GCSE Reverse percentages Practice Questions

Free GCSE Reverse percentages practice questions with full step-by-step worked solutions. Covers reverse percentage, decrease multiplier, sale price, quarter off. Practise exam-style problems and check your method.

reverse percentagedecrease multipliersale pricequarter offincrease multiplierVAT
GCSE Foundation70 questionsStep-by-step solutions
Question 1
2 markseasy
A coat is reduced by 20% in a sale. Its sale price is £48. Work out the price of the coat before the sale.
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Worked solution

  1. Decide what the given amount stands for

    £48=80% of the original\pounds 48 = 80\% \text{ of the original}

    The 20% was taken off the original, not taken off £48. So £48 is 80% of the original — this is a reverse percentage problem, and the original is the number we have to find.

  2. Divide by the multiplier

    original=£48÷0.8=£60\text{original} = \pounds 48 \div 0.8 = \pounds 60

    80% as a decimal multiplier is 0.8. The original was multiplied by 0.8 to get £48, so dividing by 0.8 undoes it and gives £60.

  3. Check by working forwards

    £60×0.8=£48\pounds 60 \times 0.8 = \pounds 48

    The round-trip check: £60 with the change applied gives £48, exactly as the question says. (Doing £48 ×\times 1.2 = £57.60 instead would be wrong — the 20% was a percentage of the original, not of £48.)

Answer
£60\pounds 60
Question 2
2 markseasy
A television is reduced by 35% in a sale. The sale price is £52. Work out the price of the television before the sale.
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Worked solution

  1. Decide what the given amount stands for

    £52=65% of the original\pounds 52 = 65\% \text{ of the original}

    The 35% was taken off the original, not taken off £52. So £52 is 65% of the original — this is a reverse percentage problem, and the original is the number we have to find.

  2. Divide by the multiplier

    original=£52÷0.65=£80\text{original} = \pounds 52 \div 0.65 = \pounds 80

    65% as a decimal multiplier is 0.65. The original was multiplied by 0.65 to get £52, so dividing by 0.65 undoes it and gives £80.

  3. Check by working forwards

    £80×0.65=£52\pounds 80 \times 0.65 = \pounds 52

    The round-trip check: £80 with the change applied gives £52, exactly as the question says. (Doing £52 ×\times 1.35 = £70.20 instead would be wrong — the 35% was a percentage of the original, not of £52.)

Answer
£80\pounds 80
Question 3
2 marksintermediate
A shirt costs £90 in a sale after a 25% discount. A student works out the original price as 90 × 1.25 = £112.50. Explain why the student is wrong and give the correct original price.
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Worked solution

  1. See what the student has done

    £90×1.25=£112.50\pounds 90 \times 1.25 = \pounds 112.50

    The student has added 25% of £90 (which is £22.50) on to £90. That treats the discount as 25% of the SALE price.

  2. See what the discount really was a percentage of

    25% of the original, not 25% of £9025\% \text{ of the original, not } 25\% \text{ of } \pounds 90

    The shop took 25% off the ORIGINAL price. The original is bigger than £90, so 25% of it is bigger than £22.50. The student used the wrong base.

  3. Write the correct relationship

    original×0.75=£90\text{original} \times 0.75 = \pounds 90

    After a 25% discount you pay 75% of the original, so £90 is 75% of the original.

  4. Divide to find the original

    original=£90÷0.75=£120\text{original} = \pounds 90 \div 0.75 = \pounds 120

    The original price was £120.

  5. Round-trip check the correct answer

    £120×0.75=£90\pounds 120 \times 0.75 = \pounds 90

    25% of £120 is £30, and £120 − £30 = £90 — the sale price in the question. The answer checks out.

  6. Round-trip check the student's answer to show it fails

    £112.50×0.75=£84.375\pounds 112.50 \times 0.75 = \pounds 84.375

    Taking 25% off the student's £112.50 gives £84.38 (to the nearest penny), not £90. The student's price fails the check, so it cannot be the original.

Answer
£120\pounds 120
Question 4
4 markshard
A dress costs £84 in a sale after a 30% discount. When the sale ends, the shop increases all sale prices by 30%. Work out how much less the dress now costs than it did before the sale.
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Worked solution

  1. Recognise a reverse percentage problem

    ?→−30%£84? \xrightarrow{-30\%} \pounds 84

    The £84 is the price AFTER the discount, and the price before the sale is unknown. The 30% was 30% of that unknown original, so we must work backwards.

  2. Name the multiplier

    100%−30%=70%=0.7100\% - 30\% = 70\% = 0.7

    A 30% discount multiplies the original price by 0.7.

  3. Divide to find the price before the sale

    original=£84÷0.7=£120\text{original} = \pounds 84 \div 0.7 = \pounds 120

    The dress cost £120 before the sale.

  4. Round-trip check the original price

    £120×0.7=£84\pounds 120 \times 0.7 = \pounds 84

    30% of £120 is £36, and £120 − £36 = £84 — the sale price in the question. Correct.

