The amount of money that can be withdrawn each month for 10 years is equal to: A. $769. 27.
Given the following data:
To determine the amount of money that can be withdrawn each month for 10 years:
First of all, we would convert the compounded interest rate into an effective monthly interest rate as follows:
[tex]r = \frac{0.032}{12} = 0.00267[/tex]
Next, we would calculate the future value for these annuities by using this formula:
[tex]A = \frac{P[(1+r)^{12t} - 1]}{r}[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]A = \frac{235.15[(1+0.00267)^{12 \times 20}\; - \;1]}{0.00267}\\\\A = \frac{235.15[(1.00267)^{240}\; - \;1]}{0.00267}\\\\A = \frac{235.15[1.8964\; - \;1]}{0.00267}\\\\A = \frac{235.15[0.8964]}{0.00267}\\\\A = \frac{210.789}{0.00267}[/tex]
A = $78,947.19
Now, we can the amount of money that can be withdrawn each month for 10 years by using this formula:
[tex]A = \frac{P[(1+r)^{12t} - 1]}{r(1+r)^{12t}}[/tex]
Making P the subject of formula, we have:
[tex]P = \frac{A[r(1+r)^{12t}]}{(1+r)^{12t} - 1}[/tex]
Substituting the given parameters into the formula, we have;
[tex]P = \frac{78,947.19[0.00267(1+0.00267)^{12 \times 10}]}{(1+0.00267)^{12 \times 10} - 1}\\\\P = \frac{78,947.19[0.00267(1.3771)]}{1.3771 - 1}\\\\P = \frac{290.278}{0.3771 }[/tex]
P = $769.27
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