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Edexcel GCSE Maths revision notes

Number

Section N
16 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 1MA1 section N

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.

How this checking works

In the exam: Formula sheet provided · Paper 1 non-calculator

Loading your tier…

Order positive and negative integers, decimals and fractions; use the symbols =, ≠, <, >, ≤, ≥

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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<−32-1.7<-\dfrac{3}{2} since −1.7<−1.5-1.7<-1.5.
  • Use == for equal values, ≠\neq for unequal values, and read ≤\leq or ≥\geq as allowing equality.
  • In an exam, show the conversion that justifies each comparison.
A number line showing that −1.85-1.85 lies to the left of −1.75-1.75.
Worked example

Write −74-\dfrac{7}{4}, −1.68-1.68, 53\dfrac{5}{3} and 1.71.7 in ascending order.

  1. 1.Convert the fractions: −74=−1.75-\dfrac{7}{4}=-1.75 and 53=1.666…\dfrac{5}{3}=1.666\ldots.
  2. 2.Compare the negative values first: −1.75<−1.68-1.75<-1.68.
  3. 3.Compare the positive values: 1.666…<1.71.666\ldots<1.7, then write the complete ordered list.

Answer: −74, −1.68, 53, 1.7-\dfrac{7}{4},\ -1.68,\ \dfrac{5}{3},\ 1.7.

Common mistakes

  • Don't write −8>−3-8>-3 because 8>38>3, ignoring that −8-8 lies farther left.
  • Don't treat ≤\leq 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.

Tier 1 · Easy

ORIGINAL

1

Insert either <<, >> or == between −0.62-0.62 and −35-\dfrac{3}{5}.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Mia says that −6>−2-6>-2 because 6>26>2. Explain why Mia is wrong.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

An integer kk satisfies −2.4<k3≤1.7-2.4<\dfrac{k}{3}\leq1.7. Write down every possible value of kk.

(3)

(Total for Question 1 is 3 marks)

Your progress and exam materials

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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.

Explanation

  • Written methods score for the method, then for the answer. In a decimal division such as 818.4÷1.2818.4\div1.2, multiply both numbers by the same power of 1010 so the divisor is an integer, divide, then place the decimal point by comparing the size of the numbers: 818.4÷1.2818.4\div1.2 is a little less than 818818, so 682682 is sensible and 68.268.2 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 −214÷35-2\dfrac{1}{4}\div\dfrac{3}{5}. Give your answer as a mixed number.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Convert the mixed number to an improper fraction: −214=−94-2\dfrac{1}{4}=-\dfrac{9}{4}.
  2. 2.Divide by multiplying by the reciprocal of 35\dfrac{3}{5}: −94×53-\dfrac{9}{4}\times\dfrac{5}{3}.
  3. 3.Cancel the common factor 33 and multiply: −3×54×1=−154-\dfrac{3\times5}{4\times1}=-\dfrac{15}{4}.
  4. 4.Write as a mixed number: −154=−334-\dfrac{15}{4}=-3\dfrac{3}{4}.

Answer: −334-3\dfrac{3}{4}

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for conversion to an improper fraction, eg −94-\dfrac{9}{4}M1Need not be shown with operators
(dep) for a method to divide by multiplying by the reciprocal, eg −94×53-\dfrac{9}{4}\times\dfrac{5}{3} or −4512-\dfrac{45}{12}M1oe
for −334-3\dfrac{3}{4}A1Accept −154-\dfrac{15}{4} only if the mixed number is not required

Common mistakes

  • Don't add fraction denominators as well as numerators, writing ab+cd=a+cb+d\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{a+c}{b+d}.
  • Don't divide by a fraction without inverting it.
  • Don't align the last digits of decimals instead of aligning their decimal points.

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 682682 many wrote 68.268.2 or 6.826.82, and that in the division of mixed numbers some flipped the wrong fraction, flipped both, or multiplied whole numbers and fractions separately, scoring nothing.

