Every Edexcel GCSE Maths 1MA1 specification point in N, read straight through: 16 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.
Checked against Edexcel 1MA1 section N. Review basis: the qualification registry sourced from the Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics (1MA1) specification; registry verification recorded 9 July 2026.
N1 · Order positive and negative integers, decimals and fractions; use the symbols =, ≠, <, >, ≤, ≥
Integers, decimals and fractions can be ordered once they are written in a comparable form.
Convert fractions to decimals, or use a common denominator, then place the values on a number line: values increase from left to right.
This is especially important for negatives, because the number with the greater distance below zero is smaller; for example, −1.7<−23 since −1.7<−1.5.
Use = for equal values, = for unequal values, and read ≤ or ≥ as allowing equality.
In an exam, show the conversion that justifies each comparison.
A number line showing that −1.85 lies to the left of −1.75.
Worked example
Write −47, −1.68, 35 and 1.7 in ascending order.
1.Convert the fractions: −47=−1.75 and 35=1.666….
2.Compare the negative values first: −1.75<−1.68.
3.Compare the positive values: 1.666…<1.7, then write the complete ordered list.
Answer: −47,−1.68,35,1.7.
Common mistakes
•Don't write −8>−3 because 8>3, ignoring that −8 lies farther left.
•Don't treat ≤ as meaning strictly less than and reject the equal endpoint.
•Don't round a recurring decimal too early and change the intended order.
Exam tip. For a 2-mark ordering question, show decimal or common-denominator conversions before the final list.
N2 · Apply the four operations, including formal written methods, to integers, decimals and simple fractions (proper and improper), and mixed numbers, positive and negative; understand and use place value
Definition. Specification N2: apply the four operations, including formal written methods, to integers, decimals and simple fractions (proper and improper) and mixed numbers, all both positive and negative, and understand and use place value.
Written methods score for the method, then for the answer. In a decimal division such as 818.4÷1.2, multiply both numbers by the same power of 10 so the divisor is an integer, divide, then place the decimal point by comparing the size of the numbers: 818.4÷1.2 is a little less than 818, so 682 is sensible and 68.2 is not.
For fraction addition or subtraction, write both fractions over a common denominator and keep the answer in the form requested, usually a mixed number. For multiplication or division of mixed numbers, convert every mixed number to an improper fraction first; to divide, multiply by the reciprocal of the second fraction only.
A show-that question must display each stage, including the unsimplified product.
Track signs separately: like signs give a positive result, unlike signs a negative one.
Cancelling common factors before multiplying keeps the arithmetic small.
Worked example
1
Work out −241÷53. Give your answer as a mixed number.
(3)
(Total for Question 1 is 3 marks)
Working
1.Convert the mixed number to an improper fraction: −241=−49.
2.Divide by multiplying by the reciprocal of 53: −49×35.
3.Cancel the common factor 3 and multiply: −4×13×5=−415.
4.Write as a mixed number: −415=−343.
Answer: −343
Mark scheme
Mark scheme for question 1
Question
Answer
Mark
Additional guidance
1
for conversion to an improper fraction, eg −49
M1
Need not be shown with operators
(dep) for a method to divide by multiplying by the reciprocal, eg −49×35 or −1245
M1
oe
for −343
A1
Accept −415 only if the mixed number is not required
Mark-scheme wording to use
•“M1 for a start to a method, eg 8184 ÷ 12 (or 818.4 ÷ 1.2) that leads to 6 as the first digit or for a complete method with no more than one arithmetic error … A1 for digits 682 A1 ft (dep M1) for correct placement of the decimal point into their final answer”— Pearson 1MA1, November 2024 Paper 1H, Q1
•“M1 for conversion to improper fractions, eg 21/4 or 7/3 or 9/4 M1 (dep) for method to divide by a fraction, eg 21/4 × 3/7 or 63/12 ÷ 28/12 C1 for complete work showing each stage as far as 9/4 or 2 7/28 Must see an intermediate step, eg 63/28”— Pearson 1MA1, November 2024 Paper 1H, Q3(b)
•“M1 for finding two fractions with a correct common denominator (multiple of 20), with at least one correct corresponding numerator, eg 12/20, 5/20 or 32/20, 45/20 … A1 for 3 17/20 or an equivalent mixed number”— Pearson 1MA1, November 2022 Paper 1H, Q2(a)
What the examiner wants. Sensible place value and a visible reciprocal step. The November 2024 Paper 1H report notes that after correctly finding 682 many wrote 68.2 or 6.82, and that in the division of mixed numbers some flipped the wrong fraction, flipped both, or multiplied whole numbers and fractions separately, scoring nothing.
