Growth and decay (Edexcel GCSE Maths): Revision Notes
Growth and decay
Growth and decay problems use repeated percentage changes to model real-world situations where values increase or decrease by the same percentage over multiple time periods.
Compound interest
Compound interest occurs when money left in a bank account earns interest not only on the original amount invested, but also on any interest already accumulated from previous periods.
When calculating compound interest, you need to apply the same percentage increase repeatedly for each time period. For example, if someone invests £40,000 at 3% compound interest per year, the calculation works as follows:
| Year | Starting balance (£) | Interest earned (£) | New balance (£) |
|---|---|---|---|
| 1 | 40,000 | 1,200 | 41,200 |
| 2 | 41,200 | 1,236 | 42,436 |
| 3 | 42,436 | 1,273.08 | 43,709.08 |
The total interest earned after 3 years would be £3,709.08, giving a final balance of £43,709.08.
Using multipliers and index notation
Rather than calculating each year separately, you can use multipliers to find the result more efficiently.
Key Multiplier Rules:
For any percentage change:
- Growth multiplier =
- Decay multiplier =
The general formula for repeated percentage change is:
Where n is the number of time periods the change occurs.
For the compound interest example above:
- Multiplier =
- Final amount =
Growth examples
Worked Example: Cell Growth in a Petri Dish
A petri dish contains 5,000 cells initially. The cell population increases by 20% each day. Find the number of cells after 4 days.
Step 1: Calculate the growth multiplier Growth multiplier =
Step 2: Apply the formula Final number = cells
Answer: After 4 days, there will be 10,368 cells in the petri dish.
This shows how exponential growth works - the population doesn't just increase by the same amount each day, but by the same percentage, leading to accelerating growth.
Repeated decrease (decay)
Depreciation and decay problems involve repeated percentage decreases. The same formula applies, but the multiplier will be less than 1.
Worked Example: Car Depreciation
A car worth £15,000 loses 8% of its value each year. Find its value after 3 years.
Step 1: Calculate the decay multiplier
Decay multiplier =
Step 2: Apply the formula Final value =
Answer: After 3 years, the car will be worth £11,620.32.
Notice how the car doesn't lose £1,200 (8% of £15,000) each year - it loses 8% of whatever its current value is, which becomes a smaller amount as the car's value decreases.
Key formula summary
Essential Formula for Growth and Decay:
Where:
- Multiplier for growth =
- Multiplier for decay =
- n = number of time periods
Exam tips
Exam Success Strategies:
- Always identify whether the problem involves growth (multiplier > 1) or decay (multiplier < 1)
- Write down the multiplier clearly before calculating
- Use index notation (powers) rather than repeated multiplication
- Check your answer makes sense - growth should give a larger final amount, decay should give a smaller one
- Round appropriately for the context (money to 2 decimal places, population to whole numbers)
Remember!
Key Points to Remember:
- Compound interest means earning interest on interest as well as the original investment
- Multipliers greater than 1 represent growth, multipliers less than 1 represent decay
- The formula works for all repeated percentage change problems
- Index notation makes calculations much more efficient than working year by year
- Always check if your final answer is reasonable for the context of the problem