Percentage Change & Growth/Decay
Percentage Change & Growth/Decay
Percentage Increase and Decrease
To increase or decrease by a percentage, use a multiplier.
- Increase by 15% → multiply by 1.15
- Decrease by 15% → multiply by 0.85
- Increase by 3.5% → multiply by 1.035
General rule: Multiplier = 1 + (percentage/100) for increase, or 1 − (percentage/100) for decrease.
Worked Example: Increase £460 by 8%.
- 460 × 1.08 = £496.80
Worked Example: Decrease 750 by 12%.
- 750 × 0.88 = 660
Percentage Change
Percentage change = (change ÷ original) × 100
Worked Example: A house price rises from £240,000 to £276,000. Find the percentage increase.
- Change = 276,000 − 240,000 = 36,000
- Percentage increase = (36,000 ÷ 240,000) × 100 = 15%
Reverse Percentages
When given the value AFTER a percentage change, find the ORIGINAL.
Worked Example: After a 30% discount, a jacket costs £56. Find the original price.
- £56 represents 70% (since 100% − 30% = 70%)
- 70% = £56
- 1% = £56 ÷ 70 = £0.80
- 100% = £0.80 × 100 = £80
Using multipliers: Original × 0.7 = 56, so Original = 56 ÷ 0.7 = £80
Worked Example: A population increases by 12% to 67,200. What was the original population?
- 67,200 ÷ 1.12 = 60,000
Repeated Percentage Change (Higher)
For the same percentage change applied multiple times, use the multiplier raised to a power.
Formula: Final value = Original × (multiplier)ⁿ
Worked Example: A town's population is 50,000 and grows by 2% each year. Find the population after 5 years.
- Population = 50,000 × (1.02)⁵
- = 50,000 × 1.10408...
- = 55,204 (to nearest whole number)
Exponential Growth and Decay
Growth: The multiplier is greater than 1. Examples include population growth, compound interest, bacterial growth.
Decay: The multiplier is between 0 and 1. Examples include depreciation, radioactive decay, cooling.
Worked Example: A car worth £18,000 depreciates by 20% each year. After how many years will it be worth less than £5,000?
| Year | Value |
|---|---|
| 0 | £18,000 |
| 1 | £18,000 × 0.8 = £14,400 |
| 2 | £14,400 × 0.8 = £11,520 |
| 3 | £11,520 × 0.8 = £9,216 |
| 4 | £9,216 × 0.8 = £7,372.80 |
| 5 | £7,372.80 × 0.8 = £5,898.24 |
| 6 | £5,898.24 × 0.8 = £4,718.59 |
- After 6 years the car is worth less than £5,000.
Or using the formula: 18000 × 0.8ⁿ < 5000 → 0.8ⁿ < 5/18 → solving by trial or logarithms gives n > 5.7, so 6 years.
Simple vs Compound Interest
Simple interest: The same fixed amount is added each period (based on original only).
- Simple interest = principal × rate × time
Compound interest: Interest is added to the running total, so you earn interest on interest.
- Final amount = P × (1 + r/100)ⁿ
Worked Example: £2,000 at 5% for 3 years.
- Simple: Interest = 2000 × 0.05 × 3 = £300. Total = £2,300
- Compound: 2000 × 1.05³ = 2000 × 1.157625 = £2,315.25
Compound interest gives £15.25 more because interest earns interest.
Exam Tips
- In reverse percentage questions, divide by the multiplier — do NOT find the percentage of the new amount
- For repeated change, always use the multiplier to the power of n, not n separate calculations
- If a question says a value "has increased by 20%" and asks for the original, divide by 1.2 (not by 0.2)
- "By what percentage" questions always use: (change ÷ ORIGINAL) × 100