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 laptop is bought for £600. Its value decreases by 20% after 1 year. Work out the value of the laptop after 1 year.
- 2.Work out 10% of 30% of £200.
- 3.A company had 8000 employees. The number of employees decreased by 5% in the first year, and then increased by 5% in the second year. Work out the number of employees at the end of the second year, to the nearest whole number.
- 4.Grace's mean mark in her maths tests rises from 70 to 84. Work out the percentage increase in her mean mark.
- 5.Work out 35% of 180.
- 6.Work out 2/5 of 45.
- 7.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.
- 8.After a price increase of 10%, a laptop costs £330. Work out the original price.
- 9.Work out 1/2 of 1/4 of 80.
- 10.A water tank holds 80 litres when full. It currently contains 60 litres. Work out what fraction of the tank is empty.
- 11.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.
- 12.Work out 50% of 60.
- 13.A car is bought for £9000. Its value decreases by 8% each year. Work out its value after 2 years.
- 14.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.
- 15.A box holds 80 chocolates. 75% of them are milk chocolates. Work out how many milk chocolates are in the box.
- 16.The value of a rare coin increases by 12% each year. The coin is currently worth £270. Work out the value of the coin after 2 years, giving your answer to the nearest penny.
- 17.Work out 3/8 of 96.
- 18.A laptop costs £800 when new. Its value decreases by 25% of its value at the start of each year. Work out how much value the laptop loses in the second year.
- 19.A savings account pays 3.5% compound interest each year. Bilal invests £1200. Work out how much interest, in total, he earns after 2 years, to the nearest penny.
- 20.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 21.The value of a motorbike falls by 12% each year. The motorbike is worth £3200 now. Write down the calculation that gives its value after 3 years.
- 22.A jumper is reduced by 15% in a sale to a price of £42.50. Work out the original price.
Answer key
- (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) £6 — First find 30% of £200, which is £60, then find 10% of that: £60 × 0.1 = £6. Adding the two percentages together instead of applying them one after the other, 10% + 30% = 40%, and finding 40% of £200 gives £80. Finding 30% of £200 = £60 correctly but stopping before applying the second percentage leaves £60 as the final answer. Finding only 10% of the original £200, ignoring the 30% entirely, gives £20.
- (b) 7980 — After the first year: 8000 × 0.95 = 7600. After the second year: 7600 × 1.05 = 7980. 8000 comes from assuming a 5% decrease followed by a 5% increase returns exactly to the starting number — it does not, because the increase acts on the smaller, already-reduced number. 8400 comes from applying only the second year's 5% increase to the original number: 8000 × 1.05 = 8400. 7600 comes from applying only the first year's 5% decrease and stopping there, without applying the second year's increase.
- (a) 20% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is 84 − 70 = 14 marks, and 14 ÷ 70 = 0.2, so 0.2 × 100 = 20. Answer: an increase of 20%. The distractors: 14% comes from quoting the 14 mark increase as though marks and per cent were the same thing; 17% comes from dividing the 14 by the new mean 84 instead of by the original 70, which gives 17% to the nearest per cent; 120% is the new mean written as a percentage of the old one, which is the whole of the new mean rather than the increase.
- (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.
- (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.
- (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.
- (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.
- (b) 10 — First find 1/4 of 80, which is 20, then find 1/2 of that: 20 ÷ 2 = 10. Adding the two fractions together instead of applying them one after the other, 1/2 + 1/4 = 3/4, and finding 3/4 of 80 gives 60. Finding 1/4 of 80 = 20 correctly but stopping before applying the second fraction leaves 20 as the final answer. Finding 1/2 of 80 = 40 first but forgetting to then find 1/4 of that leaves 40 as the final answer.
- (a) 1/4 — The empty part of the tank is 80 − 60 = 20 litres. As a fraction of the full capacity, this is 20/80, which simplifies to 1/4. Finding the fraction of the tank that is FULL instead of empty, 60/80, simplifies to 3/4 — the wrong quantity for the question asked. Writing the empty amount over the amount remaining instead of over the full capacity, 20/60, simplifies to 1/3. Comparing the empty amount to 100 instead of to the tank's actual capacity of 80, 20/100, gives 1/5.
- (b) 80% — Percentage = (40 ÷ 50) × 100 = 80%.
