A machine bought for £12 000 loses 20% of its value each year. What is it worth after 3 years?
- A loss of 20% leaves 80%, so the multiplier is 0.8.
- Value = 12 000 × 0.83, and 0.83 = 0.8 × 0.8 × 0.8 = 0.512.
- 12 000 × 0.512 = 6144.
Answer: £6144
Growth and decay
Maths · Ratio, proportion and rates of change. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
In growth and decay a quantity changes by the same percentage in each time period. Each change is worked out on the latest value, not the starting value, so the amount of change is different every time.
This means you multiply by the same multiplier once for each period. Three years of 4% growth is × 1.04 × 1.04 × 1.04, which is × 1.043. Compound interest is growth. Depreciation, where something loses value each year, is decay with a multiplier below 1.
An iterative process uses a rule again and again: the answer from one step is the starting value for the next step.
Rules to remember
- Growth: value after n periods = starting value × (1 + r/100)n
- Decay: value after n periods = starting value × (1 − r/100)n
- Doubling: value = starting value × 2n, where n = number of doubling periods.
- To find the rate: multiplier = nth root of (final value ÷ starting value).
- Iteration Vn+1 = aVn + b: put V0 in to get V1, then V1 in to get V2.
- Compound interest gives more than simple interest after the first year at the same rate.
Common mistake
Multiplying the percentage by the number of years, for example treating 15% a year for 2 years as 30%. Raise the multiplier to a power instead: 0.852.
Worked example
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