22 questions on percentage change, reverse percentages, simple and compound interest, and depreciation.
💷 Percentages and compound interest
Percentages appear on all three papers and in more contexts than any other single topic — prices, wages, interest, VAT, population, depreciation. This sheet covers the whole of it in order. It begins with percentage of an amount and percentage change, including finding the percentage rather than applying it; moves to multipliers, which make everything after this easier; then to reverse percentages, where you are given the value after the change and asked for the original, and where subtracting instead of dividing is the commonest error in the topic; then to simple interest, compound interest and depreciation, where the multiplier is raised to a power rather than multiplied by the number of years. The last few questions mix the types without labelling them, because on the paper they are not labelled either.
- 1.A recipe uses 160 g of flour. Sam wants to increase the amount by 1/4. Work out the new amount of flour.
- 2.Work out 8% of £250.
- 3.A cycle route is 350 m long. A footpath runs alongside it for 3/7 of that length. Work out the length of the footpath.
- 4.The price of a share falls by 10% on Monday and then rises by 10% on Tuesday. Work out the overall percentage change from Monday's starting price.
- 5.A gym increases its membership price by 1/5. The original price is £60. Work out the new price.
- 6.A charity raffle sells 240 tickets at £1.85 each. 40% of the money raised is given to a local hospital. Work out how much money the hospital receives.
- 7.Leah puts £4000 into a savings account paying 3% compound interest each year. At the end of 2 years she takes out all of the money and spends £1500 of it on a laptop. Work out how much of the money she has left.
- 8.Maya buys a vintage car for £15000. Its value is expected to increase by 4% each year for 2 years. She then plans to sell the car and use £2000 of the money from the sale to buy a trailer. Work out how much money she will have left after selling the car and buying the trailer.
- 9.A laptop costs £720. A carrying case for it costs 1/9 of the price of the laptop. Work out the cost of the carrying case.
- 10.In a class, 1/3 of the pupils are girls. There are 12 girls in the class. Work out how many pupils are in the class.
- 11.A jumper costs £45 at Shop A, where it is reduced by 20%. The same jumper costs £34 at Shop B, where a further 10% reduction is then applied. Work out the difference between the two reduced prices.
- 12.Work out 2/5 of 45.
- 13.A charity shop buys a coat for £24 and sells it for a profit that is 3/8 of the buying price. Work out the selling price.
- 14.Before a charity campaign, donations were £6400 per month. Donations grew by 8% in the first year after the campaign, and then fell by 3% in the second year as interest faded. Work out the amount donated per month at the end of the second year, to the nearest pound.
- 15.A colony of bacteria has 400 bacteria. The number increases by 15% each hour. Work out the number of bacteria after 1 hour.
- 16.Work out 35% of 180.
- 17.A car is bought for £17,500. Its value decreases by 12% in the first year, and by a further 10% of its reduced value in the second year. Work out the value of the car at the end of the second year, giving your answer to the nearest pound.
- 18.The same jumper is sold at two shops. Shop A charges £40 and Shop B charges £50. Write down the price at Shop A as a percentage of the price at Shop B.
- 19.2/3 of an amount of money is £30. Work out the amount.
- 20.A textbook is reduced from £60 to £45. Work out the percentage reduction.
- 21.Work out the difference between 45% of 70 and 35% of 80.
- 22.After a price increase of 10%, a laptop costs £330. Work out the original price.
Answer key
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
- (d) £20 — 1% of £250 = £2.50, so 8% = 8 × £2.50 = £20. A candidate who misplaces the decimal point and finds 0.8% instead gets £2. A candidate who confuses 8% with 80% gets £200. A candidate who rounds 8% up to the nearby 10% gets £25.
- (c) 150 m — Method: a fraction acts as an operator, so finding 3/7 of a length means dividing by the denominator and multiplying by the numerator. Working: 350 ÷ 7 = 50, so one seventh of the route is 50 m, and three sevenths is 50 × 3 = 150 m. Answer: 150 m. The distractors: 50 m comes from finding one seventh and stopping there instead of multiplying by 3; 1050 m comes from multiplying by the numerator without dividing by the denominator, giving 350 × 3 = 1050; 200 m comes from working out the stretch of the route the footpath does not run alongside, which is 4/7 of 350 m, instead of the stretch it does.
- (b) −1% — Method: write each change as a multiplier and multiply them. A 10% fall is × 0.9 and a 10% rise is × 1.1. Working: 0.9 × 1.1 = 0.99, so the final price is 99% of the original, which is 1% less. Answer: an overall change of −1%. The distractors: 0% comes from assuming a 10% fall and a 10% rise cancel — they do not, because the rise is 10% of a smaller amount; +1% has the size right but the sign wrong, from reading the multiplier 0.99 as 1% above 1 instead of 1% below it; −2% comes from finding the 1% fall and then counting it once for each of the two changes.
