Question

7100 dollars is placed in an account with an annual interest rate of 7.75%. how much will be in the account after 30 years, to the nearest cent

Answer 1

Given:

7100 dollars is placed in an account with an annual interest rate of 7.75%.

To find:

How much will be in the account after 30 years, to the nearest cent

Solution:

The sum of money placed in the account, P = 7100 dollars

The annual rate of interest, R = 7.75%

The no. of years, n = 30 years

We know,

$\boxed{\bold{A = P [1 + \frac{R}{100} ]^n}}$

Now, on substituting the values of P, R and n in the formula above, we get

The amount present in the account after 30 years will be,

= $7100 [1 + \frac{7.75}{100} ]^3^0$

= $7100 [1 + 0.0775 ]^3^0$

= $7100 [1.0775 ]^3^0$

= $\bold{\ \:66646.4028}$

rounding off to the nearest cents

= $\bold{\ \:66646.40}$

Thus, $\underline{\bold{\ \:66646.4028}}$ will be there in the account after 30 years.

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