OCR GCSE Mathematics: Number Operations and Integers — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Number Operations and Integers revision notes

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Section A

1. Calculate: (a) −7 + 12 (b) −6 × −4 (c) 15 ÷ −3 (d) (−2)³ [4]

2. Evaluate 4 + 3 × (8 − 5)² using the correct order of operations. [2]

Section B

3. Work out, showing your method:

(a) 2⅔ + 1¾ [3] (b) 3⅕ ÷ 1⅗ [3]

4. Estimate the value of (39.6 × 5.12) ÷ 0.196, showing the rounded values used. [3]

5. A number n satisfies 2 < n ≤ 9 and n is prime.

(a) List all possible values. [2] (b) Explain why 1 is not a prime number. [2]

6. Two lighthouses flash every 12 seconds and every 18 seconds. They flash together at 9:00 pm.

(a) Calculate the next time they flash together. [3] (b) State which mathematical idea you used and why. [2]

7. Round 3.04958 to (a) 3 decimal places (b) 2 significant figures (c) the nearest whole number. [3]

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Answers

1. (a) 5 [1]. (b) 24 [1]. (c) −5 [1]. (d) −8 [1].

2. Brackets first: (8 − 5) = 3; then 3² = 9 [1]; 4 + 3 × 9 = 4 + 27 = 31 [1].

3. (a) Convert to improper fractions: 8/3 + 7/4 [1]; common denominator 12: 32/12 + 21/12 = 53/12 [1] = 4 5/12 [1]. (b) 16/5 ÷ 8/5 [1]; multiply by the reciprocal: 16/5 × 5/8 [1] = 80/40 = 2 [1].

4. Round to 1 significant figure: 40 × 5 ÷ 0.2 [1]; 200 ÷ 0.2 [1] = 1000 [1].

5. (a) 3, 5, 7 [1] [1]. (b) A prime number has exactly two distinct factors [1]; 1 has only one factor — itself [1].

6. (a) LCM of 12 and 18: 12 = 2² × 3, 18 = 2 × 3², so LCM = 2² × 3² = 36 [1] [1]; they flash together again after 36 seconds, at 9:00:36 pm [1]. (b) The lowest common multiple [1], because it is the first time that is a whole number of intervals for both lighthouses [1].

7. (a) 3.050 [1]. (b) 3.0 [1]. (c) 3 [1].

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Section C — additional questions

8. A price of £65 is increased by 20%, and the new price is then decreased by 20%. Calculate the final price and explain why it is not equal to £65. [3]

9. Find the HCF and LCM of 84 and 126 using prime factorisation. [4]

10. Calculate 40% of 65% of £500, giving your method clearly. [3]

Answers to additional questions

8. Increase: 65 × 1.2 = £78 [1]. Decrease: 78 × 0.8 = £62.40 [1]. This is not £65 because the 20% decrease is calculated on the larger, already-increased price (£78), not on the original £65, so the absolute amount removed is larger than the absolute amount added [1].

9. 84 = 2² × 3 × 7; 126 = 2 × 3² × 7 [2]. HCF = product of the lowest power of each common prime = 2 × 3 × 7 = 42 [1]. LCM = product of the highest power of every prime present = 2² × 3² × 7 = 252 [1].

10. 65% of £500 = 500 × 0.65 = £325 [1]; 40% of £325 = 325 × 0.4 = £130 [1]. (Equivalently, 40% × 65% = 26% of £500 = £130, since multiplying successive percentage multipliers together gives the same overall result [1].)

A note on prime factorisation for HCF and LCM

Question 9 illustrates the general method worth memorising: write both numbers as a product of prime factors, then build the HCF by taking the lowest power of each prime that appears in both factorisations, and build the LCM by taking the highest power of every prime that appears in either factorisation. This method scales to any pair of numbers, however large, and is far more reliable under exam conditions than attempting to spot common factors by inspection, particularly once the numbers involved are no longer small and familiar.

A note on order of operations

Question 2 is a reminder that BIDMAS – Brackets, Indices, Division and Multiplication, Addition and Subtraction – has two pairs that rank equally rather than a strict six-step hierarchy: division and multiplication are worked left to right as they appear, and the same applies to addition and subtraction. A common error is treating multiplication as always ranking above division, or addition as always ranking above subtraction, rather than working strictly left to right once brackets and indices have been resolved. Applying brackets and indices first, exactly as in question 2, before moving to the remaining operations in left-to-right order, avoids the ambiguity that a rigid six-step reading of the acronym can otherwise introduce.

Where marks are usually lost

• Getting the sign wrong when multiplying two negatives.
• Adding whole parts and fraction parts separately without a common denominator.
• Rounding to a convenient number rather than 1 significant figure in estimation.
• Dropping the trailing zero in 3.050, which is needed to show three decimal places.
• Treating multiplication as always ranking above division, or addition as always ranking above subtraction, instead of working left to right within each equal-ranking pair.