involving angle FAH. This was however, dependent on the processes for any values used in this step being correct, which was normally not the case.
ABCDEFGH is a cuboid with AD = 9 cm, face diagonal FD = 13 cm and angle GHF = 49°. Triangle ADF is right-angled at A, triangle FGH is right-angled at G with GH = AD, and triangle AFH is right-angled at F.
Edexcel 1MA1 · November 2022 · Paper 3F · Q11 · 2 marks
2 marks
Examiner report
This question was answered well. Common errors were to find the surface area or the perimeter of the shape. A significant number of candidates seemed to confuse volume with area formulae (of triangles or trapeziums) as they halved a perfectly good volume and thereby ruined their answer. Candidates are advised to be more aware of the dimensionality of area and volume and how this relates to the units used for these different concepts.
Here is a cuboid. 8 cm 4 cm 5 cm Work out the volume of the cuboid. cm³
Edexcel 1MA1 · November 2022 · Paper 3F · Q13 · 3 marks
3 marks
Examiner report
In part (a)(i), the answer of 40 was most often correct. The most common incorrect answer for this part was 80, the misconception being that the angles at C and D were equal. In part (ii), most candidates were able to give the standard response. Some omitted either “angles” or “quadrilateral” from their answer, which was a requirement for the mark. In part (b) a high proportion of candidates were able score the C mark. The most common responses included: stating that the angles in a triangle should sum to 180° or that the angles of the given triangle incorrectly sum to 190°. Unfortunately, there were a number of responses seen that either stated incorrectly or spoiled a good response by asserting that the triangle should be equilateral or isosceles, which of course was an incorrect statement. Some also chose to change the 60° angle to 50°; while this answer may imply the correct angle sum it is not a correct statement as that angle did not need to be that size.
The four interior angles of quadrilateral ABCD are 120°, 120°, 80° and x°.
Edexcel 1MA1 · November 2022 · Paper 3F · Q18 · 2 marks
2 marks
Examiner report
This question posed some difficulty for candidates to complete correctly. Candidates who were able to rotate the shape 90° anticlockwise into quadrant four were rewarded with at least one mark if they were unable to position it or draw it correctly. While uncommon, if candidates were able to rotate it correctly clockwise into quadrant two with centre the origin were also rewarded with one mark. More often than not candidates scored no marks due to a variety of errors. Some candidates rotated 180° rather than 90° or were unable to rotate it anticlockwise by the correct centre (often resulting in the shape not being fully in quadrant four). Candidates would benefit from a better understanding of this assessment objective and perhaps using tracing paper.
A shaded quadrilateral has vertices (−1,−2), (−4,−2), (−2,−4) and (−1,−4). Rotate it 90° anticlockwise about (0,0).
Edexcel 1MA1 · November 2022 · Paper 3F · Q24 · 4 marks
4 marks
Examiner report
Nearly all responses scored zero marks. Responses that scored all 4 marks were very rare. Very few candidates used the correct formula for the area of the cross-section or for the whole cylinder, many using the diameter rather than the radius. Most candidates failed to find a volume, but simply found the area of a rectangle by multiplying 80 by 160 and worked with 12800 as if it was the volume of one cylinder. The most common way of scoring a mark was for finding the amount of fertiliser for each tank. Where marks were gained this was generally for the process of working out the volume of a single tank or for a process to find the volume of water per tank. A small minority of candidates got 1 mark for showing 8000 in their working. The ratio of 100:1 equalling 101 parts evaded nearly all candidates, resulting in minimal full mark answers.
Karina has 4 tanks on her tractor. Each tank is a cylinder with diameter 80 cm and height 160 cm. 160 cm 80 cm The 4 tanks are to be filled completely with a mixture of fertiliser and water. The fertiliser has to be mixed with water in the ratio 1 : 100 by volume. Karina has 32 litres of fertiliser. 1 litre = 1000 cm³ Has Karina enough fertiliser for the 4 tanks? You must show how you get your answer.
Edexcel 1MA1 · November 2022 · Paper 3F · Q25 · 4 marks
4 marks
Examiner report
There were far more correct answers than similar shape questions in previous years with many candidates getting 2 or 4 marks. If candidates were awarded two marks for the first part, they frequently went on to get two marks in the second part too. Candidates either knew how to calculate a scale factor and got the answers correct or made wild attempts at finding relationships. A few candidates fell into the classic misconception of adding something on to find the missing sides, rather than finding the ratio between the corresponding lengths. In part (a), a common error was to correctly find the scale factor of 4, but then incorrectly multiply by 5 to get 20, whilst in part (b) a common mistake was to divide by the 5.
Triangles ABC and DEF are similar, with A corresponding to D, B to E and C to F. In triangle ABC, AC = 5 cm and BC = 4 cm. In triangle DEF, DE = 22 cm and DF = 20 cm.
Edexcel 1MA1 · November 2022 · Paper 3F · Q30 · 4 marks
4 marks
Examiner report
Part (a)(i) was surprisingly well answered. A common error was for the numbers to be reversed in the vector. But in contrast part (ii) was not well answered. While many candidates were able to show a substituted expression, they often failed to multiply both the 2 and the 3 of vector a by 2 before doing the subtraction. Some candidates forgot to account for the sign and added the vectors instead. Many of the candidates failed to attempt part (b). Most of those who did make an attempt did not know (or failed to recognise) that 2d would be parallel to d, just drawing a random diagonal line. Some of the candidates drew a correct vector but not from point P. Copying the vector without doubling it was another common error. Summary Based on their performance on this paper, candidates should: • take care when reading figures from questions, and transcribing their own work • ensure that when giving explanations that responses are focused on the demand of the question • write their figures legibly, avoiding any possible confusion when writing 1s and 7s, when writing 4s and 9s, and even when writing 0s and 6s. • show their methods clearly • ensure they know how to scale numbers, and use scaling on axes of graphs • practice algebraic manipulation, particularly in relation to finding the subject of a formula • understand better how to change between units, including compound units. • practice using ratio in real life contexts
a = (2, 3), b = (-1, 2) and c = (4, 1), written as column vectors. Vector d on the grid moves 3 squares right and 2 squares down from its tail.
Edexcel 1MA1 · November 2022 · Paper 3H · Q4 · 4 marks
4 marks
Examiner report
A good discriminator, most candidates were able to score at least 2 marks for showing how to get the total volume of 4 tanks. Only the lowest attaining candidates failed to use a correct formula for the volume of a tank. A few candidates used the diameter instead of the radius to work out the volume, in effect working out the value of πd²h or sometimes πdh instead of πr²h. Some candidates who left their volumes as a multiple of π, for example 256000π, then lost the π in the next step of their working. Correct attempts to use the ratio 1 : 100 were usually restricted to answers from higher attaining candidates and many candidates used 100 where they should have used 101 in their calculations. There was a wide range of possible approaches to answering the question and responses seen were very varied. The most common successful approach was to compare the volume of the 4 tanks with the amount of mixture that could be made with the available fertilizer.
Karina has 4 tanks on her tractor. Each tank is a cylinder with diameter 80 cm and height 160 cm. 160 cm 80 cm The 4 tanks are to be filled completely with a mixture of fertiliser and water. The fertiliser has to be mixed with water in the ratio 1 : 100 by volume. Karina has 32 litres of fertiliser. 1 litre = 1000 cm³ Has Karina enough fertiliser for the 4 tanks? You must show how you get your answer.
Edexcel 1MA1 · November 2022 · Paper 3H · Q5 · 4 marks
4 marks
Examiner report
This question involving similar triangles was well answered by a majority of candidates. Often there was little or no working seen. Where working was seen, the most common approach was to find the scale factor ( 5 ) of the enlargement, mapping the smaller triangle onto the bigger triangle, then use it to find the lengths required. This approach was usually successful. It was rare to see the alternative method of using the ratio of two known sides, for example 𝐴𝐶 𝐵𝐶 = 4. Some candidates tried to solve the problem by using “Pythagoras’ rule”. The most common error seen in part (a) was for candidates to use a scale factor of 4 . In part (b) the most frequently incorrect answer seen was 4.4.
Triangles ABC and DEF are similar, with A corresponding to D, B to E and C to F. In triangle ABC, AC = 5 cm and BC = 4 cm. In triangle DEF, DE = 22 cm and DF = 20 cm.
Edexcel 1MA1 · November 2022 · Paper 3H · Q16 · 4 marks
4 marks
Examiner report
The majority of candidates achieved some credit for their answers to this question by showing they could interpret the ratio given or by showing they understood that opposite angles in a cyclic quadrilateral sum to 180°. Many candidates showed they understood the ratio by marking the angles EAC and ACE as 2x and x respectively on the diagram. Examiners also accepted values for the size of these two angles, in the given ratio, to award this mark. Similarly, values for the opposite angles in quadrilateral ABDE which added to 180 were accepted as evidence for the use of the result that opposite angles of a cyclic quadrilateral add up to 180. A relatively small proportion of candidates could find a further connection between the angles in the quadrilateral and the angles in the triangle and so score more than 2 marks. Despite the guidance that all working needed to be shown, a number of candidates wrote down 40 as their answer but with a lack of supportive and persuasive working. These candidates could not be given full credit. In particular, examiners were concerned to see some candidates using results incorrectly, for example “angles on a straight line sum to 180°” to state that angles EAB and DBC sum to 180 or that angles ABD and BDC were alternate angles.
ABDE is a cyclic quadrilateral. Points A, B and C lie on a straight line, and points E, D and C lie on a straight line. Angle DBC = 60°. Given that angle EAB : angle BCD = 2 : 1, work out angle BCD. You must show all your working.
Edexcel 1MA1 · November 2022 · Paper 3H · Q18 · 2 marks
2 marks
Examiner report
It is unfortunate that very few candidates used the working space for this question to draw a diagram to help them. Consequently, only a small number could identify the correct single transformation. Instead, many candidates suggested that the single transformation equivalent to two reflections would be another reflection or, bizarrely an enlargement, scale factor 3. Of those candidates who did identify the transformation as a translation, some gave the vector (4 0) to describe the translation instead of (8 0).
Shape A is reflected in the line with equation x = 2 to give shape B. Shape B is reflected in the line with equation x = 6 to give shape C. Describe fully the single transformation that maps shape A onto shape C.
Edexcel 1MA1 · November 2022 · Paper 3H · Q20 · 3 marks
3 marks
Examiner report
This question was not well done and there were few full proofs seen. Many candidates who attempted the question realised that they needed to consider the case of “SAS”. However, they more often than not failed to give sufficient justification to accompany the pairings of two sides or two angles. In particular, examiners could not accept “AE = DE” without any justification as this was merely writing a detail from the question. Instead, they needed to see “AE = DE (given)” or equivalent. The pairing AC with DB needed to be accompanied by some justification related to the ratio 1:2:1 given in the question and the pairing of angle EAC with angle EDB needed to be accompanied with a justification related to the base angles of isosceles triangle ADE. A substantial number of candidates made progress with reasons but failed to explicitly say that this led to the equality of the lengths or angles involved. For example, it was not uncommon to see part of an explanation as to why the ratio lead to the fact that the lengths of AC and DE were the same without actually stating that AC = DE.
In triangle ADE, points B and C lie on AD in the order A–B–C–D. Also AE = DE and AB : BC : CD = 1 : 2 : 1. Prove that triangle ACE is congruent to triangle DBE.
Edexcel 1MA1 · November 2022 · Paper 3H · Q24 · 4 marks
4 marks
Examiner report
Many of the more high attaining candidates gained either partial or full credit for their answers to this question on vectors. The vector algebra seen was generally clearly presented with accurate simplification of expressions found. Most sudents who attempted the question gave a correct expression for either 𝐶𝐸⃗⃗⃗⃗⃗ and/or 𝐶𝐹⃗⃗⃗⃗⃗ in terms of a and b. Of those candidates who gained marks but who were not successful in proving that CF is parallel to DP, it was often just the last stage which went awry. Candidates are advised that they should show examiners how their working leads to a proof that the required result is true. In this case, a statement that 𝐷𝑃⃗⃗⃗⃗⃗ = 2 𝐶𝐹⃗⃗⃗⃗⃗ or equivalent and a conclusion that therefore CF and DP are parallel was expected.
CDEF is a quadrilateral whose vertices occur in the order C–D–E–F. The vectors CD = 2a, ED = b and FE = a. Point P lies on the straight line C–E–P and CE = EP.
Edexcel 1MA1 · November 2022 · Paper 3H · Q26 · 5 marks
5 marks
Examiner report
This question, not surprisingly attracted only a small proportion of fully correct answers. However, it is encouraging to report that a much higher proportion of candidates scored some credit for their attempts. Many candidates identified the need to use the angles of the heptagon and scored a mark for a correct process to find the size of each interior angle or the size of each exterior angle. A small number of candidates worked out the size of an angle but wrongly assigned it, for example marking 51.4° on the diagram as an interior angle. A good number of candidates also realised that they needed to form an equation by using the area of the triangle, given as 30 cm² in order to make progress towards finding the length of GB. Some candidates did this successfully but few candidates were able to get any further. Summary Based on their performance on this paper, candidates are offered the following advice: • carry out a common sense check on the answers to calculations • read questions where a written explanation is expected very carefully and check your answer matches the question asked • practise less straightforward problems which involve working with a combination of algebra or geometry or probability with ratio • improve your skills in dealing with the conversion of units • ensure you know how to write down an error interval when a number has been truncated • practice writing proofs of congruence in geometry
ABCDEFG is a regular heptagon. A E D F C G B The area of triangle ABG is 30 cm² Calculate the length of GB. Give your answer correct to 3 significant figures. You must show all your working. cm
Edexcel 1MA1 · June 2022 · Paper 1F · Q3 · 1 mark
1 mark
Examiner report
Very well answered with most students gaining the mark. It was pleasing to see many using a ruler resulting in accurately drawn diagrams. Where the mark was not given, it was usually due to an inaccurate positioning of the reflected vertex of the triangle.
On an 8-by-7 square grid, a vertical mirror line is x=4. The triangle has vertices (4,2.5), (4,5.5), and (7,4). Reflect the triangle in the mirror line.
Edexcel 1MA1 · June 2022 · Paper 1F · Q22 · 3 marks
3 marks
Examiner report
This question was answered poorly by all but a few students. Many recognized the radius as being 3 cm but were unable to find an expression for the area. The height of the cylinder was often taken as 6 cm instead of 5 cm. A common error was to correctly find the area of the plan, 9π, and then multiply it by the area of the front elevation. Some students used numerical values for π and this was acceptable in assessing the process but not in the final answer. Whilst many students did realise that they needed to use 𝜋𝑟2, or even 𝜋𝑟2ℎ they lacked the confidence to evaluate this with the given information. Weaker students did attempt to count squares in working out the area of the plan but this was usually accompanied by counting squares to find the area of the front elevation as well.
A cylinder's plan is a circle of radius 3 cm; its front elevation is a rectangle 6 cm wide and 5 cm high.
Edexcel 1MA1 · June 2022 · Paper 1F · Q27 · 3 marks
3 marks
Examiner report
Predictably in this question, many students were confused between interior and exterior angles in regular polygons. Many times the interior angles of a regular pentagon and a regular hexagon were quoted (or worked out to be) 72o and 60o respectively. No marks were available after this major error. Contradictions between diagram and working were often seen. A significant number of students simply divided 360 by 3 or measured the angle with a protractor. Students who found the correct interior or exterior angles usually went on to score well, arithmetical errors again preventing full marks at times; 18 as the answer to 5 being frequently seen. It was good to see that so many students know how to find the total number of degrees in the sum of the interior angles of a polygon.
A regular hexagon and regular pentagon share a side. x is the exterior angle between their next sides at one end of the shared side.
Edexcel 1MA1 · June 2022 · Paper 1F · Q29 · 3 marks
3 marks
Examiner report
Only the more able students, appreciating the need to find the volumes of the given cubes, made any progress in this question. Some multiplied the mass by the volume and then tried to simplify their ratio. Some did divide the masses by the volumes but left the un-simplified ratio 27 : The most common error was to divide the masses by the length of an edge, leaving a ratio of 27:32. Also common was to multiply the 81 and 128 by 3 and 4 to get 243:512
Cube A has side 3 cm and mass 81 g. Cube B has side 4 cm and mass 128 g.
Edexcel 1MA1 · June 2022 · Paper 1F · Q30 · 1 mark
1 mark
Examiner report
This final question was very poorly answered by students, many leaving the solution blank. Summary: On the evidence of their performance on this paper, students need to: • take more care when reading the questions. There were a considerable number of misreads which prevented complete solutions. • write clearly so that correct values quoted are not altered in subsequent working. • be more careful when performing simple arithmetic calculations, particularly in build-up methods. • think about the reasonableness of their answers. • set their working out clearly, crossing out working that is being replaced. • avoid any temptation to merely offer an answer only without showing their working. • be able to convert fractions of an hour into minutes. • know the difference between an interior angle and an exterior angle of a regular polygon.
Write down sin 30°.
Edexcel 1MA1 · June 2022 · Paper 1H · Q5 · 3 marks
3 marks
Examiner report
It was pleasing that many students gained full marks for finding the value of x, often making use of the diagram as part of their working. The first step for many was to use (n – 2) × 180° to find the sum of the interior angles of the hexagon or the pentagon and then divide by the number of sides to find the size of an interior angle. Students who used this approach to find the interior angles often went on to show a complete method. Arithmetic errors, particularly when dividing 540 by 5, were quite common. However, these students were still credited with the methods marks. The division of 540 by 5 sometimes resulted in 18 or 180 yet neither of these values is a sensible size for the interior angle of the pentagon and should have alerted students to a problem. Alternatively, some students started by working out 360 ÷ 6 = 60 and 360 ÷ 5 = 72. Those who realised that these calculations give the sizes of the exterior angles usually went on to get full marks but a significant number of students used them as interior angles and gained no marks. Contradictions on the diagram were common.
A regular hexagon and regular pentagon share a side. x is the exterior angle between their next sides at one end of the shared side.
Edexcel 1MA1 · June 2022 · Paper 1H · Q7 · 3 marks
3 marks
Examiner report
The majority of students gained the first mark for a process to find a volume and many went on to give a correct answer. The final mark was sometimes lost because of arithmetic errors, these occurred most often when dividing 81 by 27 or dividing 128 by 64. Having found the volume of each cube some students gained no more marks because they were unable to use density = mass ÷ volume. It was common to see mass × volume and volume ÷ mass being used. Some students did not realise that it was necessary to find the volume of each cube and scored no marks at all. Often these students simply used the figures in the question, dividing 81 by 3 and dividing 128 by 4 or multiplying 81 by 3 and 128 by 4. There were also some students who calculated the surface area of the cubes and attempted to divide the mass by the surface area and gained no credit. Some students included units in their final answer but they were not penalised.
Cube A has side 3 cm and mass 81 g. Cube B has side 4 cm and mass 128 g.
Edexcel 1MA1 · June 2022 · Paper 1H · Q9 · 3 marks
3 marks
Examiner report
This question was well answered with many students gaining at least two of the three marks. It was pleasing that most students worked with area and attempted to find the total surface area of the solid. Many students showed a correct process to find the total surface area of at least five faces for each solid and gained the first two marks. Those who showed a complete process to find the total surface area of the solid often gained full marks but some solutions were spoilt by arithmetic errors. The main stumbling block to a correct final answer was not dealing correctly with the parts of the cube and cuboid that are hidden. Many students worked out the total surface area of the cube and the total surface area of the cuboid and added them together, leading to the common incorrect answer of 310. Some added the total surface area of five faces of the cube to the total surface area of the cuboid and got an answer that was 16 cm2 too big. Students who made the mistake of including four 5 cm by 6 cm faces or four 7 cm by 5 cm faces in the surface area calculation for the cuboid gained one mark only. Some students did not read the question with enough care and worked with volume instead of with area. This question benefited from a systematic approach and there were some very good solutions from students who clearly identified what they were calculating. However, there were many solutions that had calculations dotted around the page making the working difficult to follow.
A cube of side 4 cm sits on top of a 7 cm by 6 cm by 5 cm cuboid, with its square base fully in contact.
Edexcel 1MA1 · June 2022 · Paper 1H · Q13 · 3 marks
3 marks
Examiner report
This was an unfamiliar type of question but even so it was not answered as well as might have been expected. Many students recognised that the question required the use of Pythagoras and wrote down Pythagoras’ theorem, often as it appears on the formula sheet. This alone was not sufficient for the first mark – the statement of Pythagoras needed to be linked in some way to the question. Many students did this by labelling the sides of the triangle in the diagram and using these letters in Pythagoras’ theorem, some students explained what the letters they used represented. Some students labelled the sides as a2, b2 and c2 and used these incorrectly or used the labels from the diagram to form their Pythagoras equation which was not worthy of any marks unless A, B and C were either clearly linked to the diagram or clearly defined as being the diameters of the semi-circles. It was very disappointing that relatively few students scored the second mark for writing correct expressions for the areas of at least two of the three semicircles. It was very common to see the diameters being used instead of the radii in 2r and for expressions to be given for the areas of circles instead of semicircles. Students who labelled the sides 2a, 2b and 2c found it easier to write correct expressions for the areas but harder to write a Pythagoras statement. Those who did write correct expressions for the areas were then often not able to complete the chain of reasoning to give a fully correct answer. Students should be encouraged to use brackets correctly. The omission of brackets was rarely recovered and resulted in a loss of marks. Students who used numerical values for the sides of the triangle were still able to access the first two marks but a surprising number of these students were unable to write correct expressions for the areas and gained the first mark only.
The diameters of three semicircles A, B and C form the sides of a right-angled triangle, with A on the hypotenuse.
Edexcel 1MA1 · June 2022 · Paper 1H · Q15 · 5 marks
5 marks
Examiner report
In part (a) the first mark was awarded for indicating that 15a + 20b is 5 times 3a + 4b and a good proportion of students managed to do this. Some students wrote “×5” between the two vectors but did not write a mathematical statement linking AB and AC . For the award of both marks it was necessary to see a statement such as AC = 5 × AB or 15a + 20b = 5(3a + 4b) and a correct reason. Many reasons mentioned multiples and some students also referred to the vectors being parallel or having a point in common. Some very good reasoning was seen but on the other hand a large number of students had difficulty formulating a correct reason or gave no reason at all. Some students worked with BC and AB and showed that BC = 4 × AB or 12a + 16b = 4(3a + 4b). A significant number of students, however, thought that simply working out BC as 12a + 16b was sufficient to show that A, B and C lie on a straight line. Incorrect working such as 15a ÷ 3a = 5a or (15a + 20b) ÷ (3a + 4b) = 5a + 5b was very common. In part (b) many students realised that they needed to find the vector DF and gained the first mark for 3e + 6f + –10.5e – 21f. Arithmetic errors during this first step were common. Many students then gave –7.5e –15f : 3e + 6f as the final answer and gained one mark only. Those students who knew that they needed to do more in order to find a ratio of lengths generally gained the second mark for a multiplicative relationship such as DF = –2.5 DE or, more commonly, for writing a ratio such as –7.5 : 3 or –2.5 : 1. In many cases the accuracy mark was lost because the final ratio contained a negative sign. Students could also work with DE and EF but students who took this approach were usually unsuccessful in gaining full marks.
Vectors AB=3a+4b and AC=15a+20b. Separately, collinear D,E,F have DE=3e+6f and EF=−10.5e−21f.
Edexcel 1MA1 · June 2022 · Paper 1H · Q21 · 5 marks
5 marks
Examiner report
This proved to be a challenging question with very few students showing any understanding of what was required to find the total area of the two shaded regions. Most students failed to find a successful strategy and working out was often messy and difficult for examiners to follow. Fully correct answers were seen only rarely but nevertheless it was pleasing to see some excellent solutions from the most able students taking this paper. The students who scored marks had generally drawn appropriate triangles on the diagram to help formulate their approach. Those who used 2 absinC to find the area of a triangle were usually able to recall the exact value of sin60o or sin120o. Some students made a good start and gained the first two marks by finding either the area of a triangle or the area of a sector but often they did not find both and were unable to make any further progress. It was pleasing to see many students attempting this question but often they simply found the area of one circle as 16or the area of three circles and then made incorrect assumptions about the proportion shaded in searching for a solution. Summary Based on their performance on this paper, students should: ● take care when carrying out arithmetic operations and check their working to avoid careless errors ● consider whether or not an answer is reasonable and of a sensible size ● read each question carefully and ensure that their final answer matches the question asked ● show their methods clearly when using graphs ● ensure that they know the difference between surface area and volume ● practise working out an estimate for the mean from a grouped frequency table ● practise answering probability questions that require the use of algebra and ratio
Three circles each have radius 4 cm. Centres A,B,C are collinear with AB=BC=4. The two shaded regions are the parts of the middle circle lying outside both the left and right circles.
Edexcel 1MA1 · June 2022 · Paper 2F · Q7 · 3 marks
3 marks
Examiner report
Part (a) was not well answered with only approximately half the cohort able to correctly name the shape. The most common incorrect answer was pentagon. The second part of the question was answered correctly more often than part (a). The vast majority of students were able to give a correct answer for the parallel line. The last part of this question was more challenging and less than half the cohort could effectively describe a perpendicular line. The students seemed to lack knowledge of the meaning of the word perpendicular rather than the ability to describe a line on the shape. Some blank scripts were seen for this part of the question even when letters had been used for a correct answer to part (b).
Polygon ABCDEF is drawn on a square grid. Taking the bottom-left grid intersection as (0,0), its vertices are A=(2,1), B=(2,4.5), C=(3.5,6), D=(6.5,6), E=(8,4.5), and F=(8,1), joined in order.
Edexcel 1MA1 · June 2022 · Paper 2F · Q8 · 4 marks
4 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2022 · Paper 2F · Q11 · 3 marks
3 marks
Examiner report
The modal mark for this question was two marks for calculating the angle as 39 degrees. However, many students lost the final communication mark for failing to give the correct geometric reason. A few students lost the communication mark by contradicting their working. For example, ‘Angles in a triangle add up to 360’. Students should be encouraged to give their geometric reasons as they work through the question and centres should emphasise the difference between giving reasons and showing working.
Triangle ABC has angle BAC=116° and ABC=25°. ACD and BCE are straight lines; x is angle ECD.
Edexcel 1MA1 · June 2022 · Paper 2F · Q22 · 2 marks
2 marks
Examiner report
Describing transformations is a standard type of question and should have been accessible to most students. However, the most common mark awarded was zero. Students did not answer this question well, with the most common response describing the movement of the shape using left, right, up and down commands rather than using a column vector and failing to state ‘translation’. Some students used incorrect transformations, such as rotation or reflection. If a vector was given, it was often incorrect using the inverse vector instead, implying that they had gone from T to S (direction of translated shape was switched). Relatively few students gave answers with multiple transformations which was pleasing to note. Centres could encourage students to use a path on the diagram provided to connect a corresponding vertex of each of the shapes, as this seemed to be a successful approach to identifying the correct vector, and to learn the correct terminology for transformations.
On a coordinate grid, shape S has vertices (1,−3), (4,−3), (4,−4), (3,−4), (3,−5), (1,−5). Shape T has corresponding vertices (−4,3), (−1,3), (−1,2), (−2,2), (−2,1), (−4,1). T is the image of S under a single transformation.
