N Number — revision question pack

16 specification points · notes, questions, answers and worked methods

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

Loading your tier…

Answer ALL questions.

Write your answers in the spaces provided.

You must write down all the stages in your working.

N1 · Order positive and negative integers, decimals and fractions; use the symbols =, ≠, <, >, ≤, ≥

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, $-1.7<-\dfrac{3}{2}$ since $-1.7<-1.5$ .
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

lies to the left of

-1.75

Worked example

Write $-\dfrac{7}{4}$ , $-1.68$ , $\dfrac{5}{3}$ and $1.7$ in ascending order.

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

Answer: $-\dfrac{7}{4},\ -1.68,\ \dfrac{5}{3},\ 1.7$ .

Common mistakes

Don't write $-8>-3$ because $8>3$ , ignoring that $-8$ 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

1
Insert either $<$ , $>$ or $=$ between $-0.62$ and $-\dfrac{3}{5}$ .
(1)
(Total for Question 1 is 1 mark)
2
Write $-0.47$ , $-\dfrac{9}{20}$ , $-0.52$ and $-\dfrac{1}{2}$ in ascending order.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
Mia says that $-6>-2$ because $6>2$ . Explain why Mia is wrong.
(2)
(Total for Question 1 is 2 marks)
2
Write $-1.2$ , $-\dfrac{7}{6}$ and $-1.15$ in descending order.
(2)
(Total for Question 2 is 2 marks)
3
The box in $-0.6\square$ is filled with one digit. The completed decimal is greater than $-0.64$ and less than $-0.58$ . Work out all the possible digits.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
An integer $k$ satisfies $-2.4<\dfrac{k}{3}\leq1.7$ . Write down every possible value of $k$ .
(3)
(Total for Question 1 is 3 marks)
2
Sam writes $-\dfrac{13}{20}<-0.66<-\dfrac{2}{3}$ . Is Sam correct? Give a reason.
(2)
(Total for Question 2 is 2 marks)
3
Four cards show $-\dfrac{11}{18}$ , $-0.605$ , $-61\%$ and $-0.59$ . Work out the difference between the greatest value and the smallest value. Give your answer as a fraction.
(4)
(Total for Question 3 is 4 marks)
4
Points $P$ , $Q$ , $R$ and $S$ have coordinates $-\dfrac{3}{4}$ , $-0.72$ , $-73\%$ and $-\dfrac{7}{10}$ respectively. Which points lie in the interval $-0.74<x\leq-0.71$ ? Write their coordinates in descending order.
(3)
(Total for Question 4 is 3 marks)
5
The integer $k$ satisfies $-0.56<-\dfrac{k}{40}<-0.54$ . Work out $k$ . You must show all your working.
(3)
(Total for Question 5 is 3 marks)

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

Explanation

The four operations must work with positive and negative integers, decimals, proper and improper fractions, and mixed numbers. For written decimal addition or subtraction, align decimal points so digits of equal place value share a column.
When adding fractions, create a common denominator; when multiplying, cancel common factors where useful.
Convert a mixed number to an improper fraction before multiplication or division, and divide by a fraction by multiplying by its reciprocal.
Track signs separately: equal signs give a positive product or quotient and different signs give a negative one.
Examiners award method marks for these conversions even if the final arithmetic slips.

Worked example

Work out $-2\dfrac{1}{4}\div\dfrac{3}{5}$ .

1.Convert the mixed number: $-2\dfrac{1}{4}=-\dfrac{9}{4}$ .
2.Multiply by the reciprocal: $-\dfrac{9}{4}\times\dfrac{5}{3}$ .
3.Cancel the factor $3$ and evaluate to get $-\dfrac{15}{4}$ .

Answer: $-\dfrac{15}{4}=-3\dfrac{3}{4}$ .

Common mistakes

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

Exam tip

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

Tier 1 · Easy

1
Work out $-17+29$ .
(1)
(Total for Question 1 is 1 mark)
2
Work out $42.7-18.95$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Work out $7.2\div0.06$ .
(2)
(Total for Question 1 is 2 marks)
2
Work out $2\dfrac{1}{3}+1\dfrac{5}{6}$ .
(2)
(Total for Question 2 is 2 marks)
3
A roll contains $8\dfrac{3}{5}$ metres of ribbon. Six pieces, each $0.75$ metres long, are cut from the roll. Work out the length of ribbon left.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Work out $3.6-(-1.75)+\dfrac{5}{6}\div\left(-\dfrac{5}{9}\right)$ . Give your answer as a decimal.
(3)
(Total for Question 1 is 3 marks)
2
A container holds $12.5$ litres. Then $2\dfrac{3}{4}$ litres are removed and $1.6$ litres are added. Work out the final volume.
(3)
(Total for Question 2 is 3 marks)
3
An account has a balance of $-\pounds18.40$ . A payment of $\pounds75.75$ is added. A fee of $2\dfrac{2}{5}$ pounds and three payments of $\pounds16.85$ are then taken from the account. Work out the final balance.
(4)
(Total for Question 3 is 4 marks)
4
Work out $8.03-2.476+0.095\times16$ .
(3)
(Total for Question 4 is 3 marks)
5
Work out $\left(4\dfrac{1}{5}-2.85\right)\div\dfrac{3}{8}$ . Give your answer as a decimal.
(4)
(Total for Question 5 is 4 marks)

N3 · Recognise and use relationships between operations, including inverse operations; use conventional notation for priority of operations, including brackets, powers, roots and reciprocals

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

Worked example

A positive number is squared, $11$ is subtracted, and the reciprocal of the result is $\dfrac{1}{14}$ . 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: $\sqrt{25}=5$ .

Answer: $5$ .

Common mistakes

Don't add before multiplying in $18-3^2\times2$ .
Don't say the reciprocal of $\dfrac{a}{b}$ is $-\dfrac{a}{b}$ .
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

1
Work out $18-3^2\times2$ .
(1)
(Total for Question 1 is 1 mark)
2
Write down the reciprocal of $8$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Work out $\left(\sqrt{81}-5\right)^2\div2$ .
(2)
(Total for Question 1 is 2 marks)
2
Insert one pair of brackets into $18-4\times2+3$ to make the value $31$ .
(1)
(Total for Question 2 is 1 mark)
3
Luca says that $728\div14=54$ . Use an inverse operation to show that Luca is wrong. Work out the correct answer.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A positive number is squared, $11$ is subtracted, and the reciprocal of the result is $\dfrac{1}{14}$ . Find the original number.
(3)
(Total for Question 1 is 3 marks)
2
Work out $2^3+36\div\sqrt{9}$ , then subtract the reciprocal of $\dfrac{1}{4}$ .
(3)
(Total for Question 2 is 3 marks)
3
Ravi works out $60\div5\times3$ by first calculating $5\times3$ and gets $4$ . Explain Ravi's error and work out the correct value.
(3)
(Total for Question 3 is 3 marks)
4
Work out $96\div\left[2\left(3^2-1\right)\right]+\sqrt{49}$ .
(3)
(Total for Question 4 is 3 marks)
5
A number machine multiplies its input by $5$ and then subtracts $8$ . The output is $47$ . Ben says to reverse the machine by adding $8$ and then dividing by $5$ . Layla says to divide by $5$ and then add $8$ . Who is correct? Work out the input.
(3)
(Total for Question 5 is 3 marks)

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

Explanation

A prime number has exactly two positive factors, $1$ and itself; $1$ is not prime. Every integer greater than $1$ has a unique prime factorisation apart from the order of its factors.
Produce it with a factor tree or repeated division, then write repeated factors using powers. For two or more numbers, the highest common factor uses every shared prime with its smallest exponent.
The lowest common multiple uses every prime present with its largest exponent.
This method avoids double-counting factors and also supports problems about making a product into a square or cube.
Examiners require the factorisation to contain primes only.

Worked example

$180n$ is a cube number, where $n$ is a positive integer. Find the smallest possible $n$ .

1.Prime factorise: $180=2^2\times3^2\times5$ .
2.A cube needs exponents in multiples of $3$ , so supply $2^1\times3^1\times5^2$ .
3.Evaluate $n=2\times3\times25=150$ and check $180n=27000=30^3$ .

Answer: $n=150$ .

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.

Tier 1 · Easy

1
Write $756$ as a product of its prime factors.
(2)
(Total for Question 1 is 2 marks)
2
Write down all the prime factors of $42$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Find both the HCF and the LCM of $84$ and $126$ .
(3)
(Total for Question 1 is 3 marks)
2
Find the smallest multiple of both $12$ and $18$ that is greater than $100$ .
(2)
(Total for Question 2 is 2 marks)
3
A club has $96$ red counters, $144$ blue counters and $168$ green counters. The counters are put into the greatest possible number of identical bags with none left over. Work out the total number of counters in each bag.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
$180n$ is a cube number, where $n$ is a positive integer. Find the smallest possible value of $n$ .
(4)
(Total for Question 1 is 4 marks)
2
$n$ is a positive multiple of $15$ less than $100$ . The HCF of $60$ and $n$ is $15$ . Work out all the possible values of $n$ .
(3)
(Total for Question 2 is 3 marks)
3
$n$ is a positive integer. The HCF of $n$ and $72$ is $18$ . The LCM of $n$ and $72$ is $360$ . Work out $n$ . Your method must be shown.
(4)
(Total for Question 3 is 4 marks)
4
A whole number leaves a remainder of $5$ when it is divided by $18$ and also when it is divided by $24$ . Work out the smallest three-digit number with this property.
(4)
(Total for Question 4 is 4 marks)
5
The number $2^7\times3^5\times5^2$ has a factor that is a cube number. Work out its greatest factor that is a cube number.
(4)
(Total for Question 5 is 4 marks)

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)

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 $m$ ways and a following independent choice in $n$ ways, the product rule gives $m\times n$ combined outcomes.
Examiners award completeness only when the structure of the list is evident.