Exam tip

Keep an exact fraction until the final line and show the reciprocal explicitly for the division method mark.

Tier 1 · Easy

ORIGINAL

1

Work out −17+29-17+29.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

Work out 7.2÷0.067.2\div0.06.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

Work out 3.6−(−1.75)+56÷(−59)3.6-(-1.75)+\dfrac{5}{6}\div\left(-\dfrac{5}{9}\right). Give your answer as a decimal.

(3)

(Total for Question 1 is 3 marks)

Recognise and use relationships between operations, including inverse operations; use conventional notation for priority of operations, including brackets, powers, roots and reciprocals

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 ab\dfrac{a}{b} is ba\dfrac{b}{a}.
  • Examiners expect each priority stage to be visible when the question carries more than one mark.
Worked example

A positive number is squared, 1111 is subtracted, and the reciprocal of the result is 114\dfrac{1}{14}. Find the number.

  1. 1.Reverse the reciprocal: the value before taking it was 1414.
  2. 2.Undo subtracting 1111: 14+11=2514+11=25.
  3. 3.Undo the square using the positive square root: 25=5\sqrt{25}=5.

Answer: 55.

Common mistakes

  • Don't add before multiplying in 18−32×218-3^2\times2.
  • Don't say the reciprocal of ab\dfrac{a}{b} is −ab-\dfrac{a}{b}.
  • 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.

Tier 1 · Easy

ORIGINAL

1

Work out 18−32×218-3^2\times2.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

Work out (81−5)2÷2\left(\sqrt{81}-5\right)^2\div2.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A positive number is squared, 1111 is subtracted, and the reciprocal of the result is 114\dfrac{1}{14}. Find the original number.

(3)

(Total for Question 1 is 3 marks)

Prime numbers, factors (divisors), multiples, common factors and multiples, highest common factor, lowest common multiple, prime factorisation with product notation and unique factorisation theorem

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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.

Explanation

  • A prime has exactly two factors, so 11 is not prime. Every integer above 11 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=22×53500=2^2\times5^3. 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

180n180n is a cube number, where nn is a positive integer. Work out the smallest possible value of nn.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Write 180180 as a product of prime factors: 180=22×32×5180=2^2\times3^2\times5.
  2. 2.A cube number has every index a multiple of 33, so the product needs one more 22, one more 33 and two more 55s.
  3. 3.The smallest such nn is 2×3×52=1502\times3\times5^2=150.
  4. 4.Check: 180×150=27 000=303180\times150=27\,000=30^3.

Answer: n=150n=150

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for a complete method to find the prime factors of 180180, eg 22×32×52^2\times3^2\times5 or 2,2,3,3,52,2,3,3,5M1Condone the inclusion of 1 for this mark
for identifying the factors needed, eg 2×3×5×52\times3\times5\times5 or 2×3×522\times3\times5^2M1oe
for 150150A1cao

Common mistakes

  • Don't include 11 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.

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 66; the Paper 3H report notes a significant proportion gave the LCM of 315315 when the HCF of 2121 was asked for.

Exam tip

For HCF or LCM questions, write both prime-power decompositions first so the method marks remain available.

Tier 1 · Easy

ORIGINAL

1

Write 756756 as a product of its prime factors.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Find both the HCF and the LCM of 8484 and 126126.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

180n180n is a cube number, where nn is a positive integer. Find the smallest possible value of nn.

(4)

(Total for Question 1 is 4 marks)

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)

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 mm ways and a following independent choice in nn ways, the product rule gives m×nm\times 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 22, 44, 55 and 77 to make an even three-digit number. How many numbers are possible?

  1. 1.Fix the final digit as 22: choose and order two of the remaining three digits, giving 3×2=63\times2=6 numbers.
  2. 2.Fix the final digit as 44: again there are 3×2=63\times2=6 numbers.
  3. 3.Add the two disjoint cases: 6+6=126+6=12.

Answer: 1212 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.

