Question

U Do It Centers deposited $3,200 in an account two years ago and is depositing another $5,000 today. A final deposit of $3,500 will be made one year from now. What will the account balance be three years from now if the account pays 4.85 percent interest, compounded annually?

Answer 1

The account balance three years from now will be the sum of the future values of three separate deposits, each calculated using the formula for compound interest with an annual rate of 4.85%, and accounting for the different times each deposit will have been compounding.

Step-by-step explanation:

Calculating Compound Interest for Multiple Deposits

To determine the account balance three years from now with given details on multiple deposits and annual compound interest, we'll treat each deposit separately and sum the future values.

First deposit: $3,200, made two years ago, will have compounded for a total of 3 years at 4.85% interest when we reach the 3-year mark from now. Its future value (FV) is calculated using the formula FV = P(1 + r)^n, where P is the principal amount, r is the annual interest rate, and n is the number of compounding periods.

FV for the first deposit = $3,200 * (1 + 0.0485)^3

Second deposit: $5,000, made today, will have compounded for 3 years at 4.85% interest. Its future value is:

FV for the second deposit = $5,000 * (1 + 0.0485)^3

Third deposit: $3,500, will be made one year from now, and will have compounded for 2 years at 4.85% interest. Its future value is:

FV for the third deposit = $3,500 * (1 + 0.0485)^2

Adding the future values of all three deposits will give us the total account balance three years from now.

Answer 2

Total= $13,221.52

Step-by-step explanation:

Giving the following information:

Deposited $3,200 in an account two years ago and is depositing another $5,000 today.

A final deposit of $3,500 will be made one year from now.

Interest expense= 4.85% compounded annually.

We need to calculate the final value of each deposit using the following formula:

FV= PV*(1+i)^n

First deposit= 3,200*(1.0485^5)= $4,055.01

Second deposit= 5,000*(1.0485^2)= $5,496.76

Third deposit= 3,500*(1.0485)= $3,669.75

Total= $13,221.52