Question

Aria paid $75,000 for her house its property value increased by 22% per year. When Aria sold her house after eleven years, how much was it worth, to the nearest hundred dollars?

Answer 1

`$668,400`

Explanation:

This is a compound interest problem. An amount is increased by a fixed amount over a set time interval, and the new amount becomes the basis for the next interval increase calculation. The formula for compound interest is:
`A = P*(1 + (r/n)^(nt))`

Where A is the amount, P is the principal (initial amount), r is the rate of interest (as a decimal, not percentage) and n is the number of times the interest is calculated during the interval (t) intervals (years, months), and t is the number of years.

For this problem we have P = 75000, r = 0.22 , and n = 1 and t = 11
`A = 75000*(1 + (0.22/1)^(1*11))`; `A = 75000*(1.22)^(11))`

`A = 75000*(8.912)`
`A = 668373` or `$668,400` to the nearest hundred dollars.

As an example, if we took ½ the rate with twice the compounding frequency, would she have more or less value? N = 2. r =0.11
`A = 75000*(1 + (0.11/2)^(2*11))`; `A = 75000*(1.055)^(22)`
`A = 75000*(3.25)`
`A = 243565` Higher interest is better than more frequent compounding.