Edexcel 1MA1 · June 2022 · Paper 2F · Q24 · 5 marks
5 marks
Examiner report
It was pleasing to see that the many students were able to at least begin this multi-step problem involving area and proportion, with many gaining marks on this question. Most students were able to work out the area of the field at festival B and score the first mark, with many then continuing to use proportion correctly. Two or three marks were often awarded to students who were able to find the available area per person for each field, with many then continuing to complete the solution and gain full marks. However, a larger than expected number of students were unable to progress any further once they had found the area available at Festival B, with common mistakes being to subtract or divide the two areas rather than using proportionality. The final mark was occasionally lost due to arithmetic errors but the amount of working shown was pleasing to see. Part (b) of this question was demanding for most students. This type of question is a specification requirement and, as with previous years, the conversion of units of area proved not to be well known. For this example some students did have the notion that the squared term was involved but mistakenly wanted to square the 3 as well giving the popular incorrect value of 90000 or 9. Others said that you had to use 100 ×100 but said this was to multiply by 1000. This is a contradiction and as such so no marks were awarded. Students should be encouraged to check for contradictions in their explanations.
Festival A area 80,000 m² for 425 people. Festival B is 700 m by 2000 m for 6750 people.
Edexcel 1MA1 · June 2022 · Paper 2F · Q25 · 4 marks
4 marks
Examiner report
This was another question that was not well answered. The majority of students did not understand the need to work out distances using the coordinates given in the question. Both positive and negative values were accepted for these distances. There were many students who gave the answer of (11,17) implying that the differences of 7 and 8 were correctly calculated and added on to (4,9). However, in most of these responses, there was no working shown. When working was shown, students often tried to draw axes and plot points. The use of a diagram did allow some students to progress though this question and there were a small number of students who managed to find the y-coordinate.
L=(−3,1), M=(4,9), and LM:MN=2:3 on a straight line.
Edexcel 1MA1 · June 2022 · Paper 2H · Q2 · 2 marks
2 marks
Examiner report
This description of a translation was, unfortunately, not answered as well as would be expected. Many students did not properly read the question and assumed the transformation was going from left to right, and hence had the wrong vector of − . It was also disappointing to see how many students were unable to give the correct transformation of translation.
On a coordinate grid, shape S has vertices (1,−3), (4,−3), (4,−4), (3,−4), (3,−5), (1,−5). Shape T has corresponding vertices (−4,3), (−1,3), (−1,2), (−2,2), (−2,1), (−4,1). T is the image of S under a single transformation.
Edexcel 1MA1 · June 2022 · Paper 2H · Q4 · 5 marks
5 marks
Examiner report
Part (a) was the first problem question on the paper and was looking for students to compare the people per unit area at two festivals. To be able to do so students had to first find the area at one of the festivals. However, students could gain the second and third marks even if they were unsuccessful in finding this area. These two marks were for finding the area per person at one or both festivals. Part (b) required students to understand the mistake made in the unit conversion. To gain the mark students needed to explain that, for the conversion from cm2 to m2, the conversion factor is 10000 or 1002. This was done with varying degrees of success with a good number able to work with the correct conversion factor, often showing the correct conversion for 3m2 to cm2 or 300cm2 to m2. The most common incorrect answer was to show that 300 × 300 = 90,000.
Festival A area 80,000 m² for 425 people. Festival B is 700 m by 2000 m for 6750 people.
Edexcel 1MA1 · June 2022 · Paper 2H · Q5 · 4 marks
4 marks
Examiner report
This problem required students to work with coordinates and a ratio to determine the coordinates of another point. A good number of students were able to gain credit on this question, with many gaining all 4 marks. Those who didn’t score fully often only went as far as finding the vertical and horizontal distances, but then didn’t add these onto the coordinates of L or M. Another common error was to use the ratio incorrectly, typically dividing by 3 and multiplying by 2, as was finding the equation of the line LMN, which could still gain P1 for the values of 8 and 7 within the gradient.
L=(−3,1), M=(4,9), and LM:MN=2:3 on a straight line.
Edexcel 1MA1 · June 2022 · Paper 2H · Q13 · 2 marks
2 marks
Examiner report
A higher tier enlargement, and one generally completed well. Many were able to deal with the negative scale factor and correctly complete the transformation. There were a significant number of students who were unable to and either mistook the negative for a fractional enlargement and made the shape smaller or ignored the negative altogether and enlarged the shape by a scale factor of 2.
A quadrilateral with vertices (−2,−4),(−3,−2),(−5,−2),(−6,−4) is enlarged by scale factor −2 about (0,0).
Edexcel 1MA1 · June 2022 · Paper 2H · Q17 · 5 marks
5 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2022 · Paper 2H · Q18 · 4 marks
4 marks
Examiner report
This trigonometry question was again a familiar style and provided access to marks for a good number of students. It required students to first use the Cosine rule to find a missing length, and then the Sine rule to find another missing length. If a student didn’t score the first mark for the use of the Cosine Rule, they were then unable to gain any further credit. Many who scored this first mark, didn’t fully complete this process, often forgetting to square root. However, these students could then pick up the third mark (and therefore a total of 2 marks) for correctly using the Sine Rule for their incorrect value for QR. Common incorrect methods seen were attempts to use Pythagoras and SOHCAHTOA.
Triangles PQR and QRS share QR. PQ=11 cm, PR=9.4 cm, angle QPR=27°, angle QRS=88°, angle QSR=41°.
Edexcel 1MA1 · June 2022 · Paper 2H · Q20 · 4 marks
4 marks
Examiner report
This circle theorem question proved to be challenging to many. There were 3 routes through this problem. The most common start was to find BAD and then BCD before moving into triangle CDE. Alternatively, a number of students found BAD and then ABE using angles in a triangle before finding ADC. The third method was to find the reflex BOD before moving to BCD. All of these methods were seen regularly, and successfully. However, many students struggled with remembering their circle theorems and either thought opposite angles in a cyclic quadrilateral are equal, or incorrectly thought that OBCD was a cyclic quadrilateral. Reasons were frequently not accurately stated; correct geometric terms must be used.
A,B,C,D lie on a circle centre O. ADE and BCE are straight. Angle BOD=132° and angle CED=16°.
Edexcel 1MA1 · June 2022 · Paper 3F · Q6 · 2 marks
2 marks
Examiner report
Very few students scored 2 marks by correctly drawing a parallelogram. Students usually understood the definition of a quadrilateral (although there were a number of triangles, and polygons with more than 4 sides). Although students were able to draw a quadrilateral, a significant number were unable to use properties of quadrilaterals to draw a parallelogram. The majority of students who gained one mark for this question drew quadrilaterals without any lines of symmetry. Another common shape for 1 mark was a rectangle where students had understood the question but couldn’t find a shape that matched both criteria so went for one of the criteria to secure 1 mark. Where 0 marks were scored for a quadrilateral the most commonly seen were a square or an isosceles trapezium. From this question it was clear that students, on the whole, understood how to draw a shape with no lines of symmetry but it was the rotational symmetry that most did not understand.
On a square grid, draw a quadrilateral with no lines of symmetry and rotational symmetry of order 2.
Edexcel 1MA1 · June 2022 · Paper 3F · Q9 · 2 marks
2 marks
Examiner report
In part (a) most students were able to correctly give the number of faces of the triangular prism but sometimes a response of 6 was seen (maybe confusion with a cuboid). Part (b) was less successfully answered than part (a). The most common incorrect answer was 6, possibly due to the face that the diagram showed 2 triangles, 6 bold edges or 6 vertices.
A solid triangular prism.
Edexcel 1MA1 · June 2022 · Paper 3F · Q13 · 4 marks
4 marks
Examiner report
Part (a) was well attempted by most students who were able to measure accurately and convert to metres. It was pleasing to see that most showed the calculations required for the conversion. The majority of these went on to gain full marks correctly, adding the appropriate two distances and subtracting the other to find the difference in the overall total. There were a few who used their measurements to calculate the difference of 2 and then multiply by 150 which is a very efficient method. Those that did not measure accurately often went on to gain one mark for showing clear working for conversion and then the difference between the distances, but it was disappointing to find students using measurements that were more than 2 cm different to the diagram, leading to the suspicion that they either did not have a ruler, or did not know how to use it accurately. In part (b) the vast majority were unable to tackle this bearings question. Of those who did attempt this question it was common to see the bearing of C from A given. There were very few correct answers in the range with quite a few answers of 285 seen, which was just out of range.
Text-complete scale-map adaptation: on a coordinate grid measured in centimetres, plot C=(0,0), A=(−6.66,2.16) and B=(−3.58,−1.78), with north in the positive y-direction. Use scale 1 cm=150 m. Join A to B, B to C, and A to C.
Edexcel 1MA1 · June 2022 · Paper 3F · Q16 · 2 marks
2 marks
Examiner report
In most cases students knew what an isosceles triangle was and were able to draw one. However, students struggled with the concept of the base multiplied by the height is twice the area of the triangle; a common mistake was to draw an isosceles triangle which had an area of 6 cm2 or one of 24 cm2. Students need to be aware that accuracy of the drawing is important in this question: it was the drawing on the grid that was marked, and not working or labels that accompanied it.
On a centimetre grid, draw an isosceles triangle of area 12 cm².
Edexcel 1MA1 · June 2022 · Paper 3F · Q20 · 5 marks
5 marks
Examiner report
Full marks on this question were rare. It was clear that students knew a lot about angle rules and were able to apply some of this knowledge in the question, but unfortunately, they were frequently let down by their understanding of angle notation and their ability to accurately state the angle facts they were using. It was rare to see students using three letter angle notation so working out under the diagram was often likely correct but unable to score marks as it wasn't clear which angle a student was working out. Students who wrote their answers on the diagram tended to fare better. Students underestimated how to explain their reasons for their working out although those giving reasons usually gained one mark for “vertically opposite” or “opposite angles”. However, phrases like "opposite equal" and "triangles add up to 180" were common, missing the key words necessary to gain the communication marks. Reasons using angles on parallel lines were either often not given or were incorrect. Students who did not use an angle rule relating to parallel lines sometimes did 180 − 62 (PQR) and then divided their answer by 2 to incorrectly work out CQP. Others didn’t seem to understand which angle was actually CQR: it was not uncommon to see, using angles about point Q, the incorrect calculation 180 – 62 = 118.
PQR is isosceles with PQ=PR. APR and CQD are parallel; BPQ is straight; angle APB=56°.
Edexcel 1MA1 · June 2022 · Paper 3F · Q22 · 2 marks
2 marks
Examiner report
Many did not recognise the need to use Pythagoras’ theorem and therefore gained no credit. Some found the difference between the two sides given and produced an answer of 4.5 which attracted 0 marks. Some students wrote the difference of the squares the wrong way around, which gained no marks, unless they corrected this by giving the correct answer. A common incorrect method was for students to assume the perimeter was 20, and then subtract 8.5 and 4 to give an answer of 7.5, which could not be awarded any marks since it was from an incorrect method.
A right triangle has hypotenuse 8.5 cm and one leg 4 cm.
Edexcel 1MA1 · June 2022 · Paper 3F · Q28 · 3 marks
3 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2022 · Paper 3H · Q1 · 2 marks
2 marks
Examiner report
This question provided a good start to the paper. Most students gained the full 2 marks. However, a significant number of students did not apply Pythagoras’ rule correctly and instead added 4² and 8.5². These students could not be awarded any marks. It was surprising to see many students forget to take a square root and give the answer 56.25. A brief recap of the diagram might have alerted some students to there being an error in their working. In a small number of cases, trigonometry was attempted but the method was often incomplete and so no marks could be awarded.
A right triangle has hypotenuse 8.5 cm and one leg 4 cm.
Edexcel 1MA1 · June 2022 · Paper 3H · Q7 · 3 marks
3 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2022 · Paper 3H · Q12 · 3 marks
3 marks
Examiner report
In answering this question, most students showed a good understanding of how to apply trigonometrical ratios to this 2 stage problem. However, a significant proportion of them made errors in manipulating the ratio used in the second stage of the problem. It was envisaged that students would use the trigonometry of right angled triangles to solve this problem and most did. However, a large number of students used the sine rule to get their answer. Examiners were surprised how few students realised that triangle ACD was isosceles and so they were not able to write down the length of AC without further calculation. However, they usually used tan 45 = 𝐴𝐶 8 correctly to get 8 cm for the length of AC. Most students were then able to write down a relationship which would enable them to find the length of AB, and this was awarded some credit. However, too many of these students did not rearrange their equation correctly and instead of getting AB = sin20, wrote down AB = 8 × sin20, leading to an incorrect answer.
Right triangles ABC and ACD share AC; angles ACB and ACD are right angles. DC=8 cm, angle ADC=45°, angle ABC=20°.
Edexcel 1MA1 · June 2022 · Paper 3H · Q13 · 3 marks
3 marks
Examiner report
This question was successfully answered by just under a half of all students who took this paper. Nearly all students scored a mark for finding 3a as a column vector. However, students often failed to write their responses in a clear, logical manner. Few students manipulated the equation to get −2b = ( 8 17) − 3a or 2b = 3a − ( 8 17) before substituting for a. This might have led to a clearer approach and the avoidance of errors with signs. As it was, examiners were often left not knowing whether the student was trying to find the value of −2b or 2b with a consequent loss of marks for those students who did not get a fully correct final answer. Students who got one correct component of b were given the benefit of the doubt and rewarded with two of the three marks available.
a=(2,−3) and 3a−2b=(8,−17), as column vectors.
Edexcel 1MA1 · June 2022 · Paper 3H · Q15 · 3 marks
3 marks
Examiner report
There were some excellent complete, concise and clear proofs to this question involving the application of circle theorems. However, these were only seen in a minority of cases and students need to be aware of the need to state results clearly using the statements indicated on the mark scheme as guidance. There were many students who could not give reasons or who gave them with a lack of clarity. These students often gained some credit for matching up at least two pairs of equal angles. A clear indication of this on the diagram was accepted by examiners. Of students who completed the proof successfully, most used statements of the same segment theorem together with a pair of vertically opposite angles. Students should note it was not enough to state “same segment theorem” – they needed to say what the theorem states, for example “angles in the same segment are equal”. Some students used the result that angles in a triangle sum to 180° as a reason to support the equality of their third pair of angles. This was, of course, acceptable. A small number of students used the result that the angle at the centre of a circle is twice the angle at the circumference to show that angle PQX = angle PRS and that angle QPR = angle QSR, another acceptable approach. About one in ten students obtained full marks for their proof.
P,Q,R,S lie on a circle. Chords PR and SQ intersect at X, so P-X-R and S-X-Q are straight.
Edexcel 1MA1 · June 2022 · Paper 3H · Q18 · 5 marks
5 marks
Examiner report
It is encouraging to report that the majority of students gained some credit for their attempt at this question. The length of TC was often found successfully and a good proportion of students were also able to find the length of at least one of SD and TD. Only the higher attaining students were able to identify the angle which was required and put everything together in order to find the size of this angle, that is the size of angle DTS. Errors commonly seen included, substituting 12 and 14 incorrectly into the formula for the area of the cross section and making an error in the use of Pythagoras’ rule to find the length of DT.
Prism base ABCD is a 14 cm square. T lies on BC with BT:TC=4:3. Cross-section SRCD is a trapezium of area 147 cm² with CR=12 cm. ST rises from base point T to S above D.
Edexcel 1MA1 · November 2021 · Paper 1F · Q6 · 2 marks
2 marks
Examiner report
Part (b) was usually answered well; cube, cuboid and sphere being the most common errors. Part (a) less well done with ‘parallelogram’ seen far too often.
Two unlabelled diagrams are shown. (a) A four-sided shape has a longer horizontal top edge and a shorter horizontal bottom edge centred below it. Matching single-arrow marks on those two edges show they are parallel; the other two sides slope inward. (b) A 3-D solid is drawn with an elliptical top, a matching elliptical base, and two straight vertical side edges; the hidden back half of the base is dashed.
Edexcel 1MA1 · November 2021 · Paper 1F · Q16 · 4 marks
4 marks
Examiner report
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.
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.
Edexcel 1MA1 · November 2021 · Paper 1F · Q25 · 3 marks
3 marks
Examiner report
This question was not well answered. Many students gained some reward for dividing the volume (750) by the length (25) but were then unable to equate their result to the area of the cross-section. It was rare to find any formula given for the area of the cross-section or volume. An answer of 6 was awarded one mark for those students incorrectly working with a cuboid.
A prism has length 25 cm and volume 750 cm^3. Its constant cross-section is a right-angled triangle with base 5 cm.
Edexcel 1MA1 · November 2021 · Paper 1H · Q6 · 3 marks
3 marks
Examiner report
Many of the attempts to work out the height of the triangular prism failed because students treated the shape as a cuboid and worked with length × width × height. This meant that 6 was a very common incorrect answer. Some of the students who gained the first mark for dividing 750 by 25 to find the area of the cross section made no further progress but most went on to make an attempt at finding the height of the triangle. These attempts were often not successful because students failed to take into account the fact that the cross section was a triangle. Some students wrote an equation such as ½ × 5 × h × 25 = 750 and were awarded the first two marks but they were not always able to solve their equation correctly. For those students who had gained no process marks yet had only omitted division by 2 there was a special case mark for a final answer of 6.
A prism is 25 cm long and has volume 750 cm^3. Its cross-section is a right triangle with base 5 cm.
Edexcel 1MA1 · November 2021 · Paper 1H · Q7 · 4 marks
4 marks
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.
A cube has edge x cm. A sphere has radius 3 cm. Their surface areas are equal. Sphere surface area = 4πr^2.
Edexcel 1MA1 · November 2021 · Paper 1H · Q14 · 3 marks
3 marks
Examiner report
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.
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.
Edexcel 1MA1 · November 2021 · Paper 1H · Q18 · 5 marks
5 marks
Examiner report
This proved to be a challenging question and finding the length of AB proved to be beyond most students. Some of those who attempted the question identified that they need to find the perpendicular height of the trapezium and gained the first mark for writing a correct trigonometric statement such as sin30 = h/6. Some students could not recall the value of sin30 and did not complete the process to find the height. After finding the height many students struggled to make any further progress, failing to realise that they should use the formula for the area of a trapezium. It was common to see the trapezium split into two triangles and a rectangle and attempts at using Pythagoras. A good proportion of those that did substitute correctly into the formula went on to show a complete process to find the length of AB. A small number of students gave 5 as the final answer and lost the accuracy mark.
ABCD is a trapezium with parallel sides AB and CD. A sloping side of length 6 cm makes a 30° angle with a parallel side. Area=66 cm^2 and AB:CD=2:3.
Edexcel 1MA1 · November 2021 · Paper 2F · Q5 · 2 marks
2 marks
Examiner report
This question was well answered, with most students scoring the 2 marks available. The most common incorrect responses were either to draw a 6cm by 3cm rectangle, seemingly from counting the grid lines rather than the squares, or to draw either a 5cm by 1cm rectangle or a 5cm by 5cm square. When full marks were not awarded, a mark was often achieved for engaging with the scale given by either drawing a diagram with one dimension correct or by calculating both dimensions correctly.
A rectangle is 35 m by 20 m.
Edexcel 1MA1 · November 2021 · Paper 2F · Q11 · 2 marks
2 marks
Examiner report
Many students scored both marks on this question correctly drawing the reflection in the mirror line. Only a few students drew a correct reflection using an incorrect mirror line to score the method mark only. It was disappointing to see responses that scored no marks, the most common error was to attempt a rotation of the shape or a translation of the original shape to the bottom left-hand side of the grid. These incorrect transformations scored no marks.
On a square grid, a horizontal mirror line is shown. Triangle A has vertices at horizontal grid positions 1, 1 and 3, lying respectively 4, 3 and 2 squares above the mirror line.
Edexcel 1MA1 · November 2021 · Paper 2F · Q13 · 3 marks
3 marks
Examiner report
The majority of students scored both marks in part (i) calculating the angle to be 21 degrees. A small number of students used 360 in their calculation instead of 180 and this resulted in an incorrect answer of 201 degrees. The other error most often seen was arithmetic, again suggesting that students may have been reluctant to use their calculator for this question. Part (ii) was less well answered. Students who did score the communication mark often wrote ‘angles on a straight line add up to 180’. The mark was most commonly lost for missing out the word angles, other incorrect response included, ‘half a circle is 180’, here neither angle nor line were mentioned. However, the most common error was for students to state or re state their numerical calculation without mention of geometrical properties. Centres should encourage students to write short succinct explanations.
RST is a straight line. Three adjacent angles on the same side of the line are x°, 75° and 84°.
Edexcel 1MA1 · November 2021 · Paper 2F · Q25 · 4 marks
4 marks
Examiner report
This multi-stage problem proved challenging for the majority of students, with very few scoring any marks. A small proportion recognised that the application of Pythagoras’ Theorem was required, with a common incorrect method being to only perform addition or multiplication of the 2 lengths given. Partial marks were occasionally awarded for showing that squaring and adding the 2 side lengths was needed but many failed to continue correctly to square root to find the hypotenuse. Of those who were able to find the missing side length, it was rare to be able to break down the stages needed to find the total perimeter. The main challenge appeared to be recognising that one part of the perimeter was the difference between the length of the hypotenuse and the side of length 10cm. Some students found the total perimeter of one of the triangles and multiplied by 2.
A right triangle has perpendicular sides 8 mm and 10 mm. A shaded composite shape is made from two such triangles; its outside boundary lengths are 8, 8, 10, h and h - 10 mm, where h is the triangle's hypotenuse.
Edexcel 1MA1 · November 2021 · Paper 2F · Q26 · 4 marks
4 marks
Examiner report
Neither part of this question was well answered ad there were a disappointing number of blank responses. A few students indicated the recognition of the need to use trigonometric ratios. Of these, some wrote down SOHCAHTOA or drew “formula triangles” but were not able to select the most appropriate ratio to use. Students who correctly identified they needed to use the tangent ratio in part (a) normally scored both marks. When only 1 mark was awarded, the final mark was often lost due to poor application of the calculator for example the calculator was not in degree mode. The most common incorrect approach seen was to use the sum of the angles in a triangle by subtracting 90°and 34° from 180°, despite the question requiring a missing length to be calculated. Some attempted to use Pythagoras, squaring the 34 and the 12, mixing angles and side lengths was seen. Part (b) was very similar to part (a) in the quality of response. For those who realised the need to use trigonometry many did not identify the correct ratio to use or know that the inverse function was needed in order to find an angle. Although 18 = 0.83 was sometimes calculated and formed part of the working, this was insufficient for credit to be given as a correct equation involving cos x was required. Some students gave an answer of 34 which had evidently come from assuming the angles in the diagram from part (a) were the same as 90 - 56 was also seen as the working. Less often seen was an answer of 33 from adding the side lengths. Another common incorrect answer of 45 came from assuming the triangle is isosceles. Some attempted to use Pythagoras’s Theorem but were unable to complete the alternative approach far enough to be credit worthy.
ABC is right-angled at C, with AC=12 cm and angle A=56°. PQR is right-angled at Q, with hypotenuse PR=18 cm, adjacent side RQ=15 cm and angle R=x.
Edexcel 1MA1 · November 2021 · Paper 2H · Q5 · 4 marks
4 marks
Examiner report
The first real problem question on the paper and it was good to see the majority of students were able to spot the need to use Pythagoras’ Theorem to find the length of the hypotenuse. It was also really pleasing to see that once that had been identified, most were then able to score 2 marks for completing the process to find the missing length of the triangle. At this stage many understood the need to add the sides to find the perimeter, but a significant number were unable to complete this process successfully. A common misconception was to find the perimeter of one triangle and then double it, thus not dealing with the overlapping sides.
A right triangle has legs 8 mm and 10 mm. A composite made from two copies has boundary lengths 8,8,10,h,h-10, where h is the hypotenuse.
Edexcel 1MA1 · November 2021 · Paper 2H · Q6 · 4 marks
4 marks
Examiner report
It was evident that students had spent some time studying trigonometry, but again the learning did not seem to be secure. Many students were not able to identify the correct ratio for either part. Of those who did, many either substituted incorrectly, or they gained the first mark but were unable to gain the second as they struggled with the rearrangement. This was most evident in part (b) but was still seen in part (a). Of those who attempted an alternative method, a good number were again able to evidence understanding of Pythagoras’ Theorem, but were again unable to choose the correct ratio to gain the first mark.
ABC is right-angled at C with AC=12 cm and angle A=56°. PQR is right-angled at Q with hypotenuse PR=18 cm, adjacent side RQ=15 cm and angle R=x.
Edexcel 1MA1 · November 2021 · Paper 2H · Q11 · 5 marks
5 marks
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.
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.
Edexcel 1MA1 · November 2021 · Paper 2H · Q14 · 4 marks
4 marks
Examiner report
To access any marks on this question students had to understand the ratio and be able to use this with a circle theorem to find BCD or BAD. Unfortunately, many were not able to do this and thus scored no marks. Those who did, typically scored at least two marks as they normally stated the correct circle theorem. As is often the case, many students struggled to apply (or name) the ‘alternate segment theorem’ and therefore struggled to find the correct value for SBA. Some gained the second method mark for using angles in a triangle to find BDA or a small number recognised the relationship between BCD and SBD but this was rare. Very few were able to correctly find the required angle and provide both correct circle theorems.
A,B,C,D lie on a circle. SBT is tangent at B. Angle ABD=20°. Angle BAD:angle BCD=3:1.
Edexcel 1MA1 · November 2021 · Paper 2H · Q15 · 5 marks
5 marks
Examiner report
The use of more complex trigonometry proved very challenging for the bulk of the students. Many attempted to use the standard ratios, not realising they only apply to right-angles triangles. Another significant proportion had obviously studied these formulae but were unable to accurately quote them and/or substitute into them. This applied to both parts (a) and (b). In part (b) it was seen on a number of occasions, that students used the value found in part (a) in their substitution (normally 11.4 in place of 11) and therefore gained no marks. Although various alternative methods were seen in both parts, they were very rarely successful.
Triangle ABC has AB=8 cm, AC=11 cm and included angle A=72°.
Edexcel 1MA1 · November 2021 · Paper 2H · Q19 · 3 marks
3 marks
Examiner report
This ratio problem required students to work with scale factors and ratios between 3 spheres. The most common error was to work with the 27 and the ratio 1 : 2 and arrive at 54. The first mark was for the ability to cube root to find 5 and 3, and at this stage the values did not need to be given in a ratio. Many students who succeeded this far then got confused and failed to work with the value of 3 and the ratio 1 : 2. Of those who got to the values of 3 and 6, many then failed to link the 6 from sphere C and the 5 from sphere A. Even fewer then knew to square to find the ratio of surface areas. Quite a few saw the question as having to find the radius for the sphere and had difficulty rearranging the correct formula or even trying to use an incorrect formula.
Spheres A,B have volumes 125 and 27 cm^3. Radius B:radius C=1:2.
Edexcel 1MA1 · November 2021 · Paper 3F · Q5 · 1 mark
1 mark
Examiner report
A well answered question. It is not clear whether inaccurate marks were as a result of ruler misuse or from not having a ruler.
AB is a straight line segment.
Edexcel 1MA1 · November 2021 · Paper 3F · Q15 · 1 mark
1 mark
Examiner report
Many students gained the mark here, usually by explaining that the angles did not add up to 360. There were some confused answers, with some students believing that the angles could not be duplicated (eg there cannot be 2 angles of 23) or not making a choice and suggesting that the angles sum might be 180 or 360.
Jenna measures all the angles around a point as 23°, 145°, 23° and 69°.
Edexcel 1MA1 · November 2021 · Paper 3F · Q16 · 2 marks
2 marks
Examiner report
Students began to struggle from this point on in the paper. There were hardly any correct answers in this question. Those who attempted the question frequently gave “enlargement” as the transformation, but to gain marks this had to be accompanied by a further feature, and it was here that students had difficulty. The scale factor given was usually 3 instead of 4, and few regarded it necessary to give a centre point for the enlargement.