Worked example

Higher tier: Three different digits are chosen from $2$ , $4$ , $5$ 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 $3\times2=6$ numbers.
2.Fix the final digit as $4$ : again there are $3\times2=6$ 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.

Tier 1 · Easy

1
A badge uses one of the letters A, B or C and one of the numbers $1$ , $2$ or $3$ . List every possible badge code in a systematic order.
(2)
(Total for Question 1 is 2 marks)
2
Use two different digits from $1$ , $2$ and $3$ to make a two-digit number. List all the possible numbers.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
A three-digit number is made using three different digits from $2$ , $4$ , $5$ and $7$ . List all the possible even numbers.
(3)
(Total for Question 1 is 3 marks)
2
A route uses gate A or gate B, followed by lane $1$ , $2$ or $3$ . Lane $3$ is closed after gate B. List all the possible routes.
(2)
(Total for Question 2 is 2 marks)
3
An appointment can start at $10{:}00$ , $10{:}30$ or $11{:}00$ and can last $30$ minutes or $60$ minutes. It must finish by $11{:}30$ . List all the possible start-time and duration pairs.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Two different numbers are selected from $1$ , $2$ , $3$ , $4$ and $5$ . The order of selection does not matter. List every pair whose product is even and whose sum is greater than $5$ .
(3)
(Total for Question 1 is 3 marks)
2
One spinner is labelled $1$ , $2$ , $4$ and another is labelled $1$ , $3$ . Both spinners are spun once. List all the possible totals. Write each total once. You must show all your working.
(2)
(Total for Question 2 is 2 marks)
3
An outfit has one jacket, one pair of trousers and one pair of shoes. The jackets are black and grey. The trousers are navy, cream and khaki. The shoes are boots and trainers. The grey jacket is never worn with the khaki trousers, and the navy trousers are never worn with the trainers. List all the possible outfits.
(4)
(Total for Question 3 is 4 marks)
4
Higher only: A light signal uses one of four colours, one of three patterns and one of two durations. A red signal can use only the steady pattern. The other colours can use any pattern. Work out the total number of different signals.
(4)
(Total for Question 4 is 4 marks)
5
A robot reaches its target by making exactly three moves east and two moves north. It can make the five moves in any order. Work out the number of different routes. You must show a systematic method.
(4)
(Total for Question 5 is 4 marks)

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

Explanation

A positive integer power represents repeated multiplication: $a^n$ contains $n$ factors equal to $a$ .
An associated root reverses that power, so $\sqrt{81}=9$ , $\sqrt[3]{125}=5$ , and $\sqrt[4]{16}=2$ .
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, $3^3<40<4^3$ shows $3<\sqrt[3]{40}<4$ .
In an exam, distinguish an exact evaluation from an estimate and clearly state the bounding powers used.

Worked example

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

1.Use nearby cubes: $5^3=125$ and $6^3=216$ , so $5<\sqrt[3]{200}<6$ .
2.The rounding threshold is $5.5^3=166.375$ ; since $200>166.375$ , the cube root is above $5.5$ .
3.Round the estimate to the nearest whole number.

Answer: $\sqrt[3]{200}\approx6$ .

Common mistakes

Don't calculate $3^4$ as $3\times4$ 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.

Tier 1 · Easy

1
Work out $2^5$ .
(1)
(Total for Question 1 is 1 mark)
2
Write down the two consecutive whole numbers between which $\sqrt{30}$ lies.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Work out $\sqrt[3]{343}$ .
(1)
(Total for Question 1 is 1 mark)
2
Write $625$ as a power of $5$ .
(1)
(Total for Question 2 is 1 mark)
3
Write $4096$ as a power of $4$ . Hence work out $\sqrt[6]{4096}$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Work out $\sqrt[4]{1296}+\sqrt[3]{512}$ .
(3)
(Total for Question 1 is 3 marks)
2
Work out $\sqrt[3]{1000}\times\sqrt{196}\div\sqrt[3]{8}$ .
(3)
(Total for Question 2 is 3 marks)
3
$a$ and $b$ are positive integers. $a^4=625$ and $\sqrt[3]{b}=a+3$ . Work out the value of $b$ .
(3)
(Total for Question 3 is 3 marks)
4
Higher only: The positive number $x$ satisfies $x^4=580$ . Given that $4.9^4=576.4801$ , use a suitable comparison to work out $x$ to the nearest tenth.
(4)
(Total for Question 4 is 4 marks)
5
$n$ is a positive integer that is both a square number and a cube number. It satisfies $\sqrt[3]{n}+\sqrt{n}=150$ . Work out the value of $n$ .
(4)
(Total for Question 5 is 4 marks)

N7 · Calculate with roots, and with integer and fractional indices

Explanation

Integer indices include positive, zero and negative powers. For non-zero $a$ , $a^0=1$ and $a^{-n}=\dfrac{1}{a^n}$ ; the negative index creates a reciprocal, not a negative answer.
Roots reverse powers, including odd roots of negative numbers such as $\sqrt[3]{-216}=-6$ .
Higher-tier questions also use fractional indices: $a^{1/n}=\sqrt[n]{a}$ and $a^{m/n}=\sqrt[n]{a^m}$ .
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 $81^{3/4}$ .

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

Answer: $27$ .

Common mistakes

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

1
Work out $4^{-2}$ .
(1)
(Total for Question 1 is 1 mark)
2
Work out $10^0$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Work out $\sqrt[3]{-216}$ .
(1)
(Total for Question 1 is 1 mark)
2
Work out $2^{-3}+(-2)^3$ .
(2)
(Total for Question 2 is 2 marks)
3
$5^n=\dfrac{1}{625}$ . Work out the value of $n$ and then work out $(-n)^2$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Work out $\sqrt{144}+5^{-2}$ . Give an exact answer.
(3)
(Total for Question 1 is 3 marks)
2
Work out $(-3)^4\times3^{-2}\div\sqrt{81}$ .
(3)
(Total for Question 2 is 3 marks)
3
Work out $(-4)^3\times2^{-5}-9^0$ .
(3)
(Total for Question 3 is 3 marks)
4
Higher only: Work out $81^{-3/4}+32^{2/5}$ . Give your answer as a fully simplified fraction.
(4)
(Total for Question 4 is 4 marks)
5
Higher only: The positive integer $n$ satisfies $n^{3/2}=125$ . Work out the value of $n$ .
(3)
(Total for Question 5 is 3 marks)

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

Explanation

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

Worked example

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

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

Answer: $3\sqrt3$ .

Common mistakes

Don't replace $\pi$ with $3.14$ when an exact value is required.
Don't write $\sqrt{a+b}=\sqrt a+\sqrt b$ .
Don't collect unlike surds, for example treating $\sqrt2+\sqrt3$ as $2\sqrt5$ .

Exam tip

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

Tier 1 · Easy

1
Work out $\dfrac{3}{4}+\dfrac{5}{6}$ . Give an exact answer.
(2)
(Total for Question 1 is 2 marks)
2
Work out $\dfrac{7}{10}-\dfrac{1}{4}$ . Give an exact answer.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
A circle has radius $7$ cm. Work out its exact area.
(2)
(Total for Question 1 is 2 marks)
2
Simplify $5\pi-\dfrac{7\pi}{4}$ .
(2)
(Total for Question 2 is 2 marks)
3
A ribbon is $5\pi+\dfrac34$ metres long. A length of $2\pi-\dfrac16$ metres is used. Work out the exact length left.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Work out $\dfrac{3\pi}{5}+\dfrac{7\pi}{8}-\dfrac{\pi}{4}$ . Give an exact answer.
(3)
(Total for Question 1 is 3 marks)
2
Work out $\left(\dfrac{5\pi}{6}+\dfrac{7\pi}{9}\right)\div\dfrac{2}{3}$ . Give an exact answer.
(3)
(Total for Question 2 is 3 marks)
3
A wheel has radius $0.35$ metres. It makes $24$ complete rotations. Work out the exact distance travelled.
(3)
(Total for Question 3 is 3 marks)
4
Higher only: Work out $\left(\sqrt{50}-\sqrt{8}\right)\left(\sqrt{18}+\sqrt{2}\right)$ . Give an exact answer.
(4)
(Total for Question 4 is 4 marks)
5
Higher only: Rationalise the denominator and simplify $\dfrac{4}{\sqrt{7}-1}-\sqrt{7}$ .
(4)
(Total for Question 5 is 4 marks)

N9 · Calculate with and interpret standard form A × 10^n, where 1 ≤ A < 10 and n is an integer

Explanation

Standard form writes a number as $A\times10^n$ , where $1\leq A<10$ and $n$ is an integer. A positive index moves the decimal point right to make a large ordinary number; a negative index moves it left to make a small one.
For multiplication, multiply the $A$ values and add the indices; for division, divide the $A$ values and subtract the indices.
For addition or subtraction, first rewrite the terms with the same power of $10$ .
Finally normalise the coefficient so it lies in the required interval.
Examiners can award method marks for correct index work before normalisation.

Worked example

Work out $\dfrac{3.6\times10^{-4}+7.5\times10^{-5}}{1.5\times10^3}$ in standard form.

1.Use a common power: $3.6\times10^{-4}+0.75\times10^{-4}=4.35\times10^{-4}$ .
2.Divide coefficients and subtract indices: $(4.35\div1.5)\times10^{-4-3}$ .
3.Evaluate and check the coefficient: $2.9\times10^{-7}$ .

Answer: $2.9\times10^{-7}$ .

Common mistakes

Don't leave $24\times10^4$ 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 $1\leq A<10$ before submitting any standard-form answer.

