Josue is saving money and plans on making quarterly contributions into an account earning a quarterly interest rate of $1.875 \%$ . If Josue would like to end up with $\$ 95,000$ after $15$ 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. Josue is saving money and plans on making quarterly contributions into an account earning a quarterly interest rate of

1.875 \%

. If Josue would like to end up with

\$ 95,000

after

15

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:

Identify Given Values: Identify the given values from the problem. $\newline$ $A$ (future value of the account) = $\$95,000$ $\newline$ $i$ (interest rate per period) = $1.875\%$ or $0.01875$ when converted to decimal $\newline$ $n$ (number of periods) = $15$ years $*$ $4$ quarters/year = $60$ quarters $\newline$ Now we can use these values in the formula provided.
Substitute Values into Formula: Substitute the given values into the formula. $\newline$ We have the formula $A = d \times \left(\frac{(1 + i)^{n} - 1}{i}\right)$ , where $A$ is the future value, $d$ is the amount invested each period, $i$ is the interest rate per period, and $n$ is the number of periods. $\newline$ Let's plug in the values: $\newline$ $\$95,000 = d \times \left(\frac{(1 + 0.01875)^{60} - 1}{0.01875}\right)$
Calculate Compound Factor: Calculate the compound factor. $\newline$ First, calculate $(1 + i)^n$ : $\newline$ $(1 + 0.01875)^{60}$ $\newline$ Now, use a calculator to find the value.
Continue Calculation: Continue the calculation from the previous step. $\newline$ Using a calculator, we find: $\newline$ $(1 + 0.01875)^{60} \approx 2.45489$ $\newline$ Now we can update our equation: $\newline$ $\$95,000 = d \times \left(\frac{2.45489 - 1}{0.01875}\right)$
Simplify Equation: Simplify the equation further. $\newline$ Subtract $1$ from $2$ . $45489$ : $\newline$ $2.45489 - 1 = 1.45489$ $\newline$ Now, divide by the interest rate: $\newline$ $\frac{1.45489}{0.01875} \approx 77.59413$ $\newline$ Update the equation with this value: $\newline$ $\$95,000 = d \times 77.59413$
Solve for Amount Invested: Solve for $d$ , the amount invested each period. $\newline$ To find $d$ , divide the future value by the compound factor: $\newline$ $d = \frac{\$95,000}{77.59413}$ $\newline$ Use a calculator to find $d$ .
Calculate and Round Value: Calculate the value of $d$ and round to the nearest dollar. $\newline$ Using a calculator, we find: $\newline$ $d \approx \$(95,000) / 77.59413 \approx \$(1224.14)$ $\newline$ Since we need to round to the nearest dollar, $d \approx \$(1224)$ .