GCSE Maths: Multi-step and reverse fraction problems (Higher)
Part of the GCSE Maths fractions chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Multi-step and reverse fraction problems
Many exam questions describe a story in several steps. The key is to track the fraction of the original that is left at each stage, and to work backwards when the final amount is known.
Key ideas
- Fractional change: increasing by 15 means multiplying by 65. Decreasing by 14 means multiplying by 34.
- Successive changes: multiply the multipliers. Do not add the fractions.
- Reverse problems: write final = multiplier × original, then divide the final amount by the multiplier.
- Part and remainder: “13 of the rest” means 13 of the remaining amount, not of the original.
Method. Turn each change into a multiplier, multiply the multipliers, then work backwards from the final price.
- Find the first multiplier. Reducing by 15 leaves 45 of the price.1 − 15 = 45
- Find the second multiplier. The second reduction is 14 of the new price, so 34 of the new price is left.1 − 14 = 34
- Find the overall multiplier. The second change applies to the result of the first, so multiply the multipliers (do not add the fractions). Cancel the 4s.45 × 34 = 35
- Link the final price to the original. The final price is 35 of the original. 3 parts are worth £90, so one part (one fifth) is worth £90 ÷ 3.35 of original = £90
1 fifth = £90 ÷ 3 = £30 - Find the original price. The whole is 5 fifths.5 × £30 = £150
Check. Work forwards from £150. The first reduction leaves 4/5 of £150 = £120. The second leaves 3/4 of £120 = £90. This matches. ✓
Watch out. Do not add the fractions. A reduction of 15 then 14 is not a reduction of 920, because the second reduction is of a smaller price.
Method. Find the fraction of the whole box that is white, then use the 30 white chocolates to find the whole.
- Find the fraction that is not milk. The rest means everything except the milk chocolates.1 − 38 = 58
- Find the fraction that is plain. Plain is 13 of the rest, which is 13 of 58. “Of” means multiply.13 × 58 = 524
- Find the fraction that is white. White is what is left after the milk and the plain. Use a common denominator of 24.58 − 524 = 1524 − 524 = 1024 = 512
- Link the white chocolates to the whole box. 512 of the box is 30 chocolates. 5 parts are 30, so one twelfth is 30 ÷ 5.512 of the box = 30
1 twelfth = 30 ÷ 5 = 6 - Find the whole box. The whole is 12 twelfths.12 × 6 = 72 chocolates
Check. Work forwards from 72. Milk: 38 of 72 = 27. The rest is 45. Plain: 13 of 45 = 15. White: 45 − 15 = 30. This matches. ✓
Watch out. “Of the rest” means of the amount that remains after the first group, not of the whole box.
Method. For work-rate problems, find the fraction of the job each does in one hour, add them, then turn the combined rate into a time.
- Find how much A fills in one hour. A fills the whole tank in 6 hours, so in 1 hour it fills one sixth.A: 16 of the tank per hour
- Find how much B fills in one hour. B fills the whole tank in 3 hours, so in 1 hour it fills one third.B: 13 of the tank per hour
- Add the rates. Working together they fill both amounts every hour. Use a common denominator of 6.16 + 26 = 36 = 12 of the tank per hour
- Turn the combined rate into a time. If half the tank is filled each hour, the whole tank takes 1 ÷ 12 = 2 hours.1 ÷ 12 = 2
- Write the answer with units. The time is in hours.2 hours
Check. In 2 hours A fills 2 × 16 = 13 and B fills 2 × 13 = 23. 13 + 23 = 1 whole tank. ✓
Watch out. Do not just average the two times, and do not add 6 + 3. Add the rates (fractions per hour), then take the reciprocal.
Method. Turn each change into a multiplier and apply them one after the other. The second change is a fraction of the new price.
- Find the multiplier for the increase. Increasing by 15 means adding a fifth, so the price becomes 1 + 15 of what it was.1 + 15 = 65
- Apply the increase. Divide by the denominator 5, then multiply by 6.£100 ÷ 5 = £20
£20 × 6 = £120 - Find the multiplier for the decrease. Decreasing by 15 of the new price leaves 1 − 15 of it.1 − 15 = 45
- Apply the decrease. Work on the new price, £120, not on the original £100.£120 ÷ 5 = £24
£24 × 4 = £96
Check. A different route: the overall multiplier is 65 × 45 = 2425. 2425 of £100 = £96. ✓
Watch out. The price does not return to £100. The decrease is of a bigger price, so it takes off more than the increase added.
Method. Reverse problem: write sale price = original × multiplier, then divide the sale price by the multiplier.