Triangle ABC has A=(1,1), B=(1,2), C=(2,1). Triangle ADE has A=(1,1), D=(1,5), E=(5,1).
Edexcel 1MA1 · November 2021 · Paper 3F · Q19 · 4 marks
4 marks
Examiner report
Most students found this question a challenge. A significant minority of students chose to estimate the length of the side of the shaded square, guessing it was about 8 or 9cm, with some even measuring it on the diagram. Of those who gained some marks, they chose either a solution using area calculation, or Pythagoras. The former approach was more successful; many gained a mark for 8 × 8 alone. Those who chose to use Pythagoras and remembered the relevant formula to use, usually went on to gain the first 2 marks, but then became unstuck, sometimes leaving their answer as √34 or 5.83. Those prematurely rounding their calculations when processing Pythagoras frequently lost the accuracy mark as their answer was then outside the required range. The easiest mark to obtain in this question was the independent units mark. Yet many failed to get it, either because they used cm as their unit, or most commonly because they failed to state units at all with their answer.
A large square has side 8 cm. A tilted square inside it leaves four congruent right triangles at the corners, each with perpendicular sides 3 cm and 5 cm.
Edexcel 1MA1 · November 2021 · Paper 3F · Q27 · 4 marks
4 marks
Examiner report
Few students were able to recall the formula for either the area of a triangle or the area of a circle, and as a result could not access any marks in this question. Predictably the most common incorrect answer for the triangle was 64. Students were more successful with the area of the circle. When the formula for the area of a circle was correct, they were normally successful although some forgot to square 8, some squaring the π and then multiplying by 8. A small number of students incorrectly used 16 as their radius or failed to divide the area of the circle by 4 to give a quarter circle. The right-angled triangle confused some students who used Pythagoras in an attempt to calculate the length of the hypotenuse rather than calculate the area.
OPR is a right-angled triangle with OP=OR=8 cm. The quarter-circle sector OPR has centre O and radius 8 cm. The segment between chord PR and arc PR is shaded.
Edexcel 1MA1 · November 2021 · Paper 3H · Q7 · 4 marks
4 marks
Examiner report
This question was done well by a majority of students who were awarded all 4 marks. Some students found the area of the triangle and/or the area of the quadrant but could not complete the solution to get full marks. It was encouraging to see that nearly all students could recall how to find the area of a circle.
A sector has centre O, OP=OR=8 cm and angle POR=90°. The chord PR divides the quarter-circle sector into right triangle OPR and the shaded segment between PR and arc PR.
Edexcel 1MA1 · November 2021 · Paper 3H · Q10 · 4 marks
4 marks
Examiner report
This question involving similar right-angled triangles was well answered by a minority of students. There were two main routes used by students who scored full marks. Some students used a scale factor together with Pythagoras’ rule while other students used trigonometry to find the size of angle B then went on to use trigonometry again to find the length of side CB. Of those students who could only be awarded partial credit for their attempts, most scored the mark for using Pythagoras’ rule correctly to find the length of EB. Unfortunately, they often then used a scale factor of 1.5 instead of 2.5 or failed to make any further progress.
Right triangles ACB and DEB share angle B, with C and E right angles. D lies on AB and E lies on CB. AD=9 cm, DB=6 cm and DE=2 cm.
Edexcel 1MA1 · November 2021 · Paper 3H · Q13 · 5 marks
5 marks
Examiner report
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.
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.
Edexcel 1MA1 · November 2021 · Paper 3H · Q18 · 4 marks
4 marks
Examiner report
There were many students who gained some credit for their answers to this question on vectors and a significant number of answers were fully correct and concisely presented. Of those students who gained some marks but who were not successful in getting a correct expression for 𝐷𝐸⃗⃗⃗⃗⃗ , many were awarded 2 marks for a correct expression for either 𝑂𝐷⃗⃗⃗⃗⃗⃗ , 𝐷𝐵⃗⃗⃗⃗⃗⃗ or 𝐵𝐸⃗⃗⃗⃗⃗ in terms of a and b. Less successful attempts at answering the question often involved the incorrect use of the ratios given, for example quarters and thirds were often seen instead of fifths in expressions for 𝑂𝐷⃗⃗⃗⃗⃗⃗ , 𝐷𝐵⃗⃗⃗⃗⃗⃗ and 𝐵𝐸⃗⃗⃗⃗⃗ .
OABC is a trapezium with AB parallel to OC. Vector OA=a, vector AB=b and vector OC=3b. D lies on OB with OD:DB=2:3. E lies on BC with BE:EC=1:4.
Edexcel 1MA1 · November 2021 · Paper 3H · Q23 · 4 marks
4 marks
Examiner report
There were a small number of excellent answers to this question which was aimed at the best students taking this paper. These students invariably use the sine rule together with the result that sin y = sin(180 − y) to prove the given result and examiners rarely saw the alternative approach using areas of triangles. An error seen in many responses was to treat the triangles ADB and ADC as similar or congruent triangles. This is clearly not the case. Many other students tried to use the sine rule on two different triangles and wrote down incorrect statements such as 𝐵𝐷 sin 𝑥 = 𝐷𝐶 sin 𝑥 In Summary Based on their performance on this paper, students should: • carry out a common sense check on the answers to calculations, so for example in question 12 you should expect 15 people to take less time than 12 people to clean the same number of cars. • practise less straightforward problems which involve working with ratio • read questions very carefully in order to fully understand what is being asked for • improve your skills in dealing with units of time and conversion between these units • take care in interpreting the scales used on the axes in questions which involve the use and interpretation of graphs, for example question 21(a) on this paper
ABC is a triangle. D lies on BC and AD bisects angle BAC, so angle BAD=angle DAC=x°.
Edexcel 1MA1 · November 2020 · Paper 1F · Q11 · 2 marks
2 marks
The mirror line is x = 0. A polygon has vertices (4, 5), (4, 1), (3, 1), (3, 2) and (2, 2), joined in that order.
Edexcel 1MA1 · November 2020 · Paper 1F · Q13 · 4 marks
4 marks
On an accurately drawn map, B is 4.53 cm north and 2.11 cm east of A. The map scale is 1:25 000.
Edexcel 1MA1 · November 2020 · Paper 1F · Q23 · 1 mark
1 mark
Triangle A has a 10 cm side between angles 55 degrees and 45 degrees. Triangle B has sides 8 cm and 10 cm, with the 45-degree angle between the 8 cm side and the third side. Triangle C has a 10 cm side opposite a 55-degree angle and an 8 cm side adjacent to that angle. Triangle D has a 10 cm side between angles 45 degrees and 55 degrees (its third angle is 80 degrees).
Edexcel 1MA1 · November 2020 · Paper 1F · Q25 · 5 marks
5 marks
ADC is a triangle. 63 degrees 148 degrees A B C D E AED and ABC are straight lines. EB is parallel to DC. Angle EBC = 148 degrees Angle ADC = 63 degrees Work out the size of angle EAB. You must give a reason for each stage of your working.
Edexcel 1MA1 · November 2020 · Paper 1F · Q27 · 3 marks
3 marks
A prism of height 3 m stands on a horizontal floor. Its volume is 18 m^3 and the pressure it exerts on the floor is 75 N/m^2. Use pressure = force/area.
Edexcel 1MA1 · November 2020 · Paper 1H · Q4 · 1 mark
1 mark
Triangle A has a 10 cm side between angles 55 degrees and 45 degrees. Triangle B has sides 8 cm and 10 cm, with the 45-degree angle between the 8 cm side and the third side. Triangle C has a 10 cm side opposite a 55-degree angle and an 8 cm side adjacent to that angle. Triangle D has a 10 cm side between angles 45 degrees and 55 degrees (its third angle is 80 degrees).
Edexcel 1MA1 · November 2020 · Paper 1H · Q6 · 5 marks
5 marks
ADC is a triangle. 63 degrees 148 degrees A B C D E AED and ABC are straight lines. EB is parallel to DC. Angle EBC = 148 degrees Angle ADC = 63 degrees Work out the size of angle EAB. You must give a reason for each stage of your working.
Edexcel 1MA1 · November 2020 · Paper 1H · Q8 · 3 marks
3 marks
A prism of height 3 m stands on a horizontal floor. Its volume is 18 m^3 and the pressure it exerts on the floor is 75 N/m^2. Use pressure = force/area.
Edexcel 1MA1 · November 2020 · Paper 1H · Q21 · 4 marks
4 marks
DEF is a triangle. D Q E F P P is the midpoint of FD. Q is the midpoint of DE. FD -> = a and FE -> = b Use a vector method to prove that PQ is parallel to FE.
Edexcel 1MA1 · November 2020 · Paper 1H · Q22 · 5 marks
5 marks
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.
Edexcel 1MA1 · November 2020 · Paper 2F · Q8 · 3 marks
3 marks
ABC is a straight line. The adjacent angles at B are x degrees and 150 degrees. A separate incorrect diagram shows a right angle and an angle labelled 280 degrees.
Edexcel 1MA1 · November 2020 · Paper 2F · Q11 · 3 marks
3 marks
Here is a cuboid. 10 cm 15 cm 4 cm Work out the volume of the cuboid.
Edexcel 1MA1 · November 2020 · Paper 2F · Q18 · 2 marks
2 marks
Triangle A has vertices (-5, 3), (-5, 5) and (-2, 3). Triangle B has vertices (0, -3), (3, -3) and (3, -5).
Edexcel 1MA1 · November 2020 · Paper 2F · Q25 · 2 marks
2 marks
ABC is right-angled at B, with angle ACB = 34 degrees, AC = 178 mm and AB = x mm. Find x to 1 decimal place.
Edexcel 1MA1 · November 2020 · Paper 2F · Q26 · 2 marks
2 marks
Given column vectors a = (3, 4) and b = (5, -2), find 2a - 3b.
Edexcel 1MA1 · November 2020 · Paper 2F · Q27 · 4 marks
4 marks
ABC is right-angled at B, with AC = 9 m and AB = 6 m. A quarter circle has centre C and radius CB.
Edexcel 1MA1 · November 2020 · Paper 2F · Q28 · 2 marks
2 marks
Each exterior angle of a regular polygon is 15 degrees Work out the number of sides of the polygon.
Edexcel 1MA1 · November 2020 · Paper 2H · Q5 · 2 marks
2 marks
ABC is right-angled at B, with angle ACB = 34 degrees, AC = 178 mm and AB = x mm. Find x to 1 decimal place.
Edexcel 1MA1 · November 2020 · Paper 2H · Q6 · 2 marks
2 marks
Given column vectors a = (3, 4) and b = (5, -2), find 2a - 3b.
Edexcel 1MA1 · November 2020 · Paper 2H · Q7 · 4 marks
4 marks
ABC is right-angled at B, with AC = 9 m and AB = 6 m. A quarter circle has centre C and radius CB.
Edexcel 1MA1 · November 2020 · Paper 2H · Q13 · 3 marks
3 marks
In triangle ABC, angle ACB = 34 degrees, angle ABC = 26 degrees and BC = 23.8 cm.
Edexcel 1MA1 · November 2020 · Paper 2H · Q15 · 2 marks
2 marks
Triangle A has vertices (3, 3), (5, 3) and (5, 5). Triangle B has vertices (-2, -2), (-5, -2) and (-5, -5).
Edexcel 1MA1 · November 2020 · Paper 2H · Q18 · 4 marks
4 marks
A cone has base diameter 20 cm and slant height 25 cm. A circle is drawn around the cone at a slant distance of 10 cm above the base; the curved surface above this circle is painted grey. Use curved surface area of a cone = pi*r*l, where r is the radius and l is the slant height.
Edexcel 1MA1 · November 2020 · Paper 3F · Q11 · 3 marks
3 marks
PQRS is a quadrilateral. PST is a straight line. Q R T S P 130 degrees 65 degrees y degrees 95 degrees Find the value of y. y = ................................................
Edexcel 1MA1 · November 2020 · Paper 3F · Q13 · 2 marks
2 marks
An L-shaped rectilinear shape has overall width 10 cm and height 7 cm. Its top horizontal side is 4 cm and its inner vertical side is 5 cm.
Edexcel 1MA1 · November 2020 · Paper 3F · Q17 · 2 marks
2 marks
Draw accurately an isosceles triangle with sides of length 8 cm, 6 cm and 6 cm. One side of the triangle has been drawn for you. 8 cm
Edexcel 1MA1 · November 2020 · Paper 3F · Q22 · 2 marks
2 marks
A square-based pyramid sits exactly on the 6 cm by 6 cm top face of a cube, with its square base aligned edge-for-edge with that face. The apex is vertically above the centre of the common square face and is joined to all four top corners. Every edge of the solid is 6 cm, including the four apex-to-corner edges.
Edexcel 1MA1 · November 2020 · Paper 3F · Q26 · 5 marks
5 marks
Maisie knows that she needs 3 kg of grass seed to make a rectangular lawn 5 m by 9 m. Grass seed is sold in 2 kg boxes. Maisie wants to make a rectangular lawn 10 m by 14 m. She has 5 boxes of grass seed.
Edexcel 1MA1 · November 2020 · Paper 3F · Q29 · 3 marks
3 marks
A solid triangular prism has a right-triangular cross-section with perpendicular sides 4 cm and 6 cm, length 10 cm, and density 0.8 g/cm^3.
Edexcel 1MA1 · November 2020 · Paper 3H · Q9 · 5 marks
5 marks
A prism is 20 cm long. Its cross-section is a rectangle 10 cm wide and 12 cm high, topped by an isosceles triangle of base 10 cm. Each base angle of the triangular roof is 40 degrees.
Edexcel 1MA1 · November 2020 · Paper 3H · Q11 · 4 marks
4 marks
Triangle P has vertices (1, 2), (1, 5) and (2, 5). It is rotated 180 degrees about (0, 0) to give triangle Q, then Q is translated by vector (5, -2) to give triangle R.
Edexcel 1MA1 · November 2020 · Paper 3H · Q14 · 4 marks
4 marks
B, D, E and F lie on a circle; ABC is tangent to the circle at B; angle BDF = 40 degrees and angle DEF = 100 degrees. Find angle ABD, giving a reason at each stage.
Edexcel 1MA1 · November 2020 · Paper 3H · Q18 · 4 marks
4 marks
In a cube, A and H are opposite vertices, so AH is a body diagonal through the cube. AH=11.3 cm correct to the nearest millimetre.
Edexcel 1MA1 · November 2020 · Paper 3H · Q19 · 4 marks
4 marks
ABCDEF is a regular hexagon with side length x. It is enlarged, centre F, by scale factor p to give regular hexagon FGHIJK. The shaded region is the part of the enlarged hexagon outside ABCDEF.
Edexcel 1MA1 · November 2019 · Paper 1F · Q9 · 2 marks
2 marks
Examiner report
Most students scored full marks for this question. The most common errors seen were either in the subtraction of 135 from 180 or in only subtracting 35 from 180. Students making an error with the subtraction could usually be awarded one mark. Perhaps it was surprising that students making errors rarely used a common-sense check to rule out the possibility of x being over 100°.
PQR is a straight line. P R Q x 35 degrees 100 degrees Work out the size of angle x. ....................................................... degrees
Edexcel 1MA1 · November 2019 · Paper 1F · Q23 · 2 marks
2 marks
Examiner report
There were few accurately constructed lines seen in answer to this question. More commonly, where students did gain some credit for their response, it was one mark for drawing a perpendicular line from the point P to the line CD within the tolerance allowed but without any evidence of it being constructed with a ruler and compasses. Some students gave an accurate construction of the perpendicular bisector of the line CD. They could not be awarded any marks.
Use a ruler and compasses to construct the line from the point P perpendicular to the line CD. You must show all construction lines. P C D
Edexcel 1MA1 · November 2019 · Paper 1F · Q24 · 4 marks
4 marks
Examiner report
This multi-step question was not answered well. Many students mistakenly used the result that “angles on a straight line sum to 180̊” with angles ABC, BDC and BAC to get an answer of 54̊ for the size of angle BDC. Of those students who did make a correct first step to find that angle BCA was 54°, few went on to split the angle correctly in the ratio 2 : 1 and then to complete the problem successfully.
In triangle ABC, D lies on AB between A and B, so A, D and B are collinear in that order. Angle BAC=75 degrees and angle ABC=51 degrees. The internal line CD splits angle ACB so that angle DCB : angle ACD = 2 : 1.
Edexcel 1MA1 · November 2019 · Paper 1F · Q27 · 5 marks
5 marks
Examiner report
Fully correct answers to this question were hardly ever seen. Only about one third of students were able to get at least one mark for drawing a correct bearing and/or for calculating the distance travelled by the boat and using the scale of the diagram. Examiners were able to award three marks to students who plotted the position of Q accurately on the diagram. Those who did plot Q correctly were often able to use the scale to give an accurate distance from L to Q but were unable to measure the bearing, often giving the angle measured anticlockwise from the north line. Weaker students often started their working by multiplying 90 by 12 but seemingly with little purpose evident. There was a disappointing number of students who measured the length of one or more of the north lines on the diagram.
A scale drawing uses 1 cm to represent 4 km. On it, lighthouse L is 7.5 cm east and 2.7 cm south of port P. Aleena sails from P on a bearing of 070 degrees for 1.5 hours at 12 km/h, arriving at port Q.
Edexcel 1MA1 · November 2019 · Paper 1F · Q28 · 3 marks
3 marks
Examiner report
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.
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.
Edexcel 1MA1 · November 2019 · Paper 1F · Q29 · 4 marks
4 marks
Examiner report
Answers to part (a) of this question revealed several approaches. A number of students showed a good understanding of similarity and gave a correct answer which scored full marks, but many students used differences leading to the incorrect working 15 − 10 = 5, 9 − 5 = 4. They could not be awarded any credit. Some students made little headway with the question. A minority of students obtained one mark for finding a relevant scale factor but were then unable to complete the question successfully. A greater proportion of students gained some credit for their responses to part (b) than in part (a). There were a good number of fully correct answers. Where two marks were not scored, students often made some progress, for example by marking the lengths 4cm and 10 cm for HG and HK respectively and then working out that the length of GK was 6cm, but getting no further. It was not uncommon to see EG marked on the diagram as 10 cm and/or FG marked in as 6 cm, leading to an incorrect answer of 4 cm for the length of EF. Summary Based on their performance on this paper, students should: consider carefully the facts concerning angles, for example when “angles on a line add to 180” applies and when it does not practice solving problems using the mean, paying particular attention to how many pieces of data are to be added before dividing to find the mean carry out a common sense check on the answers to calculations, so for example you should expect the angle x in question 9 to be less than 90° check all calculations for arithmetic errors particularly when completing a paper where the use of a calculator is not allowed.
ABC and PQR are similar right-angled triangles. C P A 9 cm 15 cm 10 cm B R Q angle ABC = angle PQR
Edexcel 1MA1 · November 2019 · Paper 1H · Q4 · 2 marks
2 marks
Examiner report
Many students did not understand the meaning of perpendicular and were therefore unable to score on this question. Of those who did, the perpendicular was often simply drawn with no sensible attempt at construction; stray arcs were often seen. It was pleasing, however, to see a significant number of students gaining full marks, usually with arcs centre D radius DP and centre C radius CP. Quite often the construction of a perpendicular bisector of CD was seen; this gained no marks. A significant number of students seemed not to understand the definition of perpendicular.
Use a ruler and compasses to construct the line from the point P perpendicular to the line CD. You must show all construction lines. P C D
Edexcel 1MA1 · November 2019 · Paper 1H · Q5 · 4 marks
4 marks
Examiner report
180 – 75 – 51 = 54 was the common first step seen but many saw this as a method to find the required angle BDC incorrectly using ‘angles on a straight line’. Some correctly found angle ACB = 54o but could go no further and for those that did arithmetic errors were often seen preventing a complete and correct solution. Some students were unable to deal with the ratio and often 54 was simply halved. Others assumed that angle ADC or BDC was 90 degrees.
In triangle ABC, D lies on AB between A and B, so A, D and B are collinear in that order. Angle BAC=75 degrees and angle ABC=51 degrees. The internal line CD splits angle ACB so that angle DCB : angle ACD = 2 : 1.
Edexcel 1MA1 · November 2019 · Paper 1H · Q8 · 5 marks
5 marks
Examiner report
It was rare to see a fully correct solution to this question. Bearings remains a topic which many students find difficult to understand. Those students able to correctly draw a bearing of 070o usually went on to score well. The ability to use distance = speed × time did enable many students to gain at least one mark for a distance of 18 km and often an additional mark for correct use of scale. When a correct position for Q was found, the distance from L was usually correct but the bearing of Q from L, a reflex angle, was usually not correct.
A scale drawing uses 1 cm to represent 4 km. On it, lighthouse L is 7.5 cm east and 2.7 cm south of port P. Aleena sails from P on a bearing of 070 degrees for 1.5 hours at 12 km/h, arriving at port Q.
Edexcel 1MA1 · November 2019 · Paper 1H · Q14 · 4 marks
4 marks
Examiner report
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.
A right-angled triangle has perpendicular legs of lengths (x-2) cm and (x+4) cm. Its area is 27.5 cm^2.
Edexcel 1MA1 · November 2019 · Paper 1H · Q17 · 4 marks
4 marks
Examiner report
This question was very poorly answered, many simply using a scale factor of 2.25 (9 ÷ 4) in an attempt to make progress. Some looked as though they wanted to use the formulae for surface area and volume of a cylinder, but never really made any meaningful start. A few students correctly found the ratio of lengths but could go no further.
A and B are two similar cylindrical containers. A B the surface area of container A : the surface area of container B = 4 : 9 Tyler fills container A with water. She then pours all the water into container B. Tyler repeats this and stops when container B is full of water. Work out the number of times that Tyler fills container A with water. You must show all your working.
Edexcel 1MA1 · November 2019 · Paper 1H · Q22 · 4 marks
4 marks
Examiner report
Fully correct solutions were very rare indeed. Equal pairs of angles were sometimes identified but often with no reasons given. The most common award of one mark was for showing triangles AED and EBC to be equilateral with all angles of 60o shown. One mark for BC being common to both triangles was also awarded on a number of occasions. Summary Based on their performance on this paper, students should: read questions carefully. ensure that they know the difference between factors and multiples. re-visit the four rules of fractions. understand that constructions and scale drawings require accuracy of measurement. take care in interpreting the scale on the axes in graphical questions. ensure that specific questions posed in problems are answered, eg. “Is Donna correct” requires an answer of “yes” or “no” together with full explanation.
A, B, C and D lie on a circle. Chords AC and DB intersect at E, so A, E, C are collinear and D, E, B are collinear. Triangle AED is equilateral.
Edexcel 1MA1 · November 2019 · Paper 2F · Q13 · 2 marks
2 marks
Examiner report
Drawing a horizontal line and a vertical line both 6 squares in length should have been a straightforward start to the enlargement but a surprising number of students failed to gain at least one mark for drawing two of the sides correctly. Many of those who did make a correct start could not complete the enlargement correctly as the diagonal lines were often incorrectly drawn. It was noted that some students did not appear to have a ruler as their answer to this question was drawn freehand.
Shape X is a quadrilateral on a square grid. Taking its bottom-left vertex as (0,0), its vertices in order are (0,0), (0,2), (3,3), (2,0).
Edexcel 1MA1 · November 2019 · Paper 2F · Q17 · 4 marks
4 marks
Examiner report
This question was not answered as well as might have been expected. Many students could not make the correct first step and identify the size of angle ADB as 64º. A common mistake was to identify the two equal angles in triangle ADB as angles DAB and ABD or as angles ABD and ADB. Those who did find ADB = 64º often failed to make any further progress. Even some of those who got as far as finding that angle ABD = 52º and angle DBC = 128º could not complete the method to find the size of the angle marked x. Students that annotated the diagram tended to be more successful than those that tried a more formal labelling angles approach. It was disappointing that relatively few students achieved the C mark for giving two correct reasons appropriate to their method. When reasons were given they were often incomplete and did not include a reference to angles. Reasons such as ‘triangles add up to 180º’ or ‘a straight line is 180º’ are insufficient. Many gave no reasons at all or incorrect reasons based on parallel lines. A significant number of students misunderstood ‘angles on a straight line add to 180˚’ and applied it to angles that were at separate points along the line ABC.
ABC is a straight line. Point D is joined to A, B and C. AB = BC = BD. Angle DAB = 64°. Angle DCB is x.
Edexcel 1MA1 · November 2019 · Paper 2F · Q29 · 5 marks
5 marks
Examiner report
This question proved to be beyond most students on this Foundation tier paper and it was frequently not attempted. Many that did make an attempt were unable to do anything other than find the volume of the cuboid. On its own, this was insufficient to gain any marks. Many of the students who appreciated the need to work out the surface area of the cuboid could not show a complete process to do so. A common mistake was to include either four 6 by 18 faces or four 8 by 18 faces in the surface area calculation. Relatively few students were able to progress beyond a surface area calculation to find the side length of the cube but it was pleasing that some who did so were able to carry on and give a fully correct solution.
A cuboid measures 18 cm by 8 cm by 6 cm. A cube has the same total surface area as the cuboid. Janet says the cube and cuboid have equal volumes.
Edexcel 1MA1 · November 2019 · Paper 2F · Q30 · 3 marks
3 marks
Examiner report
This question was answered poorly with relatively few students showing an understanding of vector arithmetic. Some students scored the first method mark for 5 – 2 × 3 or 2 – 2 × – 1. When 5 was seen it was often not simplified correctly with errors usually occurring in the y component. Students that scored the second mark for simplifying to were rarely able to achieve the final mark for drawing the vector. Some students attempted to find the vector a – 2b by drawing but most could not manage even the first step. At all stages of the question the drawing of vectors was extremely poor and very few students drew any kind of correct vector. Summary Based on their performance on this paper, students should: ● practise working with time, converting from minutes to hours and vice versa ● learn how to read and interpret scales on graphs, particularly those where 1 small square is not 1 or 0.1 ● ensure that they know how to use their calculator ● practise finding missing angles with isosceles triangles in various orientations ● understand the difference between simple interest and compound interest ● practise working with algebra in formulating expressions or formulae and in substituting into formulae ● practise using and drawing column vectors
Two column vectors are a = (5, 2) and b = (3, -1).
Edexcel 1MA1 · November 2019 · Paper 2H · Q9 · 5 marks
5 marks
Examiner report
It was disappointing to find that so many students did not know how to find the surface area. Many only found volumes, which resulted in no marks being awarded. Of those who attempted to find the surface area, many found two of 6×8 and 4 of 6×18, but this still earned credit once they showed that they then wanted to add. Any remaining credit was only awarded for correctly processing their total surface area figure, usually by dividing their total surfaced area by 6 and then taking the square root to get their volume of the cube. Some students assumed that it was the volumes that were equal or that the cube was half the cuboid.
A cuboid is 18 cm by 8 cm by 6 cm. A cube has the same total surface area. Janet claims their volumes are equal.
Edexcel 1MA1 · November 2019 · Paper 2H · Q12 · 4 marks
4 marks
Examiner report
This was not well done with many not attempting any trigonometry. Those that did often got the first mark for the process of making a correct start to trigonometry, but then spoilt their working. Others, having calculated 5.52… went on to use Pythagoras’s Theorem but then lost their way. It was not uncommon to see students using 5.52.. as the length of ED. Quite a few students over complicated the process of solution by using the sine rule or even cosine rule in an attempt to solve, with few of these showing any success due to error in manipulation. Some students, who showed a completely correct process, rounded or truncated their interim values to the extent that their final answer was outside the range required for the final mark.