- (d) 30 — Method: 50% is one half, so 50% of a quantity is the quantity divided by 2. Working: 60 ÷ 2 = 30. Answer: 30. The distractors: 120 comes from multiplying by 2 instead of dividing; 3000 comes from multiplying by 50 without dividing by 100; 6 comes from finding 10% instead of 50%.
- (c) £7617.60 — To decrease by 8% each year, multiply by 0.92 (100% − 8%) twice. £9000 × 0.92 × 0.92 = £7617.60. £7560.00 comes from treating the two 8% decreases as a single flat 16% decrease applied once instead of compounding: £9000 × 0.84 = £7560.00. £8280.00 comes from applying the 8% decrease only once, for 1 year instead of 2: £9000 × 0.92 = £8280.00. £10497.60 comes from multiplying by 1.08 twice, increasing the value instead of decreasing it: £9000 × 1.08 × 1.08 = £10497.60.
- (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.
- (d) 60 — Method: 75% is three quarters, so divide by 4 to find one quarter and then multiply by 3. Working: 80 ÷ 4 = 20 for one quarter, and 3 × 20 = 60. Answer: 60 milk chocolates. The distractors: 20 is one quarter of 80, which is 25% and not 75%; 55 comes from subtracting 25 from 80 rather than taking 25% of 80 away from it; 75 comes from quoting the percentage itself as a number of chocolates.
- (c) £338.69 — To increase by 12% each year, multiply by 1.12 twice. £270 × 1.12 × 1.12 = £338.688, which rounds to £338.69 (nearest penny, since the third decimal place is 8). £334.80 comes from treating the two 12% increases as a single flat 24% increase applied once instead of compounding: £270 × 1.24 = £334.80. £302.40 comes from applying the 12% increase only once, for 1 year instead of 2: £270 × 1.12 = £302.40. £338.68 comes from rounding £338.688 down to the nearest penny instead of up.
- (d) 36 — To find 3/8 of 96, divide by the denominator and multiply by the numerator: 96 ÷ 8 = 12, and 12 × 3 = 36. Dividing 96 by the numerator instead of the denominator, 96 ÷ 3 = 32, uses the wrong part of the fraction. Using 3/4 instead of 3/8, perhaps misreading the denominator, gives 96 × 3/4 = 72. Finding 96 ÷ 8 = 12 correctly but forgetting to multiply by the numerator 3 leaves just 12 as the final answer.
- (a) £150 — Value after year 1: £800 × 0.75 = £600. Value after year 2: £600 × 0.75 = £450. The loss during the second year alone is £600 − £450 = £150. £450 comes from giving the value remaining after 2 years, not the amount lost during the second year. £200 comes from working out the loss during the first year instead of the second: £800 − £600 = £200. £350 comes from working out the total loss over both years instead of just the second year's loss: £800 − £450 = £350.
- (a) £85.47 — Value after 2 years: £1200 × 1.035 × 1.035 = £1285.47 (nearest penny). Interest earned = £1285.47 − £1200 = £85.47. £1285.47 is the total value of the account, not the interest earned on top of the original £1200. £84.00 comes from using simple interest instead of compound interest: £1200 × 0.035 × 2 = £84.00. £42.00 comes from working out only the first year's interest and stopping there: £1200 × 0.035 = £42.00.
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (d) 3200 × 0.88³ — A fall of 12% leaves 88% of the value, because 100 − 12 = 88, and 88% written as a decimal multiplier is 0.88. Decay repeats that multiplier once for each year, so over 3 years it is applied three times: 0.88 × 0.88 × 0.88, which is written 0.88³. The calculation is therefore 3200 × 0.88³. Adding the percentages to make a single fall of 36% would be wrong, because each year's fall is taken from a smaller value than the year before.
- (d) £50.00 — The sale price is 85% of the original, so the original price = £42.50 ÷ 0.85 = £50.00. 15% of £42.50 is £6.375. A candidate who finds 15% of £42.50 and subtracts it from the sale price gets £42.50 − £6.375 = £36.125, which is £36.13 to the nearest penny. A candidate who adds 15% of £42.50 instead of reversing the decrease gets £42.50 + £6.375 = £48.875, which is £48.88 to the nearest penny. A candidate who divides by 0.15 instead of 0.85 gets £283.33.
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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