Tier 1 · Easy

ORIGINAL

1

A badge uses one of the letters A, B or C and one of the numbers 11, 22 or 33. List every possible badge code in a systematic order.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A three-digit number is made using three different digits from 22, 44, 55 and 77. List all the possible even numbers.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Two different numbers are selected from 11, 22, 33, 44 and 55. The order of selection does not matter. List every pair whose product is even and whose sum is greater than 55.

(3)

(Total for Question 1 is 3 marks)

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A positive integer power represents repeated multiplication: ana^n contains nn factors equal to aa.
  • An associated root reverses that power, so 81=9\sqrt{81}=9, 1253=5\sqrt[3]{125}=5, and 164=2\sqrt[4]{16}=2.
  • Learn common powers of 22, 33, 44 and 55 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<433^3<40<4^3 shows 3<403<43<\sqrt[3]{40}<4.
  • In an exam, distinguish an exact evaluation from an estimate and clearly state the bounding powers used.
Worked example

Higher tier: Estimate 2003\sqrt[3]{200} to the nearest whole number without a calculator.

  1. 1.Use nearby cubes: 53=1255^3=125 and 63=2166^3=216, so 5<2003<65<\sqrt[3]{200}<6.
  2. 2.The rounding threshold is 5.53=166.3755.5^3=166.375; since 200>166.375200>166.375, the cube root is above 5.55.5.
  3. 3.Round the estimate to the nearest whole number.

Answer: 2003≈6\sqrt[3]{200}\approx6.

Common mistakes

  • Don't calculate 343^4 as 3×43\times4 instead of four factors of 33.
  • 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.

Tier 1 · Easy

ORIGINAL

1

Work out 252^5.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

Work out 3433\sqrt[3]{343}.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 3 · Hard

ORIGINAL

1

Work out 12964+5123\sqrt[4]{1296}+\sqrt[3]{512}.

(3)

(Total for Question 1 is 3 marks)

Calculate with roots, and with integer and fractional indices

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Integer indices include positive, zero and negative powers. For non-zero aa, a0=1a^0=1 and a−n=1ana^{-n}=\dfrac{1}{a^n}; the negative index creates a reciprocal, not a negative answer.
  • Roots reverse powers, including odd roots of negative numbers such as −2163=−6\sqrt[3]{-216}=-6.
  • Higher-tier questions also use fractional indices: a1/n=ana^{1/n}=\sqrt[n]{a} and am/n=amna^{m/n}=\sqrt[n]{a^m}.
  • 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 813/481^{3/4}.

  1. 1.Interpret the denominator as a root: 813/4=(814)381^{3/4}=(\sqrt[4]{81})^3.
  2. 2.Evaluate 814=3\sqrt[4]{81}=3.
  3. 3.Calculate 33=273^3=27.

Answer: 2727.

Common mistakes

  • Don't write 4−2=−164^{-2}=-16 instead of taking the reciprocal.
  • Don't treat a1/2a^{1/2} as a÷2a\div2 rather than a\sqrt a.
  • Don't evaluate (−3)2(-3)^2 and −32-3^2 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.

Tier 1 · Easy

ORIGINAL

1

Work out 4−24^{-2}.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Work out −2163\sqrt[3]{-216}.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 3 · Hard

ORIGINAL

1

Work out 144+5−2\sqrt{144}+5^{-2}. Give an exact answer.

(3)

(Total for Question 1 is 3 marks)

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An exact answer keeps fractions, multiples of π\pi 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\pi.
  • Higher-tier surds are simplified by extracting square factors: 12=4×3=23\sqrt{12}=\sqrt{4\times3}=2\sqrt3.
  • Like surds can then be collected, and a denominator such as 5\sqrt5 is rationalised by multiplying numerator and denominator by 5\sqrt5.
  • Foundation questions do not require surd manipulation, but do require exact fractions and multiples of π\pi.
  • When the command says “exact”, no rounded decimal should appear in the final answer.
Worked example

Higher tier: Simplify 75−12\sqrt{75}-\sqrt{12} and give an exact answer.