Tier 1 · Easy

1
Write $0.000072$ in standard form.
(1)
(Total for Question 1 is 1 mark)
2
Write $4\,530\,000$ in standard form.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Work out $(6\times10^7)(4\times10^{-3})$ . Give your answer in standard form.
(2)
(Total for Question 1 is 2 marks)
2
Work out $\dfrac{8.4\times10^6}{2.1\times10^2}$ . Write your answer in the form $A\times10^n$ , where $1\leq A<10$ .
(2)
(Total for Question 2 is 2 marks)
3
A parcel has mass $3.75\times10^4$ grams. Another parcel has mass $8600$ grams. Work out their total mass in standard form.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Work out $\dfrac{3.6\times10^{-4}+7.5\times10^{-5}}{1.5\times10^3}$ . Give your answer in standard form.
(4)
(Total for Question 1 is 4 marks)
2
Machine A makes $4.5\times10^5$ components and machine B makes $0.62\times10^6$ components. Work out how many more components machine B makes. Give your answer in standard form.
(3)
(Total for Question 2 is 3 marks)
3
Each tray contains $2.4\times10^3$ screws. There are $3.5\times10^2$ trays. After inspection, $6.7\times10^4$ screws are rejected. Work out the number of screws that are not rejected. Give your answer in standard form.
(4)
(Total for Question 3 is 4 marks)
4
A survey stores $4.2\times10^7$ readings. Each reading uses $1.5\times10^{-3}$ MB. After $1.1\times10^4$ MB is deleted, the remaining data is shared equally between $8\times10^2$ drives. Work out the amount on each drive. Give your answer in standard form.
(4)
(Total for Question 4 is 4 marks)
5
Factory X makes $3.2\times10^4$ clips each day for $18$ days. Factory Y makes $5.5\times10^5$ clips in total. Work out how many more clips factory X makes than factory Y. Give your answer in standard form.
(4)
(Total for Question 5 is 4 marks)

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

Explanation

A terminating decimal has finitely many decimal places, so write it over the matching power of $10$ and simplify: $0.375=\dfrac{375}{1000}=\dfrac38$ . 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 $x$ , 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 $0.\dot2\dot7=0.272727\ldots$ to a fraction.

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

Answer: $\dfrac3{11}$ .

Common mistakes

Don't write $0.375$ as $\dfrac{375}{100}$ 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.

Tier 1 · Easy

1
Write $0.375$ as a fraction in its simplest form.
(1)
(Total for Question 1 is 1 mark)
2
Write $2\dfrac{9}{25}$ as a decimal.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Write $\dfrac{37}{80}$ as a decimal.
(1)
(Total for Question 1 is 1 mark)
2
A bottle is $0.74$ full. Write the fraction of the bottle that is empty in its simplest form.
(2)
(Total for Question 2 is 2 marks)
3
Two of these three values are equal: $0.55$ , $\dfrac{11}{20}$ and $\dfrac{14}{25}$ . Write down the value that is not equal to the other two. Show your working.
(2)
(Total for Question 3 is 2 marks)

Tier 3 · Hard

1
Which is greater, $0.56$ or $\dfrac{9}{16}$ ? Work out the difference as a fraction.
(3)
(Total for Question 1 is 3 marks)
2
Work out $0.875+\dfrac{3}{20}$ . Write the result as a fully simplified fraction.
(3)
(Total for Question 2 is 3 marks)
3
A number is added to $0.36$ to make $\dfrac78$ . Work out the number. Write it as a fully simplified fraction and as a decimal.
(3)
(Total for Question 3 is 3 marks)
4
Higher only: $x=0.\dot{1}\dot{8}$ and $y=0.2\dot{7}$ . Work out $x+y$ as a fully simplified fraction. You must show all your working.
(5)
(Total for Question 4 is 5 marks)
5
Higher only: Write $\dfrac{29}{24}$ as a recurring decimal. Show enough working to identify the recurring digit.
(3)
(Total for Question 5 is 3 marks)

N11 · Identify and work with fractions in ratio problems

Explanation

Fractions and ratios describe the same partition in different forms. If a group is $\dfrac35$ of the whole, the remainder is $\dfrac25$ , so the part-to-part ratio is $3:2$ .
Conversely, in a ratio $a:b$ , the first share is $\dfrac{a}{a+b}$ of the whole and the second is $\dfrac{b}{a+b}$ .
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.

Worked example

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

1.There are $11$ parts, so one part is $330\div11=30$ .
2.Imran gets $120$ and keeps $\dfrac58\times120=75$ ; Jo gets $210$ and keeps $\dfrac57\times210=150$ .
3.Add the amounts kept: $75+150=225$ .

Answer: $\pounds225$ .

Common mistakes

Don't say the remainder of $\dfrac35$ is $\dfrac55$ instead of $\dfrac25$ .
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

1
$\dfrac{3}{5}$ 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)
2
The ratio of silver counters to gold counters is $4:9$ . Write the fraction of the counters that are gold.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
The ratio of Ava's tokens to Ben's tokens is $5:7$ . Ben has $84$ tokens. Work out $\dfrac{3}{4}$ of Ava's number of tokens.
(3)
(Total for Question 1 is 3 marks)
2
$\dfrac{2}{5}$ of some tiles are yellow. One quarter of the remaining tiles are green and the rest are blue. Write the ratio yellow : green : blue.
(3)
(Total for Question 2 is 3 marks)
3
The ratio of adults to children at an event is $7 : 4$ . Three eighths of the children wear hats. Work out the fraction of all the people at the event who are children wearing hats.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Red, blue and green beads are in the ratio $2:5:3$ . Half of the red beads are removed. Work out the fraction of the remaining beads that are blue.
(3)
(Total for Question 1 is 3 marks)
2
The ratio $A : B$ is $4 : 9$ . The amount $A$ is $\dfrac27$ of the amount $C$ . Work out the fraction of the total amount $A+B+C$ that is $B$ .
(4)
(Total for Question 2 is 4 marks)
3
The ratio of adults to children at a show is $7:5$ . One third of the adults and one fifth of the children leave. There are then $78$ people at the show. Work out the number of people who were at the show before anyone left.
(4)
(Total for Question 3 is 4 marks)
4
Two positive amounts are $A$ and $B$ . Two fifths of $A$ is equal to three sevenths of $B$ . Work out the fraction of $A+B$ that is $A$ .
(3)
(Total for Question 4 is 3 marks)

N12 · Interpret fractions and percentages as operators

Explanation

A fraction or percentage acts as an operator, meaning it multiplies a quantity. To find $\dfrac ab$ of an amount, divide by $b$ and multiply by $a$ , or multiply directly by $\dfrac ab$ .
Convert a percentage to a fraction over $100$ or to its decimal multiplier; for example, $17.5\%=0.175$ .
A percentage decrease of $p\%$ leaves the multiplier $1-\dfrac p{100}$ , while an increase uses $1+\dfrac p{100}$ .
When several operators occur, apply them to the correct intermediate amount in the stated order.
Examiners award method marks for a correct multiplier even if the evaluation is wrong.

Worked example

A machine costs $\pounds640$ . Its price falls by $15\%$ , then a customer pays $\dfrac38$ of the reduced price as a deposit. Find the balance.

1.Apply the reduction multiplier: $640\times0.85=544$ .
2.Find the deposit: $544\times\dfrac38=204$ .
3.Subtract the deposit from the reduced price: $544-204=340$ .

Answer: $\pounds340$ .

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.

Tier 1 · Easy

1
Work out $\dfrac{3}{5}$ of $70$ .
(1)
(Total for Question 1 is 1 mark)
2
Work out $30\%$ of $90$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Work out $17.5\%$ of $240$ .
(2)
(Total for Question 1 is 2 marks)
2
A museum ticket costs $\pounds35$ . Its price is increased by $18\%$ . Work out the new price.
(2)
(Total for Question 2 is 2 marks)
3
$35\%$ of a number is $84$ . Work out the number.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A machine costs $\pounds640$ . Its price is reduced by $15\%$ , then a customer pays $\dfrac{3}{8}$ of the reduced price as a deposit. Work out the balance still to pay.
(4)
(Total for Question 1 is 4 marks)
2
$\dfrac{3}{5}$ of $250$ students travel by bus. Of these students, $40\%$ buy lunch at school. Work out how many bus students buy lunch at school.
(3)
(Total for Question 2 is 3 marks)
3
A charity receives $480$ entries. Of these entries, $35\%$ are made online. Three eighths of the online entries are incomplete. Work out the number of complete online entries.
(4)
(Total for Question 3 is 4 marks)
4
Three fifths of a positive number is $42$ greater than $25\%$ of the number. Work out the number.
(4)
(Total for Question 4 is 4 marks)
5
A tank loses $15\%$ of its water. It then loses one quarter of the water that remains. Zoya says the tank has lost $40\%$ of its original amount. Is Zoya correct? Give a reason.
(4)
(Total for Question 5 is 4 marks)

N13 · Use standard units of mass, length, time, money and other measures (including standard compound measures) using decimal quantities where appropriate

Explanation

Standard units cover mass, length, time, money and other measures; compound measures combine units, such as kilometres per hour or grams per cubic centimetre. Choose units appropriate to the context, then convert all quantities to compatible units before calculating.
Length factors are squared for area and cubed for volume: because $1\text{ m}=100\text{ cm}$ , $1\text{ m}^2=10\,000\text{ cm}^2$ .
Time is base $60$ , not base $100$ , so $1.5$ hours is $1$ hour $30$ minutes.
Give money to the nearest penny when appropriate.
Examiners expect a numerical value and a correctly converted unit.

Worked example

A car travels $54$ km using $7.5$ litres per $100$ km. Fuel costs $\pounds1.68$ per litre. Find the journey’s fuel cost.

1.Fuel used $=54\times\dfrac{7.5}{100}=4.05$ litres.
2.Cost $=4.05\times1.68=6.804$ pounds.
3.Round money to the nearest penny.

Answer: $\pounds6.80$ .

Common mistakes

Don't treat $1.5$ hours as $1$ hour $50$ minutes.
Don't use the factor $100$ rather than $100^2$ 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.

