Compound Interest and Depreciation

Compound interest and depreciation use almost exactly the same method. The main difference is whether the amount goes up or down.

1. Compound interest 📈

Compound interest is when money increases by a percentage each year, and you earn interest on the interest you’ve already gained.

For example, if you put £500 in a bank account with 4% compound interest, you gain 4% in the first year. In the second year, you gain 4% of the new total, not just the original £500.

The formula

$$\text{Final amount}=\text{Original amount}\times {(1+\frac{\text{Rate}}{100})}^{\text{Number of years}}$$Final amount=Original amount×(1+100Rate​)Number of years

Here’s how to use it:

• Original amount: How much money you start with.
• Rate: The percentage increase.
• Number of years: How long the money grows for.
• The multiplier: Add the percentage to 100%, then divide by 100.

For example, 4% interest gives a multiplier of $1.04$

Worked example

A student invests £500 at 4% compound interest for 3 years. Calculate the final amount.

Step 1: Find the multiplier

$1+\frac{4}{100}=1.04$

Step 2: Substitute into the formula

$$500\times {1.04}^3$$Step 3: Calculate

$$500\times 1.124864=562.432$$Final answer, to the nearest penny £562.43
Remember: the power of 3 means you multiply by 1.04 three times.
You can explore how the starting amount, interest rate and number of years affect the final amount below.

2. Depreciation 📉

Depreciation is when something loses value by a percentage each year.
Common examples include cars, phones and machinery.
The formula is almost the same as compound interest, but instead of adding the percentage, you subtract it.

$$\text{Final value}=\text{Original value}\times {(1-\frac{\text{Rate}}{100})}^{\text{Number of years}}$$Final value=Original value×(1−100Rate​)Number of years

Worked example

A car costs £12,000 and depreciates by 15% each year. Calculate its value after 3 years.

Step 1: Find the percentage remaining
$100\mathrm{\%}-15\mathrm{\%}=85\mathrm{\%}$

Step 2: Convert to a multiplier$100\mathrm{\%}-15\mathrm{\%}=85\mathrm{\%}$

$$\frac{85}{100}=0.85$$Step 3: Use the formula

$$12000\times {0.85}^3$$

$$=12000\times 0.614125$$Final answer £7,369.50

The car loses 15% of its value each year, so you keep 85% each year.

3. The most important thing to remember

The key to GCSE questions is finding the correct multiplier.

An easy way to remember it:

• Compound interest: multiplier is bigger than 1.
• Depreciation: multiplier is smaller than 1.

Both use the same structure:

$$\boxed{{\displaystyle \text{Original amount}\times {\text{multiplier}}^{\text{time}}}}$$

5. GCSE exam tips

• Read the question carefully. Look for words such as increases, interest, decreases and depreciates.
• Use brackets around the multiplier and the power. For example, $800\times (1.05{)}^2$800×(1.05)2.
• Don’t use simple interest by mistake. Compound interest means the percentage is applied to the new total each year.
• Don’t round too early. Keep the full calculator value until the end, then round money to two decimal places if needed.
• Show your working. Writing the multiplier and the formula can help you earn method marks even if your final answer is wrong.

The biggest thing to practise is recognising whether to use $1+\frac{r}{100}$1+100r​ or $1-\frac{r}{100}$1−100r​. Once you can do that, most GCSE compound interest and depreciation questions follow the same steps.