Hard GCSE Percentage change Questions

Challenging, exam-style GCSE Percentage change questions with worked solutions. Stretch yourself on the hardest successive percentage change, combined multiplier, percentage profit, mark-up problems.

successive percentage changecombined multiplierpercentage profitmark-upbuilding up cost and revenuecomparing percentage changes
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
A hotel room costs £80\pounds 80 per night. In August the price rises by 40%40\%. In October the August price falls by 30%30\%. In January the October price falls by a further 25%25\%. Work out the overall percentage change from the original £80\pounds 80.
Show worked solution

Worked solution

  1. Write down the original price

    original=£80\text{original} = \pounds 80

    The overall percentage change is measured against this £80\pounds 80.

  2. August: find the multiplier

    100%+40%=140%=1.4100\% + 40\% = 140\% = 1.4

    A 40%40\% rise means multiplying by 1.41.4.

  3. August: work out the price

    £80×1.4=£112\pounds 80 \times 1.4 = \pounds 112

    The August price is £112\pounds 112.

  4. October: find the multiplier

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

    A 30%30\% fall means multiplying by 0.70.7.

  5. October: apply it to the AUGUST price

    £112×0.7=£78.40\pounds 112 \times 0.7 = \pounds 78.40

    The 30%30\% comes off £112\pounds 112 (that is £33.60\pounds 33.60), not off the original £80\pounds 80. The October price is £78.40\pounds 78.40.

  6. January: find the multiplier

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

    A further 25%25\% fall means multiplying by 0.750.75.

  7. January: apply it to the OCTOBER price

    £78.40×0.75=£58.80\pounds 78.40 \times 0.75 = \pounds 58.80

    The January price is £58.80\pounds 58.80.

  8. Multiply the three multipliers together

    1.4×0.7×0.751.4 \times 0.7 \times 0.75

    Successive percentage changes combine by multiplying their multipliers, never by adding the percentages.

  9. Work out the product in stages

    1.4×0.7=0.98,0.98×0.75=0.7351.4 \times 0.7 = 0.98, \quad 0.98 \times 0.75 = 0.735

    The single equivalent multiplier is 0.7350.735.

  10. Check the single multiplier against the price

    £80×0.735=£58.80\pounds 80 \times 0.735 = \pounds 58.80

    The single multiplier gives the same January price. ✓

  11. Compare the multiplier with 11

    0.735<10.735 < 1

    The multiplier is below 11, so the room is cheaper in January than it was originally.

  12. Turn the multiplier into a percentage change

    (0.735−1)×100=−26.5(0.735 - 1) \times 100 = -26.5

    The overall change is a 26.5%26.5\% decrease.

  13. Confirm from the prices

    80−58.8080×100=21.2080×100=26.5%\frac{80 - 58.80}{80} \times 100 = \frac{21.20}{80} \times 100 = 26.5\%

    The price has fallen by £21.20\pounds 21.20, and £21.20\pounds 21.20 out of the original £80\pounds 80 is 26.5%26.5\%. ✓

  14. Watch the common error

    +40%−30%−25%=−15%≠−26.5%+40\% - 30\% - 25\% = -15\% \ne -26.5\%

    Adding the percentages gives a 15%15\% fall, which is badly wrong. Notice too that +40%+40\% then −30%-30\% is not +10%+10\%: 1.4×0.7=0.981.4 \times 0.7 = 0.98, already a 2%2\% fall.

  15. State the overall percentage change

    26.5% decrease26.5\% \text{ decrease}

    The January price is 26.5%26.5\% below the original £80\pounds 80.

Answer
26.5% decrease26.5\% \text{ decrease}
Question 2
5 markschallenging
A tank holds 250250 litres of water. On Monday 18%18\% of the water is used. On Tuesday 25%25\% of the water that is left is used. Work out what percentage of the original 250250 litres is still in the tank.
Show worked solution

Worked solution

  1. Write down the original amount of water

    original=250 litres\text{original} = 250 \text{ litres}

    Everything at the end must be compared back to this 250250 litres.

  2. Find the multiplier for Monday

    100%−18%=82%=0.82100\% - 18\% = 82\% = 0.82

    If 18%18\% is used, 82%82\% is left, so multiply by 0.820.82.

  3. Work out the water used on Monday

    18% of 250=45 litres18\% \text{ of } 250 = 45 \text{ litres}

    4545 litres are used on Monday.