Tier 1 · Easy

1
Change $2.75$ kg to grams.
(1)
(Total for Question 1 is 1 mark)
2
Change $4500$ millilitres to litres.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A workshop starts at $09{:}38$ and lasts for $1$ hour $47$ minutes. Work out the finishing time.
(2)
(Total for Question 1 is 2 marks)
2
A rectangular sheet is $0.27$ m long and $46$ cm wide. Work out its area in square centimetres.
(2)
(Total for Question 2 is 2 marks)
3
A delivery contains $8$ cartons with mass $750$ g each and $3$ cartons with mass $1.2$ kg each. Work out the total mass of the cartons in kilograms.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A car travels $54$ km and uses fuel at a rate of $7.5$ litres per $100$ km. Fuel costs $\pounds1.68$ 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)
2
A tap delivers $18$ litres of water in $2$ minutes $24$ seconds. Work out the average rate in millilitres per second.
(3)
(Total for Question 2 is 3 marks)
3
A printer produces $825$ labels in $2$ minutes $30$ seconds. It continues at the same average rate for $7$ minutes. Work out the number of labels produced in the $7$ minutes.
(4)
(Total for Question 3 is 4 marks)
4
A roll holds $2.35$ km of cable. A technician cuts $48$ pieces, each $37.5$ m long, from the roll. Work out the length of cable left. Give your answer in metres.
(3)
(Total for Question 4 is 3 marks)
5
A runner covers $7.5$ km at an average speed of $12$ km/h and then rests for $8$ minutes. Work out the total time for the run and the rest. Give your answer in minutes.
(4)
(Total for Question 5 is 4 marks)

N14 · Estimate answers; check calculations using approximation and estimation, including answers obtained using technology

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 $10$ 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 $\dfrac{598.4\times0.0317}{0.204}$ . Use an estimate to decide whether it is reasonable.

1.Round inputs to convenient values: $598.4\approx600$ , $0.0317\approx0.03$ , $0.204\approx0.2$ .
2.Estimate $\dfrac{600\times0.03}{0.2}=90$ .
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 $\approx$ 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.

Tier 1 · Easy

1
Estimate the value of $19.8\times0.49$ .
(1)
(Total for Question 1 is 1 mark)
2
By rounding each number to $1$ significant figure, estimate the value of $72.4+26.3$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Estimate $\dfrac{48.7\times0.203}{0.098}$ .
(2)
(Total for Question 1 is 2 marks)
2
Estimate the value of $9.8^2+51$ by rounding each number to $1$ significant figure.
(2)
(Total for Question 2 is 2 marks)
3
By rounding each number to $1$ significant figure, estimate the value of $\dfrac{31.2\times197}{0.62}$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A calculator display gives $931.24$ for $\dfrac{598.4\times0.0317}{0.204}$ . Use an estimate to decide whether this display is reasonable. Give a reason.
(3)
(Total for Question 1 is 3 marks)
2
Each pack of seed covers $2.9$ m $^2$ . A garden has area $118$ m $^2$ . Use an estimate to decide whether $35$ packs will cover the garden. Give a reason.
(3)
(Total for Question 2 is 3 marks)
3
The value of $\dfrac{498\times19.6}{0.204}$ lies in one of these intervals. A: $400$ to $600$ . B: $4000$ to $6000$ . C: $40\,000$ to $60\,000$ . Use an estimate to choose the correct interval. Write down the letter of the interval.
(3)
(Total for Question 3 is 3 marks)
4
Replace all three numbers by their $1$ -significant-figure approximations to estimate $\dfrac{76.2\times0.384}{2.17}$ . State whether the estimate is an overestimate or an underestimate, and give a reason.
(4)
(Total for Question 4 is 4 marks)
5
A packing line seals $293$ cartons in $8.2$ minutes. It has $60$ minutes to seal $2050$ cartons. Use an estimated sealing rate to decide whether the available time should be enough.
(4)
(Total for Question 5 is 4 marks)

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

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: $5$ 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 $\dfrac u2$ ; 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

12.6

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 $4.37\leq y<4.38$ .
2.Multiply the whole interval by $100$ : $437\leq100y<438$ .
3.The greatest integer in this interval is $437$ .

Answer: $4.37\leq y<4.38$ 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.

Tier 1 · Easy

1
Write $0.007846$ correct to $2$ significant figures.
(1)
(Total for Question 1 is 1 mark)
2
Write $53\,749$ correct to $2$ significant figures.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A number $x$ is $12.6$ correct to $1$ decimal place. Write the error interval for $x$ .
(2)
(Total for Question 1 is 2 marks)
2
A number $n$ is $730$ correct to the nearest $10$ . Write the error interval for $n$ .
(2)
(Total for Question 2 is 2 marks)
3
Write $12.4867$ correct to $3$ significant figures and correct to $2$ decimal places.
(2)
(Total for Question 3 is 2 marks)

Tier 3 · Hard

1
A positive number $y$ is truncated to $4.37$ at $2$ decimal places. Write its error interval and find the greatest possible integer value of $100y$ .
(3)
(Total for Question 1 is 3 marks)
2
A length is $155$ mm correct to the nearest $5$ mm. Salma writes $152.5<x\leq157.5$ . Write the correct error interval and explain both changes to Salma's inequality signs.
(3)
(Total for Question 2 is 3 marks)
3
A positive number $x$ is $2.4$ correct to $1$ decimal place. Work out all the possible integer values of $20x$ .
(3)
(Total for Question 3 is 3 marks)
4
A positive number $x$ is $0.008$ correct to $1$ significant figure and $0.0076$ correct to $2$ significant figures. Write the error interval for $x$ that satisfies both statements.
(4)
(Total for Question 4 is 4 marks)
5
A positive integer is $68\,000$ correct to $2$ significant figures. The integer is a multiple of $400$ . Work out the least and the greatest possible values of the integer.
(4)
(Total for Question 5 is 4 marks)

N16 · Apply and interpret limits of accuracy, including upper and lower bounds

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.

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.

Tier 1 · Easy

1
A length is recorded as $12$ cm to the nearest centimetre. Write down the maximum possible rounding error.
(1)
(Total for Question 1 is 1 mark)
2
Higher only: Each of $6$ identical tubs has a mass of $260$ g correct to the nearest $10$ g. Work out the upper bound for their total mass.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
Higher only: A parcel has mass $7.6$ kg correct to the nearest $0.1$ kg. Could its exact mass be $7.66$ kg? Give a reason.
(2)
(Total for Question 1 is 2 marks)
2
A jar is marked as having mass $500$ g. The maximum possible rounding error is $5$ g. Write down the unit to which the mass was rounded and write the error interval for its exact mass $m$ .
(3)
(Total for Question 2 is 3 marks)

Tier 3 · Hard

1
Higher only: A scale records the mass of each of $8$ identical boxes as $2.4$ kg to the nearest $0.1$ kg. Could the exact total mass of the boxes be $20$ kg? Justify your answer.
(3)
(Total for Question 1 is 3 marks)
2
Higher only: A laser measure records a length as $1.372$ m to the nearest $0.001$ m. A tape measure records the same length as $137$ cm to the nearest centimetre. Which measure has the smaller maximum possible error? Give a reason.
(3)
(Total for Question 2 is 3 marks)
3
Higher only: An alloy bar has mass $1.8$ kg correct to the nearest $0.1$ kg. It is cut into $7$ blocks of equal mass. Work out the upper bound for the mass of one block. Give your answer exactly in grams.
(4)
(Total for Question 3 is 4 marks)
4
Higher only: Two boards have recorded lengths of $8.4$ m and $3.7$ m, each correct to the nearest $0.1$ m. A shelf design requires the first board to be at least $4.65$ m longer than the second. Is this guaranteed? Justify your answer using bounds.
(4)
(Total for Question 4 is 4 marks)
5
Higher only: A machine's output is recorded as $840$ parts, correct to the nearest $10$ parts. Its running time is recorded as $3.2$ hours, correct to the nearest $0.1$ hour. Could its exact output rate be greater than $270$ parts per hour? You must show all your working.
(4)
(Total for Question 5 is 4 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

N1 · Order positive and negative integers, decimals and fractions; use the symbols =, ≠, <, >, ≤, ≥

Tier 1 · Easy

Mark scheme for N1 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$-0.62<-\dfrac{3}{5}$	1	Convert $-\dfrac{3}{5}$ to $-0.6$ . Since $-0.62$ is farther left on the number line than $-0.6$ , $-0.62<-\dfrac{3}{5}$ .
2	$-0.52,\ -\dfrac{1}{2},\ -0.47,\ -\dfrac{9}{20}$	2	$-\dfrac{9}{20}=-0.45$ and $-\dfrac{1}{2}=-0.5$ . From smallest to greatest, the values are $-0.52,-0.5,-0.47,-0.45$ .

Tier 2 · Standard

Mark scheme for N1 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$-6<-2$ because $-6$ lies farther to the left on a number line.	2	For negative numbers, the number with the greater distance below zero is smaller. Since $-6$ is farther left than $-2$ , the correct comparison is $-6<-2$ .
2	$-1.15,\ -\dfrac{7}{6},\ -1.2$	2	$-\dfrac{7}{6}=-1.1666\ldots$ . For negative numbers, the value closest to zero is greatest, so the descending order is $-1.15,-1.1666\ldots,-1.2$ .
3	$0,1,2,3$	3	The completed decimal must lie between $-0.64$ and $-0.58$ . The values $-0.60$ , $-0.61$ , $-0.62$ and $-0.63$ all do this, whereas $-0.64$ is not greater than the lower endpoint. The possible digits are $0,1,2,3$ .