  5. Now go forwards for the second part

    £84×1.3\pounds 84 \times 1.3

    The shop increases the SALE price by 30%. This part is a forward calculation: the starting amount (£84) is known, so multiply by 1.3. Notice how the same question contains one reverse step and one forward step — deciding which is which is the skill being tested.

  6. Work out the new price

    £84×1.3=£109.20\pounds 84 \times 1.3 = \pounds 109.20

    After the sale ends, the dress costs £109.20.

  7. Compare with the price before the sale

    £109.20<£120\pounds 109.20 < \pounds 120

    The dress does NOT return to £120. A 30% rise on the reduced price adds less money than the 30% cut took away, because it is 30% of a smaller amount.

  8. Show the two changes in money

    30% of £120=£36,30% of £84=£25.2030\% \text{ of } \pounds 120 = \pounds 36, \quad 30\% \text{ of } \pounds 84 = \pounds 25.20

    The sale took off £36 but the increase only put back £25.20. The difference between those two amounts is exactly what is missing.

  9. Answer the question that was asked

    £120−£109.20=£10.80\pounds 120 - \pounds 109.20 = \pounds 10.80

    The dress now costs £10.80 less than it did before the sale.

  10. Check with the combined multiplier

    0.7×1.3=0.91,£120×0.91=£109.200.7 \times 1.3 = 0.91, \quad \pounds 120 \times 0.91 = \pounds 109.20

    The two changes together multiply the original by 0.91 — a 9% fall overall, not a return to the start. And 9% of £120 is exactly £10.80, confirming the answer.

Answer
£10.80\pounds 10.80
Question 5
6 markschallenging
A shop advertises "25% OFF, THEN A FURTHER 20% OFF AT THE TILL". A customer pays £96 for a coat and says "so I have saved 45%". Work out the actual percentage saved on the original price.
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Worked solution

  1. Read what the £96 is

    ?→−25%?→−20%£96? \xrightarrow{-25\%} ? \xrightarrow{-20\%} \pounds 96

    £96 is the price after BOTH discounts. The original price is unknown, and neither discount was a percentage of £96, so this is a reverse percentage problem.

  2. Name the first multiplier

    100%−25%=75%=0.75100\% - 25\% = 75\% = 0.75

    The first discount multiplies the original price by 0.75.

  3. Name the second multiplier

    100%−20%=80%=0.8100\% - 20\% = 80\% = 0.8

    The further 20% is taken off the already-reduced price, so it multiplies that price by 0.8.

  4. Combine the multipliers

    0.75×0.8=0.60.75 \times 0.8 = 0.6

    Applying one multiplier and then the other is the same as multiplying by 0.6.

  5. Say what the combined multiplier means

    0.6=60%0.6 = 60\%

    The customer pays 60% of the original price, so the saving is 40% — not the 45% the customer claims, and not the 25+20=4525 + 20 = 45% the advert seems to promise.

  6. Write the relationship

    original×0.6=£96\text{original} \times 0.6 = \pounds 96

    The original price, multiplied by 0.6, gives the £96 paid.

  7. Divide to find the original price

    original=£96÷0.6=£160\text{original} = \pounds 96 \div 0.6 = \pounds 160

    The coat was £160 before any discount.

  8. Round-trip check the first discount

    £160×0.75=£120\pounds 160 \times 0.75 = \pounds 120

    Taking 25% off £160 gives £120.

  9. Round-trip check the second discount

    £120×0.8=£96\pounds 120 \times 0.8 = \pounds 96

    Taking a further 20% off £120 gives £96, exactly what the customer paid. The original of £160 is confirmed.

  10. Find the saving in money

    £160−£96=£64\pounds 160 - \pounds 96 = \pounds 64

    The customer saved £64 altogether.

  11. Work out the actual percentage saved

    64160×100=40%\frac{64}{160} \times 100 = 40\%

    The saving is 40% of the original price. The customer's claim of 45% is wrong.

  12. Explain why the customer is wrong

    25% of £160=£40,20% of £120=£2425\% \text{ of } \pounds 160 = \pounds 40, \quad 20\% \text{ of } \pounds 120 = \pounds 24

    The first discount saved £40, but the second saved only £24, because it was 20% of the reduced £120 — not 20% of the original £160 (which would be £32). Adding the percentages, 25+20=4525 + 20 = 45, assumes both were taken from £160, and they were not.

  13. Check the two savings add up

    £40+£24=£64\pounds 40 + \pounds 24 = \pounds 64

    The two separate savings total £64, which matches the overall saving. Everything is consistent.

  14. Check the claimed 45% fails

    £160×0.55=£88≠£96\pounds 160 \times 0.55 = \pounds 88 \ne \pounds 96

    A genuine 45% saving on £160 would have left the customer paying £88, not £96. The customer paid £8 more than a 45% saving would give.

  15. State the answer

    40%40\%

    The actual saving is 40% of the original price of £160, not 45%.

Answer
40%40\%

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