- Find the multiplier. A reduction of 27 leaves 57 of the original price.1 − 27 = 57
- Write the relationship. The sale price is the multiplier times the original price.57 of original = £150
- Find one part. 5 parts are £150, so one seventh is £150 ÷ 5.£150 ÷ 5 = £30 for one seventh
- Find the whole. The original price is 7 sevenths.£30 × 7 = £210
Check. Work forwards: 27 of £210 = £60 reduction, and £210 − £60 = £150. ✓
Watch out. Do not increase £150 by 27. The reduction was a fraction of the original, so you must divide by the multiplier.
Watch out
Increasing a price by 14 and then decreasing by 14 does not return to the start: 54 × 34 = 1516, a loss of 116.Exercise 3B Multi-step and reverse problems
Non-calculatorMethod. Increasing by a fraction: multiply by (1 + the fraction).
- Find the multiplier. Adding 16 of the amount leaves 1 + 16 of it.1 + 16 = 66 + 16 = 76
- Find one part: divide by the denominator 6. The denominator says to split the amount into 6 equal parts. One part is 16 of the amount.48 ÷ 6 = 8
- Multiply by the numerator 7. The multiplier 76 says we need 7 of those parts.8 × 7 = 56
- Write the answer. The amount has been increased.56
Check. A different route: the change is 16 of 48 = 8 × 1 = 8. Then 48 + 8 = 56. ✓
Watch out. The multiplier is 76, not 16. You must add the change to the whole (1).
Method. Decreasing by a fraction: multiply by (1 − the fraction).
- Find the multiplier. Taking away 29 of the amount leaves 1 − 29 of it.1 − 29 = 99 − 29 = 79
- Find one part: divide by the denominator 9. The denominator says to split the amount into 9 equal parts. One part is 19 of the amount.90 ÷ 9 = 10
- Multiply by the numerator 7. The multiplier 79 says we need 7 of those parts.10 × 7 = 70
- Write the answer. The amount has been decreased.70
Check. A different route: the change is 29 of 90 = 10 × 2 = 20. Then 90 − 20 = 70. ✓
Watch out. The multiplier is 79, not 29. You must take the change away from the whole (1).
Method. Increasing by a fraction: multiply by (1 + the fraction).
- Find the multiplier. Adding 34 of the amount leaves 1 + 34 of it.1 + 34 = 44 + 34 = 74
- Find one part: divide by the denominator 4. The denominator says to split the amount into 4 equal parts. One part is 14 of the amount.36 ÷ 4 = 9
- Multiply by the numerator 7. The multiplier 74 says we need 7 of those parts.9 × 7 = 63
- Write the answer. The amount has been increased.63
Check. A different route: the change is 34 of 36 = 9 × 3 = 27. Then 36 + 27 = 63. ✓
Watch out. The multiplier is 74, not 34. You must add the change to the whole (1).
Method. Decreasing by a fraction: multiply by (1 − the fraction).
- Find the multiplier. Taking away 310 of the amount leaves 1 − 310 of it.1 − 310 = 1010 − 310 = 710
- Find one part: divide by the denominator 10. The denominator says to split the amount into 10 equal parts. One part is 110 of the amount.250 ÷ 10 = 25
- Multiply by the numerator 7. The multiplier 710 says we need 7 of those parts.25 × 7 = 175
- Write the answer. The amount has been decreased.175
Check. A different route: the change is 310 of 250 = 25 × 3 = 75. Then 250 − 75 = 175. ✓
Watch out. The multiplier is 710, not 310. You must take the change away from the whole (1).
Method. Reverse problem: final amount = original × multiplier, so original = final ÷ multiplier.
- Find the multiplier. Increasing by 15 gives a multiplier of 1 + 15 = 65.65
- Write the relationship. The final amount is the multiplier times the original amount.original × 65 = 60
- Find one part: divide by the numerator 6. 6 parts of the original make 60, so one part is 60 ÷ 6. Dividing by 65 is the same as dividing by 6 and multiplying by 5.60 ÷ 6 = 10
- Multiply by the denominator 5. The original is 5 parts altogether.10 × 5 = 50
- Write the answer. This is the original amount.50
Check. Work forwards: 50 × 65 = 10 × 6 = 60. ✓
Watch out. Do not just reduce or increase the final amount by the same fraction. You must divide by the multiplier.
Method. Reverse problem: final amount = original × multiplier, so original = final ÷ multiplier.
