Hard GCSE Reverse percentages Questions

Challenging, exam-style GCSE Reverse percentages questions with worked solutions. Stretch yourself on the hardest successive reverse percentages, combined multiplier, round-trip check, reverse percentage problems.

successive reverse percentagescombined multiplierround-trip checkreverse percentagedepreciationrepeated multiplier
GCSE Foundation34 questionsStep-by-step solutions
Question 1
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\%
Question 2
5 markschallenging
A recipe is scaled down so that every ingredient is reduced by 40%. The scaled-down recipe uses 0.36 kg of flour. Work out how much flour the original recipe used. Give your answer in grams.
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Worked solution

  1. Read what the 0.36 kg is

    ?→−40%0.36 kg? \xrightarrow{-40\%} 0.36\text{ kg}

    0.36 kg is the amount AFTER the reduction. The 40% was 40% of the ORIGINAL amount of flour, so this is a reverse percentage problem.

  2. Decide: forward or reverse?

    original×0.6=0.36 kg\text{original} \times 0.6 = 0.36\text{ kg}

    We are given the final amount and asked for the original, so we must divide by the multiplier, not multiply by it.

  3. Name the multiplier

    100%−40%=60%=0.6100\% - 40\% = 60\% = 0.6

    A 40% reduction leaves 60% of the flour, so the original is multiplied by 0.6.

  4. Divide to find the original amount

    original=0.36÷0.6=0.6 kg\text{original} = 0.36 \div 0.6 = 0.6\text{ kg}

    The original recipe used 0.6 kg of flour.

  5. Round-trip check

    0.6×0.6=0.36 kg0.6 \times 0.6 = 0.36\text{ kg}

    Reducing 0.6 kg by 40% gives 0.36 kg, exactly as the question says. Correct. (Notice that 0.6×0.6=0.360.6 \times 0.6 = 0.36 is a coincidence of these numbers, not a rule.)

  6. Guard against the classic error

    0.36×1.4=0.504 kg≠0.6 kg0.36 \times 1.4 = 0.504\text{ kg} \ne 0.6\text{ kg}

    Adding 40% back on to 0.36 kg gives 0.504 kg, which fails the check: reducing 0.504 kg by 40% gives 0.3024 kg, not 0.36 kg. The 40% removed was 40% of the larger original.

  7. Recall the unit conversion

    1 kg=1000 g1\text{ kg} = 1000\text{ g}

    The answer must be in grams, so a conversion is needed.

  8. Convert the original amount to grams

    0.6×1000=600 g0.6 \times 1000 = 600\text{ g}

    The original recipe used 600 g of flour.

  9. Check the round trip in grams as well

    600×0.6=360 g600 \times 0.6 = 360\text{ g}

    0.36 kg is 360 g, and 60% of 600 g is 360 g. The check works in either unit, which confirms the conversion was done correctly.

  10. Find the amount of flour removed

    600−360=240 g600 - 360 = 240\text{ g}

    The scaling down removed 240 g of flour.

  11. Check that removal as a percentage

    240600×100=40%\frac{240}{600} \times 100 = 40\%

    240 g out of the original 600 g is 40%, matching the reduction in the question.

  12. Look at the wrong base once more

    240360×100=66.6‾%\frac{240}{360} \times 100 = 66.\overline{6}\%

    Comparing the 240 g removed with the NEW amount of 360 g gives 66.7%, which is not the reduction. Percentage change is always measured against the original.

  13. Sense check the size of the answer

    600 g>360 g600\text{ g} > 360\text{ g}

    The original recipe must use more flour than the scaled-down one, and it does.

  14. Note a quick check with fractions

    0.6=35,0.36÷35=0.36×53=0.60.6 = \frac{3}{5}, \quad 0.36 \div \frac{3}{5} = 0.36 \times \frac{5}{3} = 0.6

    A 40% reduction leaves three fifths, so dividing by three fifths (multiplying by five thirds) gives the original — the same answer by an exact fraction method.

  15. State the answer

    600 g600\text{ g}

    The original recipe used 600 g of flour.

Answer
600 g600\text{ g}
Question 3
5 markschallenging
A charity keeps 15% of all money donated to cover its costs, and passes the rest on to good causes. In one week the charity passed on £2550. Work out how much of that week's donations the charity kept for its costs.
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Worked solution

  1. Work out what percentage of the donations was passed on

    100%−15%=85%100\% - 15\% = 85\%

    If 15% is kept, then 85% is passed on. So the £2550 is 85% of the total donated — not 100% of it.

  2. Recognise a reverse percentage problem

    ?→−15%£2550? \xrightarrow{-15\%} \pounds 2550

    We know the amount AFTER the 15% was taken out, and we want the total before. That is a reverse percentage.

  3. Name the multiplier

    85%=0.8585\% = 0.85

    Keeping 15% and passing on the rest multiplies the total donations by 0.85.

  4. Write the relationship

    total donated×0.85=£2550\text{total donated} \times 0.85 = \pounds 2550

    The total donated, multiplied by 0.85, gives the £2550 passed on.

