Question

Samuel deposits $1,000 every month into an account earning a monthly interest rate of 0.45%. How many years would it be until Samuel had $86,000 in the account, to the nearest tenth of a year? Use the formula: A = ((1+i)^n - 1) / i, where A is the future value, d is the monthly deposit, i is the interest rate per period, and n is the number of periods.

A) 8.3 years
B) 7.5 years
C) 9.2 years
D) 10.1 years

Answer 1

Using the future value formula for a series of deposits, you solve for the number of periods and then convert months to years to find that it takes approximately 7.5 years for Samuel to have $86,000 in the account, which makes B) 7.5 years the correct answer.

Step-by-step explanation:

To solve this problem, you'll need to use the formula for the future value of a series of deposits with interest: A = ​((1+i)^n - 1) / i. The goal is to solve for n, which is the number of periods, and then convert that into years. Let's substitute the given values into the formula where A is the future value ($86,000), d is the monthly deposit ($1,000), and i is the interest rate per period (0.45% or 0.0045).

First, rearrange the formula to solve for n: n = ​(log(A​i + 1)) / log(1+i). After substituting values you get: n = ​(log((86000​​​​​​​​​*0.0045)/1000 + 1)) / log(1+0.0045). Calculating this you'll get the number of months. To convert the months to years, divide by 12.

After calculating, you find that n is approximately 7.5 years, so the correct answer is B) 7.5 years.