ADC and DEB are straight lines. AD = 4.4 cm, BC = 8.6 cm, E is midpoint of DB, angle CDB = 90°, and angle DCB = 40°.
Edexcel 1MA1 · November 2019 · Paper 2H · Q17 · 3 marks
3 marks
Examiner report
Students should be encouraged to mark the angles on the diagram. Many got one mark for identifying the 90˚ angle. It was disappointing to find errors were then made with basic geometry to find the angle OAB. Some students found the angle 58˚ at the top of the triangle, but failed to recognise the other triangle was isosceles, or incorrectly tried to apply the alternate segment theorem and made one of the angles in the triangle 32˚, which then led to an incorrect final answer.
A and B lie on a circle with centre O. CAD is tangent at A; B, O, D are collinear; angle ODA = 32°.
Edexcel 1MA1 · November 2019 · Paper 2H · Q19 · 2 marks
2 marks
Examiner report
Many students did not know the meaning of the word “hemisphere”, or failed to divide by 2 to find the volume of half the sphere. Too many substituted the figure for the diameter rather than the radius. Some cubed rather than squaring, even though the formula was given.
A hemisphere has diameter 8.4 cm. Use volume of sphere = (4/3)pi r^3.
Edexcel 1MA1 · November 2019 · Paper 2H · Q23 · 6 marks
6 marks
Examiner report
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.
A parallelogram has adjacent sides (2x - 1) cm and (10 - x) cm with included angle 150°. Its area is greater than 15 cm^2.
Edexcel 1MA1 · November 2019 · Paper 2H · Q24 · 2 marks
2 marks
Examiner report
A very challenging question for the vast majority. Many were able to gain one mark for stating a correct co-ordinate, the most popular of which were (-1,0) or (-1,2). There were many who left out the direction in their description and some confused their description by using enlargement, translation and vectors instead of co-ordinates.
Square ABCD has A(1,4), B(3,4), C(3,2), D(1,2). It is reflected in x=-1 then rotated; two original vertices are invariant.
Edexcel 1MA1 · November 2019 · Paper 2H · Q25 · 5 marks
5 marks
Examiner report
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.
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.
Edexcel 1MA1 · November 2019 · Paper 3F · Q10 · 2 marks
2 marks
Examiner report
Part (a) required students to draw a diameter on a circle, which the vast majority were able to do. The most common incorrect response was to draw a radius rather than a diameter, or to extend the diameter beyond the circumference of the circle. In part (b) the question demanded the student draw a segment. A significant proportion of students made the mistake of drawing a sector instead.
Two blank circles are supplied.
Edexcel 1MA1 · November 2019 · Paper 3F · Q14 · 2 marks
2 marks
Examiner report
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.
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.
Edexcel 1MA1 · November 2019 · Paper 3F · Q18 · 2 marks
2 marks
Examiner report
It was disappointing to see so many students unable to fully complete this question. Many students were able to complete a reflection, but many were either unable to draw the mirror line, or just assumed it was something else. Common mirror lines seen were x = 3, x = 2.5, x = 0 and also y = 2.5. There were also a significant number of students carrying out rotations and perhaps even more commonly, translations.
Triangle P has vertices (1,4),(1,7),(2,4) on a coordinate grid.
Edexcel 1MA1 · November 2019 · Paper 3F · Q29 · 4 marks
4 marks
Examiner report
This geometry problem was broken into three steps. A process to use angles in a parallelogram (or parallel lines) to find an angle of 63. The second was to use interior or exterior angles of a pentagon, and then finally to combine these as a process to find x. With the first two steps, if there was any contradiction by mis-assigning the correct value to an incorrect angle either in working or on the diagram, the mark was not awarded. Many students were able to gain some credit, more typically for the 63, but often no more. Those who knew how to calculate interior or exterior angles, often got confused and stated them incorrectly, and as a result gained no credit. It is clear though that many students have no understanding of how to find angles within polygons. Students should also be reminded that diagrams are not drawn to scale and also be discouraged from making assumptions based on the diagram such as x being equal to the angle in the parallelogram or that x was an angle inside an isosceles triangle if a line was extended from the parallelogram.
A regular pentagon and parallelogram share a bottom vertex. The parallelogram's upper-left interior angle is 117°; x is between its left side and the pentagon's adjacent side at the shared vertex.
Edexcel 1MA1 · November 2019 · Paper 3F · Q30 · 3 marks
3 marks
Examiner report
The final problem on the paper was one that many students were able to attempt, and in many cases gain some credit. The problem could be approached in various ways. Students could either start at one shape, find its area, use the scaling factor of 9 and then calculate the radius of the other shape. This could be done starting at either A or B. However, most students started with one shape, found its area and then the other area using the scaling factor of 9. Rather than then find the radius, students re- started with the second shape and found the area and showed these were the same. Both ways were equally acceptable and a decent proportion of correct responses were seen. Of those who didn’t gain full marks the common error was to fail to divide by 4 when finding the area of A. It was pleasing to note that very few confused formulae for area and circumference and used the correct formula in their work. Summary Based on their performance on this paper, students should: be encouraged to use their calculators to check basic calculations. remember to keep working their working inside the boxes provided or use an extra sheet to ensure working worthy of credit is seen. remember to use their calculator in percentage questions. ensure that they are able to identify between the different types of transformations. show calculations for each stage of working carried out. read questions carefully to ensure that they have used all of the information provided have further practise at questions involving a passage of time and resulting compound unit calculations ensure that they have set out their work clearly so that stages of their calculations can be easily found and used.
Shape A is a quarter-circle of radius 15 cm. Shape B is a circle. Area(A)=9×Area(B).
Edexcel 1MA1 · November 2019 · Paper 3H · Q8 · 4 marks
4 marks
A regular pentagon and parallelogram share a vertex. The parallelogram upper-left interior angle is 117°; x lies between its left side and the pentagon's adjacent side at the shared vertex.
Edexcel 1MA1 · November 2019 · Paper 3H · Q9 · 2 marks
2 marks
Triangle A has vertices (1,1),(1,3),(2,3) on a coordinate grid.
Edexcel 1MA1 · November 2019 · Paper 3H · Q13 · 4 marks
4 marks
Liquids A:B are mixed 2:13 by volume. Densities are 1.21 and 1.02 g/cm^3. A cylinder of mixture has radius 3 cm and height 25 cm.
Edexcel 1MA1 · November 2019 · Paper 3H · Q18 · 5 marks
5 marks
Triangle ABC has AB=3.4 cm, AC=6.2 cm, BC=6.1 cm. D lies on BC and angle DAC=(2/5) angle BCA.
Edexcel 1MA1 · November 2019 · Paper 3H · Q23 · 5 marks
5 marks
A sector OACB is rolled into a hollow cone by joining OA and OB; arc midpoint C maps consistently. Cone volume=56.8 cm^3, height=3.6 cm. Use V=(1/3)pi r^2h and curved area=pi rl.
Edexcel 1MA1 · November 2019 · Paper 3H · Q24 · 5 marks
5 marks
OXYZ is a parallelogram with vector OX = a and vector OY = b. P lies on OX with OP:PX = 1:2. R lies on OY with OR:RY = 1:3.
Edexcel 1MA1 · June 2019 · Paper 1F · Q12 · 5 marks
5 marks
Examiner report
Part (a) was generally answered well with the majority of students recognising that angle ABC is 90 degrees, and the others must be subtracted from it. When this method was adopted it usually scored 2 marks, with the odd exception where a calculation error occurred. Very few students created an equation to help solve this question, preferring normally to state 40 without any calculation Those not scoring any marks generally used 180 degrees or, less frequently, 360 degrees and attempted to subtract the other angles from this. Some added 25 to 25 to get 50, and gave 50 as the answer. Encouragingly, very few students attempted to measure the angle with a protractor. For part (b)(i) most students scored 1 mark or better. Recognition of the correct angle (b or d) was good. A small proportion of students attempted to label the angles using ‘PQ’ rather than ‘b’ or ‘d’, so scored 0 marks. ‘Parallel lines’ as a reason was a common misconception, along with naming an incorrect relationship such as alternate angles, and simply writing ‘opposite’ rather than ‘vertically opposite’ or ‘opposite angles’ On the whole (b)(ii) was answered well. Marks were lost when students calculated the remaining angles, but then failed show a link to 360 degrees, for example some students just stated the size of each of the angles a, b, c – but gave no reasons for their sizes. Others lost the mark when arithmetic errors led to incorrect values being given. It is worth centres noting that this is a communication mark and so the answer must not be contradicted as this will result in the mark being withheld. Students should be encouraged to give concise reasons.
AB and BC are perpendicular. At B, three adjacent angles from BA clockwise to BC are 25°, x° and 25°. Separately, parallel horizontal lines RS and TU are cut by the straight line PQ, which descends from P at the upper left to Q at the lower right. At the intersection with RS, the four angles clockwise from the upper-left quadrant are labelled a, 125°, c and b. At the intersection with TU, the upper-right angle is d and the lower-right angle is e.
Edexcel 1MA1 · June 2019 · Paper 1F · Q18 · 4 marks
4 marks
Examiner report
Many fully correct justifications were seen. There were some well organised attempts clearly earning three method marks where an arithmetic error was obviously the only mistake. The most commonly used starting points were either to find the area of the rectangle or the number of paving stones needed to cover a side of the rectangle. The second step then caused difficulty, the area of the rectangle was often divided by the length of a stone, rather than the area of one stone, or the number of stones on each side was added rather than multiplied. Students using the second approach were more successful if they worked out the cost of covering one length eg 20× (£)2.50, then said they needed 4 rows like this to find £200. This kind of approach using the diagram and drawing in the rows seemed to add scaffolding for many students. Some students tried to work with perimeter which is not a correct first step and so no marks could be awarded.
A rectangular path is 600 cm by 120 cm. Square paving stones have side 30 cm, cost £2.50 each; budget £220.
Edexcel 1MA1 · June 2019 · Paper 1F · Q25 · 2 marks
2 marks
Examiner report
Most students made an attempt at this drawing question, but some just tried to put the plan, side and front elevations into one figure. Dimensions, when included, were usually correct, but many cylinders did not have the dimensions shown, which meant the accuracy mark was often lost. Students should be encouraged to re-read questions to check they have satisfied the full requirements.
Plan/front/side elevations describe a cylinder of radius 2 cm (diameter 4 cm) and height 5 cm.
Edexcel 1MA1 · June 2019 · Paper 1F · Q26 · 3 marks
3 marks
Examiner report
This question on the whole was not very well answered, as most students did not correctly reflect shape A in the x-axis. Many students had answers of -6 and -5 (seen in either order) which does not take the reflection into account, but is a translation vector for (3,2) to (-3,-3). Students who drew the reflection on the diagram were often able to achieve two or three marks, but it is clear that students need to develop their skills in multiple step transformations, clearly showing each transformation one step at a time.
Triangle A vertices (3,2),(5,2),(3,5). Reflect in x-axis then translate by (c,d) to triangle B vertices (-3,-3),(-1,-3),(-3,-6).
Edexcel 1MA1 · June 2019 · Paper 1F · Q28 · 4 marks
4 marks
Examiner report
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.
Rectangles ABCD and PQRS: QR=10 cm; BC=PQ; perimeter ABCD=26 cm; area PQRS=45 cm^2.
Edexcel 1MA1 · June 2019 · Paper 1H · Q4 · 2 marks
2 marks
Examiner report
Having recognised that the solid shape is a cylinder the challenge for some students was to draw a sketch of a cylinder. Many students did draw an acceptable sketch and those who gave the correct dimensions of the solid on their sketch gained both marks. Some students did not include any dimensions on their sketch or made a mistake with one or both dimensions, giving the height as 4 cm instead of 5 cm for example, and gained one mark only. Labels such as “plan”, “front” and “side” were often seen on the sketches instead of the dimensions. A common mistake was to draw a shape that, at first glance, looked like a cylinder but with an extra face drawn perpendicular to the front face. These types of sketches gained 0 marks.
The plan, front elevation and side elevation describe a cylinder of radius 2 cm (diameter 4 cm) and height 5 cm.
Edexcel 1MA1 · June 2019 · Paper 1H · Q5 · 3 marks
3 marks
Examiner report
Most of the students who gave the correct values of c and d had first drawn the reflection of shape A in the x-axis. Those who did not give two correct values often gained two of the three marks for either c = –6 or d = –1 or for transposing the two correct values. One correct value, either c = –6 or d = –1, scored two marks regardless of the second value or the diagram. A surprising number of students with neither value correct spurned the opportunity of scoring a relatively easy mark by failing to draw the reflection of shape A in the x-axis, sometimes drawing the reflection in the y-axis or the reflection of shape B in the x-axis but often drawing no reflection at all. Some students drew lines between the corresponding vertices of the two triangles, clearly confused with which type of transformation they were dealing with.
Triangle A has vertices (3,2), (5,2), (3,5). It is reflected in the x-axis, then translated by vector (c,d), to triangle B with vertices (-3,-3), (-1,-3), (-3,-6).
Edexcel 1MA1 · June 2019 · Paper 1H · Q7 · 4 marks
4 marks
Examiner report
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.
Rectangles ABCD and PQRS satisfy QR=10 cm, BC=PQ, perimeter ABCD=26 cm and area PQRS=45 cm^2.
Edexcel 1MA1 · June 2019 · Paper 1H · Q14 · 2 marks
2 marks
Examiner report
Students who gave the exact value of either tan30º or sin60º gained the first mark. Various strategies were used to find these values. Some students worked them out by drawing an appropriate triangle; some used patterns in a table with angles of 0º, 30º, 45º, 60º and 90º and some simply recalled them from memory. Sometimes sin30º ÷ cos30º was used to find the value of tan30º. However, the strategies used were not always successful and incorrect values were very common. When only one of the two values was correct it was more often sin60º = 3 . In some responses the values were not labelled. If only one value was correct examiners had to assume from the calculation shown that the order was tan30º×sin60º. Some of the students who got to 1 × 3 did not go on to get full marks because they gave 3 as the final answer or simplified it to 2. The question required the answer to be given in its simplest form. Some students used incorrect values such as tan30º = 1 and sin60º = 0.5 which just happened to give an answer of 0.5. When the “correct” answer clearly came from incorrect working a mark of 0 was awarded.
Find the exact value of tan 30° × sin 60°.
Edexcel 1MA1 · June 2019 · Paper 1H · Q15 · 4 marks
4 marks
Examiner report
Most students were able to gain the first method mark for using a volume formula with the correct substitution. A common mistake at this stage was to use a radius of 6 cm instead of 3 cm. It was also not uncommon to see 32 instead of 33 substituted into the formula for the volume of a sphere. Having substituted correctly many students failed to gain the second method mark because they found the total volume of a cone and a sphere, not the total volume of a cone and a hemisphere. However, many of these students were still able to gain the third method mark for simplifying the volume of the cone to 30π. Incorrect answers of 66 or 66π, from adding 30π and the volume of the sphere were very common. Many students struggled to simplify their expressions for volume because they had difficulty dealing with the fractions within the calculations. Some left the volume of the cone as 1/3π × 90 and 4/3×27 and 1/2×4/3×27 were often evaluated incorrectly. Generally, answering in terms of π was understood well. Students who used a value for π usually got into difficulties and made little progress.
A cone of height 10 cm and base diameter 6 cm sits on a hemisphere of diameter 6 cm. Cone volume=(1/3)πr^2h; sphere volume=(4/3)πr^3. Total volume is kπ cm^3.
Edexcel 1MA1 · June 2019 · Paper 2F · Q15 · 4 marks
4 marks
Examiner report
Although a few students did work with perimeter instead of area, the vast majority gained at least one mark for 90 × 60. Many, failing to read the question carefully, then simply found the area of the garden where flowers are grown. It was however pleasing to see so many fully correct solutions. Many used non calculator methods to find 40% (or 60%). Some found 40% or 60% of both dimensions as opposed to just one before finding the area.
A rectangular garden is 90 m by 60 m. Flowers occupy 40%; the rest is grass.
Edexcel 1MA1 · June 2019 · Paper 2F · Q19 · 4 marks
4 marks
Examiner report
The vast majority of students did not understand how to draw the locus of points a given distance from a point and so it was rare to see attempts at arcs drawn with point A as the centre. Often points were marked at a distance of 6 cm from A. This gained no marks other than the M1 for using the scale (180 ÷ 30) correctly. This was the only mark gained by many students who failed to draw any loci on the diagram. Clearly a lot of students had no concept of loci either not attempting the question or simply shading an area with no defined boundaries. However, many students were able to score at least two marks for a correct line segment drawn 5 cm from the line BC. Sometimes this was achieved by drawing a ‘table’ with one side 5 cm from BC. Many students shaded a small rectangular table in the middle of the region.
A kitchen plan is the interior of rectangle ABCD, with A at the top left, B at the top right, C at the bottom right and D at the bottom left. It is drawn at scale 1:30. A table must be more than 180 cm from A and more than 150 cm from the side BC.
Edexcel 1MA1 · June 2019 · Paper 2F · Q23 · 4 marks
4 marks
Examiner report
It was pleasing to see many good solutions to this question. There was, however, some misunderstanding of what was required. Some found the number of cups required to fill the whole container (2 marks maximum) and some found the number of cups needed to fill the final third of the container (3 marks maximum). The final accuracy mark was often not gained through not rounding down to a number of completely filled cups. Incorrect conversions between millilitres and cubic centimetres were condoned if the process was correct. A number of students converted ଶ ଷ to a decimal or a percentage. This was accepted provided the conversions were correct to two significant figures. Often weaker students worked with surface area or perimeter or found 2/3 of 275 and thus gained no credit.
A 30 cm by 6 cm by 19 cm cuboid container is 2/3 full of water. Each cup holds 275 ml.
Edexcel 1MA1 · June 2019 · Paper 2F · Q24 · 2 marks
2 marks
Examiner report
This question was very poorly answered suggested that many students had not covered trigonometry as part of their course, or were insufficiently confident in the application of trigonometry. Many tried simply using the numbers in the question in a variety of incorrect calculations. Students with some knowledge of trigonometry usually selected sine as the function to use, although cosine and to a lesser extent tangent were sometimes their choice. The correct quoting of sin 38o was usually followed by a fully correct solution. Some students however, having their calculators set on an incorrect mode, were unable to score full marks. Correct answers written in the body of a script, and then rounded or truncated incorrectly, gained full credit. Less able students often worked out the missing angle and gave that as their answer rather than finding the side as required.
ABC is right-angled at B. AC=16 cm and angle ACB=38°.
Edexcel 1MA1 · June 2019 · Paper 2F · Q29 · 2 marks
2 marks
Examiner report
Many students were able to score at least one mark, and often two, in this question, showing some understanding of vector arithmetic. 4 was often seen but then simplified incorrectly, sometimes by adding, sometimes by multiplication and sometimes simply by incorrect subtraction. A number of students tried to square the vector b instead of multiplying it by 2 giving 4 . Quite a few students saw this as a fraction question, 4/5 ̶ 6/4, and proceeded with a common denominator etc. A number of students wrote out 4 but they were unable to complete this method. Summary Based on their performance on this paper, students should: show working for all questions including any calculations done using a calculator read each question carefully to ensure that their final answer matches the question asked ensure that they know the difference between area and perimeter use the correct percentage of decimal or equivalent instead of a fraction, for example, 0.3 or 30% should not be used in place of ଵ ଷ remember to draw a straight line through points generated and then plotted in order to draw the graph of an equation of the form y = mx + c practise showing inequalities on a number line, drawing loci to identify a region, finding lengths or angles using trigonometry and writing down error intervals
Column vectors a=(4,5) and b=(3,2).
Edexcel 1MA1 · June 2019 · Paper 2H · Q4 · 4 marks
4 marks
Examiner report
This was the first problem type question on the paper and was answered very well. The problem required the students to carry out three steps which could be interchangeable in order, as the mark scheme shows. Typically, students followed the path of finding the total volume, then finding 2 3 of this volume, before finally dividing by 275 and rounding down to find the number of cups. There were a number of errors that brought the loss of one or two marks. Some students worked with the total volume only and were able to score two marks at best. Others worked with ଵ ଷ rather than ଶ ଷ of the total volume. These students could gain three marks only. Finally there were the students who rounded the answer up to 9 rather than down to 8, again these lost the final mark. Alongside these problems, some students thought there was a need to convert units due to a misunderstanding of the conversion between cm3 and millilitres. This was catered for in the mark scheme and students were still able to score all the process marks despite incorrect unit changes. Some students mis-interpreted this as a bounds question and calculated the volume as 30.5 × 6.5 × 19.5. Another incorrect method seen was to calculate the surface area of the cuboid instead of the volume, gaining no marks.
A 30 cm by 6 cm by 19 cm cuboid container is 2/3 full of water. Each cup holds 275 ml.
Edexcel 1MA1 · June 2019 · Paper 2H · Q5 · 2 marks
2 marks
Examiner report
As should be expected on the higher tier, the majority of students were able to gain both marks on this simple trigonometry question for a correct answer in range. Most students completed the question using the sine ratio, however some used a combination of other trigonometric ratios and Pythagoras’s Theorem. Provided they didn’t prematurely round their interim values, they ended with an answer in range and still gained both marks. As ever some students used the wrong ratio, and this typically meant zero marks. There were a small number of students who had their calculator set in radians; in this instance if the method was shown then the method mark could be awarded.
ABC is right-angled at B. AC=16 cm and angle ACB=38°.
Edexcel 1MA1 · June 2019 · Paper 2H · Q12 · 4 marks
4 marks
Examiner report
This question required students to be able to link the area and perimeter of sectors. There were two main routes through the problem. The most popular was to find the angle of the sector using the formula for the area of a sector, and then to apply this to the arc length of a sector. This left simply the need to add 2 × 7 to finish the problem. The other common method was to work with proportion, for example 40 49 and then find this proportion of the circumference. Both these methods were regularly seen and used well. There was a simple first mark that almost all were able to achieve, for finding the area of the circle. Quite a large number of students dropped the final mark for failing to add the 2 × 7 to the arc length to complete the perimeter. Several wrong responses started with the angle being 110°, presumably from measuring. Students need to be aware of that, unless stated, diagrams are not accurately drawn.
OAB is a sector of radius 7 cm and area 40 cm^2.
Edexcel 1MA1 · June 2019 · Paper 2H · Q18 · 5 marks
5 marks
Examiner report
This circle theorem problem primarily required the knowledge of only one circle theorem that opposite angles in a cyclic quadrilateral add up to 180o – alongside some of the general geometric properties. It was split into two method marks, one accuracy mark and two communication marks. Most students gained the first mark for using the isosceles triangle to find either BAD or BDA. The second mark was for using cyclic quadrilaterals to find BCD or setting up an equation for ABC and ADC, also using cyclic quadrilaterals. The accuracy mark was then for the final two steps of method and getting to the correct value for ADE. To gain all communication marks, students had to correctly state the circle theorem as well as one other correct property used. Alongside this, any other reasons stated need not be complete in terms of their wording, but must have been appropriate for the method that the individual used. If the award of both communication marks could not be made then one mark was available for either stating “opposite angles in a cyclic quadrilateral…”, or for correctly giving all other reasons appropriate for their method. A common mistake in this question was to use a wrong circle theorem.
A,B,C,D lie on a circle; C,D,E are collinear. BA=BD, CB=CD and angle ABD=40°.
Edexcel 1MA1 · June 2019 · Paper 2H · Q19 · 4 marks
4 marks
Examiner report
3D trigonometry always poses a problem for all but the most able students, but again, it was good to see the majority of students attempting the problem, with most able to score at least one mark for either splitting DA in the correct ratio, or for using tan correctly to find BE. However, it is apparent that many students did not even know which angle they were trying to calculate. Also, several found AE which is not in the required triangle and so, on its own, was not enough for a mark. An impressive number of students were able to take the next step and have a correct process to find either MB or ME and gain two marks. It was at this stage that the vast majority of students struggled to get further, with many pupils incorrectly identifying and finding EMA. The final process mark was for a correct trigonometric statement involving angle EMB, which could be gained from numerous routes through the diagram.
A triangular prism has square base ABCD of side 15 cm. ABE and CBE are right angles, angle EAB=35°. M lies on DA with DM:MA=2:3.
Edexcel 1MA1 · June 2019 · Paper 2H · Q20 · 6 marks
6 marks
Examiner report
Even at this, the highest grade, a good number of students were not only attempting part (a) but gaining some credit. Many students scored one if not two marks in part (a) for a correct expression for 𝐹𝐸 ሬሬሬሬሬ⃗. The small number who gained one mark usually gave an answer of a–b+a+b which showed that the instruction to give the vector in its simplest form had been missed. Part (b) proved to be a significant challenge for almost all students to solve using vector algebra. Some were able to gain the first mark, typically for finding 𝐶𝐸 ሬሬሬሬሬ⃗ or 𝐹𝑀 ሬሬሬሬሬሬ⃗. Very few students were then able to gain the second mark, involving the use of the correct scale factor, either n or n n . Of the few that did produce appropriate expressions, very few knew how to equate the coefficients to form an equation, and as a result only a very small minority of students were able to gain the full four marks using an algebraic method. Even students who converted the ratio n:1 into fractional amounts such as n/(n+1) too often considered these fractions as vectors in their own right and would go on to write incorrect expressions such as a–b +n/(n+1). Summary Based on their performance on this paper, students should: be encouraged to not prematurely round values within working, as this often leads to an answer outside of the acceptable range understand how to use their calculators when substituting negative values into formulae and expressions continue to spend time with new content especially topics such as error intervals for truncation, tangents and areas under graphs learn the correct language for giving geometric reasoning, without using abbreviated words remember to keep working inside the boxes provided or use an extra sheet to ensure working worthy of credit is seen.
In quadrilateral CDEF, vectors CD=a, DE=b and FC=a-b. M is midpoint of DE. X lies on FM with FX:XM=n:1, and C,X,E are collinear.
Edexcel 1MA1 · June 2019 · Paper 3F · Q16 · 3 marks
3 marks
Examiner report
Most students earned the first mark by correctly placing 5 figures on the diagram or identifying one of the missing lengths. A high proportion got 30 as their answer since they had missed out the length of 2; many more got an incorrect answer of 36 by including the inner edge. Some worked out the perimeter of the two separate rectangles and just added the two perimeters. Those who worked through the question methodically and found all the required sides usually went on to give the correct answer for full marks.
A 6 cm by 4 cm rectangle is used twice to make a 6-sided shape. One copy stands upright (width 4 cm, height 6 cm). A second copy is rotated (width 6 cm, height 4 cm) and attached to the lower 4 cm of the upright rectangle's right side, with the two bottom edges level.
Edexcel 1MA1 · June 2019 · Paper 3F · Q20 · 2 marks
2 marks
Examiner report
It was surprising how many students failed to gain the answer of 105 in this question. Students who wrote their angles on the diagram were more successful than others. However, those angles needed to be correct to gain the method mark; this was usually gained by stating that the angle in the corner of the square was 90, or the angle in the triangle was 60, which could simply be shown by writing it on the diagram. But these were often just added to give 150 as their final answer. It was clear that some students were not aware which angle EBC represented, even though it was clearly shown on the diagram.
ABDE is a square and BCD is an equilateral triangle. The triangle lies outside the square and shares side BD with it.