Common mistakes
•Don't add fraction denominators as well as numerators, writing ba+dc=b+da+c.
•Don't divide by a fraction without inverting it.
•Don't align the last digits of decimals instead of aligning their decimal points.
Exam tip. Keep an exact fraction until the final line and show the reciprocal explicitly for the division method mark.
N3 · Recognise and use relationships between operations, including inverse operations; use conventional notation for priority of operations, including brackets, powers, roots and reciprocals
Inverse operations undo one another: addition pairs with subtraction, multiplication with division, and squaring with square root when the required value is non-negative. Use these relationships to check calculations or reverse a sequence of operations.
For a written calculation, follow conventional priority: brackets first; then powers and roots; then multiplication and division; then addition and subtraction.
Operations at the same priority are completed from left to right.
A reciprocal is also an inverse operation for multiplication: the reciprocal of ba is ab.
Examiners expect each priority stage to be visible when the question carries more than one mark.
Worked example
A positive number is squared, 11 is subtracted, and the reciprocal of the result is 141. Find the number.
1.Reverse the reciprocal: the value before taking it was 14.
2.Undo subtracting 11: 14+11=25.
3.Undo the square using the positive square root: 25=5.
Answer: 5.
Common mistakes
•Don't add before multiplying in 18−32×2.
•Don't say the reciprocal of ba is −ba.
•Don't forget that a square root symbol denotes the non-negative square root.
Exam tip. In a reverse-process question, undo the operations in the opposite order and write one inverse operation per line.
N4 · Prime numbers, factors (divisors), multiples, common factors and multiples, highest common factor, lowest common multiple, prime factorisation with product notation and unique factorisation theorem
Definition. Specification N4: use the concepts and vocabulary of prime numbers, factors (divisors), multiples, common factors, common multiples, highest common factor, lowest common multiple and prime factorisation, including using product notation and the unique factorisation theorem.
A prime has exactly two factors, so 1 is not prime. Every integer above 1 is a product of primes in exactly one way, which is why the prime factorisation is a reliable route to the highest common factor and lowest common multiple.
Find it with a factor tree or repeated division by primes, then write the answer as a product using index notation: 500=22×53. A list of primes without multiplication signs is not a product and drops the accuracy mark.
For two numbers, a Venn diagram of their prime factors puts the shared primes in the overlap: the HCF is the product of the overlap and the LCM is the product of everything in the diagram.
Given numbers already in index form, the HCF takes the lower power of each prime and the LCM the higher.
Read the question twice, because HCF and LCM are routinely swapped.
Worked example
1
180n is a cube number, where n is a positive integer. Work out the smallest possible value of n.
(3)
(Total for Question 1 is 3 marks)
Working
1.Write 180 as a product of prime factors: 180=22×32×5.
2.A cube number has every index a multiple of 3, so the product needs one more 2, one more 3 and two more 5s.
3.The smallest such n is 2×3×52=150.
4.Check: 180×150=27000=303.
Answer: n=150
Mark scheme
Mark scheme for question 1
Question
Answer
Mark
Additional guidance
1
for a complete method to find the prime factors of 180, eg 22×32×5 or 2,2,3,3,5
M1
Condone the inclusion of 1 for this mark
for identifying the factors needed, eg 2×3×5×5 or 2×3×52
M1
oe
for 150
A1
cao
Mark-scheme wording to use
•“for a complete method to find prime factors, could be shown on a complete factor tree with no more than one error or by division by prime factors with no more than one error”— Pearson 1MA1, November 2022 Paper 1H, Q1
•“for listing at least 4 correct factors (with no more than 1 incorrect) of 63 or 105, could be in factor pairs or for the prime factors of 63 (3, 3, 7) or 105 (3, 5, 7)”— Pearson 1MA1, June 2024 Paper 3H, Q1
•“M1 for a complete method to find prime factors; could be shown on a complete factor tree with no more than one error or by division by prime factors with no more than one error … A1 for 2 × 2 × 31 oe”— Pearson 1MA1, June 2022 Paper 1H, Q2
What the examiner wants. A product, and the right one of HCF or LCM. The June 2024 Paper 2H report says students who listed the prime factors rather than writing them as a product scored one mark, and that asked for an LCM from index form many gave the HCF of 6; the Paper 3H report notes a significant proportion gave the LCM of 315 when the HCF of 21 was asked for.
