Percentages, reverse percentages, compound interest & depreciation – IGCSE Maths exam questions with mark scheme

1Grade 4–5Calculator3 marks

In a sale, the normal price of a coat is reduced by 15%.

The sale price of the coat is £255

Work out the normal price of the coat. [3]

for $100 - 15 = 85$ or $0.85$ oeM1
for $255 \div 0.85$ or $\dfrac{255}{85} \times 100$ oeM1
£$300$A1
Answer: £300
Examiner tip: The 15% was taken off the ORIGINAL price, so adding 15% of £255 back on gives the wrong answer (£293.25).

2Grade 6–7HigherCalculator3 marks

Ravi buys a car for £18 500

The value of the car depreciates by 12% in the first year.

After the first year, the value of the car depreciates by 8% each year.

Work out the value of the car at the end of 4 years. Give your answer to the nearest pound. [3]

for $18\,500 \times 0.88$ $(= 16\,280)$M1
for their $16\,280 \times 0.92^3$ oe (a complete method)M1
£$12\,677$A1
Answer: £12 677
Examiner tip: The 8% applies for three years (years 2, 3 and 4), not four — the first year has its own rate.

3Grade 6–7HigherCalculator4 marks

In 2024, the price of a laptop increased by 20%.

In 2025, the new price decreased by 20%.

At the end of 2025, the price of the laptop was £1152

(a) Work out the price of the laptop at the start of 2024 [3]

(b) Work out the overall percentage change in the price. State whether it is an increase or a decrease. [1]

(a) for $1.2 \times 0.8$ $(= 0.96)$ oe, or $1152 \div 0.8$ $(= 1440)$M1
for $1152 \div 0.96$ or $1440 \div 1.2$ oeM1
£$1200$A1
(b) $4\%$ decrease (ft their (a), e.g. from $\dfrac{1152 - 1200}{1200} \times 100$)B1
Answer: (a) £1200 (b) 4% decrease
Examiner tip: A 20% rise followed by a 20% fall does not cancel out: the multiplier is 1.2 × 0.8 = 0.96.

4Grade 6–7HigherCalculator3 marks

Lena invests £5000 in an account paying 3.5% compound interest per year.

At the end of which year will the value of her investment first be more than £6000? You must show your working. [3]

for $5000 \times 1.035^n$ for any $n \geqslant 2$, or a year-by-year method using $1.035$M1
for showing values either side of £6000: $5000 \times 1.035^5$ $(= 5938.43\ldots)$ and $5000 \times 1.035^6$ $(= 6146.27\ldots)$M1
$6$ (end of year 6), supported by workingA1
Answer: Year 6
Examiner tip: Show the values for both year 5 (below £6000) and year 6 (above) — that is the evidence the method mark needs.

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