Practise 21 exam-style questions on Order of operations and inverse operations for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
21 questions
1–5 marks each
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20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Question 1
2 marks
Put brackets into each calculation to make it correct. You may use more than one pair of brackets in a calculation.
Brackets are worked out first, so they change the order of operations. Try putting brackets round an addition or subtraction so that it happens before the multiplication. In the second calculation, \((3 + 2)^2\) squares 5, not just 2.
Mark scheme
1 mark
B1 for (8 − 2) × (3 + 1) = 24
cao; correct answer with no working scores full marks
1 mark
B1 for \((3 + 2)^2 \times (5 - 1) = 100\)
Question 2
2 marks
Ravi thinks of a number. He squares the number, then subtracts 7, then divides by 3. His answer is 14.
Find the two numbers Ravi could have thought of.
Model answer
Work backwards, using inverse operations in reverse order: 14 × 3 = 42 42 + 7 = 49 \(\sqrt{49} = 7\) or \(-7\) Ravi's number was 7 or −7.
Explanation
To undo a sequence of operations, apply the inverse operations in the reverse order: × 3 undoes ÷ 3, + 7 undoes − 7, and square root undoes squaring. Remember that both 7 and −7 square to 49.
Mark scheme
1 mark
M1 for working backwards with inverse operations in the correct order, e.g. 14 × 3 + 7 (= 49)
or an equation such as \(\dfrac{x^2 - 7}{3} = 14\); correct answer with no working scores full marks
1 mark
A1 for 7 and −7
both needed
Question 3
3 marks
Work out \[\dfrac{48 \times 35 \times 0.2}{14 \times 0.6}\]
Model answer
Cancel common factors: \(\dfrac{35}{14} = \dfrac{5}{2}\) and \(\dfrac{0.2}{0.6} = \dfrac{1}{3}\) So the calculation is \(48 \times \dfrac{5}{2} \times \dfrac{1}{3} = \dfrac{240}{6} = 40\)
Explanation
Multiplication and division are inverse operations, so a factor on the top cancels with the same factor on the bottom. Cancelling first keeps the numbers small. Working it out directly also works: 48 × 35 × 0.2 = 336 and 14 × 0.6 = 8.4, and 336 ÷ 8.4 = 3360 ÷ 84 = 40.
Mark scheme
1 mark
M1 for a correct first cancellation or evaluation, e.g. \(\dfrac{35}{14} = \dfrac{5}{2}\), \(\dfrac{0.2}{0.6} = \dfrac{1}{3}\), numerator 336 or denominator 8.4
correct answer with no working scores full marks
1 mark
M1 for a complete method, e.g. \(48 \times \dfrac{5}{2} \times \dfrac{1}{3}\), or 336 ÷ 8.4
1 mark
A1 for 40
cao
Question 4
4 marks
On each of three days, Fiona spends half of the money she has and then spends another £4. At the end of the third day she has £1 left.
How much money did Fiona have at the start of the first day?
Model answer
Work backwards. To undo 'halve, then subtract 4', add 4 and then double. Start of day 3: (1 + 4) × 2 = £10 Start of day 2: (10 + 4) × 2 = £28 Start of day 1: (28 + 4) × 2 = £64 Check: 64 → 32 − 4 = 28 → 14 − 4 = 10 → 5 − 4 = 1
Explanation
Each day the operations are 'halve' then 'subtract 4', so going backwards you must undo them in the reverse order: add 4 first, then double. Doubling first and then adding 4 is the most common mistake. Always check by working forwards.
Mark scheme
1 mark
P1 for reversing one day correctly: add 4 then multiply by 2, e.g. (1 + 4) × 2 = 10
or a correct forward trial for a starting amount, showing all three days; correct answer with no working scores full marks
1 mark
P1 for £10 at the start of day 3 and (10 + 4) × 2 = £28 at the start of day 2
1 mark
P1 for (28 + 4) × 2
ft their value for the start of day 2
1 mark
A1 for £64
cao
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