GCSE Maths · Ratio, proportion & rates of change

Growth & Decay

Repeated percentage change - compound interest, depreciation and growth/decay over several periods.

200 GCSE-style practice questions

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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.

  1. 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.

  2. 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.

  3. 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.

  4. 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.

  5. 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.

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.

£2000 is invested at 3% compound interest per year. Work out the value after 4 years.

  1. £2251.02
  2. £2240.00
  3. £2185.45
  4. £2240
Show the answer and the working

Answer: £2251.02

Four years of 3% growth gives 2000 × 1.03⁴ = £2251.02.

  1. The multiplier for 3% interest is 1.03.
  2. For 4 years, use 1.03⁴.
  3. 2000 × 1.03⁴ = 2000 × 1.12550881 = £2251.02.

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