Tier 3 · Hard

Mark scheme for N1 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$k=-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5$	3	Multiply every part by positive $3$ , so the inequality signs stay unchanged: $-7.2<k\leq5.1$ . The integers in this interval are $-7$ through $5$ inclusive.
2	No; the correct order is $-\dfrac{2}{3}<-0.66<-\dfrac{13}{20}$ .	2	Use denominator $300$ : $-\dfrac{2}{3}=-\dfrac{200}{300}$ , $-0.66=-\dfrac{198}{300}$ and $-\dfrac{13}{20}=-\dfrac{195}{300}$ . Therefore $-\dfrac{2}{3}<-0.66<-\dfrac{13}{20}$ , so Sam is not correct.
3	$\dfrac{19}{900}$	4	$-\dfrac{11}{18}=-0.6111\ldots$ and $-61\%=-0.61$ . The greatest value is $-0.59$ and the smallest is $-\dfrac{11}{18}$ . Their difference is $-\dfrac{59}{100}+\dfrac{11}{18}=\dfrac{38}{1800}=\dfrac{19}{900}$ .
4	$Q$ and $R$ ; $-0.72,\ -73\%$ (or $-0.72,\ -0.73$ )	3	$P=-0.75$ , $R=-0.73$ and $S=-0.70$ . Only $Q=-0.72$ and $R=-0.73$ lie in the stated interval. Since $-0.72>-0.73$ , their coordinates in descending order are $-0.72,-0.73$ .
5	$k=22$	3	Multiplying by $40$ gives $-22.4<-k<-21.6$ . Multiplying by $-1$ reverses both inequality signs, so $21.6<k<22.4$ . The only integer in this interval is $22$ .

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

Tier 1 · Easy

Mark scheme for N2 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$12$	1	The signs are different, so find $29-17=12$ and use the sign of the number with the larger magnitude. The result is $12$ .
2	$23.75$	1	Align the decimal points and write $42.70-18.95$ . Column subtraction gives $23.75$ .

Tier 2 · Standard

Mark scheme for N2 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$120$	2	Multiply both numbers by $100$ so the divisor is an integer: $7.2\div0.06=720\div6=120$ .
2	$4\dfrac{1}{6}$ (or $\dfrac{25}{6}$ )	2	$2\dfrac{1}{3}=\dfrac{14}{6}$ and $1\dfrac{5}{6}=\dfrac{11}{6}$ . Their sum is $\dfrac{25}{6}=4\dfrac{1}{6}$ .
3	$4.1$ metres	3	$8\dfrac35=8.6$ . The six pieces use $6\times0.75=4.5$ metres, so the length left is $8.6-4.5=4.1$ metres.

Tier 3 · Hard

Mark scheme for N2 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$3.85$	3	First $\dfrac{5}{6}\div\left(-\dfrac{5}{9}\right)=\dfrac{5}{6}\times\left(-\dfrac{9}{5}\right)=-\dfrac{3}{2}=-1.5$ . Then $3.6+1.75-1.5=5.35-1.5=3.85$ .
2	$11.35$ litres	3	Convert $2\dfrac{3}{4}$ to $2.75$ . After the removal there are $12.5-2.75=9.75$ litres, and after the addition there are $9.75+1.6=11.35$ litres.
3	$\pounds4.40$	4	After the payment, the balance is $-18.40+75.75=57.35$ . The fee is $2\dfrac25=2.40$ , leaving $54.95$ . The three later payments total $3\times16.85=50.55$ , so the final balance is $54.95-50.55=\pounds4.40$ .
4	$7.074$	3	Do the multiplication first: $0.095\times16=1.52$ . Written subtraction gives $8.030-2.476=5.554$ , so the result is $5.554+1.52=7.074$ .
5	$3.6$	4	$4\dfrac15=4.2$ , so the value in brackets is $4.2-2.85=1.35=\dfrac{27}{20}$ . Dividing by $\dfrac38$ means multiplying by $\dfrac83$ : $\dfrac{27}{20}\times\dfrac83=\dfrac{18}{5}=3.6$ .

N3 · Recognise and use relationships between operations, including inverse operations; use conventional notation for priority of operations, including brackets, powers, roots and reciprocals

Tier 1 · Easy

Mark scheme for N3 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$0$	1	Evaluate the power first: $3^2=9$ . Then multiply, $9\times2=18$ , and subtract: $18-18=0$ .
2	$\dfrac{1}{8}$ (or $0.125$ )	1	The reciprocal multiplies by the original number to make $1$ . Since $8\times\dfrac{1}{8}=1$ , the reciprocal is $\dfrac{1}{8}$ .

Tier 2 · Standard

Mark scheme for N3 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$8$	2	Inside the brackets, $\sqrt{81}-5=9-5=4$ . Then $4^2=16$ and $16\div2=8$ .
2	$(18-4)\times2+3=31$	1	Place the brackets around $18-4$ . Then $(18-4)\times2+3=14\times2+3=31$ .
3	$54\times14=756\ne728$ , so Luca is wrong; $728\div14=52$	3	$14\times54=756$ , not $728$ , so the inverse operation shows Luca's answer is wrong. Since $14\times52=728$ , the correct answer is $52$ .

Tier 3 · Hard

Mark scheme for N3 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$5$	3	Undo the reciprocal first: the result before taking the reciprocal was $14$ . Add $11$ to undo the subtraction, giving $25$ . The positive square root of $25$ is $5$ .
2	$16$	3	$2^3=8$ and $\sqrt9=3$ , so $36\div\sqrt9=12$ . The reciprocal of $\dfrac14$ is $4$ , giving $8+12-4=16$ .
3	Ravi should calculate from left to right; the correct value is $36$ .	3	Multiplication and division have equal priority, so they are completed from left to right. First $60\div5=12$ , then $12\times3=36$ . Ravi incorrectly treated the expression as $60\div(5\times3)$ .
4	$13$	3	Complete the inner operations first: $3^2-1=8$ , so $2\left(3^2-1\right)=16$ . Then $96\div16=6$ and $\sqrt{49}=7$ , giving $6+7=13$ .
5	Ben is correct; the input is $11$ .	3	Inverse operations must be applied in reverse order. Add $8$ to the output, giving $55$ , and then divide by $5$ , giving $11$ . Checking: $11\times5-8=47$ , so Ben is correct.

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

Tier 1 · Easy

Mark scheme for N4 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$756=2^2\times3^3\times7$	2	Divide successively by primes: $756=2\times378=2^2\times189=2^2\times3^3\times7$ .
2	$2$ , $3$ and $7$	1	$42=2\times3\times7$ , and each of $2$ , $3$ and $7$ is prime.

Tier 2 · Standard

Mark scheme for N4 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	HCF $=42$ LCM $=252$	3	Use $84=2^2\times3\times7$ and $126=2\times3^2\times7$ . The smaller common powers give $2\times3\times7=42$ . The largest powers give $2^2\times3^2\times7=252$ .
2	$108$	2	$12=2^2\times3$ and $18=2\times3^2$ , so their LCM is $2^2\times3^2=36$ . The first multiple of $36$ greater than $100$ is $36\times3=108$ .
3	$17$ counters	3	The HCF of $96$ , $144$ and $168$ is $24$ , so there are $24$ bags. Each bag contains $96\div24=4$ red, $144\div24=6$ blue and $168\div24=7$ green counters. This is $4+6+7=17$ counters.

Tier 3 · Hard

Mark scheme for N4 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$n=150$	4	Prime factorise $180=2^2\times3^2\times5$ . A cube needs every exponent to be a multiple of $3$ , so multiply by $2\times3\times5^2$ . Therefore $n=2\times3\times25=150$ , and $180n=27000=30^3$ .
2	$n=15,45,75$	3	The multiples of $15$ below $100$ are $15,30,45,60,75,90$ . Their HCFs with $60$ are respectively $15,30,15,60,15,30$ , so the required values are $15$ , $45$ and $75$ .
3	$n=90$	4	Since the HCF is $18$ , $n$ is a multiple of $18$ . Since the LCM is $360$ , $n$ is a factor of $360$ . The multiples of $18$ that divide $360$ are $18,36,72,90,180,360$ . Checking these conditions leaves $n=90$ , because $\operatorname{HCF}(90,72)=18$ and $\operatorname{LCM}(90,72)=360$ .
4	$149$	4	Subtracting $5$ from the number must give a common multiple of $18$ and $24$ . Since $18=2\times3^2$ and $24=2^3\times3$ , their LCM is $2^3\times3^2=72$ . The first multiple, $72$ , gives $77$ , which is not three-digit; the next gives $144+5=149$ .
5	$1728$	4	Every exponent in a cube number is a multiple of $3$ . The greatest allowed exponents are therefore $6$ for the factor $2$ , $3$ for the factor $3$ and $0$ for the factor $5$ . The required factor is $2^6\times3^3=64\times27=1728$ .

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)

Tier 1 · Easy

Mark scheme for N5 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	A₁, A₂, A₃, B₁, B₂, B₃, C₁, C₂, C₃	2	Hold the letter fixed while cycling through the numbers: A₁, A₂, A₃; then B₁, B₂, B₃; then C₁, C₂, C₃. This gives all $9$ codes once each.
2	$12,13,21,23,31,32$	2	Fix the tens digit in turn. Starting with $1$ gives $12,13$ ; starting with $2$ gives $21,23$ ; starting with $3$ gives $31,32$ .

Tier 2 · Standard

Mark scheme for N5 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$452$ , $472$ , $542$ , $572$ , $742$ , $752$ , $254$ , $274$ , $524$ , $574$ , $724$ , $754$	3	List by the final digit. Ending in $2$ gives $452$ , $472$ , $542$ , $572$ , $742$ , $752$ . Ending in $4$ gives $254$ , $274$ , $524$ , $574$ , $724$ , $754$ . The fixed final digit makes the list systematic and complete.
2	A₁, A₂, A₃, B₁, B₂	2	Hold the gate fixed. Gate A gives A₁, A₂ and A₃. Gate B would give B₁, B₂ and B₃, but B₃ is unavailable, leaving B₁ and B₂.
3	$10{:}00$ and $30$ minutes; $10{:}00$ and $60$ minutes; $10{:}30$ and $30$ minutes; $10{:}30$ and $60$ minutes; $11{:}00$ and $30$ minutes	3	List both durations for $10{:}00$ and both for $10{:}30$ . At $11{:}00$ , only the $30$ -minute appointment finishes by $11{:}30$ ; a $60$ -minute appointment finishes too late. This gives five pairs.

