Compound growth and decay

Objective. Build and solve repeated multiplier models over fractional and integer periods.

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Learn: the key idea

Compound growth and decay models a quantity that changes by the same percentage every time period, so each new value is the previous value times a fixed multiplier raised to a power. For growth of r% per period, the multiplier is 1+r/100; for decay, it is 1−r/100. After n periods the amount is A = P×(multiplier)ⁿ, where P is the starting value. Periods need not be whole years: six months at an annual rate uses a fractional power, n=0.5. For example, £2000 growing at 3% for 4 years gives 2000×1.03⁴ = £2251.02.

Worked example

A car worth £18000 depreciates by 12% each year. Find its value after 3 years, to the nearest pound.

1 $Depreciation multiplier = 1 - 0.12 = 0.88$
2 $Apply the multiplier for 3 years: 18000 \times 0.88³$
3 $0.88³ = 0.681472, so 18000 \times 0.681472 = 12266.496$
4 $Round to the nearest pound$

Answer: £12266

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Common mistake. Pupils often multiply by r/100 repeatedly instead of raising the multiplier to a power, treating 5% growth for 3 years as 1.15 instead of 1.05³; always write the multiplier once and apply the power n.

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