Edexcel 1MA1 · June 2019 · Paper 3F · Q28 · 4 marks
4 marks
Examiner report
Very few marks were gained in this question. The most common approach was to consider the angle BCF, and giving this as 63 earned a mark, but they could rarely proceed beyond this point. Some earned a mark by working with the total angles of a polygon, but again regularly failed to proceed beyond the point at which this figure was found. Most incorrectly used 360 as the total angles in a hexagon or pentagon; many were unable to label angles correctly, which was particularly disappointing when BCD and CDE were found, but written the wrong way around.
ABCDEF is a hexagon with one line of symmetry. AB is the top side and ED the bottom side; symmetry pairs A with B, F with C and E with D. EF = FA and BC = CD. Angle ABC = 117°. Angle BCD is twice angle CDE.
Edexcel 1MA1 · June 2019 · Paper 3F · Q29 · 5 marks
5 marks
Examiner report
Again, very few marks were gained in this question. Having been given a picture of a cylinder many students just found the volume, which was of no use to them. The other common error was to think of the curved surface area as a rectangle 1.8 × 1.6 which again usually led to no credit. There was of course some confusion as to whether a figure of 1.6 or 0.8 should be used for the radius. Marks were sometimes awarded when a student found the area of a circle (without converting this into a volume), or less frequently when they found the curved surface area. A common mark awarded was when they found the total coverage possible (7 × 5) but this was sometime spoilt when students interpreted the 5 m2 as 25 and then incorrectly calculated 7 × 25.
Jeremy must paint 3 closed cylindrical tanks completely, including top and bottom. Each tank has diameter 1.6 m and height 1.8 m. He has 7 tins of paint and each tin covers 5 m².
Edexcel 1MA1 · June 2019 · Paper 3H · Q5 · 4 marks
4 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2019 · Paper 3H · Q6 · 5 marks
5 marks
Examiner report
This question was also a good discrminator. Unfortunately, a significant proportion of students calculated the volume of the cylinder rather than the surface area and so could not be awarded much credit for their responses. However, there were many fully correct and complete solutions usually comparing the total surface area of the 3 tanks with the total surface area which could be covered with 7 tins of paint but sometimes comparing the number of tins of paint needed to paint the three tanks with the 7 tins of paint available to Jeremy. A good proprtion of students calculated the surface area of only one end of the cylinder or tried to combine the area of one or two ends with a volume or with an incorrect “area” found by calculating 1.6 by 1.8. It was pleasing to see that students often worked with exact expressions in terms of π, thus avoiding the need to round values in intermediate working and minimizing the risk of errors when calculating and writing down decimal values from calculators.
Jeremy must paint three closed cylindrical tanks completely, including the top and bottom. Each tank has diameter 1.6 m and height 1.8 m. He has 7 tins of paint, and each tin covers 5 square metres.
Edexcel 1MA1 · June 2019 · Paper 3H · Q8 · 2 marks
2 marks
Examiner report
A large proportion of students were able to score at least one mark for their responses to this question. Responses were usually clearly expressed. Part (a) was answered well. The most common description of the mistake that Sarah made was to state that the first line should read BC² = AC² – AB² or equivalent, for example that she should be subtracting not adding the numbers. A common but unacceptable response was that the mistake was not to show the square root sign over the 100. Part (b) was also answered well with most students stating that the scale factor is 2.5 and not 1.5 as required and many other students who explained that the length of one side of the enlarged triangle was incorrect together with a statement of what the length should be. The most common unacceptable response referred to C being the wrong centre of enlargement.
Triangle ABC is right-angled at B, with AB = 6 cm and AC = 8 cm. Sarah writes BC^2 = AB^2 + AC^2. Separately, on a coordinate grid with C = (0, 0), triangle PQR has P = (2, 4), Q = (4, 4), R = (4, 0). Roy draws X = (5, 10), Y = (10, 10), Z = (10, 0), intending XYZ to be an enlargement of PQR with centre C and scale factor 1.5.
Edexcel 1MA1 · June 2019 · Paper 3H · Q14 · 4 marks
4 marks
Examiner report
A small minority of students scored full marks for their responses to this question. Weaker students often showed little or no working and if they did show working, it was often characterised by assuming that the triangles were right angled and using trigonometry to fit such a situation. More able students often did realise that references to area in the question together with the diagram showing non right angled triangles suggested use of the formula ଵ ଶ absinC. However, some of these students tried to find the length of AE instead of using the ratio AB : AE to express AE as an unknown variable, for example 3x. Students who used AB = x and AE = 3x scored at least one mark and often went on to write down and solve an equation using the given fact that the sum of the areas of the triangles was equal to the area of the rectangle. Examiners were surprised at the number of students who equated the area of one triangle to the area of the rectangle which led to the incorrect final answer of 18.
ABDE is a rectangle. Congruent triangles AFE and BCD are attached to the two sides of the rectangle. AF = BC = 24 cm, angle FAE = angle CBD = 30 degrees, and the area of rectangle ABDE equals the total area of the two triangles. Also AB : AE = 1 : 3.
Edexcel 1MA1 · June 2019 · Paper 3H · Q22 · 4 marks
4 marks
Examiner report
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.
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.
Edexcel 1MA1 · June 2019 · Paper 3H · Q23 · 5 marks
5 marks
Examiner report
This question was accessible to and discriminated well between the most able students sitting this paper. Students who had a good understanding of bearings together with applications of the sine rule and cosine rule were able to produce an accurate and concise solution to the problem. Of those students who gained partial credit for their attempts, many were only able to score either one mark for a correct process to find the size of angle ABC or two marks for getting as far as finding the length of AC. Some students who got as far as finding the size of angle BAC successfully gave this as their final answer. Finding the bearing of A from C instead of the bearing from C from A was also seen quite frequently. Summary Based on their performance on this paper, students should: check working for careless errors and to see that the answer given was specifically required by the question. practise problems involving the surface areas of solids. practise expanding brackets and collecting terms especially with quadratic functions involving fractions. ensure a good understanding of multipliers in problems concerning percentages practise solving quadratic inequalities.
The diagram shows the positions of three towns, Acton (A), Barston (B) and Chorlton (C). N B 150° N 8 km N 9 km 37° A C Barston is 8 km from Acton on a bearing of 037° Chorlton is 9 km from Barston on a bearing of 150° Find the bearing of Chorlton from Acton. Give your answer correct to 1 decimal place. You must show all your working. °
Edexcel 1MA1 · November 2018 · Paper 1F · Q14 · 4 marks
4 marks
Examiner report
Most students scored at least one mark for their responses to this question and often two marks for finding the size of angle BCD and the size of angle ADC. These were often seen marked on the diagram. The third mark for a method to find the size of angle ABC was more elusive. A significant minority of students assumed the shape was a kite and so stated that angles ABC and ADC were of equal size. This was not acceptable to examiners who expected students to use the angle sum of a quadrilateral to find angle ABC. Students who did gain all three method marks for finding the angles were often able to give geometric reasons accurately and so score the final communication mark. However, centres and students are reminded that they need to express reasons in full, for example, “straight line is 180°” is not acceptable but “angles on a straight line sum to 180°” is acceptable. Some students used spurious reasons such as “alternate angles” or “corresponding angles”.
ABCD is a quadrilateral in cyclic order, with every side extended. A, B and the extension of AB are collinear; B, C and the extension of BC are collinear; C, D and the extension of CD are collinear; D, A and the extension of AD are collinear. The interior angle DAB is 100 degrees. At D, the exterior angle between DC and the extension of AD beyond D is 75 degrees. At C, the angle between the extensions of BC and DC vertically opposite interior angle BCD is 50 degrees. Also AB = AD.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1F · Q16 · 1 mark
1 mark
Examiner report
A disappointingly large number of students reasoned that the two rectangles were similar because each side of the smaller rectangle had 4 cm added to give the sides of the larger rectangle. Other students said that Jim was correct because “the rectangle has been enlarged by 4”. Only a very few students were able to state something along the lines that in order for the two rectangles to be similar a common multiplier is needed between corresponding sides. It was disappointing to read that a significant proportion of students thought that for shapes to be similar their area and/or perimeter must be the same.
Here are two rectangles. 6 cm 10 cm 8 cm 12 cm Jim says, “The two rectangles are similar because 8 + 4 = 12 and 6 + 4 = 10” Is Jim correct? Explain your answer.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1F · Q19 · 2 marks
2 marks
Examiner report
This question proved to be a good discriminator between more able students sitting this paper and each of the 0, 1 or 2 marks available was scored by a substantial number of students. The majority of students scored one mark for a rotation of 180 degrees, demonstrated by the correct orientation of the shape, while being unable to place their image in the correct place on the grid. Of those students not able to score any marks, many had rotated the shape by 90 degrees. Students are reminded that tracing paper may be used in the examination to help them with this type of question.
On a coordinate grid, shape A is a quadrilateral with vertices (-3, 5), (-2, 5), (-2, 3) and (-3, 4).
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1F · Q26 · 4 marks
4 marks
Examiner report
This question was not done well. A small proportion of students quoted a correct formula for the area of a circle. Unfortunately, they were usually unable to use the formula to find the area of a semi-circle of radius 10 cm or the area of a quarter of a circle with radius 20 cm. Many students quoted an incorrect formula, often citing the formula for the circumference of a circle instead. Many students found the area of a square of side 20 cm but this was by itself not enough to score any marks. Very few students scored 2 marks or more for their answers.
ABCD is a square of side 20 cm, with A bottom-left, B bottom-right, C top-right and D top-left. AB is the diameter of an inward semicircle of radius 10 cm above AB. AC is the quarter-circle arc from A to C with centre B and radius 20 cm. The shaded region is inside this quarter-circle sector but outside the semicircle.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1F · Q28 · 3 marks
3 marks
Examiner report
This, the last question on the paper targeted the most able students sitting the examination. It was very rare to see a fully correct solution or indeed any responses which could be credited with part marks. Summary Based on their performance on this paper, students should: • learn and practice standard techniques involved in solving simultaneous equations paying particular attention to accuracy where negative signs are involved. • use tracing paper to help in questions involving the transformation of rotation of a shape about a point. • carry out a common sense check on the answers to calculations, so for example expecting the time taken to fill a pool to be more if less taps are used. • check all calculations for arithmetic errors particularly when completing papers where the use of a calculator is not allowed.
The size of each interior angle of a regular polygon is 11 times the size of each exterior angle. Work out how many sides the polygon has.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1H · Q7 · 4 marks
4 marks
Examiner report
Many students correctly found the area of the semicircle (radius 10cm) and then often used this to compare with the area of the square, never realising the need to find the area of the quadrant (radius 20cm). A significant number introduced their own value for pi, often 3, but were usually unable to complete the proof. The fact the given result was in terms of pi should have deterred students from this approach.
ABCD is a square of side 20 cm, with A bottom-left, B bottom-right, C top-right and D top-left. AB is the diameter of an inward semicircle of radius 10 cm above AB. AC is the quarter-circle arc from A to C with centre B and radius 20 cm. The shaded region is inside this quarter-circle sector but outside the semicircle.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1H · Q8 · 3 marks
3 marks
Examiner report
Values for the sine, cosine and tangent of common angles is clearly not well known with only a few students gaining any credit in part (a). In part (b), a correct answer of 8 cm was common if not always from a correct method; 0.5 × 4 = 8 was a common error. Some students labelled the opposite side 4 cm and then added the two sides to get 8 cm. This also gained no credit. Students should be encouraged to always quote an equation for their choice of trigonometric function; those that did so here were usually correct and so gained one of the two marks available. Another common error was writing cos(0.5) = 4/x, showing little or no understanding of the value of 0.5.
A right-angled triangle has a 60-degree angle, adjacent side 4 cm and hypotenuse x cm. You are given cos 60 degrees = 0.5.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1H · Q12 · 3 marks
3 marks
Examiner report
The knowledge and understanding of the theorems of angles in a circle was poor and many students simply treated triangle ABC as an isosceles triangle quoting angles of 55o, 55o and 70o (35o × 2) or assumed that OB bisected angle CBA. Angle OAB = 34o was not uncommon but many failed to carry on after this.
B E 35° O 56° A C D A, B and C are points on the circumference of a circle, centre O. DAE is the tangent to the circle at A. Angle BAE = 56° Angle CBO = 35° Work out the size of angle CAO. You must show all your working. °
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1H · Q13 · 2 marks
2 marks
Examiner report
Very few students got full marks in this question, largely as a result of quoting an incorrect scale factor, usually -3. However, one mark was often awarded. Many students tried to describe a combination of transformations; this gained no credit.
On a coordinate grid, triangle A has vertices (-4, -1), (-4, -4) and (2, -4). Triangle B has vertices (4, 3), (4, 4) and (2, 4).
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1H · Q15 · 4 marks
4 marks
Examiner report
Only the most able of students achieved full marks here. A mark of 2 was common, however, with ratios of 2:5 and 3:4 or their equivalences seen. Many weaker students unsuccessfully tried to combine the pairs of ratios given. It was pleasing to see some students drawing diagrams to help visualise the problem.
Three solid shapes A, B and C are similar. The surface area of shape A is 4 cm2 The surface area of shape B is 25 cm2 The ratio of the volume of shape B to the volume of shape C is 27 : 64 Work out the ratio of the height of shape A to the height of shape C. Give your answer in its simplest form.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1H · Q21 · 5 marks
5 marks
Examiner report
Very few students indeed found the correct answer by employing a correct process. Of the few that did succeed, the use of congruent and similar triangles was the usual approach. Some students were able to find vector AB and vector OM with a few gaining an additional mark for the vector AP. These however were rare. Some simply said that ON = OP = 3 and NB = 2 × PM = 4. This got no marks. Students need to be aware that if a vector is shown on the diagram they need to indicate direction.
OAB is a triangle. O, N and B are collinear in that order. M is the midpoint of AB. O, P and M are collinear, and A, P and N are collinear. Vector OA = a, vector OB = b, and OP:PM = 3:2.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 2F · Q13 · 5 marks
5 marks
Examiner report
Part (a) was answered quite poorly. As a first step, most students either found the square root of 81 or divided 81 by 4. Unfortunately, far too many took the latter approach, often giving 20.25 as the final answer, and gained no marks. Many of those who found the square root of 81 went on to find the perimeter of the square but some did not and 9 was often given as the final answer. Rather surprisingly the number of fully correct responses seen in part (b) was very similar to the number seen in part (a). Many students gained one of the three marks. Some students gained the first method mark for finding the area of the triangle but could not use the area correctly to find the area of the parallelogram. A very significant number of students worked out the area of the triangle as 9 × 16, forgetting to divide by 2. These students could still gain the second method mark for using their area of the triangle correctly to find the height of the parallelogram, generally reaching an answer of 24. A number of students forgot to multiply the area of the triangle by 5 to find the area of the parallelogram. Others did not know how to find the area of a parallelogram and couldn’t get any further after multiplying by 5.
Answer both parts.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 2F · Q16 · 2 marks
2 marks
Examiner report
This is a standard question of a type that students should be very familiar with yet it was answered surprisingly poorly. Far too many students failed to describe the transformation as a reflection. Answers such as “flipped” and “mirrored” were very common and unacceptable. Many of the students who did describe it as a reflection failed to get both marks. Some wrote “reflection” and nothing else. Many attempted to describe the reflection but could not identify the mirror line as the x-axis or the line y = 0. Some students wrote “the line x”. It was not uncommon to see “reflection” with an angle or a vector or with coordinates. Some students described the transformation as a rotation. Very few students gave more than one transformation - this has been a common failing in the past.
On a coordinate grid, parallelogram A has vertices (1, 1), (3, 1), (4, 4) and (2, 4). Parallelogram B has vertices (1, -1), (3, -1), (4, -4) and (2, -4).
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 2F · Q22 · 4 marks
4 marks
Examiner report
Many students made a start by using angles on a straight line to find that angle BEG = 45º and then failed to make any further correct inroads into this problem. For the first method mark students needed to use parallel lines to find an angle, e.g. angle ABE = 70 º. Instead of using angles on parallel lines many students made incorrect assumptions about other angles. It was common to see angle ABE = 35 º or angle EBG = 110º (thinking it was alternate to angle DEB) or angle ABE = 110º. Some students attempted to use angles on parallel lines but incorrectly identified the angles that are equal, believing angle ABE to be equal to angle BEG for example, and scored no marks. The students who used angles on parallel lines correctly were usually able to find the size of angle x. It was disappointing that relatively few students achieved the C mark for giving one reason linked to parallel lines and one other reason. Some gave no reasons at all. When reasons were given they often did not include a reason linked to parallel lines or were incomplete, e.g. “triangle = 180”, and did not include a reference to angles. Many students were not able to give clear statements relating to parallel lines with the correct naming of the types of angles. Some students incorrectly named alternate angles or co-interior angles as corresponding whilst some students used the unacceptable terms “Z angles” and “C angles” in their reasons rather than “alternate angles” and “co-interior angles” or “allied angles”.
A, B and C are collinear in that order. D, E and F are collinear in that order, and line ABC is parallel to line DEF. BEG is a triangle. At B, angle CBG = 35 degrees. At E, angle DEB = 110 degrees and angle GEF = 25 degrees. The required angle x is angle BGE.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 2F · Q24 · 5 marks
5 marks
Examiner report
Although there were relatively few fully correct solutions to this multi-step problem many students did gain at least one mark. Many students could not find the area of a trapezium correctly but even so they were still able to access three of the process marks. After finding a floor area, a common approach was to work out that each tin covers an area of 10 m2 and divide their area by 10 to find the number of tins needed. Some students divided their area by 2 to find the number of litres needed and then divided by 5 to find the number of tins needed. A common mistake was to divide by 2 or by 5 but not by both. It was disappointing to note that a small number of students used a divisor of 4 instead of 2 for paint coverage because they misunderstood the notation 2m2. After a process to find the number of tins students often went on to multiply £16.99 by a non-integer value and lost a process mark. Some students showed incorrect working to find the total cost based on finding the cost of 1 litre of paint. A small number of students used a different approach, starting with the £160 John has to spend and working out the floor area that can be covered with the tins he can buy. Many responses had little or no structure to the working with calculations scattered over the page. Students who worked logically and structured their answer generally scored better.
A trapezium-shaped area has parallel sides 10 m and 16 m and perpendicular height 7 m. Paint covers 2 square metres per litre and is sold in 5-litre tins costing GBP 16.99 each. The budget is GBP 160.
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Edexcel 1MA1 · November 2018 · Paper 2H · Q3 · 4 marks
4 marks
Examiner report
Of those who gained marks, using parallel lines to find angle ABE along with angle BEF was the most common approach, usually leading to the correct answer of 60°. But many made incorrect assumptions about the diagram, some assuming an equilateral triangle and others an isosceles triangle, neither of which were correct. There was a mark attributed to giving reasons, but this was rarely awarded either because reasons were not given, or incorrect reasoning was given. An early error was frequently to state BEF as 45°.
A, B and C are collinear in that order. D, E and F are collinear in that order, and line ABC is parallel to line DEF. BEG is a triangle. At B, angle CBG = 35 degrees. At E, angle DEB = 110 degrees and angle GEF = 25 degrees. The required angle x is angle BGE.
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Edexcel 1MA1 · November 2018 · Paper 2H · Q5 · 5 marks
5 marks
Examiner report
The area of the trapezium was incorrectly calculated by many. The following step of working (that of division by 2, 5 or 10) were usually done well, but many students failed to round up the number of tins to a whole number. Some assumed you could buy the paint by the litre, and not in 5 litre tins. Those who worked through all stages ended with a comparison of 90 and 91. This was the most successful method, but other methods were credited that resulted in a comparison of the number of tins, or even the number of litres needed.
A floor is a trapezium with parallel sides 10 m and 16 m and perpendicular height 7 m. Each 5-litre tin of paint costs GBP 16.99. One litre covers 2 m^2. John has GBP 160.
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Edexcel 1MA1 · November 2018 · Paper 2H · Q8 · 3 marks
3 marks
Examiner report
Students had some work to do on the drawing before getting to R. With many the reflection in y = -x caused difficulty, but most could reflect in x = -1. But any work on the diagram was superseded by an attempt to describe the combined transformation. Most often the only description that earned credit was mention of a rotation, with some gaining an additional mark for indicating it was a 90° clockwise turn (or 270 anticlockwise). Rarely was (-1, 1) seen. Other common answers included confusing clockwise and anticlockwise, and some who tried to describe a series of transformations.
Triangle P has vertices (-1, 3), (3, 3) and (3, 5). It is reflected in the line y = -x to triangle Q, then Q is reflected in the line x = -1 to triangle R.
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Edexcel 1MA1 · November 2018 · Paper 2H · Q13 · 3 marks
3 marks
Examiner report
This question differentiated well. Many found the radius or diameter for the first mark, but from there many seemed at a loss as to what to do next. Of the minority of students who made further progress the most common method was to use Pythagoras’s Theorem to find the height of the triangle. But having found the height it was often seen that the base used was the diameter and not the radius; there were also examples of premature rounding which placed many answer out of range.
The diagram shows a circle and an equilateral triangle. One side of the equilateral triangle is a diameter of the circle. The circle has a circumference of 44 cm. Work out the area of the triangle. Give your answer correct to 3 significant figures. cm2
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Edexcel 1MA1 · November 2018 · Paper 2H · Q20 · 5 marks
5 marks
Examiner report
It was disappointing to find how few students were able to use given formulae and perform a simple substitution into them. Use of the diameter rather than the radius was seen too often, or not halving the diameter for the top half of the cone, or even not halving to find the volume of a hemisphere. Surprisingly it was not uncommon to find students who changed the 1/3 to ½ for the volume of the cone. In dealing with the densities, there was a lot of adding since students did not realise that each one applied to each part of the shape.
A solid is made by joining a frustum to a solid hemisphere of radius 3.6 cm. The frustum is formed from a cone of radius 3.6 cm and height 6.4 cm by removing a similar cone of radius 1.8 cm and height 3.2 cm. The frustum has density 2.4 g/cm^3 and the hemisphere has density 4.8 g/cm^3. Use volume of a sphere = (4/3)pi r^3 and volume of a cone = (1/3)pi r^2 h.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 2H · Q21 · 4 marks
4 marks
Examiner report
Students failed to have any cognisance of what was needed for a proof. It was also rare to see any mention of cyclic quadrilaterals. Many assumed incorrectly that triangle ABC was equilateral or isosceles with base AB often followed by drawing a line from C to O and attempting to use a right angles triangle. There were also many references to tangents, though there were none in the diagram. Summary Based on their performance on this paper, students should: • be prepared for an extensive range of topics, sufficient that they can make a good attempt at a majority of questions on the Higher paper. • practice algebraic manipulation and derivation, percentage calculation, application of ratios and proportion, and geometric reasoning and proof, in preparing for future examinations. • practice their arithmetical skills and ensure that they can use their calculator correctly to check processes and calculations. • 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. • spend more time ensuring they read the fine detail of the question to avoid giving answers that do not answer the question.
A, O and B lie on a straight line. C, R and B lie on a straight line. A, O, R and C lie on one circle, and O and B are the centre and a point on a second circle, so OR = OB. Prove that angle CAB = angle ABC.
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Edexcel 1MA1 · November 2018 · Paper 3F · Q5 · 2 marks
2 marks
Examiner report
In part (a) most students correctly identified the solid as a cuboid. Rectangular prism was seen occasionally and gained credit. A few students got mixed up with solids and 2D shapes and incorrectly gave the answer of rectangle. In part (b) there were typically only 2 answers seen, 8 and 12. The correct answer of 12 was by far the most common, but there were a number who got mixed up between edges and vertices and thus gave the answer of 8.
A box-shaped solid has six rectangular faces; its net has six rectangles and folds to make the solid.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 3F · Q9 · 2 marks
2 marks
Examiner report
It is clear that some students are not familiar with their 2D shapes. There were a number of incorrect answers seen, such as parallelogram, rhombus and even right angled triangle. In part (b) a large proportion of students correctly identified the correct two shapes.
Five polygons A to E are drawn on a square grid. Taking the top-left grid intersection as (0, 0), with x increasing to the right and y increasing downwards, their vertices are A: (1, 1), (7, 1), (6, 5), (1, 5); B: (8, 1), (7, 5), (12, 5); C: (10, 1), (14, 1), (14, 5); D: (2, 6), (2, 10), (6, 10); E: (5, 6), (15, 6), (15, 10), (9, 10).
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Edexcel 1MA1 · November 2018 · Paper 3F · Q10 · 1 mark
1 mark
Examiner report
This simple reflection was done really well by students, with again almost all students able to draw the correct reflected shape.
A horizontal mirror line is y = 0. Above the line, a shaded L-shape is made from four unit squares: one square occupies 0 <= x <= 1, 0 <= y <= 1, and three adjacent squares occupy 0 <= x <= 3, 1 <= y <= 2.
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Edexcel 1MA1 · November 2018 · Paper 3F · Q25 · 4 marks
4 marks
Examiner report
This question proved to be a significant challenge to the vast majority of students. The only mark that was accessed with any regularity was the first mark for dealing with the ratio to find the missing angles. From that point on, only a very few students were able to correctly apply trigonometry to further their solution.
ABC is a right-angled triangle. B A C 14 cm AC = 14 cm. Angle C = 90 size of angle B : size of angle A = 3 : 2 Work out the length of AB. Give your answer correct to 3 significant figures. cm
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Edexcel 1MA1 · November 2018 · Paper 3H · Q6 · 4 marks
4 marks
Examiner report
The initial ratio part at the beginning of this question allowed many students to achieve one mark on this question. For those who could not find the angle correctly there was another opportunity to achieve marks by using ‘their’ value for angle A or angle B in a correct equation using trigonometry. Many students continue to make the use of trigonometry in a right-angled triangle more difficult than is necessary, by trying to use the cosine rule or sine rule but often misquoted incorrectly. Whilst this approach is acceptable and can lead to the correct answer few students are able to apply these rules correctly. The use of a trigonometry in a right-angled triangle is usually a more successful approach. However, the most common error seen in this method was for the student to correctly write sin 54 = 14 ÷ AB but then to rearrange incorrectly to AB = 14 sin 54.
ABC is a right-angled triangle. B A C 14 cm AC = 14 cm. Angle C = 90 size of angle B : size of angle A = 3 : 2 Work out the length of AB. Give your answer correct to 3 significant figures. cm
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Edexcel 1MA1 · November 2018 · Paper 3H · Q8 · 5 marks
5 marks
Examiner report
Some fully correct answers were seen but this problem solving question was not well answered by the majority of students. Many students confused surface area and volume at some point within their working. Some students failed to find the correct values of 9, 5, 3 for the dimensions of large cuboid, often just deciding to multiply 45, 27 and 15 together as the volume. Although a good majority of the students did score one mark for using 2.53 as the volume of the smaller cube. But they then often used this with the surface area of the larger cuboid.
The diagram shows a solid metal cuboid. The areas of three of the faces are marked on the diagram. The lengths, in cm, of the edges of the cuboid are whole numbers. 27 cm2 15 cm2 45 cm2 The metal cuboid is melted and made into cubes. Each of the cubes has sides of length 2.5 cm. Work out the greatest number of these cubes that can be made.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 3H · Q12 · 4 marks
4 marks
Examiner report
The most common error seen was to assume that the diagonal length of the square was 5 cm. However, those who used Pythagoras’s Theorem correctly to find the diagonal often then gained the second mark for the length of the half- diagonal. Some did use trigonometry appropriately but often errors were seen.