  4. Work out the water left after Monday

    250×0.82=205 litres250 \times 0.82 = 205 \text{ litres}

    250−45=205250 - 45 = 205 litres are left. ✓

  5. Read the Tuesday condition carefully

    25% of the water that is LEFT25\% \text{ of the water that is LEFT}

    Tuesday's 25%25\% is 25%25\% of the 205205 litres left, NOT 25%25\% of the original 250250 litres.

  6. Find the multiplier for Tuesday

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

    If 25%25\% of what is left is used, 75%75\% of it remains, so multiply by 0.750.75.

  7. Work out the water used on Tuesday

    25% of 205=51.25 litres25\% \text{ of } 205 = 51.25 \text{ litres}

    51.2551.25 litres are used on Tuesday — not the 62.562.5 litres that 25%25\% of 250250 would be.

  8. Work out the water left after Tuesday

    205×0.75=153.75 litres205 \times 0.75 = 153.75 \text{ litres}

    205−51.25=153.75205 - 51.25 = 153.75 litres remain in the tank.

  9. Combine the multipliers

    0.82×0.75=0.6150.82 \times 0.75 = 0.615

    The two stages together are a single multiplier of 0.6150.615.

  10. Check the combined multiplier

    250×0.615=153.75250 \times 0.615 = 153.75

    The single multiplier reproduces the 153.75153.75 litres left. ✓

  11. Read the percentage straight off the multiplier

    0.615=61.5%0.615 = 61.5\%

    A multiplier of 0.6150.615 means 61.5%61.5\% of the original water is still there.

  12. Confirm from the litres

    153.75250×100=61.5%\frac{153.75}{250} \times 100 = 61.5\%

    153.75153.75 out of the original 250250 litres is 61.5%61.5\%. ✓

  13. Work out how much has gone

    100%−61.5%=38.5%100\% - 61.5\% = 38.5\%

    So 38.5%38.5\% of the original water has been used altogether.

  14. Watch the common error

    18%+25%=43%≠38.5%18\% + 25\% = 43\% \ne 38.5\%

    Adding the percentages would suggest 43%43\% used and 57%57\% left. That is wrong, because the 25%25\% is a percentage of the reduced 205205 litres.

  15. State the percentage of the original that is left

    61.5%61.5\%

    61.5%61.5\% of the original 250250 litres is still in the tank.

Answer
61.5%61.5\%
Question 3
5 markschallenging
The number of trees in a wood rises by 12%12\% one year and falls by 12%12\% the next year. Work out the overall percentage change in the number of trees.
Show worked solution

Worked solution

  1. Choose a number of trees to work with

    start=10000 trees\text{start} = 10000 \text{ trees}

    The overall percentage change does not depend on the starting number, so pick 1000010000 to keep the arithmetic exact.

  2. Find the multiplier for the 12%12\% rise

    100%+12%=112%=1.12100\% + 12\% = 112\% = 1.12

    A 12%12\% increase means multiplying by 1.121.12.

  3. Apply the rise

    10000×1.12=1120010000 \times 1.12 = 11200

    After the first year there are 1120011200 trees. 12001200 trees were added.

  4. Find the multiplier for the 12%12\% fall

    100%−12%=88%=0.88100\% - 12\% = 88\% = 0.88

    A 12%12\% decrease means multiplying by 0.880.88.

  5. Notice what the fall acts on

    12% of 11200=134412\% \text{ of } 11200 = 1344

    The 12%12\% fall is 12%12\% of the LARGER 1120011200, so 13441344 trees are lost — more than the 12001200 that were gained.

  6. Apply the fall

    11200×0.88=985611200 \times 0.88 = 9856

    After the second year there are 98569856 trees, fewer than the 1000010000 at the start.

  7. Combine the multipliers

    1.12×0.88=0.98561.12 \times 0.88 = 0.9856

    The single equivalent multiplier is 0.98560.9856.

  8. Check the combined multiplier

    10000×0.9856=985610000 \times 0.9856 = 9856

    The single multiplier reproduces the final number of trees. ✓

  9. Compare the multiplier with 11

    0.9856<10.9856 < 1

    The multiplier is below 11, so the wood ends up with fewer trees than it started with.