  1. 1.Extract the largest square factor: 75=25×3=53\sqrt{75}=\sqrt{25\times3}=5\sqrt3.
  2. 2.Similarly, 12=4×3=23\sqrt{12}=\sqrt{4\times3}=2\sqrt3.
  3. 3.Collect like surds: 53−23=335\sqrt3-2\sqrt3=3\sqrt3.

Answer: 333\sqrt3.

Common mistakes

  • Don't replace π\pi with 3.143.14 when an exact value is required.
  • Don't write a+b=a+b\sqrt{a+b}=\sqrt a+\sqrt b.
  • Don't collect unlike surds, for example treating 2+3\sqrt2+\sqrt3 as 252\sqrt5.

Exam tip

If the question says “exact”, retain π\pi, fractions or surds and simplify them fully rather than using a calculator decimal.

Tier 1 · Easy

ORIGINAL

1

Work out 34+56\dfrac{3}{4}+\dfrac{5}{6}. Give an exact answer.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A circle has radius 77 cm. Work out its exact area.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

Work out 3π5+7π8−π4\dfrac{3\pi}{5}+\dfrac{7\pi}{8}-\dfrac{\pi}{4}. Give an exact answer.

(3)

(Total for Question 1 is 3 marks)

Calculate with and interpret standard form A × 10^n, where 1 ≤ A < 10 and n is an integer

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Specification N9: calculate with and interpret standard form A×10nA\times10^n, where 1≤A<101\leq A<10 and nn is an integer, with and without a calculator.

Explanation

  • A positive index moves the decimal point right, so 5.3×104=53 0005.3\times10^4=53\,000; a negative index moves it left, so 7.4×10−5=0.000 0747.4\times10^{-5}=0.000\,074. To multiply, multiply the AA values and add the indices; to divide, divide the AA values and subtract the indices; then normalise, because 24×10−224\times10^{-2} is not standard form until it becomes 2.4×10−12.4\times10^{-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=3.42×1079.7\times10^6+2.45\times10^7=0.97\times10^7+2.45\times10^7=3.42\times10^7.
  • Adding the AA 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 34.2×10634.2\times10^6, 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 3.6×10−4+7.5×10−51.5×103\dfrac{3.6\times10^{-4}+7.5\times10^{-5}}{1.5\times10^{3}}. Give your answer in standard form.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Write the numerator with a common power of ten: 3.6×10−4+0.75×10−4=4.35×10−43.6\times10^{-4}+0.75\times10^{-4}=4.35\times10^{-4}.
  2. 2.Divide the AA values and subtract the indices: 4.351.5×10−4−3=2.9×10−7\dfrac{4.35}{1.5}\times10^{-4-3}=2.9\times10^{-7}.
  3. 3.Check the coefficient: 1≤2.9<101\leq2.9<10, so the answer is in standard form.

Answer: 2.9×10−72.9\times10^{-7}

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for 4.35×10−44.35\times10^{-4} or 0.0004350.000435 or both numbers in a form ready for addition, eg 3.6×10−4+0.75×10−43.6\times10^{-4}+0.75\times10^{-4}M1oe
for 4.35÷1.5×10−4−34.35\div1.5\times10^{-4-3} or 2.9×10n2.9\times10^n (n≠−7n\neq-7) or 0.000 000 290.000\,000\,29M1Correct answer in incorrect form
for 2.9×10−72.9\times10^{-7}A1cao

Common mistakes

  • Don't leave 24×10424\times10^4 as the final answer even though 2424 is not between 11 and 1010.
  • Don't add the indices when dividing powers of 1010.
  • Don't add coefficients whose powers of 1010 have not first been made equal.

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 34 200 00034\,200\,000 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 9.79.7 to 2.452.45 to get 12.1512.15 and attaching a power of ten to that.

Exam tip

Circle the final coefficient and check 1≤A<101\leq A<10 before submitting any standard-form answer.