Here is a pyramid with a square base ABCD. T B C A D AB = 5 m The vertex T is 12 m vertically above the midpoint of AC. Calculate the size of angle TAC. °
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Edexcel 1MA1 · November 2018 · Paper 3H · Q16 · 5 marks
5 marks
Examiner report
Some fully correct answers were seen for this question. When attempted, the use of the sine rule or cosine rule was generally well done with students obtaining the correct value for length of OD and this gained two marks. The calculations to find the length of the arc were often incorrect as many students found the area and not the length. Students who were successful up to this stage then often added the 6 instead of subtracting it or forgot to add the 14 so did not give a full and complete method.
The shaded shape ABCD is the union of triangle AOD and the major sector AOC of a circle with centre O and radius 6 cm. The major arc from A through B to C forms the curved boundary. O, C and D are collinear in that order. AD = 14 cm, angle AOD = 140 degrees and angle OAD = 24 degrees.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 3H · Q20 · 2 marks
2 marks
Examiner report
A very small number of correct answers were seen for this question. Students who attempted the question often drew some or all of the transformations given. However they then gave the top of the triangle (-3.5 , 4) as the invariant point. Summary Based on their performance on this paper, students should · spend time practising rounding to a given number of significant figures. · learn to recognise routine questions and apply the most efficient approaches when answering these questions. · read carefully the description of lengths and values within each question and use them appropriately. · be aware that working with specific values is inappropriate when asked to carry out a proof.
Triangle A has vertices (-4,1), (-3,1), (-3.5,4). It is rotated 180 degrees about (-2,0), then translated by vector (-3,2).
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Edexcel 1MA1 · June 2018 · Paper 1F · Q9 · 2 marks
2 marks
Examiner report
Of the more sensible incorrect responses, ‘diameter’ or ‘circumference’ and ‘chord’ or ‘circumference’ were the most common incorrect answers for parts (a) and (b) respectively. Only a minority of students were able to correctly identify the tangent in part (b).
The centre of a circle is marked. A segment joins the centre to the circumference and a line touches the circle at one point.
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Edexcel 1MA1 · June 2018 · Paper 1F · Q23 · 6 marks
6 marks
Examiner report
In part (a), the most common error was an isosceles triangle of base 6cm and height 5cm (instead of 4cm). Some students offered 3-D diagrams or nets, all such attempts gained no marks. In part (b), many students tried to work out the volume of the pyramid instead of the surface area. Forgetting to divide by 2 for the area of a triangle was common. Many students left their answer without units and so failed to score this independent mark.
A square-based pyramid has base side 6 cm, perpendicular height 4 cm and face slant height 5 cm.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 1F · Q24 · 5 marks
5 marks
Examiner report
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.
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 1F · Q26 · 2 marks
2 marks
Examiner report
For the final question on this examination paper, this was answered quite well with about half of all students gaining at least one mark. The correct doubling of vector a was often seen although sometimes written as a fraction occasionally, only the 5 was doubled. Answers of , , and were the most common incorrect offerings. Summary Based on their performance on this paper, students should: practice arithmetic processes; particular short division and negative number arithmetic ensure that all working is shown in a logical, ordered fashion and can be easily read learn and be able to recall metric conversions, for example, 1 kg is 1000 g remember to divide by two when finding the area of a triangle, i.e. use ½ × base × height practice estimating values to calculations by first rounding given values consolidate work on ratio and the links between ratios and fractions
Vectors a = (5,2) and b = (-1,7).
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Edexcel 1MA1 · June 2018 · Paper 1H · Q5 · 6 marks
6 marks
Examiner report
Most students drew an isosceles triangle in part (a) and often this triangle had base 6 cm and height 4 cm. A frequent mistake was to draw the triangle with an incorrect height, most commonly this was 5 cm, which meant that only one mark was awarded. Some students did not know what was expected for a front elevation. Sometimes the isosceles triangle was drawn as part of a 3-D shape or, less commonly, as part of a net and in these cases no marks could be awarded. In part (b) many students knew that to find the total surface area they needed to find the areas of the base and the four triangular faces. Many fully correct answers of 96 were seen. A very common mistake was to use 4 cm rather than 5 cm as the height of a triangular face and students who did this were awarded one mark if they completed the rest of the method correctly and gave an answer of 84. Some students used two triangles of height 5 cm and two triangles of height 4 cm. Some forgot to halve the product of base × height; others halved both! Volume calculations using 1/3×base×height were also seen. The final mark was for giving the correct units with the total surface area and this could be awarded whether or not any value given for the total surface area was correct. There were of course those students who gave no units at all and some who wrote cm or cm3 instead of cm2.
A square-based pyramid has base side 6 cm, height 4 cm and face slant height 5 cm.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 1H · Q6 · 5 marks
5 marks
Examiner report
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.
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 · Q7 · 2 marks
2 marks
Examiner report
Most students attempted to draw shapes R and S on the grid and these were often shown in the correct positions. Errors in the positioning of shape R and shape S were usually due to students not being able to identify the lines x = –1 and y = –2 although it was not uncommon to see shape R one square to the right of its correct position with the line x = –1 drawn. In some responses it was difficult to determine how the students had come up with their images. Students who drew shapes R and S in the correct positions were usually able to recognise that the single transformation that will map shape T to shape S is a rotation (or an enlargement, which was rarely seen as an answer) although some thought that it is a reflection. Descriptions of the transformation were generally good although mistakes were sometimes made with the angle or with the centre of rotation and a few students referred to a ‘turn’ rather than a ‘rotation’. It is not acceptable to write the centre of rotation as a vector. It was apparent that some students had used tracing paper when answering this question and this was generally a successful strategy for determining the centre of rotation. Some students gave more than one transformation and got no marks for the description; these seemed to be fewer in number than in previous series. Students who failed to gain any marks for the description of the transformation were still able to get one of the two marks for a fully correct diagram.
Reflect T in x=-1 and then y=-2.
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Edexcel 1MA1 · June 2018 · Paper 1H · Q8 · 4 marks
4 marks
Examiner report
This question was generally answered very well with most students able to find the lengths of the sides of the triangle by dividing 72 in the ratio 3 : 4 : 5. A common error was to divide 72 by 3 and by 4 and by 5. Few students used the alternative method of scaling up ratios to find the lengths of the sides. Having found the lengths of the sides most students went on to identify the correct two sides for the area calculation and use a correct method to find the area of the triangle. A few students used 18 cm and 30 cm instead of 18 cm and 24 cm and some forgot to divide by 2 when finding the area. The accuracy mark was often lost because of arithmetic errors, particularly in the area calculation. Some students sensibly reduced ½ ×18×24 to 9×24 or to 12×18 but others correctly wrote ½×18×24 but then halved both 18 and 24 before multiplying. The majority chose to multiply 18 by 24 and then divide by 2. Very few students used the method of scaling area.
The perimeter of a right-angled triangle is 72 cm. The side lengths are in the ratio 3:4:5. Work out its area.
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Edexcel 1MA1 · June 2018 · Paper 1H · Q11 · 5 marks
5 marks
Examiner report
Many students started by drawing in the radius OB and marking angle ABO as x° although there was a significant number of students that took angle ABC to be x°. There was generally a good recognition of the 90° angle between a radius and a tangent with angle OBC usually marked as a right angle on the diagram. Some students were unable to make any further progress but those who realised that angle BAO = x were then in a position to use the sum of the angles in triangle ACB to find the size of angle ACB. Many students, though, decided to also work out the sizes of angles AOB and BOC. The latter caused difficulties, with students struggling to simplify 180 – (180 – 2x) or writing it incorrectly as 180 – 180 – 2x and sometimes giving the unreasonable negative answer of –2x. Students who were able to write a correct expression for angle ACB in terms of x were not always able to write the expression in its simplest from and some who did get to 90 – 2x then ‘simplified’ it to 45 – x and lost the accuracy mark. This question required students to give reasons for each stage of their working. It was encouraging to see responses with full and correctly worded reasons but unfortunately many students did not use the correct terminology. Many of the reasons given for angle OBC = 90°, for example, made no mention of ‘radius’. Statements such as “the angle between a tangent and a circle is 90º” or “the tangent is at right angles to the circumference” were very common. Students who found the size of angle ACB as 90 – 2x but gave no correct reasons were awarded 3 of the 5 marks. A small number of students drew the chord from B to the point where OC cuts the circle and then used other circle theorems, sometimes with complete success. Some students were unable to deal with the ‘x’ and worked with numbers and it was not uncommon to see angles AOB and BOC marked as right angles.
A and B lie on a circle centre O. BC is tangent at B; A, O and C are collinear with O between A and C; angle ABO=x.
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Edexcel 1MA1 · June 2018 · Paper 2F · Q15 · 2 marks
2 marks
Examiner report
For part (a) students’ responses varied considerably with about a third of students able to demonstrate an understanding of the angles facts around isosceles triangles and managing to articulate that x was not a base angle. Some students worked out the value of x to be 54 degrees as a way of showing it not to be 63. A substantial number of students attempted to explain that if x was 63 degrees and the two other angles were the base angles of the isosceles triangle and so also 63 degrees then the triangle would make 189 degrees which is too much. Three letter angle notation was seen, for example stating that it was angles CBA and CAB which were equal however some students used two letter line notation when describing angles e.g. angle AC is equal to angle BC which is not acceptable. A large number of responses related to sides rather than angles with many stating they were ‘parallel’. Some responses demonstrated confusion between the different types of triangle and their properties. A small number of students stated incorrectly that this was not an isosceles triangle. A lack of mathematical reasoning continues to prevent some students from gaining marks when explanations are required. This type of communication is a mandatory requirement of this specification and so centres are advised to practice these types of questions and the associated vocabulary. In part (b) the student could either give the correct reason William used or identify the incorrect reason given and some then attempted to correct the incorrect reason. In this question all approaches were acceptable. Unfortunately, many incorrect answers were seen, for example, many students referred incorrectly to opposite angles being equal or that corresponding angles are not equal. Also, students continue to use incorrect language for alternate angles such as alternative angles or ‘Z’ angles. Mark schemes should be used by centres to see the correct terminology expected in these types of questions.
In triangle ABC, BC = AC, angle ABC = 63 degrees and angle ACB = x. Separately, parallel lines DE and FGH are cut by EG; the acute angle at E is 57 degrees and obtuse angle FGE is y.
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Edexcel 1MA1 · June 2018 · Paper 2F · Q18 · 2 marks
2 marks
Examiner report
Students generally recognised that the transformation shown was a reflection. There were a variety of ways to describe this - reflect, reflected, reflection. Some used the incorrect statement of mirrored or flipped. Many were unsure how to describe the y-axis and a number used expressions such as y = 0, line y or even point (0,0). Some even gave column vectors, thus confusing a reflection with a translation. There was a requirement for a single transformation and many students gave more than one transformation; this did not score any marks.
Shape A has vertices (3,1), (4,1), (4,2), (3,4). Shape B has (-3,1), (-4,1), (-4,2), (-3,4).
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Edexcel 1MA1 · June 2018 · Paper 2F · Q19 · 5 marks
5 marks
Examiner report
This question proved challenging for many students and some blank responses were seen. This was a problem solving question where a fully correct solution would involve finding three costs, for the labour, the fencing on the straight edge and the fencing around the semi-circle. This final aspect was often not found. It was common to see responses that confused area and circumference for the circle. The most successful start seen was to calculate labour cost for three days i.e. £180 × 3 = £540. Also, many of these students were able to work out one cost 50 × 29.86 = 1493 and then add two costs which resulted in the award of three process marks. Other students used a structured approach to calculate the perimeter of the semicircle but forgot to add the diameter length when calculating the total perimeter of the semicircle. Overall students who attempted this question seemed to miss out one aspect of the question. Students are advised, especially on the longer questions, to continually refer back to the question to check they have dealt with all the aspects of the problem set.
A field is a semicircle of diameter 50 m. Fencing costs GBP29.86 per metre plus GBP180 for each of 3 days.
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Edexcel 1MA1 · June 2018 · Paper 2F · Q26 · 5 marks
5 marks
Examiner report
This question was not well answered with many students just working with the dimensions given and calculating 7.2×8.4×18, or sometimes going on to divide this value by 2. Those students who recognised this question required the use of Pythagoras’s Theorem had mixed success. Some students used 8.4² + 7.2² but a good number of correct processes and values were seen. Once the correct values were found students chose one of two options: some just found the volume of the cuboid using height, width and depth of the prism but others chose to find the area of the triangle, they then tended to go on to receive full marks for correctly using this to find the volume of the prism. Summary Based on their performance on this paper, students are offered the following advice: learn and be able to use metric conversion scale factors take care when entering negative numbers into a calculator particularly in conjunction with the ‘square’ button show all working and consider the final answer to check if it is reasonable for the context of the question ensure that the definitions of mathematical terms such as prime and HCF are known practice ‘explain’ and ‘give reason’ type questions
A triangular prism is 18 cm long. Its right-triangle cross-section has leg 7.2 cm and hypotenuse 8.4 cm.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 2H · Q7 · 2 marks
2 marks
Examiner report
This was a really well answered question. Almost all students were able to gain at least one mark for a shape of the correct size and orientation, and a very significant number got the second for also having it in the correct position.
Shape A has vertices (3,4), (3,10), (6,10), (6,7). Enlarge by scale factor 1/3 about (0,1).
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Edexcel 1MA1 · June 2018 · Paper 2H · Q10 · 3 marks
3 marks
Examiner report
It was clear that many students have spent little time working with diagrammatical representations of vectors. Very commonly the vector ˗a rather than ˗2a was drawn, but the most common reason for dropping a mark was by missing the arrow on the vector. When drawing vectors it is essential to indicate direction. In part (b), it was disappointing to see that even some of the more able students were unable to tackle this question. Many either failed to read off the correct column vectors from the diagram or to multiply b by 2. It was very rare to see the vector a + 2b shown on the grid.
Vectors a=(1,2), b=(1,-3).
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Edexcel 1MA1 · June 2018 · Paper 2H · Q13 · 4 marks
4 marks
Examiner report
This was one of the first questions that really separated the most able students from the others. Many were able to score a single mark for finding one correct angle, normally one of the two right angles or for BCD = 180 ˗ y. That was as far as many were then able to go, as to gain the second mark the process needed to be complete. There were a number of routes students could take but many got lost, either by assuming wrongly that BAD was an isosceles triangle, or quite commonly by stating that angle BDE was 90°. When students gained two marks, to get the third they had to correctly state the circle theorems they had used. This part was answered better with many gaining the mark, usually for recognising that either x was acute not obtuse, or that y was obtuse not reflex. Quite a few used the fact that 200 was too big for a triangle and a small number used opposite angles in a cyclic quadrilateral. Students should be careful with clarity in their answers. In this case, some explanations appeared to have the right idea but did not mention x or y.
A, B, C and D lie on a circle centre O in the cyclic order A-B-C-D; the tangent is at D. x=angle BDO and y=angle BAD.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 2H · Q18 · 4 marks
4 marks
Examiner report
Students generally find 3D trigonometry difficult, and that was the case with this question. A high number of students wrongly made the assumption that the base of the cuboid was a square, and tried to use Pythagoras’s Theorem to find AC. Although done correctly, this was an unnecessary step so did not gain any marks. Some students were awarded the first mark for unambiguously identifying the angle CAH on the diagram as the angle to be found. The students that were most successful showed clear working by identifying the correct right triangle from the cuboid and redrawing it in the working space. Centres should encourage their students to draw “intermediate” diagrams to help them get into a problem such as this. There were a significant number of students who, rather than using standard trigonometry, used the sine rule either once or twice. Some students again chose inefficient methods by including an extra step, such as using Pythagoras’ Theorem to find the length AH before using trigonometry.
Cuboid ABCDEFGH has AB=7.3 cm, vertical CH=8.1 cm and base angle BCA=48 degrees.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 2H · Q19 · 5 marks
5 marks
Examiner report
Unsurprisingly at this stage in the paper, many students were unable to gain any marks. However, due to the two part nature of the solution, many who struggled in the process to find the radius, were then often able to then score one mark for their work with surface area. With the first part of the solution most students who dropped marks did so because they didn’t work with the ଵ ସ, either by dividing the formula by 4, or for multiplying 576π by 4 OR they omitted the π from one side of the equation. Many students then struggled with the manipulation of the formula and were unable to find the value of r; in particular a number took the square root rather than cube root as a final step. Few students managed to arrive at a radius of 12, even if a correct formula was used, as the manipulation of algebra in the rearranging of the equation was generally poor. Many took the cube root too early, or did not know how to deal with the inverse of 4/3. Worryingly, many seemed to think that the inverse of multiplication is subtraction. In the second part of the solution, many gained the third process mark for substituting their value of r into ସగయ ସ Many got no further as they either forgot to find include the flat surface areas or did so incorrectly.
S is one quarter of a solid sphere and has volume 576pi cm^3.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 3F · Q12 · 5 marks
5 marks
Examiner report
This question assessed a number of key aspects where it was possible for students to pick up some marks. It was, however, disappointing that about half the candidature failed to gain any marks in this question. The lengths were measured accurately by most who had a ruler and many students were able to find the perimeter from their lengths. Use of the scale or conversion to metres was the step most likely to be missed, leading to the award of only three of the five available marks. This might have been because they thought the value in centimetres for the perimeter was in fact already in meters. A small number of students believed the side lengths to have values of 1 and 200. Some had trouble distinguishing between perimeter and area and subsequently multiplied the lengths they had measured which gained no further marks. There were a number of blank scripts which could have been from the lack of a ruler, or not realising that a single measurement in the accepted range would have scored. On completing a multi-step question students should be encouraged to re-read the question to check they have met all the requirements.
A scale drawing of a rectangular tennis court measures 11.5 cm by 5.8 cm at scale 1:200.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 3F · Q17 · 3 marks
3 marks
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.
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 · Q21 · 2 marks
2 marks
Examiner report
Very few students were able to use the formula for the area of a trapezium to calculate the area of the trapezium. It was very common to see squares being counted, which often led to an incorrect area, however credit was given if these responses then went on to give a triangle of equal area, as long as the figure for the area of the trapezium was clearly stated. The use of a compound approach, through splitting the trapezium into a triangle and parallelogram or two triangles and a rectangle, were usually successful in finding the correct area. The usual misconception seen was when students forgot to divide the product of the triangle base and its perpendicular height by 2. The other misconception observed was when responses were worked in terms of the perimeter. Students should be encouraged to use the grid provided as some responses were drawn off the grid.
A trapezium has parallel sides 2 cm and 7 cm and perpendicular height 4 cm.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 3F · Q23 · 3 marks
3 marks
Examiner report
It was not uncommon to find that part (a) was not attempted. Of those who recognised that it was trigonometry most were able to identify that it was the cosine ratio but some used 11/7 instead of 7/11. Some incorrectly used the rounded value of 0.63 which then led to an answer of 50.9o, which was outside the acceptable range for the final mark. Others found cos (7/11) rather than cos -1 (7/11). The most common incorrect solution was to use Pythagoras to try to find the answer though the area of the triangle was sometimes seen. Part (b) was very poorly answered. The vast majority of those who did attempt it stated 'decrease ' as the reason, mistakenly thinking that the question was about the size of the angle and not the cosine of it. Some assumed that the angle would stay the same. Very few actually compared the two fractions 7/11 and 7/10 but most of those who did this gave the correct reason. Other incorrect answer included explanations that that because the length decreased, cosine or the angle decreased. An extremely common misconception was that they found out what happened to the inverse cosine (which decreased) and gave this as their answer (a decrease).
Right triangle ABC has C = 90 degrees, AB = 11 cm and BC = 7 cm.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 3F · Q26 · 5 marks
5 marks
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.
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 · Q27 · 4 marks
4 marks
Examiner report
This question was attempted by most students but rarely with any success. Those who understood about scale factors usually gained full marks, but they were in the minority. The most common answers were (a) 10.6 and (b) 10.8 obtained from finding the numerical difference between 12.6 and 8.4, that is 4.2 added or subtracted from the lengths. Again, Pythagoras was seen far too often, and some seemed to think that the triangles were identical, writing down 6.4 and 15. The small number of students who understood this concept and were able to find the scale factor (1.5) were generally successful in answering both parts.
Similar triangles ABC and DEF have AB=8.4 corresponding to DE=12.6; AC=6.4 to DF; BC to EF=15.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 3H · Q3 · 2 marks
2 marks
Examiner report
A majority of students scored at least one of the two marks available for responses to this question. There were many correct triangles drawn with area 18 cm², often right-angled triangles with base 6 cm and height 6 cm. Of those students who made an error with the calculation of the area of the trapezium, a substantial number were able to use their area to construct a triangle with the same area. Examiners awarded one mark to these students. Some students relied on counting squares to find the area of the trapezium and errors were often made in these cases.
A trapezium has parallel sides 2 cm and 7 cm and perpendicular height 4 cm.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 3H · Q5 · 3 marks
3 marks
Examiner report
Students were usually able to work out the angle ABC by using the cosine function. A small number of students used Pythagoras’ rule to find the length of the unknown side. They then usually used the sine function successfully to find angle ABC. Some students built some inaccuracy into their calculation by evaluating ଵଵ as 0.63 and then finding an angle outside the range accepted for the accuracy mark. Part (b) of the question was very poorly answered across the ability range. Students often appeared not to have read the question carefully and answered the question “Will the size of angle ABC increase or decrease?” instead of the question asked. Most of the students who did give a correct answer together with an acceptable explanation compared the sizes of the fractions ଵ and ଵଵ sometimes stating that a reduction in the size of the denominator leads to an increase in the value of cos ABC.
Right triangle ABC has angle C=90 degrees, hypotenuse AB=11 cm and side BC=7 cm. AB is then reduced to 10 cm while BC remains 7 cm.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 3H · Q8 · 5 marks
5 marks
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.
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 · Q13 · 3 marks
3 marks
Examiner report
A small minority of students scored full marks for their responses to this question and when the question was completed correctly, answers were usually accurate and within the range of acceptable values. Some students could write down the ratio of lengths or linear scale factor to score the first available mark but could not make any further progress with the question. Weak students merely worked out ଷ ସ of 10 and gave their answer as 7.5 or cubed 3 and 4 to get the ratio of the volumes, thereby failing to realise the need to find the ratio of lengths first. Some very able students worked in exact terms to express the ratio of volumes as 3√3 : 8 before successfully working out the volume of shape A. Students who worked with approximations to scale factors, expressing them in decimal form, usually retained enough accuracy to obtain an answer within the acceptable range of values.
Similar solids A and B have surface-area ratio A:B = 3:4. Volume of B is 10 cm^3.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 3H · Q17 · 5 marks
5 marks
Examiner report
This question was answered well by many students who produced a clearly written and accurate solution. Where students did not score full marks, they often found the length of BD successfully by applying the sine rule. Unfortunately a good number of these students then mistakenly used Pythagoras’ rule on triangle ABD. Students who used the cosine rule for the second stage in finding the length of AD sometimes did not process their calculation using the correct order of operations, instead working out AD² = (11.4² + 7.39² − 2 × 11.4 × 7.39) × cos86. A small proportion of students did not retain enough accuracy in intermediate calculations to obtain a final answer within the acceptable range of values. Some students did not attempt this question.
Triangles DBC and DAB share DB. DC=12.5 cm, angle DBC=109, angle DCB=34; AB=11.4 cm and angle ABD=86 degrees.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2018 · Paper 3H · Q19 · 5 marks
5 marks
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.
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 · June 2018 · Paper 3H · Q21 · 5 marks
5 marks
Examiner report
This, the last question on the paper targeted the most able students sitting the examination. A complete and well reasoned proof that the two triangles are congruent was required to gain full credit in part (a) of the question. Many students showed some understanding of what they needed to do and they often scored the mark for pairing up angle ADP with angle QBC and giving justification that they are both 90° or that it was “given” that they are equal. Statements relating to the pair of sides AD and BC or the pair of angles DAB and BCQ were much less convincing so students more often scored one mark rather than the two marks possible for the complete identification of all three aspects with reasons. Reasons were often incomplete or poorly expressed. For example “AD = BC because they are parallel” was just one response which was seen quite often and could not be accepted. Some students mistakenly argued the case for a proof involving SAS rather than ASA. A logical progression of statements needed in part (b) to explain why AQ is parallel to PC was only rarely seen. Few students used the congruency of the triangles ADP and CBQ to start their explanation. Instead they often stated that APCQ was a parallelogram without any or without sufficient justification. Summary Based on their performance on this paper, students are offered the following advice: practise writing clear explanations, bearing in mind exactly what is asked in the question and what evidence you should give to support your answer. learn standard techniques involving the use of scale factors in the context of similar shapes. practise expanding brackets and collecting terms especially where negative signs are involved. carry out a common sense check on the answers to calculations, so for example you should expect the number of red counters in question 6a to be a whole number. carry out a check of your solution(s) for an equation by substituting them back into the equation. practise solving equations involving algebraic fractions, eg question 7 and question 19 in this paper.
ABCD is a parallelogram. A-B-P and Q-D-C are straight lines. Angles ADP and CBQ are both 90 degrees.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 1F · Q8 · 2 marks
2 marks
Examiner report
This question was accessible to all students and a good proportion of fully correct answers were seen. Many students drew the rectangle 8 by 4 without any working out being shown. Some students applied the ratio 2 : 1 incorrectly and drew a rectangle 3 by 6 as a result, this scored 1 mark. Other students initially struggled to find factors of 32 but resiliently kept dividing by 2 to find them which was pleasing to see.
The length of a rectangle is twice as long as the width of the rectangle. The area of the rectangle is 32 cm2. Draw the rectangle on the centimetre grid.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 1F · Q13 · 4 marks
4 marks
Examiner report
This question was not well answered and many responses were blank. Surface area is an expected skill and knowledge that a cube has six faces is also required but most students failed to realise this and so did not divide 294 by 6. Even when drawing a diagram students often still could not see the 6 faces. Some divided by 4 or 2 or even divided by 2 and divided by 4, adding the answers together, in the possible belief that this was the same as dividing by 6. Of those few that did divide by 6 most found the answer 49, although 48 was a popular incorrect figure, most did not realise this was the answer for the area of a square face and did not go on to find the square root. Some other incorrect methods included 49 × 49 × 49 and 294 × 294
The total surface area of a cube is 294 cm2. Work out the volume of the cube.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 1F · Q25 · 4 marks
4 marks
Examiner report
This question often allowed students to score part marks as many could identify at least one angle in the correct place. Others were then able to go on to show that the angles in the triangle ABF must be 35, 75 and 70. When given, the more straightforward reasons for angles were generally clear and correct for example, opposite angles, angles in a triangle and angles on a straight line were often correctly explained. However, reasons relating to parallel lines or opposite angles in a parallelogram were often not seen and so full marks could rarely be awarded. A minority of students incorrectly assumed that triangles ABF or DEF were isosceles. Centres should remind students of the need to ensure they clearly label their angles either on the diagram or using 3 letter notation. Also to give full reasons for their working and only give the reasons they actually use.
ABCD is a parallelogram. E,D,C are collinear; F lies on AD; B,F,E are collinear. Angle EFD=35 degrees and angle DCB=75 degrees. Show angle ABF=70 degrees, giving reasons.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 1F · Q26 · 4 marks
4 marks
Examiner report
This question was not well answered. A high proportion of students did not attempt the question at all. Many of the answers seen tried to compare only the radii whilst another common response was to say, incorrectly, that ‘one circle out of 3 is shaded so 1 is shaded’ with no further explanation offered. Also comments such as ‘Daisy is correct there are 3 sections to the logo and one is shaded’ were very common Few students used the area of a circle formula and of those that did, it was exceedingly rare that pi was cancelled in any comparison or working. Where marks were gained, it was for calculating the area of one circle, usually using the radius of 10. Centres should discuss with students that if a question appears to require multiple calculations with pi on a non-calculator paper that an alternative method is likely to be available. The thought process required for this question did appear to be beyond the majority of students entered at this level in this November session.
