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 washing machine costs £320 before VAT. VAT is charged at 20%. Work out the total price including VAT.
- 2.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.
- 3.A sponsored walk raised £350 for charity. 20% of the money raised is spent on equipment. Work out how much is spent on equipment.
- 4.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.
- 5.After a price increase of 10%, a laptop costs £330. Work out the original price.
- 6.£3000 is invested in a savings account that pays 4% compound interest each year. Work out the value of the investment after 2 years.
- 7.Write 1/5 as a percentage.
- 8.Work out 15% of £40, using 10% and 5%.
- 9.A laptop is bought for £600. Its value decreases by 20% after 1 year. Work out the value of the laptop after 1 year.
- 10.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 11.Work out 10% of 30% of £200.
- 12.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.
- 13.A recipe uses 160 g of flour. Sam wants to increase the amount by 1/4. Work out the new amount of flour.
- 14.A jumper is reduced by 15% in a sale to a price of £42.50. Work out the original price.
- 15.A company's turnover this year is £180,000. Last year's turnover was £120,000. Write down this year's turnover as a percentage of last year's turnover.
- 16.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.
- 17.A box holds 140 pens. 25% of the pens are red. Work out how many of the pens are red.
- 18.A water tank holds 80 litres when full. It currently contains 60 litres. Work out what fraction of the tank is empty.
- 19.Work out 50% of 60.
- 20.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.
- 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.Grace's mean mark in her maths tests rises from 70 to 84. Work out the percentage increase in her mean mark.
Answer key
- (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.
- (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.
- (c) £70 — Method: 20% of an amount is 20/100 of it; a reliable route is to find 10% by dividing by 10 and then double it. Working: 10% of £350 is £350 ÷ 10 = £35, and 20% is twice as much, £35 × 2 = £70. Answer: £70. The distractors: £35 comes from finding 10% and stopping there; £17.50 comes from reading 20% as one twentieth and working out £350 ÷ 20 = £17.50; £280 comes from taking 20% off the money raised rather than finding 20% of it, giving £350 ÷ 5 = £70 and then £350 − £70 = £280.
- (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.
- (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) £3244.80 — A 4% increase each year means the value becomes 100% + 4% = 104% of the previous year's value, and 104% = 1.04, so the multiplier is 1.04. Multiply by 1.04 for each year: £3000 × 1.04 × 1.04 = £3244.80. (£3240.00 comes from using simple interest instead of compound interest: £3000 + 2 × (£3000 × 0.04) = £3240.00. £3120.00 is the value after only 1 year, £3000 × 1.04, forgetting to apply the second year's growth. £244.80 is the interest earned, £3244.80 − £3000, not the total value of the investment.)
- (a) 20% — Method: a percentage is a number of parts per hundred, so divide the numerator by the denominator and multiply by 100. Working: 1 ÷ 5 = 0.2, and 0.2 × 100 = 20; the same result comes from scaling 1/5 to 20/100. Answer: 20%. The distractors: 5% comes from reading the denominator 5 straight off as the percentage; 25% comes from confusing 1/5 with 1/4, the fraction met most often in percentage work; 50% comes from a place value slip in the division, writing 1 ÷ 5 as 0.5 instead of 0.2.
- (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) £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.
- (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.
- (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.
- (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.
- (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) £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.
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (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.
- (a) 35 — Method: 25% is 25/100, which cancels to 1/4, so finding 25% of an amount means dividing it by 4. Working: 25% = 25/100 = 1/4, and 140 ÷ 4 = 35. Answer: 35 pens. The distractors: 70 comes from halving instead of quartering, confusing 25% with 50%; 105 comes from working out the pens that are not red, which is 75% of 140, instead of the pens that are; 25 comes from ignoring the percent sign and reading the 25% as a count of 25 pens.
- (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.
- (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%.
- (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.
- (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.
- (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.
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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