GCSE Maths revision · Ratio and proportion

Compound interest and depreciation

Foundation and Higher

The idea

Compound change applies the same percentage to the new amount each time, so you multiply by the multiplier once for each year: $\text{amount} \times \text{multiplier}^n$ .

Find the multiplier: $3\%$ growth is $1.03$ , $3\%$ decay is $0.97$ .
Raise it to the number of years: $2000 \times 1.03^5$ .
Round only at the end, to the nearest penny for money.

Watch out Compound interest is not $5 \times 3\%$ of the start: that is simple interest.

See it worked

Worked exampleCalculator allowed2 marks

Sam invests £3000 for 3 years at $2\%$ per year compound interest.

Work out the value of Sam's investment at the end of the 3 years. Give your answer to the nearest penny.

Each year the value is multiplied by $1.02$ . $3000 \times 1.02^3 = 3183.624$
To the nearest penny: £3183.62.

Answer £3183.62

Your turn

Question 1Non-calculator2 marks

There are $2000$ bacteria in a dish. The number of bacteria increases by $50\%$ each day.

Work out the number of bacteria after 3 days.

Question 2Calculator allowed3 marks

£1500 is invested at 4% compound interest per year. Interest is added at the end of each year. What is the least whole number of years before the balance exceeds £1750? You must show your working.

Question 3Non-calculator3 marks

Two accounts initially contain equal amounts. Account A gains 10% in year 1 and loses 10% in year 2. Account B gains r% in year 1 and loses the same r% in year 2, where r > 0. After year 2, B contains 32/33 of the amount in A. Find r. You must show your working.

In past papers

93 official questions in 85 papers since 2017. The latest:

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