Reverse percentages Quiz (Grade 5)

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Topic:

Score: 0 / 10

Reverse percentages — Notes & Worked Examples Show / Hide

Explanation:

Reverse percentage questions give you the new value after a percentage change and ask for the original amount (100%). Never work out the percentage of the new amount—the percentage was calculated from the original amount, not the final one.

First write the new percentage by adding to 100% (for an increase) or subtracting from 100% (for a decrease/sale). Then either divide the new amount by the decimal multiplier, or divide to find 1% and multiply by 100 to get 100%.

Example 1: In a 15% off sale, a jacket costs £68. Work out the normal price of the jacket before the sale.

Solution:

Sale price is 100% − 15% = 85% of the original price

85% = £68 (decimal multiplier = 0.85)

1% = £68 ÷ 85 = £0.80

100% = £0.80 × 100 (or £68 ÷ 0.85) = £80

Answer: £80

Example 2: After a 6% pay rise, Maya earns £25,440 per year. What was her salary before the pay rise?

Solution:

New salary is 100% + 6% = 106% of original salary

Decimal multiplier = 1.06

Original × 1.06 = £25,440

Original = £25,440 ÷ 1.06 = £24,000

Answer: £24,000

One Page Guide (PDF) Reverse percentages Worksheets

Question 1

[2 marks]

In a sale, the normal prices are reduced by 15%
Amina buys a dress in the sale for £46.75
Work out the normal price of the dress.

Worked Solution & Answer

46.75 ÷ 0.85 = £55 M1 A1

Answer: £ 55

Question Paper (PDF) • Mark Scheme (PDF)

Question 2

[2 marks]

The normal price of a mattress is reduced by 40% in a sale.
The price of the mattress in the sale is £660
Work out the normal price of the mattress.

Worked Solution & Answer

60% of normal price = £660 M1
⇒ 10% = £110
⇒ 100% = £1100 A1

Answer: £ 1100

Question Paper (PDF) • Mark Scheme (PDF)

Question 3

[2 marks]

The price of a holiday increases by 20%
This 20% increase adds £240 to the price of the holiday.
Work out the price of the holiday before the increase.

Worked Solution & Answer

20% = £240 M1
⇒ 100% = £240 × 5 = £1200 A1

Answer: £ 1200

Question Paper (PDF) • Mark Scheme (PDF)

Question 4

[2 marks]

In a sale, normal prices are reduced by 30%
The sale price of a TV is £280
Work out the normal price of the TV.

Worked Solution & Answer

70% = £280 M1
⇒ 100% = £280 ÷ 0.7 = £400 A1

Answer: £ 400

Question Paper (PDF) • Mark Scheme (PDF)

Question 5

[2 marks]

The value of Michelle’s car has decreased by 15%
The car now has a value of £13 600
Work out the value of Michelle’s car before the decrease.

Worked Solution & Answer

13 600 ÷ 0.85 = £16 000 M1 A1

Answer: £ 16 000

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 AQA Nov 2021 3H Q14

[3 marks]

Phil sells ties.
He increases the original price of each tie by 10% to £13.20
A month later he announces a sale: 10% OFF ALL TIES.

Phil says,
“The ties will be back to their original price, because each change was by 10%”
Is he correct?
Tick a box: [ ] Yes [ ] No
Show working to support your answer.

Worked Solution & Answer

Original price = £13.20 ÷ 1.10 = £12.00 M1
Sale price = £13.20 × 0.90 = £11.88 M1
Since £11.88 ≠ £12.00, Phil is incorrect (tick No). A1

Answer: No, sale price is £11.88 whereas original price was £12.00

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 IGCSE Jun 2025 1F Q22 / 1H Q5

[3 marks]

In a sale, normal prices are reduced by 28%
The sale price of a watch is 198 euros.
Work out the normal price of the watch.

euros

Worked Solution & Answer

100% – 28% = 72% M2
Normal price = 198 ÷ 0.72 = 275 euros A1

Answer: 275 euros

Question Paper (PDF) • Mark Scheme (PDF)

Question 8 IGCSE Nov 2024 1F Q24 / 1H Q7

[4 marks]

Shop A and Shop B have offers for buying the same type of laptop.

The normal price of the laptop in Shop A is different to the normal price of the laptop in Shop B
Which shop gives more money off their normal price?
Show your working clearly.

Worked Solution & Answer

Shop A discount = 0.16 × £475 = £76 M1
Shop B normal price = £408 ÷ 0.85 = £480 M1
Shop B discount = £480 − £408 = £72 M1
Therefore Shop A gives more money off. A1

Answer: Shop A (£76 off vs £72 off in Shop B)

Question Paper (PDF) • Mark Scheme (PDF)

Question 9

[4 marks]

(a) Price increased by 10% to £440. Jim says 10% of 440 is 44 so original was £396. Is Jim correct? Explain. (1)
(b) Tina sells bikes: Nov 25% fewer than Oct, Dec 50% more than Nov. Dec = 180 bikes. How many bikes in Oct? (3)

(a) Is Jim correct?
Explain your answer. [1 mark]

(b) How many bikes did Tina sell in October? [3 marks]

Worked Solution & Answer

(a)

No: Jim took 10% of the new price instead of the original price
Original price = 440 ÷ 1.1 = £400, not £396 B1

Answer (a): No with reason (10% should be of original price £400, not £440)

(b)

December = 1.5 × November ⇒ November = 180 ÷ 1.5 = 120 M2
November = 0.75 × October ⇒ October = 120 ÷ 0.75 = 160 bikes A1

Answer (b): 160

Question Paper (PDF) • Mark Scheme (PDF)

Question 10 IGCSE Jan 2022 1HR Q10 / 1FR Q24

[6 marks]

Jonty has a storage container in the shape of a cuboid, as shown in the diagram.

Jonty is going to paint the outside of his storage container, apart from the base which is shown shaded in the diagram.
He needs enough paint to cover the four sides and the top.
Each tin of paint covers an area of 15 m²
The cost of each tin of paint recently increased by 10%
After the increase, the cost of each tin of paint is £26.95
Jonty says
"Before the increase, I could have bought enough paint for less than £200"
Show that Jonty is correct.
Show your working clearly.

Worked Solution & Answer

Area = 15 + 72 + 30 = 117 m² M2
Tins = 117 ÷ 15 = 7.8 → 8 tins M2
Original price per tin = £26.95 ÷ 1.1 = £24.50 M1
Cost for 8 tins = 8 × £24.50 = £196 < £200. C1

Answer: £196 < £200 (Jonty is correct)

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 30 marks)

Reverse percentages Worksheets Next: Standard form →

Grade 5 Quizzes

25 Topics

5A - Combining ratios	5B - Direct and inverse proportion
5C - Reverse percentages	5D - Standard form
5E - Speed	5F - Changing the subject
5G - Quadratic expressions	5H - Solving quadratic equations
5J - Drawing quadratic graphs	5K - Cubic/reciprocal graphs
5L - Simultaneous equations	5M - Sector areas and arc lengths
5N - Sim. equ. graphically	5P - Midpoint of a line segment
5Q - Gradient of a line	5R - Equation of a line
5S - Spheres and cones	5T - Similar shapes (lengths)
5U - Trigonometry (SOHCAHTOA)	5V - Exact trig. values
5W - Probability trees	5X - Vectors
5Y - Venn diagrams	5Z - Density
5AA - Pressure