Question

Sally invests $4,000 in a savings account which compounds interest monthly. After 10 years, she has a balance of $4,573.23. What is the annual interest rate (as a percentage) of the account, rounded to the nearest 0.01%?

Answer 1

The problem states that the interest is compounded monthly. For uniformity, we convert 10 years to months. This is equivalent to 120 months.

The equation would be


`F=P (1+i)^(n)`

where F is the future worth = $4,573.23
P is the present worth = $4,000
n is the number of periods = 120 months


`4,573.23= 4,000(1+i)^(120)`


`(4573.23)/(4000)= (1+i)^(120)`


`1+i = \sqrt[20]{1.1433075}`


`i = 1.00672-1`


`i = 0.00672 \ (nominal \ rate)`

To convert to effective rate


`i_(eff) =(1+ (i)/(m) )^(m)-1`

where m is the number of periods in a year. There are 120 month in a year.


`i_(eff) =(1+ (0.00672)/(120) )^(120)-1`


`i_(eff) = 0.00674`

or,

The annual interest rate would be 0.67%