A logo is three concentric circles: inner radius 4 cm, shaded annulus from radius 4 cm to 7 cm, and outer radius 10 cm. Daisy says exactly 1/3 is shaded. Is she correct? Show working.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 1H · Q3 · 4 marks
4 marks
Examiner report
Students continue to struggle with geometry questions in forming complete solutions with full reasoning. There were a great number of methods evident within the responses seen, and most students were able to score some marks. For the B1 it was typically for 35 marked on the diagram at AFB, although any correct angle marked gained this mark. 2 marks was a common score, either B1C1 for an angle marked and a correct reason, or for a correct pair marked as per the mark scheme. However, many who scored B1M1 also scored C1 for an appropriate reason. It continues to be disappointing to see students with fully correct working drop marks due to incorrect reasons stated, or reasons incorrectly stated.
ABCD is a parallelogram. E,D,C are collinear; F lies on AD; B,F,E are collinear. Angle EFD=35 degrees and angle DCB=75 degrees. Show angle ABF=70 degrees, giving reasons.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 1H · Q4 · 4 marks
4 marks
Examiner report
The second problem question on the paper proved to be a huge challenge for many in this cohort. In particular students providing just a written explanation relating to 1 circle out of 3 so it is 1 . Those who started to work with πr2 typically got at least 2 marks for a method to find the shaded area, which was very pleasing to see. Of those who got this far it was about even as to whether they could then draw a correct conclusion and compare to 1 . Noticeable was the error that occurred when students believed 33 to be equal to 1 .
A logo is three concentric circles with radii 4 cm, 7 cm and 10 cm. The annulus between 4 cm and 7 cm is shaded. Is exactly 1/3 of the logo shaded? Show working.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 1H · Q18 · 2 marks
2 marks
Examiner report
A very disappointing performance on this enlargement question. Most students completed either an enlargement of scale factor 2 or 1 , and as such scored zero marks. Few had a method or understanding to arrive at a triangle in the correct orientation and size. This resulted in very few marks being awarded.
Triangle P has vertices (2,7), (2,3), (4,3). Enlarge it by scale factor -1/2 about (0,0), and label the image Q.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 1H · Q19 · 4 marks
4 marks
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.
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
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.
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 · Q22 · 5 marks
5 marks
Examiner report
It was good to see a number of students having some understanding of similar shapes and awarding 2 marks for a value of x =2 correctly found wasn’t uncommon. Almost no students were then able to deduce the second assumption that could be made and as such no further marks were awarded.
In triangle ACE, B lies on AC with AB=8 cm and BC=x cm; D lies on AE with AD=12 cm and DE=3 cm. Triangles ABD and ACE are similar. Find both possible values of x, stating the two possible side correspondences.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 1H · Q23 · 5 marks
5 marks
Examiner report
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.
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 · Q9 · 3 marks
3 marks
Examiner report
Part (a) was done well, with the most common answer being 2.75. Most showed 5.5 × 0.5; there was some evidence that students did not have a ruler, and were guessing the length. There were a few students who, having shown 5.5 × 0.5, then gave their answer as 3. Students should be encouraged to write their accurate answer and not round it. In part (b) some students appeared confused as to which angle they were giving the bearing for. But even when it was clear the correct angle was being found, the protractor was being read wrong (eg 130). There was also some evidence that students were estimating the angle (perhaps because they did not have a protractor).
An accurately drawn map joins Backley to Cremford. The measured road is 5.3-5.7 cm and the scale is 1 cm to 0.5 km. Its clockwise bearing from north at Backley is 128-132 degrees. Find the distance and bearing.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 2F · Q10 · 3 marks
3 marks
Examiner report
In part (a) is was not uncommon to find students missing off the units from their answer; this was far more frequent than those giving an incorrect unit (such as cm). There was some evidence of counting squares to get to the numerical answer, but this usually led to an inaccurate value. It was disappointing that a significant minority did not halve their value of 24. In part (b) many students had difficulty in naming the shape, giving almost any quadrilateral other than “kite”.
On a centimetre grid, triangle P has base 4 cm and perpendicular height 6 cm. Quadrilateral Q has two pairs of adjacent equal sides. (a) Find P's area. (b) Name Q.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 2F · Q18 · 3 marks
3 marks
Examiner report
Few students scored any marks in part (a), with the commonest answer being 10, from 80 . Some scored 1 mark for working with the circumference; no marks were gained from using the area formula. Part (b) was better answered with about half gaining credit from recognising that some aspects remained constant. Unfortunately, many answered “yes” because they thought the circumference had changed.
A circle has diameter 80 m and 8 marked points. (a) Initially the points are equally spaced; find the arc distance between neighbours. (b) After four points move around the same circle, does the mean neighbour-to-neighbour distance change? Explain.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 2F · Q20 · 2 marks
2 marks
Examiner report
Many students met with some success in this question. In part (a) there were some students who rotated the shape by 90° rather than 180°. Sometimes the shape was not accurately drawn in the correct position. In part (b) students were not careful enough counting squares, and sometimes positioned the shape within one square of what was needed. Some failed to take account of the minus signs in determining direction of move.
Trapezium T has vertices (2,-1),(4,-1),(5,-2),(2,-2). (a) Rotate T 180 degrees about the origin. (b) Translate T by vector (-1,-3).
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 2F · Q22 · 5 marks
5 marks
Examiner report
Only the best students were able to work through to a correct solution, but part marks were awarded to those who attempted to do something of worth with the diagram. Some started with Pythagoras on the left hand right-angled triangle, but of course only earned marks if it was of the form 7.52−62 (ie not added). It was not uncommon to find some attempting to find the area of the trapezium, which of course earned no marks. There was some (independent) credit for working with trigonometry. This could be done in the left-hand triangle (if the angle was made clear) or in the right-hand triangle (with their stated value for the base). But only a minority of students realised that trigonometry was needed.
ABCD is a trapezium. 10 cm B C 7.5 cm 6 cm A D 24 cm Work out the size of angle CDA. Give your answer correct to 1 decimal place.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 2H · Q3 · 3 marks
3 marks
Examiner report
For those students who realised that they needed to use the formula for the circumference of a circle, and could recall it, this question proved to be straightforward. It was surprising to see that a significant number of students instead opted to divide the diameter by 8 to give 10 as their answer. In part (b) students who stated that the total distance or that the number of points remained constant were awarded the mark available. It was encouraging to see that this part of the question was answered quite well, sometimes even after an incorrect response to part (a). Students generally gave a clear decision coupled with a clearly expressed reason. Where this was not the case, students often referred to larger and smaller gaps between the points. Students who thought that the mean distance would change used the reason that the points would not now be equally spaced.
A circle has diameter 80 m and 8 marked points. (a) Find the equal arc distance. (b) If four points move around the same circle, does the mean neighbour gap change? Explain.
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Edexcel 1MA1 · November 2017 · Paper 2H · Q5 · 2 marks
2 marks
Examiner report
Students found part (a) involving the rotation of the trapezium more accessible than part (b) where they had to translate the shape. However, it was disappointing to see the number of responses to part (a) where the trapezium A was placed in an incorrect quadrant and examiners were left wondering why more students had not used tracing paper to help them. Part (b) was not well done.
Trapezium T has vertices (2,-1),(4,-1),(5,-2),(2,-2). (a) Rotate 180 degrees about the origin. (b) Translate by (-1,-3).
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Edexcel 1MA1 · November 2017 · Paper 2H · Q7 · 5 marks
5 marks
Examiner report
A good number of students did realise that this problem required the use of Pythagoras’ theorem and trigonometry but were not always able to apply them correctly. Students who realised that using Pythagoras’ theorem was the most efficient way to start the problem usually gained the first 2 marks, although some students calculated 7.5² 6² instead of 7.5² െ 6². A common error was for students to subtract 10 from 24 to find the missing base length of the right angled triangle needed to find angle CDA. Students who clearly showed a correct use of trigonometry to find angle CDA were awarded a process mark even if they had used an incorrect value for this length, for example 14. It was encouraging to see a good number of fully correct solutions to this multi-step question.
ABCD is a trapezium with parallel bases 10 cm and 24 cm. The sloping side is 7.5 cm and height 6 cm. Find angle CDA to 1 decimal place.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 2H · Q12 · 3 marks
3 marks
Examiner report
Many students were able to score 1 mark for calculating the size of an exterior angle of the regular polygon (360 ÷ 12) or for calculating the size of an interior angle (1800 ÷ 12) and about a half of these students were able to complete the question correctly to find the required angle. Evidence of confusion between exterior and interior angles was relatively rare. A significant number of students gave their final answer as 150, suggesting, possibly, that they didn’t understand the angle notation used.
In a regular 12-sided polygon, RS and ST are adjacent sides and RT is a diagonal. Find angle SRT, showing working.
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Edexcel 1MA1 · November 2017 · Paper 2H · Q19 · 4 marks
4 marks
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.
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 · Q20 · 5 marks
5 marks
Examiner report
Some students were able to score one mark for calculating the area of the sector or for identifying a right angle between a radius and a tangent or two marks for both. A significant number of students wrote down a correct expression for the area of a circle of radius 10 cm but then did not work out the correct fraction of the circle. Few students were able to give a correct method to find a length in order to calculate the area of the kite. There were a relatively small number of fully correct answers.
A circle has centre O, radius 10 m and angle AOC=120 degrees. Tangents from B touch at A and C. Find the area between the two tangents and minor arc AC, to 3 s.f.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 3F · Q7 · 4 marks
4 marks
Examiner report
Although a few students tried to show that the triangle was isosceles by assuming that it was and working ‘backwards’, most students did try to find the size of the required angles BCA and CAB, using a standard approach. Quite often reasons were omitted, or those given were incomplete, usually omitting the word ‘angles’. Many students tried to use ‘angles on a straight line’ to find angle BCD, only to show incorrect working of 180 – 117 – 54 = 9 Subsequent working to find angle CAB was then incorrect and gained no credit. Having correctly found the size of the required angles, together with correct reasoning, many failed to complete the solution with a statement explaining why the triangle is isosceles. There were regular mentions of two equal sides but not angles. Many candidates just explained that the triangle is isosceles because it has two equal angles without also giving the reasons relating to how they calculated the angles. Others gave the reason that an isosceles triangle has two equal sides rather than two equal angles. Some students incorrectly assumed that angle BCA was 54o stating the reason "base angles of an isosceles are equal".
A 54 degrees 117 degrees B C D BCD is a straight line. ABC is a triangle. Show that triangle ABC is an isosceles triangle. Give a reason for each stage of your working.
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Edexcel 1MA1 · November 2017 · Paper 3F · Q8 · 2 marks
2 marks
Examiner report
30m was the modal correct answer for this question, where the height of the building was estimated at two and a half bus lengths. Many students however used the height of the bus for comparison with the height of the building and made many errors. It was not enough for the award of the method mark to simply say that there were about 6 to 7 bus heights equivalent to the height of the building. Students had first to make a sensible comparison between the length of the bus and its height, possibly using the scale of 2cm = 12m.
12 m The picture shows a bus next to a building. The bus has a length of 12 m. The bus and the building are drawn to the same scale. Work out an estimate for the height, in metres, of the building.
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Edexcel 1MA1 · November 2017 · Paper 3F · Q14 · 2 marks
2 marks
Examiner report
Many students gained the one mark for correctly describing the transformation as a reflection. Although a lot of students did not use the correct terminology instead using ‘mirrored’ or ‘flipped’ so getting no mark. However significantly fewer were able to fully describe it. Reflection in the line x or about the origin were common errors. Several appeared to refer to the x-axis as x = 0.
Shape B is the mirror image of shape A across the x-axis. Describe the transformation fully.
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Edexcel 1MA1 · November 2017 · Paper 3F · Q16 · 5 marks
5 marks
Examiner report
Both parts of this question were poorly answered. Confusion with the conversion between metres and centimetres prevented very many students completing a solution. Many found the volumes and but then divided the volume of a box by the volume of the van. A significant number of students correctly attempted to find the maximum number of boxes by finding the correct numbers, (6, 5 and 4), fitting in each dimension of the van. Unfortunately, many simply added these numbers (to get 15) or selected the lowest value and divided by 3 to find the time. Many students calculated the time by dividing the least number (in this case 4) by 3. Some students calculated surface areas rather than volumes. In part (b), very few students answered the question given. A correct response described how the time might change (greater or less) together with a coherent reason. Many students simply said that the time will change, without reason, or argued that there will be less boxes or it will be more difficult to fill the van without ever mentioning the time implication. Centres are advised to give their students more guidance on how to fully answer this type of questions.
Chloe has a van. She is going to use the van to deliver boxes. Each box is a cuboid, 40 cm by 30 cm by 35 cm. 35 cm 30 cm 40 cm The space for boxes in the van has maximum length 2.4 m maximum width 1.5 m maximum height 1.4 m The space for boxes is empty. Chloe wants to put as many boxes as possible into the van. She can put 3 boxes into the van in one minute. Assume that the space for boxes is in the shape of a cuboid. (a) Work out how many minutes it should take Chloe to put as many boxes as possible into the van. The space for boxes might not be in the shape of a cuboid. (b) Explain how this could affect the time it would take Chloe to put as many boxes as possible into the van.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2017 · Paper 3F · Q24 · 5 marks
5 marks
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.
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.
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Edexcel 1MA1 · November 2017 · Paper 3H · Q6 · 5 marks
5 marks
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.
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 · Q14 · 3 marks
3 marks
Examiner report
Relatively few students showed that they understood the relationships between lengths, areas and volumes in similar figures. Those who did recognise that they should use the ratio of the volumes, 27 : 8, to find the ratio of the lengths or the length scale factor were usually able to give a complete method to show that the surface area of cone B is 132 cm2. Many students assumed the result they were given instead of proving it.
Cone A and cone B are mathematically similar. The ratio of the volume of cone A to the volume of cone B is 27 : 8 The surface area of cone A is 297 cm2 Show that the surface area of cone B is 132 cm2
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Edexcel 1MA1 · November 2017 · Paper 3H · Q17 · 5 marks
5 marks
Examiner report
In order to start this question and work out the length of CD, students needed to recall that the area of a triangle is given by 2 1 absinC. Those that did recall this correctly often gained the first mark for writing a correct statement such as 0.5 × 11 × CD × sin105 = 56. Mistakes, though, were often made when rearranging to find CD. Those students that did not show a correct process to work out CD were still able to gain subsequent process marks - one mark for using the cosine rule to work out AC and one mark for using the sine rule to work out AB. A number of students assumed that the triangles were right-angled and tried to find CD by using base (AD) × height (CD) ÷ 2 = 56. It was not uncommon to see students attempting to use Pythagoras and SOHCAHTOA when trying to work out the length of AC.
Triangles ABC and ADC share AC. In ABC, angles B=118 and C=48. In ADC, angle ADC=105, AD=11 m and area=56 m^2. Find AB to 1 d.p.
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Edexcel 1MA1 · November 2017 · Paper 3H · Q20 · 4 marks
4 marks
Examiner report
Some of the students that started by drawing in the radius OC to give two isosceles triangles were able to go on and show that angle ACB = 90º. Some did so by introducing algebraic notation whereas others used angle notation. It was pleasing to see some very good attempts but these were few in number. Students should note that this type of geometric proof does require full and correctly worded reasons to be given. It is not enough to state, for example, that angle OAC = angle OCA, it is also necessary to give a reason why. Those students that showed that angle ACB = 90º but gave no reasons or incomplete reasons were awarded 3 of the 4 marks. Some students ignored the statement “You must not use any circle theorems in your proof” and focused on angles in a semicircle.
A,B,C lie on a circle centre O and AOB is a diameter. Prove angle ACB=90 degrees without circle theorems.
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Edexcel 1MA1 · November 2017 · Paper 3H · Q21 · 5 marks
5 marks
Examiner report
Some students scored one mark for AB = b – a or BA = a – b but few were able to make any further meaningful progress. Those that did were most likely to find a correct expression for MN. Few students wrote that AP = k(b – a ) which meant that correct expressions for MP and PN were rare. Mistakes were sometimes made with the direction signs of the vectors. Summary Based on their performance in this paper, students should: • Read the information given in each question very carefully. • Practise solving problems that require an algebraic approach. • Use a calculator to work out accurate values without rounding or truncating early. • Practice using their knowledge in different ways and in a wide variety of contexts. • Recognise which rules and theorems are associated with triangles without a 90º angle and use them appropriately. • Give correctly worded reasons when presenting a geometric proof.
O,A,N; O,M,B; and A,P,B are straight. AN=2OA, M is midpoint OB, OA=a, OB=b, and AP=kAB. Given M,P,N are collinear, find k.
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Edexcel 1MA1 · June 2017 · Paper 1F · Q15 · 4 marks
4 marks
Examiner report
Another problem solving question and another case of most students being able to gain marks. It is still surprising to see so many students being unable to find the area of a triangle correctly; this needs to be another focus point for centres. Those who failed to recall the correct formula automatically lost two marks on this question. This was because the first mark was for the process to find the area of the triangle and then if they had made a mistake there, invariably their final answer was wrong. However, the other two marks were accessed by most. In fact a very large proportion of students understood the processes required to solve this problem: area of triangle; multiply by 6; divide by 16. Again, as is the case through the whole paper, poor arithmetic caused the final mark to be rarely awarded, in particular the division by 16 causing problems for many.
Here are a triangle and a rectangle. 9 cm 16 cm 8 cm The area of the rectangle is 6 times the area of the triangle. Work out the width of the rectangle.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2017 · Paper 1F · Q18 · 5 marks
5 marks
Examiner report
This problem solving question involved estimation in a context. As with the previous problem questions most were able to get some marks for suitable processes. Typically these marks were awarded for evidence of estimation or for dividing an area by a coverage figure. Many students were unable to correctly recall the formula for the area of a circle; this is a formula that must be learnt by students. For some reason, a number of students thought that they needed to use 360o somewhere in their solution. In part (b) the common response was to discuss the rounding done in the final step, i.e. rounding to a whole number of bags, which was not the essence of the question. To gain marks students had to discuss the effect of their estimation of either π or the coverage figure. Commonly when discussing the latter, students wrongly stated that rounding up to 50 would give an overestimate, where it would actual bring about an underestimate.
Balena has a garden in the shape of a circle of radius 10 m. He is going to cover the garden with grass seed to make a lawn. Grass seed is sold in boxes. Each box of grass seed will cover 46 m2 of garden. 10 m Balena wants to cover all the garden with grass seed. (a) Work out an estimate for the number of boxes of grass seed Balena needs. You must show your working. (b) Is your estimate for part (a) an underestimate or an overestimate? Give a reason for your answer.
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Edexcel 1MA1 · June 2017 · Paper 1F · Q25 · 5 marks
5 marks
Examiner report
Another problem solving question and again many students were able to gain some credit. The less able students tended to gain just one mark for multiplying lengths by 1.5. The more able realised the need to use Pythagoras’s theorem and most who did, did so successfully. Although a good number were able to get full marks a large number of those who showed the correct processes were again let down by their arithmetic skills, or by failing to multiply the diagonal by 1.5
This rectangular frame is made from 5 straight pieces of metal. 5 m 12 m The weight of the metal is 1.5 kg per metre. Work out the total weight of the metal in the frame.
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Edexcel 1MA1 · June 2017 · Paper 1F · Q27 · 3 marks
3 marks
Examiner report
Vector geometry is new to foundation specification. Very few students understood what to do at all and as a result very few marks were scored. Summary Based on their performance on this paper students should: continue to work to master the basic skills of arithmetic, including being able to correctly choose use the four operations with integers, fractions and decimals remember to show each stage of their working, especially when using build up methods for percentages work further on strategies for attempting problem solving skills spend greater time on the topics that are new to the foundation tier in this specification learn the conversions within the metric system. E.g. 1 m = 100 cm learn formulae such as those for the area of a triangle, area and circumference of a circle
ABCD is a parallelogram whose diagonals meet at O. Vector OA = a and vector OB = b. (a) Find vector DB in terms of b. (b) Find vector AB in terms of a and b. (c) Find vector AD in terms of a and b.
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Edexcel 1MA1 · June 2017 · Paper 1H · Q5 · 5 marks
5 marks
Examiner report
A significant minority of candidates found the area of the rectangle and then multiplied this by the 1.5, gaining no marks. However the majority who used Pythagoras’s theorem were successful in at least gaining the marks for a Pythagorean approach; some were unable to state the square root of 169. Many were able to go on and complete the question, though there were arithmetic errors, whether candidates found individual masses and added, or added the lengths first and multiplied by 1.5. When applying the latter method those finding it as 47+ 23.5 were more successful than those attempting to multiply 47 by 1.5. Some divided by 1.5 instead of multiplying while others included the diagonal twice, but overall this question was very well answered. Candidates need to read the question carefully; for example, an error for some was not appreciating that this was a rectangular frame as attempts to find the area were seen.
This rectangular frame is made from 5 straight pieces of metal. 5 m 12 m The weight of the metal is 1.5 kg per metre. Work out the total weight of the metal in the frame.
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Edexcel 1MA1 · June 2017 · Paper 1H · Q15 · 4 marks
4 marks
Examiner report
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.
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 · Q18 · 4 marks
4 marks
Examiner report
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).
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 · Q19 · 4 marks
4 marks
Examiner report
Those candidates who were able to understand the problem scored the first mark for correctly adding X and D to the diagram or by making a relevant correct first step; expressing vector AD in terms of a and c. Many candidates that got this far were able to start to use a vector equation using vector CD, however some candidates expressed this incorrectly by using incorrect signs that stopped any further progress. Of those candidates who were able to show that vector CD was 2.5c, some were perplexed as to how they related this to the proportional part of the question. Of the candidates who were able to recognise the proportional connection, they often left their final solution as 1 ÷ 2.5 and lost the final accuracy mark.
OABC is a parallelogram with vectors OA=a and OC=c. X is the midpoint of AC. O,C,D are collinear and OC:CD=k:1. Given vector XD=3c-(1/2)a, find k.
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Edexcel 1MA1 · June 2017 · Paper 1H · Q21 · 4 marks
4 marks
Examiner report
Very few candidates were prepared for this question. It was common for candidates to gain two marks for attempting to use the information given in the question, that is AB = CD and ABC = BCD but they usually omitted to state BC was shared or common to both. Many used the wrong pair of triangles for their congruent pair and some thought that the equal symbols on AB and CD meant they were parallel. Candidates often omitted essential detail to complete a formal proof for congruence so failing to get the final mark.
ABCD is a quadrilateral. B C A D AB = CD. Angle ABC = angle BCD. Prove that AC = BD.
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Edexcel 1MA1 · June 2017 · Paper 1H · Q22 · 5 marks
5 marks
![Edexcel 1MA1 · June 2017 · Paper 1H · Q22 · 5 marks — ABEF and CBED are congruent parallelograms with AB=BC=x cm and angle ABC=30 degrees. P lies on AF and Q on CD, with BP=BQ=10 cm. Prove cos(angle PBQ)=1-[x^2(2-sqrt(3))]/200.](/api/paper-crops/1ma1/2017-06/higher-paper-1/q22/question-r1-81a1f3bc.webp)
Examiner report
This question was set as a differentiator for those aiming for grade 9, so it was not unexpected to find that many candidates did not answer the question at all. Others found every angle going in order to get something down on paper. Several candidates tried to use circle theorems to help them, particularly ‘the angle at the centre is twice the angle at the circumference’ even though there was no circle. Some attempted to work backwards, but unless they realised a connection with the Cosine Rule it was unlikely any marks were earned. Some thought that the 10 cm was the length covered by BAP, so that AP = 10 – x. Some candidates picked up a mark for stating cos 30 = , but not all of those who wrote down a value for cos 30 got it correct. Some didn’t know the cosine rule well enough, or couldn’t rearrange it to get cos PBQ. At this level, candidates need to either label their diagrams clearly or use proper terminology for the sides and angles in the formula in order to make their work easier to follow. Many stuck to the a, b and c in the formula a2 = b2 + c2 −2bccosA despite these letters being used for other lengths/angles on the diagram. There was some assumption that cos PBQ = cos 30. It was either substituted on the LHS of the formula they had to prove, or substituted in to the cosine rule they were using to find PQ, usually shown as PQ2 = 102 + 102 – 2 × 10 × 10 × cos 30, despite the sides of length 10 and the angle 30 being from different triangles. Despite all of the above, there were some extremely well presented solutions. Summary Based on their performance on this paper, students should: present their work logically and in an organised way on the page, sufficient that the order of the process of solution is clear and unambiguous include working out for all questions where appropriate practise their arithmetic skills and learn multiplication tables; these are essential particularly on a non-calculator paper practise algebraic manipulation and derivation, percentage calculations and application of ratios and proportion spend more time ensuring they read the fine detail of the question to avoid giving answers that do not answer the question.
ABEF and CBED are congruent parallelograms with AB=BC=x cm and angle ABC=30 degrees. P lies on AF and Q on CD, with BP=BQ=10 cm. Prove cos(angle PBQ)=1-[x^2(2-sqrt(3))]/200.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · June 2017 · Paper 2F · Q9 · 3 marks
3 marks
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”.
Angles around a point are 3x degrees, 2x degrees and 90 degrees. Find x.
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Edexcel 1MA1 · June 2017 · Paper 2F · Q13 · 4 marks
4 marks
Examiner report
Many students drew an accurate diagram showing the correct position of the quadrilateral in response to part (a) of this question. Of students who did not give a fully correct answer, some were able to draw a shape with the correct orientation but not placed in the correct position in the diagram. A small minority of students carried out the rotation anticlockwise and some students rotated the shape by 180°. Naming the type of transformation used in part (b) was straightforward for many students but only a small proportion of them could describe the detail of the reflection by stating “in the y axis” or “in x = 0”. Though most students avoided naming more than one transformation (which would have led to the award of no marks) there were many cases where they added details of a transformation other than reflection eg “in (0, 0)”. Weaker students often replaced “reflection” with terms such as “flipped over” which, as usual, was not accepted as a correct description of the transformation.
Shape A has vertices (1,-4), (1,-3), (2,-1), (4,-4). (a) Rotate A 90 degrees clockwise about the origin. (b) Triangle B has vertices (-5,-2), (-1,-2), (-1,1); triangle C has vertices (1,-2), (5,-2), (1,1). Describe fully the single transformation mapping B to C.
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Edexcel 1MA1 · June 2017 · Paper 2F · Q16 · 3 marks
3 marks
Examiner report
The use of the construction of a circle centered on A and the perpendicular bisector of BC was not often seen and this inevitably reduced the number of students being able to accurately locate at least one of the two possible positions for T. More students drew a suitable circle than drew a perpendicular bisector but there were also many answers, some successful in showing the position of T accurately, where there was no evidence of either. The mark awarded for drawing at least part of a circle radius 2.5 cm or for showing a point 2.5 cm from point A was accessed by a high proportion of students but examiners did not award the mark for T being equidistant from point B and point C as frequently.
On a map, 1 cm represents 100 m. Point T is 250 m from A and equidistant from B and C. Construct one possible position for T.