  5. Divide to find the total donated

    total donated=£2550÷0.85=£3000\text{total donated} = \pounds 2550 \div 0.85 = \pounds 3000

    The total donated that week was £3000.

  6. Round-trip check the total

    £3000×0.85=£2550\pounds 3000 \times 0.85 = \pounds 2550

    85% of £3000 is £2550, exactly the amount passed on. Correct.

  7. Answer the question that was asked

    £3000−£2550=£450\pounds 3000 - \pounds 2550 = \pounds 450

    The charity kept £450 for its costs.

  8. Check the amount kept as a percentage

    15% of £3000=£45015\% \text{ of } \pounds 3000 = \pounds 450

    15% of the £3000 total is £450, which agrees. The costs are a percentage of the TOTAL donated.

  9. Guard against the classic error

    15% of £2550=£382.50≠£45015\% \text{ of } \pounds 2550 = \pounds 382.50 \ne \pounds 450

    Taking 15% of the £2550 passed on gives £382.50 — the wrong answer, because the 15% was 15% of the total donations, not 15% of the money passed on.

  10. Show why that error fails the check

    £2550+£382.50=£2932.50≠£3000\pounds 2550 + \pounds 382.50 = \pounds 2932.50 \ne \pounds 3000

    If the charity had kept only £382.50, the total would have been £2932.50, and 15% of that is £439.88 — not £382.50. The figures do not fit together, so the method is wrong.

  11. Check the split adds up

    £450+£2550=£3000\pounds 450 + \pounds 2550 = \pounds 3000

    The money kept plus the money passed on equals the total donated, as it must.

  12. Check the ratio of the split

    450:2550=15:85=3:17450 : 2550 = 15 : 85 = 3 : 17

    The kept and passed-on amounts are in the ratio 15 : 85, which simplifies to 3 : 17 — exactly the split the charity promised.

  13. Sense check the size of the answer

    £450<£2550\pounds 450 < \pounds 2550

    The charity keeps much less than it passes on, which fits keeping only 15%.

  14. Note the shortcut for next time

    kept=£2550×1585\text{kept} = \pounds 2550 \times \frac{15}{85}

    Because kept : passed on is 15 : 85, the amount kept is 15/85 of £2550, which is £450 again — a quicker route, but only once the reverse step is understood.

  15. State the answer

    £450\pounds 450

    The charity kept £450 for its costs, out of £3000 donated.

Answer
£450\pounds 450
Question 4
5 markschallenging
Meera received a 5% pay rise, and a year later a further 4% pay rise. She now earns £27300 per year. Work out her salary before the two rises.
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Worked solution

  1. Read the chain of changes

    ?  →+5%  ?  →+4%  £27300? \; \xrightarrow{+5\%} \; ? \; \xrightarrow{+4\%} \; \pounds 27300

    £27300 is the amount after every change has happened. The original is unknown, and each percentage was a percentage of the amount at the time — never a percentage of £27300. So we must work backwards.

  2. Decide: forward or reverse?

    final=original×multipliers\text{final} = \text{original} \times \text{multipliers}

    A forward problem gives you the original and asks for the final amount, so you multiply. Here we are given the final amount and asked for the original, so we must divide. Getting this decision right is the whole skill.

  3. Name the first multiplier

    100%+5%=105%=1.05100\% + 5\% = 105\% = 1.05

    A 5% increase multiplies the original by 1.05.

  4. Name the second multiplier

    100%+4%=104%=1.04100\% + 4\% = 104\% = 1.04

    A 4% increase multiplies the original by 1.04.

  5. Combine the multipliers

    1.05×1.04=1.0921.05 \times 1.04 = 1.092

    One multiplier after another is the same as multiplying by their product.

  6. Say what the overall multiplier means

    overall multiplier=1.092\text{overall multiplier} = 1.092

    The overall multiplier is 1.092, so Meera now earns 109.2% of her old salary — a 9.2% rise overall, not the 9% you get by adding 5% and 4%. The extra 0.2% is the 4% rise being applied to the money the 5% rise had already added.

  7. Write the relationship

    original×1.092=£27300\text{original} \times 1.092 = \pounds 27300

    The original, multiplied by 1.092, gives £27300.

  8. Divide to find the original

    original=£27300÷1.092=£25000\text{original} = \pounds 27300 \div 1.092 = \pounds 25000

    Dividing by the overall multiplier undoes every change at once. The original was £25000.

  9. Round-trip check, change 1

    £25000×1.05=£26250\pounds 25000 \times 1.05 = \pounds 26250

    Applying the 5% increase to £25000 gives £26250.

  10. Round-trip check, change 2

    £26250×1.04=£27300\pounds 26250 \times 1.04 = \pounds 27300

    Applying the 4% increase to £26250 gives £27300. This is the amount given in the question, so the original of £25000 is confirmed.

