Question

Jenny invests $8500 for 3 years in a savings account.
She gets 2.3% per year compound interest.

How much money will Jenny have in her savings account at the end of 3 years?
Give your answer correct to the nearest dollar.

Answer 1

SOLUTION

GIVEN

Jenny invests $8500 for 3 years in a savings account.

She gets 2.3% per year compound interest.

TO DETERMINE

The money Jenny will have in her savings account at the end of 3 years

( correct to the nearest dollar . )

EVALUATION

Invested amount = $ 8500

Rate of interest = 2.3%

Time = 3 years

Hence the money

$\displaystyle \sf{ = 8500 \times { \bigg(1 + \frac{2.3}{100} \bigg)}^{3} }$

$\displaystyle \sf{ = 8500 \times { \bigg(1 + 0.023 \bigg)}^{3} }$

$\displaystyle \sf{ = 8500 \times { \bigg(1 .023 \bigg)}^{3} }$

= 8500 × 1.07059917

= 9100.09295

≈ 9100 ( correct to the nearest dollar )

FINAL ANSWER

The money Jenny will have in her savings account at the end of 3 years $ 9100

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Answer 2

Given:

Jenny invests $8500 for 3 years in a savings account.

She gets 2.3% per year compound interest.

To find:

How much money will Jenny have in her savings account at the end of 3 years?  (Give your answer correct to the nearest dollar.)

Solution:

The formula for calculating the amount in compound interest is as follows:

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

The sum of money, P = $ 8500

The rate of interest, R = 2.3%

The no. of years, n = 3 years

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

$A = 8500 [ 1+ \frac{2.3}{100} ]^3$

$\implies A = 8500 [ \frac{102.3}{100} ]^3$

$\implies A = 8500 [ 1.023 ]^3$

$\implies A = 9100.09$

rounding off to the nearest dollar

$\implies \bold{A = \ \:9100}$

Thus, Jenny will have $ 9100 in her savings account at the end of 3 years.

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