Questions: Question 11, 8.5.15 Part 1 of 2 a. Use the appropriate formula to determine the periodic deposit. b. How much of the financial goal comes from deposits and how much comes from interest? Periodic Deposit, Rate, Time, Financial Goal ? at the end of each month, 7.75% compounded monthly, 45 years, 1,500,000 a. The periodic deposit is 1,245.

Question 11, 8.5.15 Part 1 of 2 a. Use the appropriate formula to determine the periodic deposit. b. How much of the financial goal comes from deposits and how much comes from interest? Periodic Deposit, Rate, Time, Financial Goal ? at the end of each month, 7.75% compounded monthly, 45 years, 1,500,000 a. The periodic deposit is 1,245.
Transcript text: Question 11, 8.5.15 Part 1 of 2 a. Use the appropriate formula to determine the periodic deposit. b. How much of the financial goal comes from deposits and how much comes from interest? \begin{tabular}{|l|l|l|l|} \hline Periodic Deposit & Rate & Time & Financial Goal \\ \hline$\$$ ? at the end of each month & $7.75 \%$ compounded monthly & 45 years & $\$ 1,500,000$ \\ \hline \end{tabular} a. The periodic deposit is $\$ 1,245$.
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Solution

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Solution Steps

Solution Approach

To determine the periodic deposit needed to reach a financial goal with compound interest, we can use the future value of an annuity formula. This formula relates the periodic deposit, interest rate, number of periods, and future value. We will rearrange the formula to solve for the periodic deposit. Given the interest rate is compounded monthly, we will convert the annual rate to a monthly rate and calculate the number of periods in months.

Step 1: Calculate Monthly Rate and Total Months

Given the annual interest rate \( r = 7.75\% \), we convert it to a monthly rate: \[ r_{monthly} = \frac{7.75}{100 \times 12} = 0.0064583333 \] The total time in months for \( 45 \) years is: \[ n = 45 \times 12 = 540 \]

Step 2: Use the Future Value of Annuity Formula

We apply the future value of an annuity formula rearranged to solve for the periodic deposit \( P \): \[ FV = P \times \left( \frac{(1 + r_{monthly})^n - 1}{r_{monthly}} \right) \] Rearranging gives: \[ P = \frac{FV}{\left( \frac{(1 + r_{monthly})^n - 1}{r_{monthly}} \right)} \] Substituting \( FV = 1500000 \): \[ P = \frac{1500000}{\left( \frac{(1 + 0.0064583333)^{540} - 1}{0.0064583333} \right)} \]

Step 3: Calculate Periodic Deposit

After performing the calculations, we find: \[ P \approx 310 \] Rounding up to the nearest dollar gives: \[ P = 310 \]

Final Answer

The periodic deposit required is \\(\boxed{310}\\).

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