- Find the multiplier. Decreasing by 14 gives a multiplier of 1 − 14 = 34.34
- Write the relationship. The final amount is the multiplier times the original amount.original × 34 = 45
- Find one part: divide by the numerator 3. 3 parts of the original make 45, so one part is 45 ÷ 3. Dividing by 34 is the same as dividing by 3 and multiplying by 4.45 ÷ 3 = 15
- Multiply by the denominator 4. The original is 4 parts altogether.15 × 4 = 60
- Write the answer. This is the original amount.60
Check. Work forwards: 60 × 34 = 15 × 3 = 45. ✓
Watch out. Do not just reduce or increase the final amount by the same fraction. You must divide by the multiplier.
Method. Reverse problem: final amount = original × multiplier, so original = final ÷ multiplier.
- Find the multiplier. Increasing by 38 gives a multiplier of 1 + 38 = 118.118
- Write the relationship. The final amount is the multiplier times the original amount.original × 118 = 110
- Find one part: divide by the numerator 11. 11 parts of the original make 110, so one part is 110 ÷ 11. Dividing by 118 is the same as dividing by 11 and multiplying by 8.110 ÷ 11 = 10
- Multiply by the denominator 8. The original is 8 parts altogether.10 × 8 = 80
- Write the answer. This is the original amount.80
Check. Work forwards: 80 × 118 = 10 × 11 = 110. ✓
Watch out. Do not just reduce or increase the final amount by the same fraction. You must divide by the multiplier.
Method. Reverse problem: final amount = original × multiplier, so original = final ÷ multiplier.
- Find the multiplier. Decreasing by 25 gives a multiplier of 1 − 25 = 35.35
- Write the relationship. The final amount is the multiplier times the original amount.original × 35 = 42
- Find one part: divide by the numerator 3. 3 parts of the original make 42, so one part is 42 ÷ 3. Dividing by 35 is the same as dividing by 3 and multiplying by 5.42 ÷ 3 = 14
- Multiply by the denominator 5. The original is 5 parts altogether.14 × 5 = 70
- Write the answer. This is the original amount.70
Check. Work forwards: 70 × 35 = 14 × 3 = 42. ✓
Watch out. Do not just reduce or increase the final amount by the same fraction. You must divide by the multiplier.
Method. “A fraction of a fraction” means multiply. Write each relationship, then substitute one into the other.
- Write each relationship. “Is a fraction of” means multiply by that fraction.A = 35 of B
B = 23 of C - Substitute B. B is 2/3 of C, so A is 3/5 of something that is 2/3 of C.A = 35 × 23 of C
- Cancel and multiply. Cancel the 3s: 3 on the top and 3 on the bottom.35 × 23 = 15 × 21 = 25
- Write the answer. A as a fraction of C.A is 25 of C.
Check. Put in a number: let C = 15. B = 23 of 15 = 10, and A = 35 of 10 = 6. Then 6 ÷ 15 = 25. ✓
Watch out. Multiply the fractions. Do not add them.
Method. Use the fraction from part (a): A is 2/5 of C.
- Recall the result of part (a). A is two fifths of C.A = 25 of C
- Divide C by the denominator. Find one fifth of C = 60.60 ÷ 5 = 12
- Multiply by the numerator. Two fifths is 2 lots of one fifth.12 × 2 = 24
Check. A different route: B = 23 of 60 = 40, and A = 35 of 40 = 24. ✓
Watch out. Use the single fraction from part (a). It saves a step.
Method. A ratio 3 : 5 means the club is split into 3 + 5 = 8 equal parts, with 3 parts boys and 5 parts girls.
- Say what the ratio means. The ratio 3 : 5 compares boys to girls: for every 3 boys there are 5 girls. We can picture the club as equal parts, 3 of them boys and 5 of them girls.boys : girls = 3 : 5
- Add the parts of the ratio. This gives the total number of equal parts in the whole club.3 + 5 = 8 equal parts
- Write the boys as a fraction of the club. 3 of the 8 equal parts are boys. The total goes on the bottom.38
Check. The girls are 58, and 38 + 58 = 88 = 1 whole club. ✓
Watch out. The denominator is the total of the ratio parts (8), not the girls' part (5).
Method. Use the ratio and the number of boys to find the value of one part.
- Boys are 3 parts. We know the number of boys.3 parts = 24 boys
- Find the value of 1 part. Divide by 3.24 ÷ 3 = 8
- Girls are 5 parts. Multiply the value of one part by 5.5 × 8 = 40
Check. Total = 24 + 40 = 64, which is 8 parts × 8. And 24 : 40 = 3 : 5 (divide both by 8). ✓
Watch out. Find the value of one part first.
Method. Track the fraction of the book left after each day, then work backwards from the 36 pages.