Tier 3 · Hard

Mark scheme for N5 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$(2,4)$ , $(2,5)$ , $(3,4)$ , $(4,5)$	3	List unordered pairs in rows beginning with $1$ , then $2$ , then $3$ , then $4$ . Reject pairs with an odd product or sum at most $5$ . The pairs left are $(2,4)$ , $(2,5)$ , $(3,4)$ and $(4,5)$ .
2	$2,\ 3,\ 4,\ 5,\ 7$	2	Fix the result on the first spinner. With $1$ , the totals are $2,4$ ; with $2$ , they are $3,5$ ; with $4$ , they are $5,7$ . This records all six outcomes, so the possible totals are $2, 3, 4, 5, 7$ .
3	black-navy-boots; black-cream-boots; black-cream-trainers; black-khaki-boots; black-khaki-trainers; grey-navy-boots; grey-cream-boots; grey-cream-trainers	4	Hold the jacket and trousers fixed, then test both pairs of shoes. The black jacket gives five outfits: navy-boots, both cream outfits and both khaki outfits. The grey jacket gives three outfits: navy-boots and both cream outfits. The restrictions remove every grey-khaki outfit and each navy-trainers outfit.
4	$20$ signals	4	For each of the three non-red colours there are $3\times2=6$ pattern-duration choices, giving $3\times6=18$ signals. Red has one pattern and two durations, giving $2$ more. The total is $18+2=20$ .
5	$10$ routes	4	List routes by the positions of the two north moves: NNEEE, NENEE, NEENE, NEEEN, ENNEE, ENENE, ENEEN, EENNE, EENEN and EEENN. This systematic list has $10$ different routes.

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

Tier 1 · Easy

Mark scheme for N6 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$32$	1	$2^5=2\times2\times2\times2\times2=32$ .
2	$5$ and $6$	1	$5^2=25$ and $6^2=36$ . Since $25<30<36$ , $\sqrt{30}$ lies between $5$ and $6$ .

Tier 2 · Standard

Mark scheme for N6 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$7$	1	Since $7^3=7\times7\times7=343$ , the associated cube root is $\sqrt[3]{343}=7$ .
2	$5^4$	1	$625=5\times125=5\times5\times25=5\times5\times5\times5$ , so $625=5^4$ .
3	$4096=4^6$ and $\sqrt[6]{4096}=4$	3	$4^2=16$ , $4^3=64$ and $4^6=64^2=4096$ . Therefore the number whose sixth power is $4096$ is $4$ , so $\sqrt[6]{4096}=4$ .

Tier 3 · Hard

Mark scheme for N6 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$14$	3	Because $6^4=1296$ , $\sqrt[4]{1296}=6$ . Because $8^3=512$ , $\sqrt[3]{512}=8$ . Their sum is $6+8=14$ .
2	$70$	3	$\sqrt[3]{1000}=10$ , $\sqrt{196}=14$ and $\sqrt[3]{8}=2$ . Therefore $10\times14\div2=70$ .
3	$b=512$	3	Since $5^4=625$ and $a$ is positive, $a=5$ . Then $\sqrt[3]{b}=5+3=8$ , so $b=8^3=512$ .
4	$x=4.9$	4	$4.9^4=576.4801<580$ , so $x>4.9$ . At the rounding boundary, $4.95^2=24.5025>24.5$ , so $4.95^4>24.5^2=600.25>580$ . Hence $x<4.95$ , and $x$ rounds to $4.9$ to $1$ decimal place.
5	$n=15\,625$	4	Because $n$ is both a square and a cube, write $n=a^6$ for a positive integer $a$ . Then $\sqrt[3]{n}=a^2$ and $\sqrt n=a^3$ , so $a^2+a^3=150$ . Since $5^2+5^3=25+125=150$ , $a=5$ and $n=5^6=15\,625$ .

N7 · Calculate with roots, and with integer and fractional indices

Tier 1 · Easy

Mark scheme for N7 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$\dfrac{1}{16}$	1	A negative index means take the reciprocal: $4^{-2}=\dfrac{1}{4^2}=\dfrac{1}{16}$ .
2	$1$	1	Any non-zero number raised to the power $0$ is $1$ , so $10^0=1$ .

Tier 2 · Standard

Mark scheme for N7 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$-6$	1	The cube root is the number whose cube is $-216$ . Since $(-6)^3=-216$ , $\sqrt[3]{-216}=-6$ .
2	$-\dfrac{63}{8}$ (or $-7\dfrac{7}{8}$ or $-7.875$ )	2	$2^{-3}=\dfrac{1}{2^3}=\dfrac18$ and $(-2)^3=-8$ . Therefore $\dfrac18-8=\dfrac18-\dfrac{64}{8}=-\dfrac{63}{8}$ .
3	$n=-4$ and $(-n)^2=16$	3	$625=5^4$ , so $\dfrac1{625}=5^{-4}$ and therefore $n=-4$ . This gives $(-n)^2=4^2=16$ .

Tier 3 · Hard

Mark scheme for N7 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$\dfrac{301}{25}$	3	$\sqrt{144}=12$ and $5^{-2}=1/5^2=1/25$ . Therefore $12+\dfrac{1}{25}=\dfrac{300}{25}+\dfrac{1}{25}=\dfrac{301}{25}$ .
2	$1$	3	$(-3)^4=81$ , $3^{-2}=\dfrac19$ and $\sqrt{81}=9$ . Therefore $81\times\dfrac19\div9=9\div9=1$ .
3	$-3$	3	$(-4)^3=-64$ , $2^{-5}=\dfrac1{32}$ and $9^0=1$ . Therefore $-64\times\dfrac1{32}-1=-2-1=-3$ .
4	$\dfrac{109}{27}$	4	$81^{1/4}=3$ , so $81^{-3/4}=3^{-3}=\dfrac1{27}$ . Also $32^{1/5}=2$ , so $32^{2/5}=2^2=4$ . Hence the total is $4+\dfrac1{27}=\dfrac{109}{27}$ .
5	$n=25$	3	$n^{3/2}=(\sqrt n)^3$ . Since $125=5^3$ , $\sqrt n=5$ . Squaring gives $n=25$ ; checking, $25^{3/2}=5^3=125$ .

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

Tier 1 · Easy

Mark scheme for N8 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$\dfrac{19}{12}$ $1\dfrac{7}{12}$	2	Use denominator $12$ : $\dfrac{3}{4}=\dfrac{9}{12}$ and $\dfrac{5}{6}=\dfrac{10}{12}$ . Their sum is $\dfrac{19}{12}=1\dfrac{7}{12}$ .
2	$\dfrac{9}{20}$ (or $0.45$ )	2	Use denominator $20$ : $\dfrac{7}{10}=\dfrac{14}{20}$ and $\dfrac14=\dfrac5{20}$ . The difference is $\dfrac9{20}$ .

Tier 2 · Standard

Mark scheme for N8 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$49\pi\text{ cm}^2$	2	Use $A=\pi r^2$ . Then $A=\pi\times7^2=49\pi\text{ cm}^2$ .
2	$\dfrac{13\pi}{4}$	2	Write $5\pi$ in quarters: $5\pi=\dfrac{20\pi}{4}$ . Then $\dfrac{20\pi}{4}-\dfrac{7\pi}{4}=\dfrac{13\pi}{4}$ .
3	$3\pi+\dfrac{11}{12}$ metres	3	Subtract the used length: $5\pi+\dfrac34-(2\pi-\dfrac16)=3\pi+\dfrac34+\dfrac16$ . Using denominator $12$ gives $3\pi+\dfrac9{12}+\dfrac2{12}=3\pi+\dfrac{11}{12}$ metres.

Tier 3 · Hard

Mark scheme for N8 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$\dfrac{49\pi}{40}$	3	Use denominator $40$ : $\dfrac{3\pi}{5}=\dfrac{24\pi}{40}$ , $\dfrac{7\pi}{8}=\dfrac{35\pi}{40}$ and $\dfrac{\pi}{4}=\dfrac{10\pi}{40}$ . Therefore the result is $\dfrac{(24+35-10)\pi}{40}=\dfrac{49\pi}{40}$ .
2	$\dfrac{29\pi}{12}$	3	First $\dfrac{5\pi}{6}+\dfrac{7\pi}{9}=\dfrac{15\pi}{18}+\dfrac{14\pi}{18}=\dfrac{29\pi}{18}$ . Dividing by $\dfrac23$ means multiplying by $\dfrac32$ , giving $\dfrac{29\pi}{18}\times\dfrac32=\dfrac{29\pi}{12}$ .
3	$\dfrac{84\pi}{5}$ metres (or $16.8\pi$ metres)	3	One rotation covers the circumference $2\pi\times0.35=0.7\pi$ metres. In $24$ rotations the distance is $24\times0.7\pi=16.8\pi=\dfrac{84\pi}{5}$ metres.
4	$24$	4	$\sqrt{50}=5\sqrt2$ , $\sqrt8=2\sqrt2$ and $\sqrt{18}=3\sqrt2$ . The expression becomes $(3\sqrt2)(4\sqrt2)=12\times2=24$ .
5	$\dfrac{2-\sqrt{7}}{3}$	4	Multiply the fraction by $\dfrac{\sqrt7+1}{\sqrt7+1}$ to get $\dfrac{4(\sqrt7+1)}{7-1}=\dfrac{2\sqrt7+2}{3}$ . Subtracting $\sqrt7=\dfrac{3\sqrt7}{3}$ gives $\dfrac{2-\sqrt7}{3}$ .