  10. Turn the multiplier into a percentage change

    (0.9856−1)×100=−1.44(0.9856 - 1) \times 100 = -1.44

    The overall change is a 1.44%1.44\% decrease.

  11. Confirm from the tree counts

    10000−985610000×100=14410000×100=1.44%\frac{10000 - 9856}{10000} \times 100 = \frac{144}{10000} \times 100 = 1.44\%

    144144 trees have been lost out of the original 1000010000, which is 1.44%1.44\%. ✓

  12. Watch the common error

    +12%−12%≠0%+12\% - 12\% \ne 0\%

    The rise and fall do not cancel, because the fall is a percentage of a bigger number than the rise was.

  13. See where the 1.44%1.44\% comes from

    0.12×0.12=0.0144=1.44%0.12 \times 0.12 = 0.0144 = 1.44\%

    For an x%x\% rise followed by an x%x\% fall the multiplier is (1+x)(1−x)=1−x2(1 + x)(1 - x) = 1 - x^2, so the loss is exactly 0.122=1.44%0.12^2 = 1.44\%.

  14. Check with a different starting number

    2500×1.12×0.88=24642500 \times 1.12 \times 0.88 = 2464

    25002500 trees would become 24642464, a fall of 3636, and 36÷2500×100=1.44%36 \div 2500 \times 100 = 1.44\%. The same answer appears. ✓

  15. State the overall percentage change

    1.44% decrease1.44\% \text{ decrease}

    The wood ends up with 1.44%1.44\% fewer trees than it had at the start.

Answer
1.44% decrease1.44\% \text{ decrease}
Question 4
6 markschallenging
A shopkeeper buys 6060 T-shirts for a total of £480\pounds 480. He sells 4545 of them at £12\pounds 12 each and the other 1515 at £7\pounds 7 each. Work out the percentage profit.
Show worked solution

Worked solution

  1. Write down the total cost

    cost=£480\text{cost} = \pounds 480

    The shopkeeper pays £480\pounds 480 for all 6060 T-shirts. This is the original amount.

  2. Work out the cost of one T-shirt

    £480÷60=£8\pounds 480 \div 60 = \pounds 8

    Each T-shirt costs £8\pounds 8. (Useful for checking, but the percentage profit can be worked out from the totals.)

  3. Work out the money from the full-price T-shirts

    45×£12=£54045 \times \pounds 12 = \pounds 540

    The 4545 T-shirts sold at £12\pounds 12 bring in £540\pounds 540.

  4. Check how many T-shirts are left

    60−45=1560 - 45 = 15

    That leaves 1515 T-shirts, which matches the question.

  5. Work out the money from the cheaper T-shirts

    15×£7=£10515 \times \pounds 7 = \pounds 105

    The 1515 T-shirts sold at £7\pounds 7 bring in £105\pounds 105. Note £7\pounds 7 is BELOW the £8\pounds 8 cost, so these are sold at a loss.

  6. Work out the total money taken

    £540+£105=£645\pounds 540 + \pounds 105 = \pounds 645

    Altogether the shopkeeper takes £645\pounds 645.

  7. Compare the takings with the cost

    £645>£480\pounds 645 > \pounds 480

    He takes more than he paid, so overall there is a profit despite the 1515 sold at a loss.

  8. Work out the profit

    £645−£480=£165\pounds 645 - \pounds 480 = \pounds 165

    The overall profit is £165\pounds 165.

  9. Check the profit shirt by shirt

    45×£4−15×£1=£180−£15=£16545 \times \pounds 4 - 15 \times \pounds 1 = \pounds 180 - \pounds 15 = \pounds 165

    Each full-price shirt makes £12−£8=£4\pounds 12 - \pounds 8 = \pounds 4, and each cheap shirt loses £8−£7=£1\pounds 8 - \pounds 7 = \pounds 1. The totals agree. ✓

  10. Recall the percentage profit formula

    percentage profit=profitcost price×100\text{percentage profit} = \frac{\text{profit}}{\text{cost price}} \times 100

    The £480\pounds 480 cost is the original amount, so it is the base.

  11. Substitute the numbers

    165480×100\frac{165}{480} \times 100

    Divide the £165\pounds 165 profit by the £480\pounds 480 cost — not by the £645\pounds 645 taken.

  12. Evaluate the division

    165480=0.34375\frac{165}{480} = 0.34375

    The profit is 0.343750.34375 of the cost.

