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MathidiaMathematics Academy
Number fluency, bounds and accuracy
Lesson 3 of 75
Standard practice

Repeated percentage change

Objective. Model growth and decay over multiple periods with multipliers.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Repeated percentage change models a quantity that grows or decays by the same percentage over several equal time periods, using a multiplier raised to a power. A growth of 4% per year uses multiplier 1.04, and decay of 4% uses 0.96; after n periods the amount is original × multiplierⁿ. For example, £2000 invested at 3% compound interest for 5 years becomes 2000 × 1.03⁵. This is far quicker and more accurate than recalculating the percentage from scratch each year. Always check whether the question wants the final amount or just the total change.

Worked example

A car worth £12000 depreciates by 15% per year. Find its value after 3 years of repeated percentage decrease.

£12000.00
Money illustration for £12000
  1. 1Decay of 15% gives multiplier 0.85.
  2. 2Apply the multiplier for 3 years: 12000 × 0.85³.
  3. 30.85³ = 0.614125.
  4. 412000 × 0.614125 = £7369.50.

Answer: £7369.50

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Practice

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Common mistake. Pupils often multiply the original amount by the percentage change n times instead of raising the multiplier to the power n, e.g. using 2000 × 1.03 × 5. Always use 2000 × 1.03⁵ for repeated change.

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