OCR GCSE Mathematics: Fractions, Decimals and Percentages — 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: Fractions, Decimals and Percentages study guide | Fractions, Decimals and Percentages revision notes

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Section A

1. Convert 5/8 to a decimal, then to a percentage. [2]

2. Write 0.85 as a fraction in its simplest form. [2]

Section B

3. Work out, showing your method:

(a) 3/4 + 5/6 [3] (b) 2⅗ − 1½ [3]

4. Find 5/9 of £252. [2]

5. A coat costs £120 and is reduced by 30% in a sale. Calculate the sale price. [3]

6. A phone costs £276 after a 15% price increase. Calculate the original price. [3]

7. A savings account holds £3,600 and pays 4% interest per year, compounded annually. Calculate the value of the account after 3 years, giving your answer to the nearest penny. [4]

8. A student scored 39 out of 52 in a maths test and 28 out of 35 in a science test. In which subject did the student achieve the higher percentage? Show your working. [4]

9. Order these values from smallest to largest, showing your method: 7/10, 68%, 0.072, 0.72. [4]

Answers

1. 5 ÷ 8 = 0.625 [1]; 0.625 × 100 = 62.5% [1].

2. 0.85 = 85/100 [1]; dividing numerator and denominator by 5 gives 17/20 [1].

3. (a) Common denominator 12: 9/12 + 10/12 [1] = 19/12 [1] = 1 7/12 [1]. (b) Convert to improper fractions: 13/5 − 3/2 [1]; common denominator 10: 26/10 − 15/10 [1] = 11/10 = 1 1/10 [1].

4. £252 ÷ 9 = 28 [1]; 28 × 5 = £140 [1].

5. Multiplier for a 30% decrease = 1 − 0.30 = 0.70 [1]; £120 × 0.70 [1] = £84 [1].

6. £276 represents (100% + 15%) = 115% of the original price [1]; original price = £276 ÷ 1.15 [1] = £240 [1].

7. Multiplier = 1 + (4 ÷ 100) = 1.04 [1]; value after 3 years = £3,600 × 1.04³ [1] = £3,600 × 1.124864 [1] = £4,049.51 (2 d.p.) [1].

8. Maths: 39 ÷ 52 = 0.75 = 75% [1]. Science: 28 ÷ 35 = 0.8 = 80% [1]. The student achieved a higher percentage in science [1], since 80% is greater than 75% [1].

9. Convert all four values to decimals: 7/10 = 0.7; 68% = 0.68; 0.072 stays as 0.072; 0.72 stays as 0.72 [2]. Ordered smallest to largest: 0.072, 68%, 7/10, 0.72 [2].

A note on the reverse-percentage method

Question 6 is the question type most commonly answered incorrectly on this topic. A student who calculates 15% of £276 and subtracts it — reasoning that the price “went up by 15%, so take 15% back off” — gets £234.60, which is wrong, because £276 is 115% of the original price, not 100% of some other amount minus 15%. The reliable method is always to identify what percentage the given value represents of the original (100% plus or minus the stated change), then divide the given value by that percentage as a decimal. The same logic applies whether the change described is an increase or a decrease: a value after a decrease represents less than 100% of the original, so it is still divided, never multiplied, to recover the original amount. Practising the habit of writing out “this value represents ___% of the original” before choosing an operation removes most of the errors this question type produces.

A note on comparing percentages fairly

Question 8 tests a distinction that is easy to miss under time pressure: 39 out of 52 and 28 out of 35 cannot be compared directly as raw scores, because the totals are different. Converting each to a percentage (or an equivalent fraction with the same denominator) puts both scores on the same scale, which is the only way to compare them meaningfully. This same principle — that two quantities can only be compared directly once they are expressed in the same form or against the same total — reappears throughout GCSE maths, from comparing ratios to comparing rates, so it is worth recognising as a general technique rather than a one-off trick specific to percentages.

Where marks are usually lost

• Adding or subtracting fractions without first converting to a common denominator.
• Calculating a percentage of the wrong amount in a reverse-percentage question — working from the final value as if it were 100% of the original.
• Using the original amount rather than the already-changed amount when applying a second percentage change, or vice versa, in multi-step percentage problems.
• Forgetting to raise the multiplier to a power when compounding over more than one year, and instead multiplying by (1 + rate) only once.
• Comparing two percentages or fractions with different original totals without converting both to the same form first.

Approaching fractions, decimals and percentages questions

Before starting any percentage calculation, decide explicitly whether the number given in the question is the original value (calculate a percentage of it directly) or the final value after a change has already been applied (use the reverse-percentage method instead), since this single judgement determines the entire method and is the most heavily examined distinction on this topic. For compound-change questions, always build a single multiplier first — 1 plus or minus the rate as a decimal — and raise it to the power matching the number of time periods, rather than repeating a percentage-and-add calculation year by year, since the multiplier method is both faster and less prone to rounding errors carried across steps. When a question asks you to compare two quantities expressed as fractions of different totals, convert both to percentages or decimals before comparing, and never compare the numerators or the raw scores directly. Finally, for fraction arithmetic, check first whether the question needs a common denominator (adding or subtracting) or a reciprocal (dividing), since applying the wrong one of these two methods is one of the most common sources of lost marks on this topic even among otherwise confident students.