Compound Interest

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What is compound interest?

Compound interest is where interest is paid on the interest from the previous year (or whatever time frame is being used), as well as on the original amount
This is different from simple interest where interest is only paid on the original amount
- Simple interest goes up by the same amount each time whereas compound interest goes up by an increasing amount each time

How do you work with compound interest?

Keep multiplying by the decimal equivalent of the percentage you want (the multiplier, p)
A 25% increase (p = 1 + 0.25) each year for 3 years is the same as multiplying by 1.25 × 1.25 × 1.25
- Using powers, this is the same as × 1.25³
In general, the multiplier p applied n times gives an overall multiplier of pⁿ
If the percentages change varies from year to year, multiply by each one in order
- a 5% increase one year followed by a 45% increase the next year is 1.05 × 1.45
In general, the multiplier p₁ followed by the multiplier p₂ followed by the multiplier p₃... etc gives an overall multiplier of p₁p₂p₃...
Alternative method: A formula for the final ("after") amount is $P {(1 + \frac{r}{100})}^{n}$ where...
- ...P is the original ("before") amount, r is the % increase, and n is the number of years
- Note that $1 + \frac{r}{100}$ is the same value as the multiplier

Examiner Tip

It is easier to multiply by the decimal equivalent "raised to a power" than to multiply by the decimal equivalent several times in a row

Worked example

Jasmina invests £1200 in a savings account which pays compound interest at the rate of 2% per year for 7 years.

To the nearest pound, what is her investment worth at the end of the 7 years?

We want an increase of 2% per year, this is equivalent to a multiplier of 1.02, or 102% of the original amount

This multiplier is applied 7 times; $\times 1.02 \times 1.02 \times 1.02 \times 1.02 \times 1.02 \times 1.02 \times 1.02 = 1 . 02^{7}$

Therefore the final value after 7 years will be

$£ 1200 \times 1 . 02^{7} = £ 1378.42$

Round to the nearest pound

$£ 1378$

Alternative method
Or use the formula for the final amount $P {(1 + \frac{r}{100})}^{n}$
Substitute P is 1200, r = 2 and n = 7 into the formula

$1200 {(1 + \frac{2}{100})}^{7}$

£1378 (to the nearest pound)

Depreciation

What is meant by depreciation?

Depreciation is where an item loses value over time
- For example: cars, game consoles, etc
Depreciation is usually calculated as a percentage decrease at the end of each year
- This works the same as compound interest, but with a percentage decrease

How do I calculate a depreciation?

You would calculate the new value after depreciation using the same method as compound interest
- Identify the multiplier, p (1 - "% as a decimal")
  - 10% depreciation would have a multiplier of p = 1 - 0.1 = 0.9
  - 1% depreciation would have a multiplier of p = 1 - 0.01 = 0.99
- Raise the multiplier to the power of the number of years (or months etc)
  - $p^{n}$
- Multiply by the starting value
New value is $A \times p^{n}$
- A is the starting value
- p is the multiplier for the depreciation
- n is the number of years
Alternative method: A formula for the final ("after") amount is $P {(1 - \frac{r}{100})}^{n}$ where...
- ...P is the original ("before") amount, r is the % decrease, and n is the number of years
If you are asked to find the amount the value has depreciated by:
- Find the difference between the starting value and the new value

Worked example

Mercy buys a car for £20 000. Each year its value depreciates by 15%.

Find the value of the car after 3 full years.

Identify the multiplier

100% - 15% = 85%

p = 1 - 0.15 = 0.85

Raise to the power of number of years

0.85³

Multiply by the starting value

£20 000 × 0.85³

= £12 282.50

Alternative method
Or use the formula for the final amount $P {(1 - \frac{r}{100})}^{n}$
Substitute P is 20 000, r = 15 and n = 3 into the formula

$20 00 {(1 - \frac{15}{100})}^{3}$

£12 282.50

Compound Interest (Cambridge O Level Maths)

Revision Note