Audrey is saving money and plans on making quarterly contributions into an account earning an annual interest rate of $3.8 \%$ compounded quarterly. If Audrey would like to end up with $\$ 6,000$ after $8$ years, how much does she need to contribute to the account every quarter, to the nearest dollar? Use the following formula to determine your answer. $\newline$ $A=d\left(\frac{(1+i)^{n}-1}{i}\right)$ $\newline$ $A=$ the future value of the account after $n$ periods $\newline$ $d=$ the amount invested at the end of each period $\newline$ $i=$ the interest rate per period $\newline$ $n=$ the number of periods $\newline$ Answer:

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Grade 7

Compound interest

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Q. Audrey is saving money and plans on making quarterly contributions into an account earning an annual interest rate of

3.8 \%

compounded quarterly. If Audrey would like to end up with

\$ 6,000

after

8

years, how much does she need to contribute to the account every quarter, to the nearest dollar? Use the following formula to determine your answer.

\newline

A=d\left(\frac{(1+i)^{n}-1}{i}\right)

\newline

A=

the future value of the account after

n

periods

\newline

d=

the amount invested at the end of each period

\newline

i=

the interest rate per period

\newline

n=

the number of periods

\newline

Answer:

Identify values: Identify the given values from the problem. $\newline$ $A$ (future value of the account) = $\$6,000$ $\newline$ $i$ (interest rate per period) = $3.8\%$ annual interest rate compounded quarterly, which is $\frac{3.8\%}{4}$ per quarter $\newline$ $n$ (number of periods) = $8$ years with quarterly contributions, which is $8 \times 4$ quarters per year
Convert interest rate: Convert the annual interest rate to a quarterly interest rate. $i = \frac{3.8\% \text{ per year}}{4 \text{ quarters per year}} = \frac{0.038}{4} = 0.0095$ per quarter
Calculate total periods: Calculate the total number of periods. $n = 8$ years $* 4$ quarters per year $= 32$ quarters
Use formula for amount: Use the formula to find the amount invested at the end of each period $d$ . $A = d \times \left(\frac{(1 + i)^{n} - 1}{i}\right)$ (\newline\) $\$6,000 = d \times \left(\frac{(1 + 0.0095)^{32} - 1}{0.0095}\right)$
Calculate inside parentheses: Calculate the value inside the parentheses. $\newline$ $(1 + 0.0095)^{32} - 1 = (1.0095)^{32} - 1$
Calculate exponentiation: Calculate the exponentiation. $\newline$ $(1.0095)^{32} \approx 1.349858807576003$
Subtract from result: Subtract $1$ from the result of the exponentiation. $\newline$ $1.349858807576003 - 1 \approx 0.349858807576003$
Divide by interest rate: Divide the result by the interest rate per period. $\newline$ $0.349858807576003 / 0.0095 \approx 36.82724395326347$
Solve for d: Solve for d by dividing the future value of the account by the result from the previous step. $\newline$ $\$6,000 / 36.82724395326347 \approx 162.974$
Round result: Round the result to the nearest dollar. $d \approx \$163$