# An investment of $1200 is increased to $1800 in 6 years. What is the annual rate of increase?

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The investment of $1200 is increased to $1800 in six years.

If the amount is considered to increase in a linear fashion, which is applicable for simple interest, the rate of increase is `(((1800 - 1200)/6)/1200)*100` = `(100/1200)*100` `~~` 8.33%

If the increase is exponential, as is the case in compound interest, the rate of increase is r where:

`1200*(1+r)^6 = 1800`

=> `(1+r)^6 = 3/2`

=> r = `(3/2)^(1/6) - 1`

=> r = 0.0699

=> r `~~` 6.99%

**The required rate of interest can be either 8.33% or 6.99% for the investment increasing in a linear fashion or increasing exponentially.**