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.