Common mistakes
•Don't include 1 in a list of prime numbers.
•Don't stop a factor tree with a composite number at an endpoint.
•Don't use the larger exponents for the HCF instead of for the LCM.
Exam tip. For HCF or LCM questions, write both prime-power decompositions first so the method marks remain available.
N5 · Apply systematic listing strategies, including use of the product rule for counting (m ways of doing one task and n ways of doing another gives m × n ways in total)
A systematic listing strategy fixes one choice and cycles through every permitted second choice before moving on. Use a clear order, such as increasing digits or alphabetical letters, so you can see that no result is missing or repeated.
When order matters, AB and BA are different outcomes; when order does not matter, list each pair only once.
A table or tree can organise several stages.
Higher tier: if one task can be completed in m ways and a following independent choice in n ways, the product rule gives m×n combined outcomes.
Examiners award completeness only when the structure of the list is evident.
Worked example
Higher tier: Three different digits are chosen from 2, 4, 5 and 7 to make an even three-digit number. How many numbers are possible?
1.Fix the final digit as 2: choose and order two of the remaining three digits, giving 3×2=6 numbers.
2.Fix the final digit as 4: again there are 3×2=6 numbers.
3.Add the two disjoint cases: 6+6=12.
Answer: 12 even numbers.
Common mistakes
•Don't change two positions at once and omit a valid outcome.
•Don't count AB and BA twice when the question says order does not matter.
•Don't use the product rule even though later choices depend on earlier restrictions.
Exam tip. Group a list by one fixed position or case; an unstructured collection may not earn the method mark for systematic listing.
N6 · Use positive integer powers and associated real roots (square, cube and higher), recognise powers of 2, 3, 4, 5; estimate powers and roots of any given positive number
A positive integer power represents repeated multiplication: an contains n factors equal to a.
An associated root reverses that power, so 81=9, 3125=5, and 416=2.
Learn common powers of 2, 3, 4 and 5 because they make exact roots quickly recognisable.
Higher tier: to estimate a root that is not exact, place its radicand between nearby known powers; for example, 33<40<43 shows 3<340<4.
In an exam, distinguish an exact evaluation from an estimate and clearly state the bounding powers used.
Worked example
Higher tier: Estimate 3200 to the nearest whole number without a calculator.
1.Use nearby cubes: 53=125 and 63=216, so 5<3200<6.
2.The rounding threshold is 5.53=166.375; since 200>166.375, the cube root is above 5.5.
3.Round the estimate to the nearest whole number.
Answer: 3200≈6.
Common mistakes
•Don't calculate 34 as 3×4 instead of four factors of 3.
•Don't use a square root when the inverse operation required is a cube root.
•Don't state an estimated root as an exact equality.
Exam tip. For an estimate, write the two consecutive known powers that bound the radicand before choosing the nearer root.
N7 · Calculate with roots, and with integer and fractional indices
Integer indices include positive, zero and negative powers. For non-zero a, a0=1 and a−n=an1; the negative index creates a reciprocal, not a negative answer.
Roots reverse powers, including odd roots of negative numbers such as 3−216=−6.
Higher-tier questions also use fractional indices: a1/n=na and am/n=nam.
Foundation questions stay with roots and integer indices.
Choose the form that makes evaluation simplest, keep brackets around a negative base, and show the reciprocal or root step very clearly before giving the final value.
Worked example
Higher tier: Work out 813/4.
1.Interpret the denominator as a root: 813/4=(481)3.
2.Evaluate 481=3.
3.Calculate 33=27.
Answer: 27.
Common mistakes
•Don't write 4−2=−16 instead of taking the reciprocal.
•Don't treat a1/2 as a÷2 rather than a.
•Don't evaluate (−3)2 and −32 as though the brackets made no difference.
