Question

Cooper invested $3,100 in an account paying an interest rate of 2%

compounded quarterly. Robert invested $3,100 in an account paying an
interest rate of 3 % compounded continuously.After 12 years, how much
more money would Robert have in his account than Cooper, to the nearest
dollar?

Answer 1

Robert will have $481 more in his account than Cooper.

Explanation:

Compound interest:

The compound interest formula is given by:

`A(t) = P(1 + (r)/(n))^(nt)`

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Continuous compounding:

Similar to compound interest:

`A(t) = P(1 + r)^t`

Cooper:

Cooper invested $3,100 in an account paying an interest rate of 2% compounded quarterly. This means that
`P = 3100, r = 0.02, n = 4`

After 12 years is A(12). So

`A(t) = P(1 + (r)/(n))^(nt)`

`A(12) = 3100(1 + (0.02)/(4))^(4*12)`

`A(12) = 3938.5`

Cooper will have $3938.5 in his account.

Robert:

Robert invested $3,100 in an account paying an interest rate of 3 % compounded continuously. So
`P = 3100, r = 0.03`.

After 12 years.

`A(t) = P(1 + r)^t`

`A(12) = 3100(1 + 0.03)^(12)`

`A(12) = 4419.9`

Robert will have $4419.9 in his account.

How much more money would Robert have in his account than Cooper, to the nearest dollar?

4419.9 - 3938.5 = 481.4

To the nearest dollar, Robert will have $481 more in his account than Cooper.