For **percentage** **increase** and **decrease**, and we can use the **compound interest** formula to help solve many problems involving repeated percentage increase and decrease.

#### Increase Example 1

Suppose we are told that the population of a village increases by 2% each year, and that the current population is 2,345. What would it be in 5 years’ time?

- We are increasing the population by 2% which means we are adding 2% to 100% or using a
**multiplier**of 1.02

We have applied the same principal used in our compound interest to get this particular answer.

#### Decrease Example 2

We can use a formula that is similar to compound interest to work out values that decrease or depreciate at a constant rate over time

If we start with an amount called the principal ‘P’, a rate of depreciation ‘D’ and apply it for some number of years ‘N’, the value of the investment can be expressed as:

Using this if you were told that the value of a new car depreciates at a rate of 15% per year and that the car cost £12,000 in 2005, how much would it be worth in 2013?

We have not been told the value ‘**N**’ but we can calculate that there are 8 years between 2005 and 2013.

Applying our depreciation formula with ‘P’ equal to 12,000, ‘D’ equal to 15% and ‘N’ equal 8 years

V = 12000 x (1-0.15)^{8}

We get an answer of £3,270 to the nearest pound.