Exam tip. On Foundation, show the reciprocal for a negative integer index; on Higher, rewrite a fractional index as a root before evaluating.
N8 · Calculate exactly with fractions, surds and multiples of π; simplify surd expressions involving squares (e.g. √12 = √(4 × 3) = √4 × √3 = 2√3) and rationalise denominators
An exact answer keeps fractions, multiples of π and, on Higher tier, surds rather than replacing them with rounded decimals. Combine exact fractions with a common denominator and leave circle results as a coefficient of π.
Higher-tier surds are simplified by extracting square factors: 12=4×3=23.
Like surds can then be collected, and a denominator such as 5 is rationalised by multiplying numerator and denominator by 5.
Foundation questions do not require surd manipulation, but do require exact fractions and multiples of π.
When the command says “exact”, no rounded decimal should appear in the final answer.
Worked example
Higher tier: Simplify 75−12 and give an exact answer.
1.Extract the largest square factor: 75=25×3=53.
2.Similarly, 12=4×3=23.
3.Collect like surds: 53−23=33.
Answer: 33.
Common mistakes
•Don't replace π with 3.14 when an exact value is required.
•Don't write a+b=a+b.
•Don't collect unlike surds, for example treating 2+3 as 25.
Exam tip. If the question says “exact”, retain π, fractions or surds and simplify them fully rather than using a calculator decimal.
N9 · Calculate with and interpret standard form A × 10^n, where 1 ≤ A < 10 and n is an integer
Definition. Specification N9: calculate with and interpret standard form A×10n, where 1≤A<10 and n is an integer, with and without a calculator.
A positive index moves the decimal point right, so 5.3×104=53000; a negative index moves it left, so 7.4×10−5=0.000074. To multiply, multiply the A values and add the indices; to divide, divide the A values and subtract the indices; then normalise, because 24×10−2 is not standard form until it becomes 2.4×10−1.
To add or subtract, write both numbers with the same power of ten first, or convert to ordinary numbers: 9.7×106+2.45×107=0.97×107+2.45×107=3.42×107.
Adding the A values when the powers differ is the standard error and scores nothing.
A method mark is available for a correct value in a non-standard form such as 34.2×106, and a final answer left as an ordinary number when standard form was asked for loses the last mark.
To order numbers, put them all in the same form first.
Worked example
1
Work out 1.5×1033.6×10−4+7.5×10−5. Give your answer in standard form.
(3)
(Total for Question 1 is 3 marks)
Working
1.Write the numerator with a common power of ten: 3.6×10−4+0.75×10−4=4.35×10−4.
2.Divide the A values and subtract the indices: 1.54.35×10−4−3=2.9×10−7.
3.Check the coefficient: 1≤2.9<10, so the answer is in standard form.
Answer: 2.9×10−7
Mark scheme
Mark scheme for question 1
Question
Answer
Mark
Additional guidance
1
for 4.35×10−4 or 0.000435 or both numbers in a form ready for addition, eg 3.6×10−4+0.75×10−4
M1
oe
for 4.35÷1.5×10−4−3 or 2.9×10n (n=−7) or 0.00000029
M1
Correct answer in incorrect form
for 2.9×10−7
A1
cao
Mark-scheme wording to use
•“for 9 700 000 + 24 500 000 (= 34 200 000) or 3.42 × 10^n (n ≠ 7) oe or 3.4 × 10^7 or correct answer in incorrect form eg 34.2 × 10^6 or both in a form ready for addition, eg 9.7 × 10^6 + 24.5 × 10^6”— Pearson 1MA1, June 2024 Paper 3H, Q2(b)
•“M1 for 4×6×10^(3–5) or 0.24 oe eg 24 × 10^–2 or 2.4×10^n where n ≠ –1 A1 cao”— Pearson 1MA1, June 2022 Paper 1H, Q4(c)
•“for correct conversions to same format, condoning one error”— Pearson 1MA1, June 2020 Paper 1H, Q9
What the examiner wants. The powers dealt with before the coefficients. The June 2024 Paper 3H report says most reached 34200000 on a calculator but many could not then write it in standard form, and that those scoring nothing had ignored the differing powers of ten, adding 9.7 to 2.45 to get 12.15 and attaching a power of ten to that.