- (b) £72 — One fifth of £60 = £12. New price = £60 + £12 = £72. A candidate who gives the increase instead of the new price gets £12. A candidate who subtracts the increase instead of adding it gets £60 − £12 = £48. A candidate who uses 1/4 instead of 1/5 gets £60 + £15 = £75.
- (a) £177.60 — Total raised = 240 × £1.85 = £444.00. The hospital receives 40% of this: £444.00 × 0.4 = £177.60. A candidate who works out the remaining 60% instead of the 40% given away gets £266.40. A candidate who forgets to find the percentage and gives the full total gets £444.00. A candidate who halves 40% by mistake and uses 20% gets £88.80.
- (c) £2743.60 — Each year the balance is multiplied by 1.03. After the first year: 4000 × 1.03 = 4120. After the second year: 4120 × 1.03 = 4243.60, so that is what Leah takes out. She then spends £1500 of it, which leaves 4243.60 − 1500 = 2743.60. She has £2743.60 left.
- (a) £14224 — Value after 2 years: £15000 × 1.04 × 1.04 = £16224. Money left after buying the trailer: £16224 − £2000 = £14224. £14200 comes from treating the two 4% increases as a single flat 8% increase applied once instead of compounding: £15000 × 1.08 = £16200, and £16200 − £2000 = £14200. £13600 comes from applying the 4% increase only once, for 1 year instead of 2: £15000 × 1.04 = £15600, and £15600 − £2000 = £13600. £18224 comes from adding the £2000 instead of subtracting it: £16224 + £2000 = £18224.
- (c) £80 — Method: a unit fraction acts as an operator, so finding 1/9 of a price means dividing that price by 9. Working: £720 ÷ 9 = £80. Answer: £80. The distractors: £6480 comes from multiplying by the denominator instead of dividing by it, giving £720 × 9 = £6480; £640 comes from working out what is left of the £720 once the case is paid for, £720 − £80, instead of the cost of the case itself; £72 comes from dividing by 10 instead of 9, treating one ninth as one tenth.
- (d) 36 — Method: the fraction is acting as an operator on the whole class, so one third of the class equals 12; the operation has to be reversed, and the inverse of dividing by 3 is multiplying by 3. Working: 1/3 × (number of pupils) = 12, so the number of pupils = 12 × 3 = 36. Answer: 36 pupils. The distractors: 4 comes from applying the operator instead of reversing it, working out 12 ÷ 3 = 4; 18 comes from reading the 12 girls as two thirds of the class, giving 12 ÷ 2 × 3 = 18; 24 comes from working out the number of boys, the other two thirds, as 2 × 12 = 24 and giving that instead of the size of the class.
- (d) £5.40 — Method: work out the reduced price at each shop separately, then subtract the smaller from the larger. Working: Shop A's reduced price is £45 × 0.8 = £36, and Shop B's reduced price is £34 × 0.9 = £30.60, so the difference is £36 − £30.60 = £5.40. Answer: £5.40. £11.00 comes from comparing the two ORIGINAL prices, £45 − £34, without applying either shop's reduction at all. £1.60 comes from finding Shop A's reduced price correctly, £36, but then subtracting Shop B's original (unreduced) price of £34 instead of its reduced price. £66.60 comes from adding the two reduced prices together, £36 + £30.60, instead of subtracting them.
- (c) 18 — 45 ÷ 5 = 9, and 2 × 9 = 18. A candidate who stops after finding one fifth gets 9. A candidate who uses 3/5 instead of 2/5 gets 27. A candidate who uses 4/5 instead of 2/5 gets 36.
- (d) £33.00 — The profit is 3/8 of £24 = (£24 ÷ 8) × 3 = £3 × 3 = £9.00. Selling price = £24 + £9.00 = £33.00. A candidate who gives the profit instead of the selling price gets £9.00. A candidate who subtracts the profit instead of adding it gets £24 − £9 = £15.00. A candidate who works out one eighth of £24 and adds that on, forgetting to multiply by the numerator 3, gets £24 + £3 = £27.00.
- (d) £6705 — After the first year: £6400 × 1.08 = £6912. After the second year: £6912 × 0.97 = £6704.64, which rounds to £6705 (nearest pound). £6720 comes from treating the +8% and −3% changes as a single net +5% change applied to the original amount instead of applying each change in turn: £6400 × 1.05 = £6720. £6912 comes from applying only the first year's growth and stopping there, without applying the second year's fall. £7104 comes from adding the two percentages together as +11% and applying that to the original amount instead of applying each change to the correct starting amount in turn: £6400 × 1.11 = £7104.