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Edexcel 1MA1 · June 2017 · Paper 2F · Q19 · 4 marks
4 marks
Examiner report
Many students were able to score at least 2 of the 4 marks available for drawing accurate elevations. For these students it was usually the case that they gained both marks for a correct front elevation but then scored at most one mark for the side elevation. Sometimes students drew the front elevation only. Side elevations often consisted of rectangles with incorrect dimensions, most commonly a rectangle with height 1 cm. Those students who drew a correct 2 cm by 4 cm rectangle for the side elevation often forgot to draw in a solid line at height of 1 cm. A significant number of students drew at least one correct elevation, usually the front elevation, then added on to their diagram in an attempt to show this as part of a 3 dimensional figure.
A prism has trapezium front cross-section with base 2 m, parallel vertical sides 0.5 m and 2 m, and length 1 m. Draw front and side elevations at scale 2 cm to 1 m.
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Edexcel 1MA1 · June 2017 · Paper 2F · Q21 · 4 marks
4 marks
Examiner report
Students taking this paper seemed unfamiliar with the techniques needed to solve problems such as this involving similar triangles. It was rare to see an attempt involving the use of scale factors or ratios. Instead many students mistakenly thought they could apply Pythagoras’s theorem to one of the triangles despite there being no indication they were right angled. A sizeable proportion of students merely subtracted lengths to find AE, for example, 8.1 − 2.6 = 5.5. 2.6 was quoted as the length of AE by some students and examiners wondered whether they had confused the arrows on sides EA and DB with the notation used for showing lengths are equal. Students may have found it helpful to redraw the diagram as two separate similar triangles
A B C E D ABC and EDC are straight lines. EA is parallel to DB. EC = 8.1 cm. DC = 5.4 cm. DB = 2.6 cm. (a) Work out the length of AE. AC = 6.15 cm. (b) Work out the length of AB.
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Edexcel 1MA1 · June 2017 · Paper 2H · Q3 · 4 marks
4 marks
Examiner report
The front elevation of a trapezium was usually correctly drawn even when shown in a 3D configuration. The side elevation however was very poorly done; a 2cm by 1cm rectangle or a 5cm by 2cm rectangle were the most common errors seen. Even when a correct 4cm by 2cm rectangle was drawn the 2 cm line, 1cm above the base, was often omitted. The students who drew the diagrams side by side had a better success rate as they seemed to have a better sense that they should be the same height. Unfortunately, many students still fail to include the horizontal line in the side elevation. Some students drew nets of the prism and rarely gained any credit.
A prism has trapezium front cross-section with base 2 m, parallel vertical sides 0.5 m and 2 m, and length 1 m. Draw front and side elevations at scale 2 cm to 1 m.
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Edexcel 1MA1 · June 2017 · Paper 2H · Q5 · 4 marks
4 marks
Examiner report
This question was poorly answered even by many of the more able students. In part (a), many incorrectly used a ratio 2 : 1 (5.4 : 2.7) and many assumed the triangles to be right-angled at A and B and tried to apply Pythagoras’s theorem or trigonometric methods. In part (b), the most common error for those students gaining any credit was simply to divide 6.15 by the correct ratio 1.5 and then leave their answer as 4.1 without ever trying to find the length of AB.
A B C E D ABC and EDC are straight lines. EA is parallel to DB. EC = 8.1 cm. DC = 5.4 cm. DB = 2.6 cm. (a) Work out the length of AE. AC = 6.15 cm. (b) Work out the length of AB.
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Edexcel 1MA1 · June 2017 · Paper 2H · Q9 · 3 marks
3 marks
Examiner report
The majority of students gained at least one mark for a correct reflection of A in the x-axis. Disappointingly this was the only mark many achieved through not being able to draw the line y = x, some drawing y =−x by mistake. Even where y = x was correctly drawn, a large number of students had difficulty in reflecting in a diagonal line. More able students usually gained full marks.
The diagram shows triangle A drawn on a grid. 5 4 3 A 2 1 -1 -2 -3 -4 -5 1 2 3 4 5 O -1 -2 -3 -4 -5 Kyle reflects triangle A in the x-axis to get triangle B. He then reflects triangle B in the line y = x to get triangle C. Amy reflects triangle A in the line y = x to get triangle D. She is then going to reflect triangle D in the x-axis to get triangle E. Amy says that triangle E should be in the same position as triangle C. Is Amy correct? You must show how you get your answer.
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Edexcel 1MA1 · June 2017 · Paper 2H · Q15 · 3 marks
3 marks
Examiner report
Many students tried to prove the two triangles were congruent instead of similar and invented lengths of sides which they argued were equal. Clearly proving two triangles similar was not well known. Angles not in the required triangle were often quoted. AC and BD were often assumed to be diameters. Many assumed that the lines AB and DC were parallel and gave “alternate angles” as their reasons for what they saw as equal angles. Angles subtended by an arc being equal in size was a reason that was accepted (but not by a chord as this is ambiguous as a chord subtends in both the major and minor segments) but any reference to a “bow tie theorem” was not. Many references to cyclic quadrilaterals were seen, usually gaining no marks as the wrong angles were identified and wrong reasons stated. Many students gained their only mark for correctly identifying angles AEB and DEC as being equal because they were vertically opposite angles. Quite a few students thought these opposite angles were right-angles. Students need to learn the accepted version of the circle theorems. The use of three letters to identify angles was used by most students and some labelled their diagrams for additional clarification.
A, B, C and D are four points on the circumference of a circle. A B E C D AEC and BED are straight lines. Prove that triangle ABE and triangle DCE are similar. You must give reasons for each stage of your working.
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Edexcel 1MA1 · June 2017 · Paper 2H · Q17 · 5 marks
5 marks
Examiner report
It was pleasing to see many students give fully correct solutions to this question. However, many did not and errors included; failure to recognise an angle of 60o in an equilateral triangle, taking 7 cm as the height of the triangle of base 7 cm when finding its area, using an incorrect formula, eg. 2 (a +b) sin C, using 5.5 cm or 7 cm as the radius of the sector OQN and some students found the length of the arc rather than area of the sector. The cosine rule was sometimes used unnecessarily to find angle AOB and the height of the triangle was often found using Pythagoras’s Theorem. Some found the area of the triangle as a percentage of the whole sector. Quite a number of students found the area of the whole circle but did not find the area of the sector. In this question, premature rounding prevented the award of full marks on many occasions.
ONQ is a 60-degree sector of a circle with centre O and radius 11 cm. A lies on ON and B on OQ such that triangle AOB is equilateral with side 7 cm. Find the shaded area (sector minus triangle) as a percentage of the sector area, to 1 decimal place.
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Edexcel 1MA1 · June 2017 · Paper 2H · Q21 · 4 marks
4 marks
Examiner report
This question was the least well answered question on the paper, although a great number of students gained one mark for giving a correct length of 96 mm for the rectangle. The most common error was in thinking that the rectangle was a 96 mm by 96 mm square. Very few students recognised the need to find the height of the equilateral triangle formed from the three centres, preferring instead to find the areas of the three circles and then being unable to find the remaining areas. This question highlighted a lack of experience of enriching tasks where knowledge has to be applied in challenging situations.
The diagram shows 3 identical circles inside a rectangle. Each circle touches the other two circles and the sides of the rectangle, as shown in the diagram. The radius of each circle is 24 mm. Work out the area of the rectangle. Give your answer correct to 3 significant figures.
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Edexcel 1MA1 · June 2017 · Paper 3F · Q12 · 5 marks
5 marks
Examiner report
In part (a) many students found the volume of one cube, 2 × 2 × 2 = 8, and said that Vera is correct because 6 × 8 = 48. Those that carried out the calculations but did not state that Vera is correct scored only one of the two marks. There were also responses in which the incorrect evaluation of correct calculations led to one mark being scored. Some students did not read Vera’s statement with sufficient care and stated that she is incorrect because the volume of the cube is 8 cm3, not 48 cm3. Many of the incorrect responses seemed to be referring to surface area rather than to volume. Part (b) was poorly answered with many students unable to draw a cuboid that could be made with 6 of the cubes. Some did draw a 2 by 3 by 1 cuboid or a 1 by 6 by 1 cuboid and scored one mark and sometimes these cuboids also showed the correct dimensions. There were many cuboids drawn with either no dimensions or incorrect dimensions and many cubes drawn. Some students drew a rectangle rather than a 3D shape gaining no marks. It should be noted that even a badly drawn cuboid with appropriate dimensions would have gained a mark. It was disappointing that relatively few students were able to make a correct attempt at finding the surface area of their cuboid. Volume calculations were very common.
The diagram shows a cube of side length 2 cm. 2 cm 2 cm 2 cm Vera says, "The volume of any solid made with 6 of these cubes is 48 cm^3 " (a) Is Vera correct? You must show your working. (b) (i) Draw a cuboid that can be made with 6 of these cubes. Write the dimensions of the cuboid on your diagram. (ii) Work out the surface area of your cuboid.
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Edexcel 1MA1 · June 2017 · Paper 3F · Q13 · 5 marks
5 marks
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.
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 · Q19 · 4 marks
4 marks
Examiner report
Showing that polygon P is a hexagon proved to be beyond many students and this question was not well answered with a significant number of blank responses seen. Fully correct solutions were usually based on interior angles although some neat solutions that used exterior angles were seen. Many of the successful students started by finding the interior angle of a dodecagon and then used angles round a point to work out that the interior angle of polygon P is 120°. The final step for these students was to show that the interior angle of a hexagon is 120°. Alternatively, some students worked out the interior angles of a dodecagon and a hexagon and the final step for these students was using angles round a point to show that 150 + 120 + 90 = 360. Some students omitted the final step and scored three of the four marks. Some students worked out one angle, the exterior angle or interior angle of either a dodecagon or a hexagon, and got no further. It was not uncommon to see errors in arithmetic. There were a significant number of students who gave a worded discussion often indicating that the three sides of polygon P could be reflected to form a hexagon. Others drew a hexagon either on the diagram or in the working space.
In the diagram, AB, BC and CD are three sides of a regular polygon P. D A square polygon P C square B regular 12-sided polygon Show that polygon P is a hexagon. You must show your working.
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Edexcel 1MA1 · June 2017 · Paper 3F · Q21 · 2 marks
2 marks
Examiner report
This question was not answered as well as might have been expected. The most common approach was to show that each side of the larger triangle is 2.5 times longer than the corresponding side of the smaller triangle. Some students used a scale factor of 2.5 but did not apply it to all three sides and could only be awarded one of the two marks. A few students used trigonometry to work out the size of the smallest angle in each triangle and stated that the two triangles are similar because they have the same angles. Many students had no idea how to show that the two triangles are mathematically similar. Some worked out the area or the perimeter of each triangle and there were some who applied Pythagoras’ theorem to both triangles. Many students simply wrote a statement that referred to each triangle having a right angle. Some students thought it was enough to rotate the triangle to look like the other one to show they were similar.
One right triangle has side lengths 7.5 cm, 10 cm and 12.5 cm. Another right triangle has side lengths 3 cm, 4 cm and 5 cm. Show that the triangles are mathematically similar.
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Edexcel 1MA1 · June 2017 · Paper 3H · Q5 · 4 marks
4 marks
Examiner report
The modal score for this question was 4, however, some students did not attempt the question, or tried to explain without calculations why polygon P was a hexagon, these approaches should not be encouraged as they rarely score any marks. Most students began with the dodecagon and were generally successful in determining it’s exterior or interior angle, they then proceeded to find the interior angle of polygon P either by summing the exterior angle of the dodecagon and the interior angle of the square or by using the interior angle of the dodecagon and square and the sum of angles around a point. Students then scored full marks by using appropriate angle facts for the hexagon for comparison. Many students used the formula (n − 2) × 180° for the sum of interior angles. Students who did not score full marks usually forgot to show calculations for BOTH of the shapes. A number of other strategies were also successful and included the calculation of the exterior angle or the number of sides of polygon P. Common mistakes made were to confuse interior and exterior angles, or to just state the interior angle of a hexagon as 120° without any justification at all thus not gaining the final communication mark as explanations were either incorrect or incomplete. Another common mistake was to state that a hexagon has 5 sides. This was disappointing on a higher paper.
In the diagram, AB, BC and CD are three sides of a regular polygon P. D A square polygon P C square B regular 12-sided polygon Show that polygon P is a hexagon. You must show your working.
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Edexcel 1MA1 · June 2017 · Paper 3H · Q7 · 2 marks
2 marks
Examiner report
This question was well answered with over 75% of students gaining full marks. Most successful students used the sine ratio for right angle triangles. Some, however, decided to go the ‘long’ way around the question by use of the cosine ratio thus calculating the wrong side length and then proceeded to use Pythagoras’s theorem to find the correct length of AB. A significant number of students used the sine rule with an angle of 90 The most common errors seen were the use of the wrong trigonometric ratio eg cosine or tangent, calculating BC instead of AB or using the side length, 15 as the angle instead of 23 Students should try to use the simplest approach possible and not look for ‘a problem’ in every question.
ABC is right-angled at B. AC=15 cm and angle ACB=23 degrees. Calculate AB to 3 significant figures.
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Edexcel 1MA1 · June 2017 · Paper 3H · Q8 · 4 marks
4 marks
Examiner report
A good proportion of fully correct answers were seen. Although some students found this question challenging, most managed to score at least one mark for either writing an equation for the radius in terms of x, correctly writing the area of the circle in an equation or calculating r correctly without explicitly writing an equation eg r was often seen or used. Although most drew an appropriate diagram of a square inside a circle, many gave their radius or their diameter as their final answer for x, it was also not unusual to have r incorrectly rearranged to πr = 7 Of those who realised x was not the same as the radius or diameter generally progressed by choosing to apply Pythagoras’s theorem to the problem and a few choosing to use trigonometry. However, those who used Pythagoras’s theorem then struggled to rearrange the equation or forgot to square or square root at the appropriate time. The sine rule was seen on a few occasions as was trigonometry using a right-angled triangle. A general lack of correct algebraic manipulation held back some students.
A square of side x cm is inscribed in a circle of area 49 cm^2. Find x to 3 significant figures.
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Edexcel 1MA1 · June 2017 · Paper 3H · Q15 · 5 marks
5 marks
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.
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 · Q18 · 5 marks
5 marks
Examiner report
A respectable proportion of students scored at least one mark for recognising the right angle between a radius and the tangent. Many displayed this in their diagram. Some students began by calculating the unknown side in the right-angled triangle using Pythagoras’s theorem. Whilst on its own this scored no marks, it was sometimes used successfully by students to correctly find a suitable angle. Other students used incorrect trigonometry and so scored no further marks. Some students assumed the angle subtended by the arc as 2 × 60° which again did not merit any marks. Once a suitable angle was found some students stopped there whilst others went on to find the arc length by using an appropriate method. Unfortunately some students used the area of the circle and found the area of the sector instead of the arc length. Another error seen in the later stages of the question was to give the length of the minor arc. Students are advised to carefully read the question. Centres are advised to stress the use of standard 3 letter notation for angles and arc lengths.
A,B,C lie on a circle of radius 5 cm, centre O. DA and DC are tangents and DO=9 cm. Arc ABC is the major arc from A to C through B. Find its length to 3 significant figures.
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Edexcel 1MA1 · June 2017 · Paper 3H · Q20 · 4 marks
4 marks
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.
(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 · Q15 · 3 marks
3 marks
Carpet tiles are going to be used to cover a floor.
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Edexcel 1MA1 · sample · Paper 1F · Q23 · 3 marks
3 marks
AE, DBG and CF are parallel straight lines. DA = DB = DC. Angle EAB = angle BCF = 38°.
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Edexcel 1MA1 · sample · Paper 1F · Q26 · 2 marks
2 marks
The plan, front elevation and side elevation of a solid prism are drawn on a centimetre
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Edexcel 1MA1 · sample · Paper 1F · Q28 · 5 marks
5 marks
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 1F · Q29 · 2 marks
2 marks
Triangles A and B are drawn on a coordinate grid.
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Edexcel 1MA1 · sample · Paper 1F · Q30 · 2 marks
2 marks
a and b are column vectors.
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Edexcel 1MA1 · sample · Paper 1H · Q3 · 3 marks
3 marks
AE, DBG and CF are parallel straight lines. DA = DB = DC. Angle EAB = angle BCF = 38°.
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Edexcel 1MA1 · sample · Paper 1H · Q6 · 2 marks
2 marks
The plan, front elevation and side elevation of a solid prism are drawn on a centimetre
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Edexcel 1MA1 · sample · Paper 1H · Q8 · 5 marks
5 marks
Here is a parallelogram.
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Edexcel 1MA1 · sample · Paper 1H · Q13 · 3 marks
3 marks
Rectangle ABCD is mathematically similar to rectangle DAEF.
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Edexcel 1MA1 · sample · Paper 1H · Q18 · 3 marks
3 marks
Solid A and solid B are mathematically similar.
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Edexcel 1MA1 · sample · Paper 1H · Q20 · 3 marks
3 marks
Shape A is the quadrilateral with vertices (1, 1), (1, 2), (2, 2) and (3, 1).
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Edexcel 1MA1 · sample · Paper 1H · Q25 · 5 marks
5 marks
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 · Q13 · 4 marks
4 marks
ABE and CBD are straight lines intersecting at B. Angle BAC = 50° and angle DBE = 80°.
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Edexcel 1MA1 · sample · Paper 2F · Q14 · 5 marks
5 marks
A cuboid tank is full of oil; an empty cuboid container will receive the unspilled oil.
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Edexcel 1MA1 · sample · Paper 2F · Q15 · 2 marks
2 marks
Two towns are marked on an accurately drawn scale map.
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Edexcel 1MA1 · sample · Paper 2F · Q22 · 5 marks
5 marks
A tin of varnish costs £15. A 6 m by 11 m floor is to be covered. Helen assumes each tin covers 12 m².
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Edexcel 1MA1 · sample · Paper 2F · Q24 · 2 marks
2 marks
PQR is right-angled at R. PQ = 14 cm is the hypotenuse and PR = 5 cm.
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Edexcel 1MA1 · sample · Paper 2F · Q28 · 4 marks
4 marks
Triangles ABD and BCD are right-angled.
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Edexcel 1MA1 · sample · Paper 2H · Q2 · 2 marks
2 marks
PQR is right-angled at R.
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Edexcel 1MA1 · sample · Paper 2H · Q6 · 4 marks
4 marks
Triangles ABD and BCD are right-angled.
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Edexcel 1MA1 · sample · Paper 2H · Q11 · 4 marks
4 marks
S and T lie on a circle with centre O; PT is tangent at T; S, O and P are collinear.
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Edexcel 1MA1 · sample · Paper 2H · Q17 · 2 marks
2 marks
The diagram shows a sector of a circle of radius 7 cm.
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Edexcel 1MA1 · sample · Paper 2H · Q21 · 5 marks
5 marks
In triangle RPQ, RP = 8.7 cm PQ = 5.2 cm Angle PRQ = 32°
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Edexcel 1MA1 · sample · Paper 2H · Q22 · 5 marks
5 marks
A glass frustum is formed by removing a small cone from a similar large cone.
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Edexcel 1MA1 · sample · Paper 3F · Q4 · 3 marks
3 marks
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 · Q13 · 4 marks
4 marks
WXYZ is a quadrilateral. X, Y and V are collinear, with Y between X and V. Angle ZYV = 147°, angle WZY = 145°, and angle ZWX = angle WXY.
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Edexcel 1MA1 · sample · Paper 3F · Q18 · 4 marks
4 marks
A shaded region is formed from a trapezium by removing two identical semicircles.
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Edexcel 1MA1 · sample · Paper 3H · Q1 · 4 marks
4 marks
A shaded region is formed from a trapezium by removing two identical semicircles.
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Edexcel 1MA1 · sample · Paper 3H · Q13 · 3 marks
3 marks
ABCD is a rhombus. Points M and N lie on diagonal BD, with DN = MB.
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Edexcel 1MA1 · sample · Paper 3H · Q16 · 5 marks
5 marks
VABCD is a solid pyramid with square base ABCD of side 20 cm. Each sloping edge makes an angle of 55° with the base plane.
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Edexcel 1MA1 · sample · Paper 3H · Q18 · 5 marks
5 marks
O, M and A are collinear, with M the midpoint of OA. O, N and B are collinear. A, B and C are collinear, with B the midpoint of AC.
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Edexcel 1MA1 · specimen2 · Paper 1F · Q4 · 2 marks
2 marks
A square-based pyramid is shown.
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Edexcel 1MA1 · specimen2 · Paper 1F · Q16 · 3 marks
3 marks
In quadrilateral ABCD, the two interior angles on the same side of transversal BC are 120° at B and 57° at C.
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Edexcel 1MA1 · specimen2 · Paper 1F · Q20 · 5 marks
5 marks
A square has perimeter 16 cm, so side length 4 cm. A shaded quadrilateral has vertices at the square's bottom-left corner, a point on the top edge 3 cm left of the top-right corner, the top-right corner, and a point on the right edge 2 cm below the top-right corner.
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Edexcel 1MA1 · specimen2 · Paper 1F · Q22 · 4 marks
4 marks
Triangle ABC has perimeter 20 cm, with AB = 7 cm and BC = 4 cm.
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Edexcel 1MA1 · specimen2 · Paper 1H · Q2 · 5 marks
5 marks
A square has perimeter 16 cm, so side length 4 cm. A shaded quadrilateral has vertices at the square's bottom-left corner, a point on the top edge 3 cm left of the top-right corner, the top-right corner, and a point on the right edge 2 cm below the top-right corner.
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Edexcel 1MA1 · specimen2 · Paper 1H · Q4 · 4 marks
4 marks
Triangle ABC has perimeter 20 cm, with AB = 7 cm and BC = 4 cm.
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Edexcel 1MA1 · specimen2 · Paper 1H · Q9 · 3 marks
3 marks
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
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 · Q12 · 3 marks
3 marks
ABCD is a parallelogram. Diagonals AC and BD meet at E.
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Edexcel 1MA1 · specimen2 · Paper 1H · Q18 · 4 marks
4 marks
LMN is right-angled at L. Q lies on LN, P lies on MN, and PQ∥LM. Area(PNQ)=8 cm² and area(LPQ)=16 cm².
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Edexcel 1MA1 · specimen2 · Paper 1H · Q22 · 5 marks
5 marks
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 · Q9 · 1 mark
1 mark
The smallest angle in a triangle is 25°. The triangle is enlarged by scale factor 3. Ben says the smallest angle in the enlarged triangle is 75°.
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Edexcel 1MA1 · specimen2 · Paper 2F · Q22 · 2 marks
2 marks
Congruent coordinate shape B is obtained from shape A by moving every vertex 4 right and 3 down.
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Edexcel 1MA1 · specimen2 · Paper 2F · Q25 · 3 marks
3 marks
A regular octagon and regular hexagon share a vertex and one side; x is the remaining angle around that vertex.
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Edexcel 1MA1 · specimen2 · Paper 2F · Q28 · 3 marks
3 marks
Arc ABC is a quarter-circle with centre O and radius 4.8 cm. AC is a chord, and the shaded segment is the sector AOC minus triangle AOC.
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Edexcel 1MA1 · specimen2 · Paper 2F · Q29 · 5 marks
5 marks
ABC is isosceles BA=BC. D lies on AC. ABD is isosceles AB=AD. ∠ABD=72°.
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Edexcel 1MA1 · specimen2 · Paper 2H · Q1 · 2 marks
2 marks
Congruent coordinate shape B is obtained from shape A by moving every vertex 4 right and 3 down.
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Edexcel 1MA1 · specimen2 · Paper 2H · Q4 · 3 marks
3 marks
A regular octagon and regular hexagon share a vertex and one side; x is the remaining angle around that vertex.
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Edexcel 1MA1 · specimen2 · Paper 2H · Q7 · 3 marks
3 marks
Arc ABC is a quarter-circle with centre O and radius 4.8 cm. AC is a chord, and the shaded segment is the sector AOC minus triangle AOC.
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Edexcel 1MA1 · specimen2 · Paper 2H · Q17 · 4 marks
4 marks
A right-angled triangle has hypotenuse a = 8.3 cm, correct to the nearest millimetre, and one leg b = 6.1 cm, correct to the nearest millimetre. The other leg has length c.
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Edexcel 1MA1 · specimen2 · Paper 2H · Q20 · 5 marks
5 marks
PQRST is a regular pentagon. R, U and T lie on a circle with centre O. QR and PT are tangents to the circle, and R, S and U are collinear.
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Edexcel 1MA1 · specimen2 · Paper 3F · Q4 · 1 mark
1 mark
On the grid, draw a parallelogram.
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Edexcel 1MA1 · specimen2 · Paper 3F · Q13 · 4 marks
4 marks
A, B and C lie on a straight line. Triangle BCD has angles x at B, 35° at C and 30° at D. In quadrilateral ABDE, the interior angle at B is 180° − x; the angles at D and E are both y, and the angle at A is 95°.
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Edexcel 1MA1 · specimen2 · Paper 3F · Q18 · 1 mark
1 mark
A diagram shows that B lies 25° west of north from A.
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Edexcel 1MA1 · specimen2 · Paper 3F · Q23 · 2 marks
2 marks
The diagram represents a solid made from seven centimetre cubes. Six cubes form a complete 2 × 3 rectangular base and the seventh cube is stacked on top of one of the base cubes.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · specimen2 · Paper 3F · Q28 · 3 marks
3 marks
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 · Q29 · 6 marks
6 marks
An empty oil tank is a prism with a trapezium cross-section. The parallel sides of the trapezium are 3 m and 1.4 m, its perpendicular height is 2.4 m, and the prism is 2.5 m long. The tank must be filled to 85% of its capacity. After one minute, 300 litres of oil are in the tank; assume oil continues to flow at this rate. Use 1 m³ = 1000 litres.
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Edexcel 1MA1 · specimen2 · Paper 3H · Q2 · 2 marks
2 marks
The diagram represents a solid made from seven centimetre cubes. Six cubes form a complete 2 × 3 rectangular base and the seventh cube is stacked on top of one of the base cubes.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · specimen2 · Paper 3H · Q7 · 3 marks
3 marks
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.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · specimen2 · Paper 3H · Q8 · 6 marks
6 marks
An empty oil tank is a prism with a trapezium cross-section. The parallel sides of the trapezium are 3 m and 1.4 m, its perpendicular height is 2.4 m, and the prism is 2.5 m long. The tank must be filled to 85% of its capacity. After one minute, 300 litres of oil are in the tank; assume oil continues to flow at this rate. Use 1 m³ = 1000 litres.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · specimen2 · Paper 3H · Q13 · 2 marks
2 marks
A triangle on a coordinate grid has vertices (4, 2), (4, 4) and (6, 4). The centre of enlargement is (0, 2).
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Edexcel 1MA1 · specimen2 · Paper 3H · Q16 · 2 marks
2 marks
A sector of a circle has centre O and radius 4 cm. The minor central angle AOC is 100°. Arc ABC is the major arc from A to C through B.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · specimen2 · Paper 3H · Q20 · 4 marks
4 marks
OAB is a triangle. The position vectors are OA = 2a and OB = 2b. Point P lies on AB such that AP:PB = 5:3, and OP = k(3a + 5b), where k is a scalar.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · specimen2 · Paper 3H · Q21 · 5 marks
5 marks
ABC is a triangle and D lies on AB. AC = 4.9 cm, AD = 3.8 cm, ∠CAD = 80° and ∠DCB = 25°.
Pearson does not publish a direct link to this question paper.
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Everything here stays free. If geometry and measures is where the marks keep going, I teach it one to one from the mark schemes. Free intro call, then a free first lesson.