Common mistakes
•Don't leave 24×104 as the final answer even though 24 is not between 1 and 10.
•Don't add the indices when dividing powers of 10.
•Don't add coefficients whose powers of 10 have not first been made equal.
Exam tip. Circle the final coefficient and check 1≤A<10 before submitting any standard-form answer.
N10 · Work interchangeably with terminating decimals and their corresponding fractions (such as 3.5 and 7/2 or 0.375 or 3/8); change recurring decimals into their corresponding fractions and vice versa
A terminating decimal has finitely many decimal places, so write it over the matching power of 10 and simplify: 0.375=1000375=83. Convert a fraction to a decimal by dividing its numerator by its denominator.
On Higher tier, recurring decimals are also converted algebraically: name the decimal x, multiply by a power of 10 so the repeating blocks align, subtract to remove the recurring part, and solve for x.
A recurring dot or bar must cover the whole repeating block.
Foundation questions use terminating decimal–fraction equivalents only.
Examiners expect the resulting fraction in its simplest form.
Worked example
Higher tier: Convert 0.2˙7˙=0.272727… to a fraction.
1.Let x=0.272727…, so 100x=27.272727….
2.Subtract: 100x−x=27, giving 99x=27.
3.Solve and simplify: x=9927=113.
Answer: 113.
Common mistakes
•Don't write 0.375 as 100375 or use the wrong place-value denominator.
•Don't stop before simplifying the numerator and denominator fully.
•Don't multiply by 10 when the repeating block has two digits for a recurring decimal.
Exam tip. For a recurring decimal, align identical recurring tails before subtracting so they cancel exactly.
N11 · Identify and work with fractions in ratio problems
Fractions and ratios describe the same partition in different forms. If a group is 53 of the whole, the remainder is 52, so the part-to-part ratio is 3:2.
Conversely, in a ratio a:b, the first share is a+ba of the whole and the second is a+bb.
In multi-step problems, first use the ratio to find each share, then apply any fraction operator to the relevant share.
Keep the named order of the groups throughout.
Examiners commonly award one mark for finding one ratio part and a later accuracy mark for the required fractional amount.
Worked example
£330 is shared between Imran and Jo in the ratio 4:7. Imran spends 83 of his share and Jo spends 72 of hers. Find the total left.
1.There are 11 parts, so one part is 330÷11=30.
2.Imran gets 120 and keeps 85×120=75; Jo gets 210 and keeps 75×210=150.
3.Add the amounts kept: 75+150=225.
Answer: £225.
Common mistakes
•Don't say the remainder of 53 is 55 instead of 52.
•Don't use one ratio part as the denominator of the fraction of the whole.
•Don't apply a spending fraction as though it were the fraction left.
Exam tip. Label each share and write the total number of ratio parts before applying any later fraction.
N12 · Interpret fractions and percentages as operators
Definition. Specification N12: interpret fractions and percentages as operators. A fraction or percentage “of” a quantity means multiply by it, and a sequence of operators is applied in order, each to the current amount.
To find ba of an amount, divide by b and multiply by a, or multiply by the fraction directly: 72 of 420 is 420×2÷7=120. A percentage becomes a decimal or fraction operator, 35% of 420 being 420×0.35=147.
A percentage change is one multiplier: a 15% decrease multiplies by 0.85, an increase of 15% by 1.15.
When several operators apply, each acts on the amount that remains, not on the original: “83 of the reduced price” means multiply the reduced price by 83.
Problems that mix a fraction, a percentage and a ratio are worked one step at a time, subtracting the shares already taken before splitting the remainder, and the scheme gives a process mark for each share.
A correct answer with no working scores nothing on these questions.
Worked example
1
A machine costs £640. Its price is reduced by 15%. A customer pays 83 of the reduced price as a deposit and the rest in one further payment. Work out the amount of the further payment.
(4)
(Total for Question 1 is 4 marks)
Working
1.Apply the reduction multiplier: 640×0.85=£544.
2.Find the deposit as a fraction of the reduced price: 544×83=£204.
3.Subtract to find the balance: 544−204=£340, or directly 544×85.
