You have $$ \$ 300,000 $$ saved for retirement. Your account earns $$ 7 \% $$ interest. How much will you be able to pull out each month, if you want to be able to take withdrawals for 25 years? $$ \$ \square $$

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UpStudy AI · Accepted Answer

Let the monthly withdrawal be $$ P $$ and use the annuity formula

$$
P = \frac{r \cdot PV}{1 - (1 + r)^{-N}}
$$

where

- $$ PV = \$300,\!000 $$ (the present value),
- $$ r = \frac{0.07}{12} $$ (the monthly interest rate), and
- $$ N = 25 \times 12 = 300 $$ (the total number of withdrawals).

**Step 1. Find the monthly interest rate**

$$
r = \frac{0.07}{12} \approx 0.0058333
$$

**Step 2. Substitute the values into the formula**

$$
P = \frac{\frac{0.07}{12} \times 300,\!000}{1 - \left(1 + \frac{0.07}{12}\right)^{-300}}
$$

This simplifies to

$$
P = \frac{300,\!000 \times 0.0058333}{1 - \left(1.0058333\right)^{-300}}
$$

**Step 3. Compute the numerator**

$$
300,\!000 \times 0.0058333 \approx 1750
$$

**Step 4. Compute the denominator**

First, calculate $$\left(1.0058333\right)^{300}$$. Taking logarithms:

$$
\ln\left(1.0058333\right) \approx 0.0058166, \quad 300 \times 0.0058166 \approx 1.745
$$

Thus,

$$
\left(1.0058333\right)^{300} \approx e^{1.745} \approx 5.723
$$

Then,

$$
\left(1.0058333\right)^{-300} \approx \frac{1}{5.723} \approx 0.1748
$$

Now, the denominator is

$$
1 - 0.1748 \approx 0.8252
$$

**Step 5. Compute $$ P $$**

$$
P \approx \frac{1750}{0.8252} \approx 2120
$$

Thus, the monthly withdrawal amount is approximately

$$
\$2120
$$
You can withdraw approximately \$2,120 each month.

You have \( \$ 300,000 \) saved for retirement. Your account earns \( 7 \% \) interest. How much will you be able to pull out each month, if you want to be able to take withdrawals for 25 years? \( \$ \square \)

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