Tier 1 · Easy

ORIGINAL

1

Write 0.0000720.000072 in standard form.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Work out (6×107)(4×10−3)(6\times10^7)(4\times10^{-3}). Give your answer in standard form.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

Work out 3.6×10−4+7.5×10−51.5×103\dfrac{3.6\times10^{-4}+7.5\times10^{-5}}{1.5\times10^3}. Give your answer in standard form.

(4)

(Total for Question 1 is 4 marks)

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A terminating decimal has finitely many decimal places, so write it over the matching power of 1010 and simplify: 0.375=3751000=380.375=\dfrac{375}{1000}=\dfrac38. 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 xx, multiply by a power of 1010 so the repeating blocks align, subtract to remove the recurring part, and solve for xx.
  • 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…0.\dot2\dot7=0.272727\ldots to a fraction.

  1. 1.Let x=0.272727…x=0.272727\ldots, so 100x=27.272727…100x=27.272727\ldots.
  2. 2.Subtract: 100x−x=27100x-x=27, giving 99x=2799x=27.
  3. 3.Solve and simplify: x=2799=311x=\dfrac{27}{99}=\dfrac3{11}.

Answer: 311\dfrac3{11}.

Common mistakes

  • Don't write 0.3750.375 as 375100\dfrac{375}{100} or use the wrong place-value denominator.
  • Don't stop before simplifying the numerator and denominator fully.
  • Don't multiply by 1010 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.

Tier 1 · Easy

ORIGINAL

1

Write 0.3750.375 as a fraction in its simplest form.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Write 3780\dfrac{37}{80} as a decimal.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 3 · Hard

ORIGINAL

1

Which is greater, 0.560.56 or 916\dfrac{9}{16}? Work out the difference as a fraction.

(3)

(Total for Question 1 is 3 marks)

Identify and work with fractions in ratio problems

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Fractions and ratios describe the same partition in different forms. If a group is 35\dfrac35 of the whole, the remainder is 25\dfrac25, so the part-to-part ratio is 3:23:2.
  • Conversely, in a ratio a:ba:b, the first share is aa+b\dfrac{a}{a+b} of the whole and the second is ba+b\dfrac{b}{a+b}.
  • 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.
Another way to see this:
Worked example

£330\pounds330 is shared between Imran and Jo in the ratio 4:74:7. Imran spends 38\dfrac38 of his share and Jo spends 27\dfrac27 of hers. Find the total left.

  1. 1.There are 1111 parts, so one part is 330÷11=30330\div11=30.
  2. 2.Imran gets 120120 and keeps 58×120=75\dfrac58\times120=75; Jo gets 210210 and keeps 57×210=150\dfrac57\times210=150.
  3. 3.Add the amounts kept: 75+150=22575+150=225.

Answer: £225\pounds225.

Common mistakes

  • Don't say the remainder of 35\dfrac35 is 55\dfrac55 instead of 25\dfrac25.
  • 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.

Tier 1 · Easy

ORIGINAL

1

35\dfrac{3}{5} of the beads in a bag are red and the rest are blue. Write the ratio of red beads to blue beads.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

The ratio of Ava's tokens to Ben's tokens is 5:75:7. Ben has 8484 tokens. Work out 34\dfrac{3}{4} of Ava's number of tokens.

(3)

(Total for Question 1 is 3 marks)

Interpret fractions and percentages as operators

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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.

Explanation

  • To find ab\dfrac{a}{b} of an amount, divide by bb and multiply by aa, or multiply by the fraction directly: 27\dfrac{2}{7} of 420420 is 420×2÷7=120420\times2\div7=120. A percentage becomes a decimal or fraction operator, 35%35\% of 420420 being 420×0.35=147420\times0.35=147.
  • A percentage change is one multiplier: a 15%15\% decrease multiplies by 0.850.85, an increase of 15%15\% by 1.151.15.
  • When several operators apply, each acts on the amount that remains, not on the original: “38\dfrac{3}{8} of the reduced price” means multiply the reduced price by 38\dfrac{3}{8}.
  • 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\pounds640. Its price is reduced by 15%15\%. A customer pays 38\dfrac{3}{8} 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. 1.Apply the reduction multiplier: 640×0.85=£544640\times0.85=\pounds544.
  2. 2.Find the deposit as a fraction of the reduced price: 544×38=£204544\times\dfrac{3}{8}=\pounds204.
  3. 3.Subtract to find the balance: 544−204=£340544-204=\pounds340, or directly 544×58544\times\dfrac{5}{8}.