N9 · Calculate with and interpret standard form A × 10^n, where 1 ≤ A < 10 and n is an integer

Tier 1 · Easy

Mark scheme for N9 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$7.2\times10^{-5}$	1	Move the decimal point $5$ places right to make $7.2$ . Moving right gives a negative power, so $0.000072=7.2\times10^{-5}$ .
2	$4.53\times10^6$	1	Move the decimal point $6$ places left to make $4.53$ , so $4\,530\,000=4.53\times10^6$ .

Tier 2 · Standard

Mark scheme for N9 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$2.4\times10^5$	2	Multiply the factors and add the indices: $6\times4=24$ and $10^7\times10^{-3}=10^4$ . Thus the product is $24\times10^4=2.4\times10^5$ .
2	$4\times10^4$	2	Divide the numbers and subtract the indices: $(8.4\div2.1)\times10^{6-2}=4\times10^4$ .
3	$4.61\times10^4$ grams	3	$8600=0.86\times10^4$ . The total is therefore $(3.75+0.86)\times10^4=4.61\times10^4$ grams.

Tier 3 · Hard

Mark scheme for N9 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$2.9\times10^{-7}$	4	Express the numerator with a common power: $3.6\times10^{-4}+0.75\times10^{-4}=4.35\times10^{-4}$ . Divide the factors and subtract the indices: $(4.35/1.5)\times10^{-4-3}=2.9\times10^{-7}$ .
2	$1.7\times10^5$ components	3	Write both amounts with the same power: $0.62\times10^6=6.2\times10^5$ . The difference is $(6.2-4.5)\times10^5=1.7\times10^5$ components.
3	$7.73\times10^5$ screws	4	The trays contain $(2.4\times10^3)(3.5\times10^2)=8.4\times10^5$ screws. Write the rejected amount as $0.67\times10^5$ . The number left is $(8.4-0.67)\times10^5=7.73\times10^5$ screws.
4	$6.5\times10^1$ MB	4	The readings use $(4.2\times10^7)(1.5\times10^{-3})=6.3\times10^4$ MB. After deletion, $6.3\times10^4-1.1\times10^4=5.2\times10^4$ MB remains. Each drive receives $\dfrac{5.2\times10^4}{8\times10^2}=0.65\times10^2=6.5\times10^1$ MB.
5	$2.6\times10^4$ clips	4	Factory X makes $(3.2\times10^4)\times18=57.6\times10^4=5.76\times10^5$ clips. The difference is $5.76\times10^5-5.5\times10^5=0.26\times10^5=2.6\times10^4$ clips.

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

Tier 1 · Easy

Mark scheme for N10 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$\dfrac{3}{8}$	1	$0.375=\dfrac{375}{1000}$ . Divide numerator and denominator by $125$ to obtain $\dfrac{3}{8}$ .
2	$2.36$	1	$\dfrac{9}{25}=\dfrac{36}{100}=0.36$ , so $2\dfrac{9}{25}=2.36$ .

Tier 2 · Standard

Mark scheme for N10 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$0.4625$	1	Make the denominator a power of $10$ : $\dfrac{37}{80}=\dfrac{4625}{10000}=0.4625$ .
2	$\dfrac{13}{50}$	2	$0.74=\dfrac{74}{100}=\dfrac{37}{50}$ , so the empty fraction is $1-\dfrac{37}{50}=\dfrac{13}{50}$ .
3	$\dfrac{14}{25}$	2	$\dfrac{11}{20}=\dfrac{55}{100}=0.55$ , while $\dfrac{14}{25}=\dfrac{56}{100}=0.56$ . Therefore $\dfrac{14}{25}$ is the value that is not equal to the other two.

Tier 3 · Hard

Mark scheme for N10 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$\dfrac{9}{16}$ is greater. The difference is $\dfrac{1}{400}$ .	3	$\dfrac{9}{16}=0.5625$ , so it is greater than $0.56$ . The difference is $0.5625-0.56=0.0025=\dfrac{25}{10000}=\dfrac{1}{400}$ .
2	$\dfrac{41}{40}$ (or $1\dfrac{1}{40}$ )	3	$0.875=\dfrac78=\dfrac{35}{40}$ and $\dfrac3{20}=\dfrac6{40}$ . Therefore the sum is $\dfrac{35}{40}+\dfrac6{40}=\dfrac{41}{40}$ .
3	$\dfrac{103}{200}=0.515$	3	$\dfrac78=0.875$ , so the number is $0.875-0.36=0.515$ . As a fraction, $0.515=\dfrac{515}{1000}=\dfrac{103}{200}$ .
4	$\dfrac{91}{198}$	5	For $x=0.1818\ldots$ , subtracting $x$ from $100x$ gives $99x=18$ , so $x=\dfrac2{11}$ . For $y=0.2777\ldots$ , subtracting $10y$ from $100y$ gives $90y=25$ , so $y=\dfrac5{18}$ . Hence $x+y=\dfrac2{11}+\dfrac5{18}=\dfrac{36+55}{198}=\dfrac{91}{198}$ .
5	$1.208\dot{3}$ (or $1.208333\ldots$ )	3	Long division gives $29\div24=1.208333\ldots$ . After the first $3$ , the remainder repeats, so every following digit is also $3$ . Therefore the recurring decimal is $1.208\dot3$ .

N11 · Identify and work with fractions in ratio problems

Tier 1 · Easy

Mark scheme for N11 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$3:2$	1	The blue fraction is $1-\dfrac{3}{5}=\dfrac{2}{5}$ . Therefore red : blue is $\dfrac{3}{5}:\dfrac{2}{5}=3:2$ .
2	$\dfrac{9}{13}$	1	There are $4+9=13$ equal parts altogether and $9$ are gold, so the required fraction is $\dfrac9{13}$ .

Tier 2 · Standard

Mark scheme for N11 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$45$ tokens	3	Seven ratio parts represent $84$ , so one part is $84\div7=12$ . Ava has $5\times12=60$ tokens. Then $\dfrac{3}{4}\times60=45$ .
2	$8:3:9$	3	Use $20$ equal parts. Yellow tiles take $8$ parts, leaving $12$ . Green tiles take one quarter of the remainder, so take $3$ parts, leaving $9$ blue parts. The ratio is $8:3:9$ .
3	$\dfrac{3}{22}$	3	Children make up $\dfrac4{11}$ of all the people. The required fraction is $\dfrac38\times\dfrac4{11}=\dfrac{12}{88}=\dfrac3{22}$ .

Tier 3 · Hard

Mark scheme for N11 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$\dfrac{5}{9}$	3	Start with $2+5+3=10$ equal parts. Removing half of the red beads removes $1$ part, so $9$ parts remain. The $5$ blue parts are therefore $\dfrac59$ of the remaining beads.
2	$\dfrac13$	4	Use $A=8$ equal parts. Then $A : B=4 : 9$ gives $B=18$ parts. Since $A=\dfrac27C$ , $C=28$ parts. The total is $8+18+28=54$ parts, so the fraction that is $B$ is $\dfrac{18}{54}=\dfrac13$ .
3	$108$ people	4	Let the original numbers be $7k$ and $5k$ . The numbers remaining are $\dfrac23\times7k=\dfrac{14k}{3}$ adults and $\dfrac45\times5k=4k$ children. Thus $\dfrac{26k}{3}=78$ , so $k=9$ . Originally there were $12k=108$ people.
4	$\dfrac{15}{29}$	3	$\dfrac25A=\dfrac37B$ . Multiplying by $35$ gives $14A=15B$ , so $A:B=15:14$ . There are $29$ ratio parts altogether and $15$ belong to $A$ , so the required fraction is $\dfrac{15}{29}$ .

N12 · Interpret fractions and percentages as operators

Tier 1 · Easy

Mark scheme for N12 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$42$	1	Divide by $5$ and multiply by $3$ : $70\div5\times3=14\times3=42$ .
2	$27$	1	$10\%$ of $90$ is $9$ , so $30\%$ is $3\times9=27$ .

Tier 2 · Standard

Mark scheme for N12 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$42$	2	Use the decimal operator $17.5\%=0.175$ . Then $0.175\times240=42$ .
2	$\pounds41.30$	2	$18\%$ of $\pounds35$ is $0.18\times35=\pounds6.30$ . Adding the increase gives $\pounds35+\pounds6.30=\pounds41.30$ .
3	$240$	3	$35\%=\dfrac{35}{100}$ . If $35\%$ is $84$ , then $5\%$ is $84\div7=12$ , so $100\%$ is $12\times20=240$ .

Tier 3 · Hard

Mark scheme for N12 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$\pounds340$	4	After the reduction, the price is $0.85\times640=544$ . The deposit is $\dfrac{3}{8}\times544=204$ . Therefore the remaining balance is $544-204=340$ .
2	$60$ students	3	The number travelling by bus is $\dfrac35\times250=150$ . Then $40\%$ of $150$ is $0.4\times150=60$ , so $60$ bus students buy lunch at school.
3	$105$ entries	4	The number made online is $0.35\times480=168$ . The complete fraction is $1-\dfrac38=\dfrac58$ . Therefore the number of complete online entries is $\dfrac58\times168=105$ .
4	$120$	4	The difference between $\dfrac35=60\%$ and $25\%$ is $35\%$ . Therefore $35\%$ of the number is $42$ , so the number is $42\div0.35=120$ .
5	No; the tank has lost $36.25\%$ (or $\dfrac{29}{80}$ ) of its original amount.	4	After the first loss, $85\%$ remains. Three quarters of this then remains, so the final fraction is $0.85\times\dfrac34=0.6375$ . The fraction lost is $1-0.6375=0.3625=36.25\%=\dfrac{29}{80}$ , not $40\%$ .

