Integers, order of operations, primes, HCF/LCM, powers & roots – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator3 marks

The table shows the lowest temperature recorded one night in five cities.

City	Oslo	Cairo	Moscow	London	Madrid
Temperature (°C)	$-7$	$4$	$-12$	$0$	$3$

(a) Write the temperatures in order, starting with the lowest. [1]

(b) Work out the difference between the highest temperature and the lowest temperature. [1]

(a) $-12,\ -7,\ 0,\ 3,\ 4$ (accept city names in the order Moscow, Oslo, London, Madrid, Cairo; accept reverse order only if clearly labelled highest first — otherwise B0)B1
(b) $16$ (°C) (do not accept $-16$)B1
(c) $2$ (°C)B1
Answer: (a) $-12, -7, 0, 3, 4$ (b) $16\,^\circ$C (c) $2\,^\circ$C
Examiner tip: In (b), the difference between 4 and −12 is 4 − (−12) = 16, not 8: subtracting a negative means adding.

2Grade 1–3FoundationCalculator5 marks

Here is a list of numbers.

$$9 \quad 15 \quad 23 \quad 27 \quad 32 \quad 51 \quad 64 \quad 81$$

From the numbers in the list, write down

(a) a prime number [1]

(b) a cube number [1]

(d) a factor of 96 [1]

(e) Kai says, “51 is a prime number because it is odd and it is not in the 5 times table.”
Explain why Kai is wrong. [1]

(a) $23$B1
(b) $27$ or $64$ (accept either; B0 if any incorrect number is also given)B1
(c) $64$B1
(d) $32$B1
(e) correct explanation, e.g. $51 = 3 \times 17$ oe, or “51 is divisible by 3”, or “51 has factors other than 1 and itself, e.g. 3 (or 17)” (accept “51 is in the 3 times table”; do not accept “it is not prime” with no reason)C1
Answer: (a) 23 (b) 27 or 64 (c) 64 (d) 32 (e) $51 = 3\times17$, so it has factors other than 1 and itself
Examiner tip: 51 looks prime but its digit sum is 6, so it is divisible by 3 — always test 3, 7 and 11 before calling a number prime.

3Grade 4–5Calculator4 marks

(a) Work out $18 - 3 \times (7 - 2)^2 \div 5$ [1]

(b) Copy the calculation and insert one pair of brackets to make it correct.
$$4 + 6 \times 5 - 2 = 48$$ [1]

(i) $4.7 \times 360$ [1]

(ii) $1692 \div 0.36$ [1]

(a) $3$B1
(b) $(4 + 6) \times 5 - 2 = 48$B1
(c)(i) $1692$B1
(c)(ii) $4700$B1
Answer: (a) 3 (b) $(4+6)\times5-2$ (c)(i) 1692 (ii) 4700
Examiner tip: In (a) the bracket and the power come first: (7 − 2)² = 25, then 3 × 25 ÷ 5 = 15 before subtracting from 18.

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4Grade 4–5Calculator3 marks

The diagram shows an incomplete factor tree for 588

(a) Write the missing numbers in the two boxes. [1]

(b) Write 588 as a product of powers of its prime factors. [1]

(a) $49$ and $7$ (both required)B1
(b) $2^2 \times 3 \times 7^2$ (accept factors in any order; accept $3^1$; do not accept $2\times2\times3\times7\times7$ as the answer is required in index form)B1
(c) $3$ (ft from their product of prime factors in (b))B1
Answer: (a) 49, 7 (b) $2^2\times3\times7^2$ (c) $k = 3$
Examiner tip: “Product of powers” means index form — writing 2 × 2 × 3 × 7 × 7 loses the mark even though the primes are right.

5Grade 4–5Calculator4 marks

(a) Find the highest common factor (HCF) of 84 and 120 [2]

(b) Find the lowest common multiple (LCM) of 84 and 120 [2]

(a) for a correct method, e.g. $84 = 2^2\times3\times7$ and $120 = 2^3\times3\times5$, or listing at least 3 factors of each number (with no more than one error), or a correct Venn diagramM1
(a) $12$A1
(b) for a correct method, e.g. $2^3\times3\times5\times7$, or $\dfrac{84\times120}{12}$, or listing multiples of both numbers up to at least 840 for one of themM1
(b) $840$ (SC B1 for any common multiple of 84 and 120, e.g. 10 080, if M0 scored)A1
Answer: (a) 12 (b) 840
Examiner tip: Students often swap the two: the HCF can be no bigger than the smaller number, and the LCM can be no smaller than the larger one.