Answer: £340\pounds340

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for a process to find the reduced price, eg 640×0.85640\times0.85 or 640−640×15÷100640-640\times15\div100 (= 544)P1oe
for a process to find the deposit, eg “544” ×3÷8\times3\div8 (= 204)P1
(dep) for a complete process, eg “544” −- “204” or “544” ×5÷8\times5\div8P1
for 340340A1cao

Common mistakes

  • Don't divide by the numerator and multiply by the denominator for a fraction of an amount.
  • Don't use 0.150.15 as the multiplier for the price after a 15%15\% decrease.
  • Don't calculate the deposit from the original price instead of the reduced price.

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%88.8\%, and the Paper 3H report of the same series notes that increasing or decreasing £46.75\pounds46.75 by 15%15\% scored nothing when the operator had to be reversed.

Exam tip

Write each fraction or percentage as a multiplier beside the amount it operates on.

Tier 1 · Easy

ORIGINAL

1

Work out 35\dfrac{3}{5} of 7070.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

Work out 17.5%17.5\% of 240240.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A machine costs £640\pounds640. Its price is reduced by 15%15\%, then a customer pays 38\dfrac{3}{8} of the reduced price as a deposit. Work out the balance still to pay.

(4)

(Total for Question 1 is 4 marks)

Use standard units of mass, length, time, money and other measures (including standard compound measures) using decimal quantities where appropriate

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Specification N13: use standard units of mass, length, time, money and other measures, including with decimal quantities.

Explanation

  • Work in one unit at a time and convert before calculating: 11 kg =1000=1000 g, 11 km =1000=1000 m, 11 litre =1000=1000 cm3^3, and money in either pounds or pence throughout, converting at the end.
  • Time is the trap: 1.251.25 hours is 11 hour 1515 minutes, not 11 hour 2525 minutes, and 2.752.75 hours is 22 hours 4545 minutes, so a time from distance÷speed\text{distance}\div\text{speed} must be converted to hours and minutes when the question asks for it.
  • A rate written per 100100 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 7.57.5 litres of fuel for every 100100 km it travels. Fuel costs £1.68\pounds1.68 per litre. Work out the cost of the fuel for a journey of 5454 km.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Scale the fuel rate to the journey: 54÷100×7.5=4.0554\div100\times7.5=4.05 litres.
  2. 2.Multiply by the price per litre: 4.05×1.68=6.8044.05\times1.68=6.804.
  3. 3.Round money to the nearest penny: £6.80\pounds6.80.

Answer: £6.80\pounds6.80

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for a process to find the fuel used, eg 54÷100×7.554\div100\times7.5 (= 4.05)P1oe
for a complete process, eg “4.05” ×1.68\times1.68P1
for 6.806.80A1accept 6.8

Common mistakes

  • Don't treat 1.51.5 hours as 11 hour 5050 minutes.
  • Don't use the factor 100100 rather than 1002100^2 when converting square metres to square centimetres.
  • Don't omit the compound unit from the final answer.

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 55 miles at 3030 miles per hour ranged from fractions of a minute up to 900900 minutes, and urges checking the reasonableness of an answer; those who wrote “3030 miles in 6060 minutes” and divided both by 66 found the 1010 minutes cleanly.

Exam tip

Write the target unit before starting; every conversion should move the data towards that unit.

