Julian is saving money and plans on making quarterly contributions into an account earning a quarterly interest rate of $1.875 \%$ . If Julian would like to end up with $\$ 17,000$ after $10$ years, how much does he 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. Julian is saving money and plans on making quarterly contributions into an account earning a quarterly interest rate of

1.875 \%

. If Julian would like to end up with

\$ 17,000

after

10

years, how much does he 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:

Given Values: We are given: $\newline$ Future value of the account, $A = \$17,000$ $\newline$ Quarterly interest rate, $i = 1.875\%$ or $0.01875$ (as a decimal) $\newline$ Number of years, $t = 10$ $\newline$ Since contributions are made quarterly, there are $4$ periods per year. $\newline$ Number of periods, $n = 4$ periods/year $* 10$ years = $40$ periods $\newline$ We need to find the amount invested at the end of each period, $d$ . $\newline$ Use the formula $A = d\left(\frac{(1+i)^{n}-1}{i}\right)$ to solve for $d$ .
Convert Interest Rate: First, convert the interest rate from a percentage to a decimal by dividing by $100$ . $\newline$ $i = \frac{1.875\%}{100} = 0.01875$
Calculate Number of Periods: Next, calculate the number of periods over $10$ years with quarterly contributions. $\newline$ $n = 4$ periods/year $* 10$ years $= 40$ periods
Plug Values into Formula: Now, plug the values of $A$ , $i$ , and $n$ into the formula to solve for $d$ . $A = d\left(\frac{(1+i)^{n}-1}{i}\right)$ $\$17,000 = d\left(\frac{(1+0.01875)^{40}-1}{0.01875}\right)$
Calculate Factor: Calculate the factor $\frac{(1+i)^{n}-1}{i}$ . $\newline$ $\frac{(1+0.01875)^{40}-1}{0.01875} = \frac{(1.01875)^{40}-1}{0.01875}$ $\newline$ First, calculate $(1.01875)^{40}$ . $\newline$ $(1.01875)^{40} \approx 2.11356$
Subtract from Result: Subtract $1$ from the result of $(1.01875)^{40}$ . $2.11356 - 1 \approx 1.11356$
Divide by Interest Rate: Divide the result by $i$ ( $0.01875$ ). $\newline$ $1.11356 / 0.01875 \approx 59.3904$
Solve for d: Now, solve for d using the calculated factor. $\newline$ $17,000 = d \times 59.3904$ $\newline$ d = $\frac{17,000}{59.3904}$ $\newline$ d \approx $286.26$
Round to Nearest Dollar: Since the question asks for the nearest dollar, round the result to the nearest whole number. $\newline$ $d \approx \$286$