Answer: £340
Mark scheme
Mark scheme for question 1
Question
Answer
Mark
Additional guidance
1
for a process to find the reduced price, eg 640×0.85 or 640−640×15÷100 (= 544)
P1
oe
for a process to find the deposit, eg “544” ×3÷8 (= 204)
P1
(dep) for a complete process, eg “544” − “204” or “544” ×5÷8
P1
for 340
A1
cao
Mark-scheme wording to use
•“P1 for 160 ÷ (3+7) (= 16) or 3/(3 + 7) (= 3/10) P1 for “16” × 3 (= 48) or “3/10” × 160 (= 48) P1 for a correct step using 48 eg “48” ÷ 8 (= 6) or “48” × 25 ÷ 100 (= 12) … P1 for a complete process to find the number of petrol cars, eg “48” – “6” – “12” oe … A1 cao … Award no marks for a correct answer with no supportive working”— Pearson 1MA1, June 2022 Paper 1H, Q3
•“for a process to find the number of banana cakes, eg 420 × 0.35 oe (= 147)”— Pearson 1MA1, June 2017 Paper 3H, Q4
•“P1 Process to find number of boys walking and number of girls walking P1 Complete process to find probability A1 10/26 oe”— Pearson 1MA1, Sample Paper 3F, Q15
What the examiner wants. Each operator applied to the right amount, in order. The November 2024 Paper 3H report describes a ratio and percentage problem where students who found the quantities correctly then worked out a percentage profit for each item separately and added them, giving 88.8%, and the Paper 3H report of the same series notes that increasing or decreasing £46.75 by 15% scored nothing when the operator had to be reversed.
Common mistakes
•Don't divide by the numerator and multiply by the denominator for a fraction of an amount.
•Don't use 0.15 as the multiplier for the price after a 15% decrease.
•Don't calculate the deposit from the original price instead of the reduced price.
Exam tip. Write each fraction or percentage as a multiplier beside the amount it operates on.
N13 · Use standard units of mass, length, time, money and other measures (including standard compound measures) using decimal quantities where appropriate
Definition. Specification N13: use standard units of mass, length, time, money and other measures, including with decimal quantities.
Work in one unit at a time and convert before calculating: 1 kg =1000 g, 1 km =1000 m, 1 litre =1000 cm3, and money in either pounds or pence throughout, converting at the end.
Time is the trap: 1.25 hours is 1 hour 15 minutes, not 1 hour 25 minutes, and 2.75 hours is 2 hours 45 minutes, so a time from distance÷speed must be converted to hours and minutes when the question asks for it.
A rate written per 100 km or per kilogram is scaled to the actual distance or mass first.
Round money to the nearest penny only at the end.
Check the size of the answer against common sense; a journey time of hundreds of minutes for a few miles signals a wrong operation.
Worked example
1
A car uses 7.5 litres of fuel for every 100 km it travels. Fuel costs £1.68 per litre. Work out the cost of the fuel for a journey of 54 km.
(3)
(Total for Question 1 is 3 marks)
Working
1.Scale the fuel rate to the journey: 54÷100×7.5=4.05 litres.
2.Multiply by the price per litre: 4.05×1.68=6.804.
3.Round money to the nearest penny: £6.80.
Answer: £6.80
Mark scheme
Mark scheme for question 1
Question
Answer
Mark
Additional guidance
1
for a process to find the fuel used, eg 54÷100×7.5 (= 4.05)
P1
oe
for a complete process, eg “4.05” ×1.68
P1
for 6.80
A1
accept 6.8
Mark-scheme wording to use
•“for 30 / 24 (= 1.25) or 12 / 8 (= 1.5) … May be written in hours and/or minutes”— Pearson 1MA1, June 2020 Paper 3H, Q2
•“for finding the sum of their two times … or 3 h 15 min or 2 h 75 min”— Pearson 1MA1, June 2020 Paper 3H, Q2
•“for full process to find the number of bags sold eg 5 × 1000 ÷ 250 (= 20)”— Pearson 1MA1, June 2018 Paper 1H, Q3
What the examiner wants. A sensible size. The November 2024 Paper 1H report says answers for the time to travel about 5 miles at 30 miles per hour ranged from fractions of a minute up to 900 minutes, and urges checking the reasonableness of an answer; those who wrote “30 miles in 60 minutes” and divided both by 6 found the 10 minutes cleanly.