- (b) 460 — To increase by 15%, multiply by 1.15 (100% + 15%). 400 × 1.15 = 460. 60 comes from working out only the increase, 400 × 0.15 = 60, and forgetting to add it to the original number. 415 comes from adding 15 directly to 400 instead of 15% of 400. 340 comes from multiplying by 0.85, decreasing instead of increasing: 400 × 0.85 = 340.
- (b) 63 — 10% of 180 = 18, so 5% = 9. 35% = (3 × 18) + 9 = 54 + 9 = 63. A candidate who uses 25% instead of 35% gets 45. A candidate who doubles 35% to get 70% by mistake gets 126. A candidate who subtracts 35 from 180 instead of finding a percentage gets 145.
- (b) £13,860 — Method: apply the first year's percentage decrease, then apply the second year's percentage decrease to the new value. Working: after the first year, the car is worth £17,500 × 0.88. Multiplying this result by 0.90 gives the value at the end of the second year, £13,860. Answer: £13,860. £13,650 comes from adding the two percentages together (12% + 10% = 22%) and applying a single 22% decrease, £17,500 × 0.78 = £13,650, instead of applying the decreases one after the other. £15,750 comes from applying only the second year's 10% decrease to the original price, forgetting the first year's decrease entirely, £17,500 × 0.90 = £15,750. £15,400 comes from applying only the first year's 12% decrease and stopping there, forgetting to apply the second year's decrease at all.
- (b) 80% — Percentage = (40 ÷ 50) × 100 = 80%.
- (b) £45 — Since 2/3 of the amount is £30, one third is £30 ÷ 2 = £15, and the whole amount is three thirds: £15 × 3 = £45. Applying the fraction forwards to £30 instead of reversing it, £30 × 2/3 = £20, treats the given amount as the whole rather than as two thirds of it. Finding one third correctly as £15 but forgetting to multiply by 3 to get the whole amount leaves £15 as the final answer. Reading £30 as one third of the amount rather than as two thirds, and so multiplying straight by 3, gives £30 × 3 = £90.
- (d) 25% — Method: percentage decrease = decrease ÷ original amount × 100. Working: the reduction is £60 − £45 = £15, and 15 ÷ 60 = 0.25, so 0.25 × 100 = 25%. Answer: 25%. The distractors: 15% comes from quoting the £15 reduction as though pounds and per cent were the same thing; 33% comes from dividing the £15 by the new price £45 instead of by the original £60, which gives 33% to the nearest per cent; 75% is the new price written as a percentage of the old one, which is what is still paid rather than what has been taken off.
- (b) 3.5 — Method: work out each percentage of its number separately, then subtract the smaller result from the larger one. Working: 45% of 70 = 31.5, and 35% of 80 = 28, so the difference is 31.5 − 28 = 3.5. Answer: 3.5. 11.5 comes from pairing the percentages with the wrong numbers, working out 35% of 70 = 24.5 and 45% of 80 = 36, and finding their difference. 59.5 comes from adding the two correct results, 31.5 + 28, instead of subtracting them. 10 comes from simply subtracting the two percentages themselves, 45 − 35, without applying them to the numbers at all.
- (b) £300 — The increased price is 110% of the original, so the original price = £330 ÷ 1.1 = £300. A candidate who finds 10% of £330 and subtracts it, wrongly treating £330 as the original, gets £330 − £33 = £297. A candidate who adds 10% of £330 again instead of reversing the increase gets £330 + £33 = £363. A candidate who divides by 0.1 instead of 1.1 gets £3,300.
What is on this worksheet?
The sheet holds 22 questions drawn from the MathsUK bank — the content areas covered: Ratio, proportion and rates of change, Number (statements R9, R16, N12). It is pitched at GCSE Foundation and takes about 40 minutes to work through in full. It is built for independent practice, with full answers at the end for self-marking.
How to use the sheet well
- Print it or open it on screen — both work. Printing is A4; the screen view fits phones and tablets.
- Do all 22 questions before checking — about 40 minutes is the guide, but there is no time pressure.
- Check the answers — press “Show answers” or print the answer page separately.
- Redo the questions you got wrong — twice as effective as doing 22 fresh ones.
- “New questions” — builds a fresh sheet on the same statements, so you can practise again without repeats.
Why this sheet helps
MathsUK worksheets use questions graded by difficulty and a fair spread of correct-answer positions (the answer is not always (a)) — so the student really has to think about each question rather than guess a pattern. Every question is tagged to a DfE content statement and checked before it enters the bank. The answers come with a step-by-step explanation, not just a value — so a wrong answer becomes a lesson.
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