AQA GCSE Mathematics: Number — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Number revision notes

Tier guide: parts marked (Higher) use content from the Higher content only column of the AQA 8300 specification — fractional indices (N7), surds (N8), recurring decimals (N10) and upper and lower bounds (N16). All other parts use Foundation content and suit both tiers.

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Section A

1. Evaluate without a calculator: (a) 5⁻² (b) (Higher) 16^(3/4) (c) (2/3)⁻¹ [3]

2. Write 0.000 037 in standard form, and 4.2 × 10⁵ as an ordinary number. [2]

Section B

3. (Higher, N8) Simplify fully:

(a) √50 + √18 [2] (b) (3 + √5)(3 − √5) [2] (c) rationalise 6 ÷ √3 [2]

4. (Higher, N16) A length is measured as 8.4 cm to the nearest 0.1 cm. (At Foundation tier, part (a) would be asked as an error interval, under N15; part (b), calculating with bounds, is Higher only.)

(a) State the upper and lower bounds. [2] (b) A rectangle has this length and a width of 5.2 cm, also to the nearest 0.1 cm. Calculate the upper bound of its area. [3]

5. A jacket costs £84 after a 30% reduction.

(a) Calculate the original price. [3] (b) An investment of £2000 earns 3.5% compound interest per year. Calculate its value after 4 years. [3]

6. (Higher, N10) Write 0.4̇5̇ (0.454545…) as a fraction in its simplest form. [3]

7. Express 360 as a product of prime factors, and hence find the HCF and LCM of 360 and 84. [4]

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Answers

1. (a) 1/25 [1]. (b) (⁴√16)³ = 2³ = 8 [1]. (c) 3/2 [1].

2. 3.7 × 10⁻⁵ [1]; 420 000 [1].

3. (a) √50 = 5√2, √18 = 3√2 [1]; total = 8√2 [1]. (b) 9 − 3√5 + 3√5 − 5 [1] = 4 [1]. (c) Multiply top and bottom by √3: 6√3 ÷ 3 [1] = 2√3 [1].

4. (a) Lower bound 8.35 cm, upper bound 8.45 cm [1] [1]. (b) Upper bounds are 8.45 and 5.25 [1]; area = 8.45 × 5.25 [1] = 44.3625 cm² [1].

5. (a) £84 represents 70% [1]; 1% = £1.20 [1]; original = £120 [1]. (b) 2000 × 1.035⁴ [1] [1] = £2295.05 (to the nearest penny) [1].

6. Let x = 0.454545… Then 100x = 45.4545… [1]; subtracting, 99x = 45 [1]; x = 45/99 = 5/11 [1].

7. 360 = 2³ × 3² × 5 [1]; 84 = 2² × 3 × 7 [1]. HCF = 2² × 3 = 12 [1]; LCM = 2³ × 3² × 5 × 7 = 2520 [1].

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Section C — additional questions

8. Evaluate without a calculator: (a) (Higher) 27^(2/3) (b) 4⁻³ [2]

9. A car’s value depreciates by 15% per year. If it is bought for £18,000, calculate its value after 3 years, giving your answer to the nearest pound. [3]

10. (Higher, N8) Simplify √72 − √8, giving your answer in the form k√2. [2]

11. (Higher as worded, N16) A field is measured as 120 m long, to the nearest 5 m. Give the upper and lower bounds of this measurement. (At Foundation tier the same skill is asked as an error interval, under N15.) [2]

Answers to Section C

8. (a) 27^(2/3) = (³√27)² = 3² = 9 [1]. (b) 4⁻³ = 1/4³ = 1/64 [1].

9. After year 1: 18000 × 0.85 = £15,300; after year 2: ×0.85 again; after year 3: 18000 × 0.85³ [1] [1] = 18000 × 0.614125 = £11,054 (nearest pound) [1].

10. √72 = 6√2, √8 = 2√2 [1]; 6√2 − 2√2 = 4√2 [1].

11. Lower bound 117.5 m, upper bound 122.5 m [1] [1].

A note on compound percentage change

Questions 5(b) and 9 both use the same underlying principle – repeated percentage change is multiplicative, not additive, so a value increasing by 3.5% each year for 4 years is found by multiplying by 1.035 four times (equivalently, raising 1.035 to the power 4), not by adding 3.5% four times over. The same logic applies to repeated percentage decrease: multiplying by 0.85 (100% − 15%) for each year of depreciation, rather than subtracting 15% of the original value in each year, is what correctly accounts for depreciation being calculated on an already-reduced value in later years. Confusing compound change with simple, additive change is one of the most common sources of lost marks across this whole topic area.

A note on surd simplification (Higher)

Questions 3(a) and 10 both rely on the same technique: finding the largest square-number factor of the number under the root, since this is what allows a surd to be simplified to the form k√n. For 72, the largest square factor is 36 (72 = 36 × 2), so √72 simplifies to 6√2; for 8, the largest square factor is 4 (8 = 4 × 2), so √8 simplifies to 2√2. Once two surds share the same value under the root, they can be added or subtracted by combining their coefficients directly, in exactly the way like terms are combined in algebra – which is why √72 − √8 becomes 6√2 − 2√2 = 4√2 rather than requiring any further simplification once both terms share the same root.

A note on fractional and negative indices

Question 1 and question 8 both test the two index rules that most often trip candidates up: a fractional index such as ^(2/3) means take the root indicated by the denominator (here, a cube root) and raise the result to the power indicated by the numerator, so 27^(2/3) is the cube root of 27, squared, giving 9. Squaring first and then taking the cube root (729, then 9) is not wrong: for any positive base the two orders always give the same value, perfect cube or not. Taking the root first is simply the practical choice, because it keeps the intermediate numbers small (3 rather than 729) and makes a non-calculator evaluation much easier. A negative index, by contrast, does not make the answer negative — it means “one over” the positive-index version, so 4⁻³ equals 1 divided by 4³, not negative 4³. Keeping these two rules distinct, rather than treating a negative index and a fractional index as variations on the same idea, avoids a common and entirely avoidable source of lost marks.

Where marks are usually lost

• Treating a negative index as a negative answer.
• (Higher) Using the lower bound of the width when finding an upper bound of an area.
• Calculating 30% of £84 instead of treating £84 as 70%.
• (Higher) Multiplying by 10 instead of 100 for a two-digit recurring block.