Growth & Decay
Repeated percentage change - compound interest, depreciation and growth/decay over several periods.
Watch the lesson
Every point below, worked through in one video, and each point links straight to its own part.
Learn
Growth & Decay, explained point by point
Everything the GCSE specification expects you to be able to do, and how to actually do it - the same lesson a signed-in student studies from.
Use multipliers for repeated percentage increase and decrease
Each period multiplies by the same factor, so apply the multiplier repeatedly rather than adding percentages. Two years of 10% growth is × 1.1 × 1.1 = × 1.21 - a 21% rise, not 20%. Percentages compound; they never simply add.
Calculate compound interest
Compound interest pays interest on the interest. £2000 at 3% for 4 years is 2000 × 1.03⁴ ≈ £2251.02. If the question asks for the INTEREST alone, subtract the original amount at the end.
Solve depreciation problems
Depreciation is the same machinery with a multiplier below 1. A £12 000 car losing 15% a year is worth 12000 × 0.85³ ≈ £7369 after 3 years. The value falls quickly at first and more slowly later - that curve is the shape of decay.
Set up and use expressions of the form P × (multiplier)ⁿ for growth and decay
Every repeated-change situation compresses to P × mⁿ: starting amount, multiplier, number of periods. A population of 5000 growing 8% a year is 5000 × 1.08ⁿ after n years. Write the expression first - it turns the whole problem into one calculator step.
Find how many periods growth or decay takes to pass a target value
Try increasing powers of the multiplier until you cross the target. How many years for £3000 at 5% to pass £4000? 3000 × 1.05⁶ ≈ £4020 - so 6 years. Show the value either side of the crossing to justify your answer.
Common mistakes in Growth & Decay
The slips that cost marks year after year, and what to do instead.
Treating compound interest as simple interest, so £2000 at 5% a year for 3 years grows to £2300.
Each year earns interest on the new total, so multiply by 1.05 once a year: 2000 × = £2315.25.
Putting the number of years in as a multiplier, as in 500 × 1.04 × 2.
The years go in the power: 500 × = £540.80. Multiplying by 2 instead gives £1040, which should look far too big.
Giving the final amount when the question asks for the interest earned.
Take the starting amount away at the end: £2315.25 − £2000 = £315.25 of interest.
Taking 15% of the original price off every year, so a £12 000 car loses £1800 each year.
Multiply by 0.85 for each year of 15% depreciation: 12 000 × = £7369.50. Each year loses 15% of a smaller value.
Stopping a year short when finding how long growth takes to pass a target.
Show the values either side of the target: 3000 × ≈ £3828.84 and 3000 × ≈ £4020.29, so £3000 at 5% takes 6 years to pass £4000.
Practice
Try a Growth & Decay question
A GCSE-style original question from this topic. Have a go before you open the working - deciding on an answer first is what makes the working stick.
Answer: £2251.02
Four years of 3% growth gives 2000 × 1.03⁴ = £2251.02.
- The multiplier for 3% interest is 1.03.
- For 4 years, use 1.03⁴.
- 2000 × 1.03⁴ = 2000 × 1.12550881 = £2251.02.
Ready to practise Growth & Decay?
Sign up free to practise this topic - questions that adapt to how you're doing, with instant encouraging feedback on every answer. No card needed.
Already have an account? Log in