Tier 1 · Easy

ORIGINAL

1

Change 2.752.75 kg to grams.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

A workshop starts at 09:3809{:}38 and lasts for 11 hour 4747 minutes. Work out the finishing time.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A car travels 5454 km and uses fuel at a rate of 7.57.5 litres per 100100 km. Fuel costs £1.68\pounds1.68 per litre. Work out the fuel cost for the journey, giving your answer to the nearest penny.

(4)

(Total for Question 1 is 4 marks)

Estimate answers; check calculations using approximation and estimation, including answers obtained using technology

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 ≈\approx, 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 1010 or 100100, suspect a place-value entry error.
  • Examiners require both the estimate and a clear comparison when asked to comment.
Worked example

A calculator gives 931.24931.24 for 598.4×0.03170.204\dfrac{598.4\times0.0317}{0.204}. Use an estimate to decide whether it is reasonable.

  1. 1.Round inputs to convenient values: 598.4≈600598.4\approx600, 0.0317≈0.030.0317\approx0.03, 0.204≈0.20.204\approx0.2.
  2. 2.Estimate 600×0.030.2=90\dfrac{600\times0.03}{0.2}=90.
  3. 3.Compare 931.24931.24 with 9090: 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 ≈\approx between an expression and its estimate.
  • Don't round a small decimal such as 0.03170.0317 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.

Tier 1 · Easy

ORIGINAL

1

Estimate the value of 19.8×0.4919.8\times0.49.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

Estimate 48.7×0.2030.098\dfrac{48.7\times0.203}{0.098}.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A calculator display gives 931.24931.24 for 598.4×0.03170.204\dfrac{598.4\times0.0317}{0.204}. Use an estimate to decide whether this display is reasonable. Give a reason.

(3)

(Total for Question 1 is 3 marks)

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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: 55 or more rounds up, while 44 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 uu, subtract and add u2\dfrac u2; 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 12.612.6 correct to one decimal place.
Worked example

A positive number yy is truncated to 4.374.37 at two decimal places. Write its error interval and find the greatest integer value of 100y100y.

  1. 1.Truncation gives 4.37≤y<4.384.37\leq y<4.38.
  2. 2.Multiply the whole interval by 100100: 437≤100y<438437\leq100y<438.
  3. 3.The greatest integer in this interval is 437437.

Answer: 4.37≤y<4.384.37\leq y<4.38 and the greatest integer value is 437437.

Common mistakes

  • Don't count leading zeros as significant figures in 0.0078460.007846.
  • 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.

Tier 1 · Easy

ORIGINAL

1

Write 0.0078460.007846 correct to 22 significant figures.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

A number xx is 12.612.6 correct to 11 decimal place. Write the error interval for xx.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A positive number yy is truncated to 4.374.37 at 22 decimal places. Write its error interval and find the greatest possible integer value of 100y100y.

(3)

(Total for Question 1 is 3 marks)

Apply and interpret limits of accuracy, including upper and lower bounds

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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.
Another way to see this:
Worked example

Higher tier: Two lengths are 4.24.2 cm and 3.73.7 cm, each correct to the nearest 0.10.1 cm. Could their exact total be less than 7.87.8 cm?

  1. 1.Write the lower limits: the lengths are at least 4.154.15 cm and 3.653.65 cm.
  2. 2.Find the smallest possible total: 4.15+3.65=7.804.15+3.65=7.80 cm.
  3. 3.Compare with 7.87.8 cm: no exact total can be smaller.

Answer: No; the exact total is at least 7.807.80 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.

Tier 1 · Easy

ORIGINAL

1

A length is recorded as 1212 cm to the nearest centimetre. Write down the maximum possible rounding error.

(1)

(Total for Question 1 is 1 mark)

Tier 3 · Hard

ORIGINAL

1

Higher only: A scale records the mass of each of 88 identical boxes as 2.42.4 kg to the nearest 0.10.1 kg. Could the exact total mass of the boxes be 2020 kg? Justify your answer.

(3)

(Total for Question 1 is 3 marks)

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