  13. Turn it into a percentage

    0.34375×100=34.375%0.34375 \times 100 = 34.375\%

    The percentage profit is 34.375%34.375\%, which is exact.

  14. Check with a multiplier

    £480×1.34375=£645\pounds 480 \times 1.34375 = \pounds 645

    A 34.375%34.375\% profit has multiplier 1.343751.34375, and £480×1.34375=£645\pounds 480 \times 1.34375 = \pounds 645. ✓

  15. State the percentage profit

    34.375% profit34.375\% \text{ profit}

    Dividing by the £645\pounds 645 takings instead would give 25.58%25.58\%, which is not the percentage profit.

Answer
34.375% profit34.375\% \text{ profit}
Question 5
6 markschallenging
In a sale, every price is reduced by 20%20\%. On the final day, a further 30%30\% is taken off the sale price. A customer says: "That is 50%50\% off altogether." Work out the actual overall percentage reduction.
Show worked solution

Worked solution

  1. Write down what the customer claims

    20%+30%=50%  (claim)20\% + 30\% = 50\% \; \text{(claim)}

    The customer has added the two percentages. We will test this.

  2. Choose a price to work with

    original=£100\text{original} = \pounds 100

    The overall percentage reduction is the same for every price, so use £100\pounds 100 to make it clear.

  3. Find the multiplier for the first reduction

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

    A 20%20\% reduction means multiplying by 0.80.8.

  4. Apply the first reduction

    £100×0.8=£80\pounds 100 \times 0.8 = \pounds 80

    The sale price is £80\pounds 80. So far £20\pounds 20 has come off.

  5. Find the multiplier for the second reduction

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

    A further 30%30\% reduction means multiplying by 0.70.7.

  6. Notice what the 30%30\% is taken from

    30% of £80=£2430\% \text{ of } \pounds 80 = \pounds 24

    The extra 30%30\% comes off the £80\pounds 80 SALE price, not off the original £100\pounds 100. Only £24\pounds 24 comes off, not £30\pounds 30.

  7. Apply the second reduction

    £80×0.7=£56\pounds 80 \times 0.7 = \pounds 56

    The final-day price is £56\pounds 56.

  8. Combine the multipliers into one

    0.8×0.7=0.560.8 \times 0.7 = 0.56

    The single equivalent multiplier is 0.560.56.

  9. Check the combined multiplier

    £100×0.56=£56\pounds 100 \times 0.56 = \pounds 56

    The single multiplier gives the same final price. ✓

  10. Read the reduction from the multiplier

    100%−56%=44%100\% - 56\% = 44\%

    A multiplier of 0.560.56 means the customer pays 56%56\% of the original price, so 44%44\% has come off.

  11. Confirm with the money

    £100−£56=£44\pounds 100 - \pounds 56 = \pounds 44

    £44£44 has come off the original £100£100, and 44÷100×100=44%44 \div 100 \times 100 = 44\%. ✓

  12. Test the claim of 50%50\% off

    £100×0.5=£50≠£56\pounds 100 \times 0.5 = \pounds 50 \ne \pounds 56

    A genuine 50%50\% off would give £50\pounds 50, but the shopper actually pays £56\pounds 56. The claim is wrong.

  13. Explain the error

    20% of £100=£20,30% of £80=£2420\% \text{ of } \pounds 100 = \pounds 20, \quad 30\% \text{ of } \pounds 80 = \pounds 24

    The two percentages are taken from different amounts — £100\pounds 100 and then £80\pounds 80 — so they cannot be added.

  14. Check with a different price

    £250×0.8×0.7=£140\pounds 250 \times 0.8 \times 0.7 = \pounds 140

    £250£250 falls to £140£140, a fall of £110£110, and 110÷250×100=44%110 \div 250 \times 100 = 44\%. The same 44%44\% appears whatever the price. ✓

  15. State the actual overall reduction

    44% reduction44\% \text{ reduction}

    The true overall reduction is 44%44\%, not the 50%50\% the customer claimed.

Answer
44% decrease44\% \text{ decrease}

Unlock 29 more Percentage change questions

Create a free account to work through every GCSE Percentage change question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Percentage change practice

Practice QuestionsWorked Solutions

Related Ratio & Proportion topics

Algebraic proportionCompound measuresDirect and inverse proportionDividing in a ratioRatios, fractions and linear functionsInstantaneous rate of change