- Find the fraction left after Monday. Ali has read 25, so the rest is left.1 − 25 = 35
- Find the fraction read on Tuesday. He reads 13 of the remaining pages, which is 13 of 35. “Of” means multiply.13 × 35 = 15
- Find the fraction left after Tuesday. Take Tuesday's reading away from what was left.35 − 15 = 25
- Use the 36 pages. 25 of the book is 36 pages, so one fifth is 36 ÷ 2.1 fifth = 18 pages
- Find the whole book. The book is 5 fifths.5 × 18 = 90 pages
Check. Work forwards from 90. Monday: 2/5 of 90 = 36, leaving 54. Tuesday: 1/3 of 54 = 18, leaving 36. This matches. ✓
Watch out. “Of the remaining pages” means of what is left after Monday, not of the whole book.
Method. The extra 12 litres changes the tank from 3/5 full to 4/5 full, so 12 litres fills the difference.
- Find the fraction of the tank that 12 litres fills. Subtract the starting fraction from the final fraction.45 − 35 = 15
- Write what we know. One fifth of the tank is 12 litres.15 of the tank = 12 litres
- Find the whole tank. The tank is 5 fifths, so multiply by 5.5 × 12 = 60 litres
Check. 35 of 60 = 36 and 45 of 60 = 48. 48 − 36 = 12. ✓
Watch out. Do not use 3/5 or 4/5 on their own. It is the difference that matches the 12 litres.
Method. For work-rate problems, find the fraction of the job each pipe does in one hour, add them, then turn the combined rate into a time.
- Find how much pipe A fills in one hour. It fills the pool in 4 hours, so in 1 hour it fills one quarter.A: 14 per hour
- Find how much pipe B fills in one hour. It fills the pool in 12 hours, so in 1 hour it fills one twelfth.B: 112 per hour
- Add the rates. Working together they fill both amounts each hour. The LCM of 4 and 12 is 12.312 + 112 = 412 = 13 per hour
- Find the time. If one third of the pool is filled each hour, the whole pool takes 1 ÷ 13 hours.1 ÷ 13 = 3 hours
Check. In 3 hours A fills 3 × 14 = 34 and B fills 3 × 112 = 14. 34 + 14 = 1 whole pool. ✓
Watch out. Do not add the times (4 + 12). Add the rates, which are fractions per hour.
Method. To test a claim about successive changes, work out the overall multiplier.
- Find the multiplier for the increase. Increasing by 15 means the amount becomes 1 + 15 of what it was.1 + 15 = 65
- Find the multiplier for the decrease. Decreasing by 15 (of the new amount) means 1 − 15 of it is left.1 − 15 = 45
- Combine them. The second change applies to the result of the first, so multiply the multipliers.65 × 45 = 2425
- Compare with 1. A multiplier of 1 means no change. A multiplier less than 1 means the amount is smaller.2425 < 1
- Write the conclusion. The amount ends up smaller than the start.Mia is wrong.
Check. Try 100: increase by 15 gives 120, then decrease by 15 of 120 (24) gives 96, not 100. ✓
Watch out. The second change is a fraction of the new amount, not of the original. That is why they do not cancel.
Method. Combine the two multipliers, then work backwards from the final price.
- Find the first multiplier. A reduction of 13 leaves 23.1 − 13 = 23
- Find the second multiplier. A further reduction of 14 of the sale price leaves 34.1 − 14 = 34
- Find the overall multiplier. Multiply the multipliers. Cancel the 3s and the 2 and 4.23 × 34 = 12
- Use the final price. The final price is 12 of the original price.12 of the original = £60
- Find the original price. The original is 2 halves.2 × £60 = £120
Check. Work forwards: 23 of £120 = £80, and 34 of £80 = £60. ✓
Watch out. Do not add the reductions: 13 + 14 is not the total reduction, because the second one is of the smaller price.
Method. Multiply the multipliers for each step, then work backwards from the amount left.
- Find the multiplier for each step. Spending 1/4 leaves 3/4. Spending 2/5 of what is left leaves 3/5. Spending 1/3 of what is left after that leaves 2/3.34, 35, 23
- Find the overall multiplier. Multiply the three multipliers, cancelling as you go.34 × 35 × 23 = 1860 = 310
- Use the £18. 310 of the starting money is £18. 3 parts are £18, so one tenth is £18 ÷ 3.1 tenth = £6
- Find the starting amount. The whole is 10 tenths.10 × £6 = £60
Check. Work forwards from £60. After the first spend: 3/4 of 60 = 45. After the second: 3/5 of 45 = 27. After the third: 2/3 of 27 = 18. This matches. ✓
Watch out. Each fraction is of what is left at that moment, not of the starting amount.