  11. Guard against the classic error

    5%+4%=9%⇒£27300÷1.09≈£25045.87≠£250005\% + 4\% = 9\% \Rightarrow \pounds 27300 \div 1.09 \approx \pounds 25045.87 \ne \pounds 25000

    Adding the percentages gives 9% and a salary of about £25045.87, which fails the round-trip check: £25045.87 × 1.092 is about £27350.09, not £27300. The second rise was 4% of the already-increased salary.

  12. State the overall percentage rise

    1.092=109.2%⇒9.2% rise1.092 = 109.2\% \Rightarrow 9.2\% \text{ rise}

    The two rises together are worth 9.2%, not 9%. The extra 0.2% is the 4% rise being paid on the money the 5% rise had already added.

  13. Check the rise in money

    £27300−£25000=£2300\pounds 27300 - \pounds 25000 = \pounds 2300

    Meera is £2300 a year better off. As a percentage of her old salary that is 2300/25000=9.22300/25000 = 9.2%, which matches the combined multiplier exactly.

  14. Check each rise separately

    £25000×1.05=£26250,4% of £26250=£1050\pounds 25000 \times 1.05 = \pounds 26250, \quad 4\% \text{ of } \pounds 26250 = \pounds 1050

    The first rise added £1250 and the second added £1050 — a total of £2300, as above. The second rise is worth more than 4% of £25000 (£1000) because it was calculated on the higher salary.

  15. State the answer

    £25000\pounds 25000

    Meera earned £25000 before the two rises.

Answer
£25000\pounds 25000
Question 5
6 markschallenging
A student says: "To undo a 20% increase, you just decrease by 20%." A price is £120 after a 20% increase. Show that the student is wrong, and work out the percentage decrease that really would undo a 20% increase.
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Worked solution

  1. Find the true original price

    original×1.2=£120\text{original} \times 1.2 = \pounds 120

    The £120 is the price AFTER the 20% increase, so this is a reverse percentage problem.

  2. Divide by the multiplier

    £120÷1.2=£100\pounds 120 \div 1.2 = \pounds 100

    The price before the increase was £100.

  3. Round-trip check the original

    £100×1.2=£120\pounds 100 \times 1.2 = \pounds 120

    20% of £100 is £20, and £100 + £20 = £120. So £100 is definitely the original.

  4. Now test the student's method

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

    Decreasing £120 by 20% gives £96, not £100. The student's method has lost £4.

  5. Say why the student's method fails

    20% of £100=£20,20% of £120=£2420\% \text{ of } \pounds 100 = \pounds 20, \quad 20\% \text{ of } \pounds 120 = \pounds 24

    The increase added 20% of £100, which is £20. The student then removes 20% of £120, which is £24 — a bigger amount, because it is a percentage of a bigger number. Taking away more than was added lands below the start.

  6. Look at the multipliers

    1.2×0.8=0.96≠11.2 \times 0.8 = 0.96 \ne 1

    To undo a change, the two multipliers must multiply to 1. Here they give 0.96, so the price ends at 96% of the original — a 4% fall overall, not back to the start.

  7. Work out the multiplier that DOES undo the increase

    11.2=0.8333…=56\frac{1}{1.2} = 0.8333\ldots = \frac{5}{6}

    The undoing multiplier is the reciprocal of 1.2, which is five sixths.

  8. Turn that multiplier into a percentage decrease

    1−56=161 - \frac{5}{6} = \frac{1}{6}

    A multiplier of five sixths means removing one sixth.

  9. Write the percentage decrease

    16×100=16.6‾%≈16.7%\frac{1}{6} \times 100 = 16.\overline{6}\% \approx 16.7\%

    A decrease of one sixth, about 16.7%, is what really undoes a 20% increase.

  10. Check the correct decrease

    £120×56=£100\pounds 120 \times \frac{5}{6} = \pounds 100

    Five sixths of £120 is exactly £100, the true original. The round-trip check passes.

  11. Check with the multipliers

    1.2×56=11.2 \times \frac{5}{6} = 1

    The two multipliers now multiply to exactly 1, which is what "undoing" means.

  12. See the general rule

    undo ×m by ÷m\text{undo } \times m \text{ by } \div m

    You undo a percentage change by DIVIDING by its multiplier, never by applying the opposite percentage. The opposite percentage is measured from the wrong base.

  13. Note the size of the error

    £100−£96=£4\pounds 100 - \pounds 96 = \pounds 4

    The student's method is £4 out on this price — and the error grows with the size of the percentage.

  14. Note when the student's idea nearly works

    small percentages⇒small error\text{small percentages} \Rightarrow \text{small error}

    For a 1% increase, undoing with a 1% decrease is out by only 0.01%. That is why the mistake feels reasonable — but it is still wrong.

  15. Choose the correct statement

    16.6‾%16.\overline{6}\%

    The student is wrong: £120 decreased by 20% is £96, not £100. The decrease that undoes a 20% increase is one sixth, about 16.7%.

Answer
16.6‾% (one sixth)16.\overline{6}\% \text{ (one sixth)}

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