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.
- 1.2/3 of an amount of money is £30. Work out the amount.
- 2.After a price increase of 10%, a laptop costs £330. Work out the original price.
- 3.In a test, Amelia answered 18 of the 24 questions correctly. Work out the percentage of the questions she answered correctly.
- 4.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 5.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.
- 6.Priya invests £750 in a savings account that pays simple interest. After 3 years, the account contains £840. Work out the annual rate of simple interest.
- 7.A washing machine costs £320 before VAT. VAT is charged at 20%. Work out the total price including VAT.
- 8.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.
- 9.Work out 25% of 200.
- 10.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.
- 11.A concert hall has 300 seats. 20% of the seats are in the balcony. Work out how many of the seats are in the balcony.
- 12.A car is bought for £12000. Its value falls by 15% in the first year and by 10% in each year after that. Work out the value of the car 3 years after it was bought.
- 13.A sponsored walk is 36 km long. Aisha has completed 8/12 of the walk. Work out how far she has walked.
- 14.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.
- 15.There are 2500 electric cars registered in a town. The number is predicted to increase by 6% each year. Work out the predicted number of electric cars after 3 years, to the nearest whole number.
- 16.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 17.A laptop is bought for £600. Its value decreases by 20% after 1 year. Work out the value of the laptop after 1 year.
- 18.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.
- 19.The number of members of a running club increases from 45 to 54. Work out the percentage increase.
- 20.Work out 15% of £40, using 10% and 5%.
- 21.The rent on a flat increases by 10% one year and by a further 10% the following year. Work out the overall percentage increase over the two years.
- 22.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.
Answer key
- (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.
- (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.
- (a) 75% — Method: to express one quantity as a percentage of another, divide the part by the whole and multiply by 100. Working: 18 ÷ 24 = 0.75, and 0.75 × 100 = 75. Answer: 75%. The distractors: 25% is the percentage she got wrong, 6 out of 24; 133% comes from dividing the whole by the part, 24 ÷ 18; 18% comes from writing the number of correct answers with a percent sign.
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (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) 4% — Method: find the total interest earned, share it equally across the number of years to find one year's interest, then write it as a percentage of the amount invested. Working: total interest = £840 − £750 = £90, so one year's interest is £90 ÷ 3 = £30, and £30 as a percentage of £750 is (£30 ÷ £750) × 100 = 4%. Answer: 4%. 12% comes from treating the total interest of £90 as if it were earned in a single year, (£90 ÷ £750) × 100 = 12%, forgetting to divide by 3 years. 0.04% comes from finding the correct decimal, £30 ÷ £750 = 0.04, but forgetting to multiply by 100 to convert it into a percentage. 112% comes from writing the final amount, £840, as a percentage of the amount invested, £750, without first subtracting the £750 to find the interest alone.
- (c) £384.00 — Adding 20% VAT means multiplying the price by 1.2: £320 × 1.2 = £384.00. Treating the 20% as a flat £20 rather than a percentage of the price, £320 + £20, gives £340.00. Working out the VAT amount alone, £320 × 0.2 = £64.00, and stopping there without adding it back to the original price gives just the VAT, not the total price. Misplacing the decimal point and using 2% instead of 20%, £320 × 1.02, gives £326.40.
- (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.
- (a) 50 — Method: 25% is one quarter, so 25% of a quantity is the quantity divided by 4. Working: 200 ÷ 4 = 50. Answer: 50. The distractors: 25 comes from writing the percentage itself as the answer; 100 comes from halving, which is 50% not 25%; 800 comes from multiplying by 4 instead of dividing.
- (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.
- (d) 60 — Method: a percentage acts as an operator, so finding 20% of an amount means multiplying it by 20/100, which cancels to 1/5. Working: 20% = 20/100 = 1/5, and 300 ÷ 5 = 60. Answer: 60 seats. The distractors: 15 comes from reading 20% as one twentieth and working out 300 ÷ 20 = 15; 30 comes from finding 10% of 300 and stopping there instead of doubling it; 6 comes from converting 20% to 0.02 rather than 0.2, giving 0.02 × 300 = 6.
- (d) £8262 — A fall of 15% is a multiplier of 0.85 and a fall of 10% is a multiplier of 0.9, and each multiplier acts on the value at the start of its own year. After year 1: 12000 × 0.85 = 10200. After year 2: 10200 × 0.9 = 9180. After year 3: 9180 × 0.9 = 8262. The value 3 years after the car was bought is £8262. Adding the percentages to make a single fall of 35% would be wrong, because the later falls are taken from smaller values.
- (d) 24 km — Method: a fraction acts as an operator, so finding 8/12 of a distance means dividing by the denominator and multiplying by the numerator. Working: 36 ÷ 12 = 3, so one twelfth of the walk is 3 km, and eight twelfths is 3 × 8 = 24 km. Answer: 24 km. The distractors: 12 km comes from working out the part of the walk still left, the other four twelfths, instead of the part already completed; 288 km comes from multiplying by the numerator without dividing by the denominator, giving 36 × 8 = 288; 4.5 km comes from dividing by the numerator instead of multiplying by it, giving 36 ÷ 8 = 4.5.
- (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) 2978 — A rise of 6% is a multiplier of 1.06, applied once for each year. After year 1: 2500 × 1.06 = 2650. After year 2: 2650 × 1.06 = 2809. After year 3: 2809 × 1.06 = 2977.54, which is 2978 to the nearest whole number. Multiplying by 1.18 in one go would be wrong, because the second and third years grow from larger numbers than the first.
- (c) a decrease of 25% — Method: use multipliers. An increase of 50% is × 1.5 and a decrease of 50% is × 0.5. Working: 1.5 × 0.5 = 0.75, so the final price is 75% of the original. Answer: a decrease of 25%. The distractors: no change comes from assuming +50% and −50% cancel; a decrease of 50% comes from applying only the second change; an increase of 25% has the direction wrong.
- (c) £480.00 — To decrease by 20%, multiply by 0.80 (100% − 20%). £600 × 0.80 = £480.00. £120.00 comes from working out only the decrease (£600 × 0.20) and forgetting to subtract it from the original value. £580.00 comes from subtracting 20 directly instead of 20% of £600. £720.00 comes from multiplying by 1.20, adding the percentage instead of subtracting it.
- (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) 20% — Method: percentage increase = (increase ÷ original) × 100. Working: the increase is 54 − 45 = 9, and 9 ÷ 45 = 0.2, so the percentage increase is 0.2 × 100 = 20. Answer: 20%. The distractors: 9% comes from writing the actual increase as a percentage; 16.7% comes from dividing by the new value 54 instead of the original 45; 120% is the multiplier 1.2 written as a change rather than the change itself.
- (a) £6 — 10% of £40 is £4, and 5% of £40 is half of that, £2. Adding these gives 15% of £40 = £4 + £2 = £6. Finding only the 10% part and stopping there gives £4. Finding only the 5% part and stopping there gives £2. Multiplying 40 by 15 without dividing by 100 gives £600, which treats the percentage as if it were a whole number multiplier.
- (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
- (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.
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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