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Answer :
Using the formula for monthly compound interest, we can calculate the balance after one month. To solve this problem, we can use the formula for the withdrawal from an account with monthly compounding interest:
P = D * (((1 + r)^n - 1) / r)
Where:
P = Present value of the account ($59,251.76)
D = Monthly withdrawal ($500)
r = Monthly interest rate (6%/12 months = 0.5% = 0.005)
n = Number of withdrawals (in months)
Rearrange the formula to solve for n:
n = ln((D/P * r) + 1) / ln(1 + r)
Now plug in the given values:
n = ln((500/59,251.76 * 0.005) + 1) / ln(1 + 0.005)
n ≈ 162.34 months
Since we need to find the number of years, we will divide the number of months by 12:
162.34 months / 12 months = 13.53 years
The closest answer to 13.53 years among the given options is 12 years 6 months (option c). Therefore, you will be withdrawing for approximately 12 years and 6 months.
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It will take approximately 181.18 months to exhaust the account at the current withdrawal rate. This is equivalent to about d) 15 years and 1 month (since there are 12 months in a year). So the answer is (d) 15 years.
To calculate the number of years it will take to exhaust the account while withdrawing 500 at the end of each month, we need to use the formula for the future value of an annuity:
[tex]FV = PMT x [(1 + r)^n - 1] / r[/tex]
where:
FV = future value
PMT = payment amount per period
r = interest rate per period
n = number of periods
In this case, PMT = 500, r = 6%/12 = 0.5% per month, and FV = 59,251.76.
We can solve for n by plugging in these values and solving for n:
[tex]59,251.76 = 500 x [(1 + 0.005)^n - 1] / 0.005[/tex]
Multiplying both sides by 0.005 and simplifying, we get:
[tex]296.26 = (1.005^n - 1)[/tex]
Taking the natural logarithm of both sides, we get:
ln(296.26 + 1) = n x ln(1.005)
n = ln(296.26 + 1) / ln(1.005)
n ≈ 181.18
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