Common mistakes
•Don't treat 1.5 hours as 1 hour 50 minutes.
•Don't use the factor 100 rather than 1002 when converting square metres to square centimetres.
•Don't omit the compound unit from the final answer.
Exam tip. Write the target unit before starting; every conversion should move the data towards that unit.
N14 · Estimate answers; check calculations using approximation and estimation, including answers obtained using technology
An estimate replaces the original inputs with nearby values that are easy to calculate, often by rounding each to one significant figure. Carry out the operation on those rounded inputs and use ≈, because the result is not exact.
Estimation checks whether a calculator answer has a sensible size, sign and decimal position; it should be an independent calculation, not a rounding of the calculator display.
Choose compatible approximations, especially for division, so mental arithmetic stays simple.
If the exact answer differs by about a factor of 10 or 100, suspect a place-value entry error.
Examiners require both the estimate and a clear comparison when asked to comment.
Worked example
A calculator gives 931.24 for 0.204598.4×0.0317. Use an estimate to decide whether it is reasonable.
1.Round inputs to convenient values: 598.4≈600, 0.0317≈0.03, 0.204≈0.2.
2.Estimate 0.2600×0.03=90.
3.Compare 931.24 with 90: the display is roughly ten times too large.
Answer: The display is not reasonable.
Common mistakes
•Don't round only the final calculator answer instead of estimating from the inputs.
•Don't use = rather than ≈ between an expression and its estimate.
•Don't round a small decimal such as 0.0317 to zero.
Exam tip. In a “check using estimation” question, state whether the given answer is reasonable and support the decision with your rounded calculation.
N15 · Round numbers and measures to an appropriate degree of accuracy (decimal places or significant figures); use inequality notation to specify simple error intervals due to truncation or rounding
Decimal places count digits after the decimal point; significant figures begin at the first non-zero digit. Identify the deciding digit immediately after the required place: 5 or more rounds up, while 4 or less leaves the retained digit unchanged.
Choose a degree of accuracy appropriate to the context, such as money to the nearest penny.
A rounded value also represents an error interval.
For rounding to a unit u, subtract and add 2u; include the lower boundary but exclude the upper boundary because that endpoint rounds to the next value.
Truncation instead keeps values from the stated number up to the next truncation step.
The error interval for 12.6 correct to one decimal place.
Worked example
A positive number y is truncated to 4.37 at two decimal places. Write its error interval and find the greatest integer value of 100y.
1.Truncation gives 4.37≤y<4.38.
2.Multiply the whole interval by 100: 437≤100y<438.
3.The greatest integer in this interval is 437.
Answer: 4.37≤y<4.38 and the greatest integer value is 437.
Common mistakes
•Don't count leading zeros as significant figures in 0.007846.
•Don't include the upper boundary of a rounding interval.
•Don't use half a rounding unit for a truncation interval.
Exam tip. State the rounding or truncation unit first; it determines both interval endpoints and which endpoint is excluded.
N16 · Apply and interpret limits of accuracy, including upper and lower bounds
A measured or rounded value stands for a range of possible true values. The lower and upper limits are usually half a rounding unit below and above the stated value, with the upper limit excluded.
Use these limits to find a maximum possible error or to decide whether a claimed result is possible. For a sum, the smallest total uses all lower limits and the greatest possible total approaches all upper limits.
Higher-tier bounds calculations also choose numerator and denominator limits deliberately, but Foundation questions can require interpreting accuracy and maximum error.
Never treat rounded inputs as exact.
Examiners expect the chosen limits to be written before the calculation.
Worked example
Higher tier: Two lengths are 4.2 cm and 3.7 cm, each correct to the nearest 0.1 cm. Could their exact total be less than 7.8 cm?
1.Write the lower limits: the lengths are at least 4.15 cm and 3.65 cm.
2.Find the smallest possible total: 4.15+3.65=7.80 cm.
3.Compare with 7.8 cm: no exact total can be smaller.
Answer: No; the exact total is at least 7.80 cm.
Common mistakes
•Don't use the displayed rounded values as though they were exact.
•Don't include an upper limit even though that endpoint rounds to the next displayed value.
•Don't use upper limits when the question asks for the smallest possible total.
Exam tip. Write “lower” or “upper” beside each substituted value so the examiner can see why it gives the required extreme.