6Grade 4–5Calculator5 marks

(a) Simplify $7^5 \times 7^3$ Give your answer as a power of 7 [1]

(b) Simplify $5^9 \div 5^{-2}$ Give your answer as a power of 5 [1]

(d) $$\frac{2^5 \times 2^{-3}}{2^6} = 2^n$$ Find the value of $n$. [2]

(a) $7^8$B1
(b) $5^{11}$B1
(c) $1$B1
(d) for $2^2$ seen as the numerator, or $5 + (-3) - 6$ oeM1
(d) $n = -4$A1
Answer: (a) $7^8$ (b) $5^{11}$ (c) 1 (d) $n=-4$
Examiner tip: Dividing by $5^{-2}$ means subtracting −2, so the index goes up: 9 − (−2) = 11, not 7.

7Grade 4–5Calculator3 marks

(a) Use your calculator to work out the value of $$\sqrt[3]{\frac{4.2^2 + 17.3}{2.6 \times 0.85}}$$ Write down all the figures on your calculator display. [2]

(b) Give your answer to part (a) correct to 3 significant figures. [1]

(a) for $34.94$ or $2.21$ seen, or $15.81\ldots$ seenM1
(a) $2.509825\ldots$ (accept $2.5098$ or better)A1
(b) $2.51$ (ft their answer to (a) if it has more than 3 sf)B1
Answer: (a) 2.509825586… (b) 2.51
Examiner tip: Use brackets round the whole fraction (or the fraction key) before taking the cube root — otherwise only the numerator gets cube-rooted.

8Grade 6–7HigherCalculator3 marks

Maya is buying food for a barbecue.

Sausages are sold in packs of 12
Bread rolls are sold in packs of 20
Paper napkins are sold in packs of 45

Maya wants to buy exactly the same number of sausages, bread rolls and napkins.

Work out the smallest number of packs of each item that she can buy. [3]

for a correct method to find a common multiple, e.g. $12 = 2^2\times3$, $20 = 2^2\times5$, $45 = 3^2\times5$, or listing multiples of at least two of the numbers (at least 3 multiples each)M1
for identifying $180$ as the LCM (may be implied by a correct number of packs)A1
sausages 15 packs, rolls 9 packs, napkins 4 packs (all three, clearly linked to the items)A1
Answer: 15 packs of sausages, 9 packs of rolls and 4 packs of napkins (180 of each)
Examiner tip: Finding 180 is not the end — the question asks for the number of packs, so divide 180 by each pack size.

9Grade 6–7HigherCalculator4 marks

$A = 2^3 \times 3^2 \times 5$ and $B = 2 \times 3^4 \times 7$

(a) Find the highest common factor (HCF) of $A$ and $B$. [1]

(b) Find the lowest common multiple (LCM) of $A$ and $B$.
Give your answer as a product of powers of its prime factors. [1]

(c) $n$ is a positive integer.
The HCF of $A$ and $n$ is 12
The LCM of $A$ and $n$ is 2520
Find the value of $n$. [2]

(a) $18$ (accept $2\times3^2$)B1
(b) $2^3\times3^4\times5\times7$ (any order)B1
(c) for $\dfrac{12\times2520}{360}$ oe, or $A = 360$ with $2520 = 2^3\times3^2\times5\times7$ and $12 = 2^2\times3$ used to identify that $n$ contains $2^2$, $3$ and $7$M1
(c) $84$A1
Answer: (a) 18 (b) $2^3\times3^4\times5\times7$ (c) $n = 84$
Examiner tip: For the HCF take the lowest power of each shared prime; for the LCM take the highest power of every prime that appears.

10Grade 8–9HigherCalculator4 marks

A school hall has fewer than 400 chairs.

When the chairs are put in rows of 4, there is 1 chair left over.
When the chairs are put in rows of 6, there is 1 chair left over.
When the chairs are put in rows of 10, there is 1 chair left over.
When the chairs are put in rows of 7, there are no chairs left over.

Work out the number of chairs in the hall.
You must show all your working. [4]

for finding $60$ as the LCM of 4, 6 and 10, or a common multiple of 4, 6 and 10 (e.g. 120)M1
for adding 1 to multiples of 60 (or of their common multiple), e.g. $61, 121, 181, \ldots$, at least three valuesM1
for testing their values for divisibility by 7 (at least two tested), or $60k + 1 = 7m$ oeM1
$301$ (with working; $301 = 7\times43$)A1
Answer: 301 chairs
Examiner tip: Working with 60k + 1 (not 60k) is the key step; a bare answer of 301 with no working may score nothing because working is demanded.

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