N13 · Use standard units of mass, length, time, money and other measures (including standard compound measures) using decimal quantities where appropriate

Tier 1 · Easy

Mark scheme for N13 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$2750$ g	1	There are $1000$ grams in a kilogram, so $2.75\times1000=2750$ g.
2	$4.5$ litres	1	There are $1000$ millilitres in a litre, so $4500\div1000=4.5$ litres.

Tier 2 · Standard

Mark scheme for N13 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$11{:}25$	2	Adding $1$ hour gives $10{:}38$ . Adding $47$ minutes gives $11{:}25$ because $22$ minutes reach $11{:}00$ and $25$ minutes remain.
2	$1242$ cm $^2$	2	$0.27$ m is $27$ cm. The area is therefore $27\times46=1242$ cm $^2$ .
3	$9.6$ kg	3	The eight smaller cartons have mass $8\times750=6000$ g, which is $6$ kg. The other cartons have mass $3\times1.2=3.6$ kg. The total is $6+3.6=9.6$ kg.

Tier 3 · Hard

Mark scheme for N13 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$\pounds6.80$	4	The fuel used is $54\times\dfrac{7.5}{100}=4.05$ litres. The cost is $4.05\times1.68=6.804$ , which rounds to $\pounds6.80$ to the nearest penny.
2	$125$ ml/s	3	$18$ litres is $18\,000$ ml and $2$ minutes $24$ seconds is $2\times60+24=144$ seconds. The average rate is $18\,000\div144=125$ ml/s.
3	$2310$ labels	4	$2$ minutes $30$ seconds is $150$ seconds, so the rate is $825\div150=5.5$ labels per second. Seven minutes is $420$ seconds. The number produced is $5.5\times420=2310$ labels.
4	$550$ m	3	Convert the original length: $2.35$ km $=2350$ m. The pieces use $48\times37.5=1800$ m, so the length left is $2350-1800=550$ m.
5	$45.5$ minutes (or $45$ minutes $30$ seconds)	4	The running time is $7.5\div12=0.625$ hours. This is $0.625\times60=37.5$ minutes. Including the rest gives $37.5+8=45.5$ minutes.

N14 · Estimate answers; check calculations using approximation and estimation, including answers obtained using technology

Tier 1 · Easy

Mark scheme for N14 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$10$	1	Use convenient one-significant-figure values: $19.8\approx20$ and $0.49\approx0.5$ . Then $20\times0.5=10$ .
2	$100$	1	Round each number to $1$ significant figure: $72.4\approx70$ and $26.3\approx30$ . Then $70+30=100$ .

Tier 2 · Standard

Mark scheme for N14 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$100$	2	Round to convenient values: $48.7\approx50$ , $0.203\approx0.2$ and $0.098\approx0.1$ . Then $\dfrac{50\times0.2}{0.1}=\dfrac{10}{0.1}=100$ .
2	$150$	2	Use $9.8\approx10$ and $51\approx50$ . Then $10^2+50=100+50=150$ .
3	$10\,000$	3	$31.2\approx30$ , $197\approx200$ and $0.62\approx0.6$ . The estimate is $\dfrac{30\times200}{0.6}=\dfrac{6000}{0.6}=10\,000$ .

Tier 3 · Hard

Mark scheme for N14 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	The display is not reasonable. An estimate is $90$ , so the displayed answer is about ten times too large.	3	Use $598.4\approx600$ , $0.0317\approx0.03$ and $0.204\approx0.2$ . This gives $\dfrac{600\times0.03}{0.2}=\dfrac{18}{0.2}=90$ . Since $931.24$ is near $900$ rather than $90$ , it is not reasonable and likely has a decimal-place error.
2	No; about $40$ packs are needed, so $35$ packs will not cover the garden.	3	Use $118\approx120$ and $2.9\approx3$ . The estimated number of packs is $120\div3=40$ . Since $35<40$ , $35$ packs will not cover the garden.
3	C $40\,000$ to $60\,000$	3	$498\approx500$ , $19.6\approx20$ and $0.204\approx0.2$ . The estimate is $\dfrac{500\times20}{0.2}=50\,000$ , which lies in interval C.
4	$16$ ; it is an overestimate.	4	$76.2\approx80$ , $0.384\approx0.4$ and $2.17\approx2$ , giving $\dfrac{80\times0.4}{2}=16$ . Both numerator factors were rounded up while the denominator was rounded down, so these changes all increase the value. The estimate is therefore an overestimate.
5	Yes; the estimated capacity is $2250$ cartons, so the time should be enough.	4	Use $293\approx300$ and $8.2\approx8$ , giving an estimated rate of $300\div8=37.5$ cartons per minute. In $60$ minutes the line should seal about $37.5\times60=2250$ cartons. This exceeds the target by $200$ cartons, about $9.8\%$ of $2050$ , so the estimate supports the decision that the time should be enough.

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

Tier 1 · Easy

Mark scheme for N15 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$0.0078$	1	The first two significant digits are $7$ and $8$ . The next digit is $4$ , so the $8$ stays unchanged and the rounded value is $0.0078$ .
2	$54\,000$	1	The first two significant digits are $5$ and $3$ . The next digit is $7$ , so round the $3$ up to $4$ to get $54\,000$ .

Tier 2 · Standard

Mark scheme for N15 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$12.55\leq x<12.65$	2	The rounding unit is $0.1$ , so half a unit is $0.05$ . Subtract and add $0.05$ to get the boundaries $12.55$ and $12.65$ ; include the lower boundary only.
2	$725\leq n<735$	2	Half of the rounding unit $10$ is $5$ . The lower boundary is $730-5=725$ and is included; the upper boundary is $730+5=735$ and is excluded.
3	$12.5$ to $3$ significant figures and $12.49$ to $2$ decimal places	2	For $3$ significant figures, keep $1,2,4$ and use the next digit $8$ , giving $12.5$ . For $2$ decimal places, keep $12.48$ and use the next digit $6$ , giving $12.49$ .

Tier 3 · Hard

Mark scheme for N15 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$4.37\leq y<4.38$ Greatest possible integer value of $100y$ is $437$ .	3	Truncation to $2$ decimal places keeps every value from $4.37$ up to but not including $4.38$ , so $4.37\leq y<4.38$ . Multiplying by $100$ gives $437\leq100y<438$ , whose greatest possible integer value is $437$ .
2	$152.5\leq x<157.5$ ; $152.5$ is included because it rounds to $155$ mm, but $157.5$ is excluded because it rounds to $160$ mm.	3	Half of the rounding unit $5$ mm is $2.5$ mm. The lower bound $152.5$ mm rounds to $155$ mm, so it is included. The upper bound $157.5$ mm rounds to $160$ mm, so it is excluded. Hence $152.5\leq x<157.5$ .
3	$47$ and $48$	3	$2.35\leq x<2.45$ . Multiplying the interval by $20$ gives $47\leq20x<49$ . The integer values in this interval are $47$ and $48$ .
4	$0.00755\leq x<0.00765$	4	The first statement gives $0.0075\leq x<0.0085$ . The second gives the narrower interval $0.00755\leq x<0.00765$ , which lies wholly inside the first interval. Their intersection is therefore $0.00755\leq x<0.00765$ .
5	Least $=67\,600$ ; greatest $=68\,400$ .	4	The rounding interval is $67\,500\leq n<68\,500$ . The first multiple of $400$ in this interval is $67\,600$ . The last is $68\,400$ , because the next multiple, $68\,800$ , is outside the interval.

N16 · Apply and interpret limits of accuracy, including upper and lower bounds

Tier 1 · Easy

Mark scheme for N16 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$0.5$ cm	1	The rounding unit is $1$ cm, so the true length can differ from the recorded value by half of this: $1\div2=0.5$ cm.
2	$1590$ g	2	The upper bound for one tub is $260+5=265$ g. The upper bound for the total mass is $6\times265=1590$ g.

Tier 2 · Standard

Mark scheme for N16 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	No; the exact mass is less than $7.65$ kg.	2	The upper bound is $7.6+0.05=7.65$ kg and is not included. Since $7.66>7.65$ , the exact mass cannot be $7.66$ kg.
2	Rounded to the nearest $10$ g; $495\leq m<505$	3	The rounding unit is twice the maximum possible error, so it is $2\times5=10$ g. The exact mass can be $5$ g below $500$ but must be less than $5$ g above it, giving $495\leq m<505$ .

Tier 3 · Hard

Mark scheme for N16 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	No. The exact total must be less than $19.6$ kg.	3	A displayed mass of $2.4$ kg means one box has mass less than $2.45$ kg. Therefore $8$ boxes have total mass less than $8\times2.45=19.6$ kg. Since $20>19.6$ , a total of $20$ kg is impossible.
2	The laser measure; its maximum possible error is $0.5$ mm, compared with $5$ mm for the tape measure.	3	The laser measure has maximum error $0.001\div2=0.0005$ m, which is $0.5$ mm. The tape measure has maximum error $1\div2=0.5$ cm, which is $5$ mm. Therefore the laser measure has the smaller maximum possible error.
3	$\dfrac{1850}{7}$ g (or $264\dfrac{2}{7}$ g)	4	The upper bound for the bar's mass is $1.85$ kg, which is $1850$ g. Dividing this upper bound equally between $7$ blocks gives $1850\div7=\dfrac{1850}{7}$ g for the upper bound of one block.
4	No; the lower bound for the difference is $4.60$ m, which is $0.05$ m below $4.65$ m.	4	Use the lower bound of the first board and the upper bound of the second board. The lower bound for the difference is $8.35-3.75=4.60$ m. This is $0.05$ m less than the required $4.65$ m, so the required difference is not guaranteed.
5	No; $270\times3.15=850.5>845$ .	4	The greatest possible output is less than $845$ parts and the least possible time is $3.15$ hours. A rate of $270$ parts per hour for $3.15$ hours would require $270\times3.15=850.5$ parts. Since $850.5>845$ , even the greatest possible output is too small, so the exact rate cannot be greater than $270$ parts per hour.