To determine how much you should deposit at the end of each month into an IRA that pays 5% interest compounded monthly, so that you can earn $110,000 a year from interest alone after 50 years, we can follow these steps:
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Determine the amount needed in the IRA at retirement: Since you want to earn $110,000 per year purely from interest, and the interest rate is 5% compounded monthly, we first need to find out the total principal amount required to achieve this.
The effective annual interest rate (with monthly compounding) for 5% is given by: \[ \text{Effective Annual Rate} = (1 + \frac{0.05}{12})^{12} - 1 \approx 0.051161 \text{ or } 5.1161% \]
However, for simplicity, we can continue to work with the nominal rate for the calculations regarding withdrawals.
If you want to earn $110,000 per year from the interest: \[ \text{Principal needed} = \frac{\text{Annual Withdrawal}}{\text{Annual Interest Rate}} = \frac{110,000}{0.05} = 2,200,000 \]
So, you need $2,200,000 in the IRA at the end of 50 years.
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Calculate the monthly deposit amount: Now we need to determine how much to deposit monthly to accumulate $2,200,000 in 50 years at a 5% annual interest rate compounded monthly.
The future value of a series of monthly deposits (ordinary annuity) can be calculated with the formula: \[ FV = P \times \frac{(1 + r)^n - 1}{r} \] where:
- \(FV\) = future value ($2,200,000)
- \(P\) = monthly deposit
- \(r\) = monthly interest rate (annual rate divided by 12)
- \(n\) = total number of deposits (number of years multiplied by 12)
Plugging in the values:
- \(r = \frac{0.05}{12} = \frac{0.05}{12} = 0.0041667\)
- \(n = 50 \times 12 = 600\)
Now, rearranging the formula to solve for \(P\): \[ P = \frac{FV \cdot r}{(1 + r)^n - 1} \]
Plugging in for \(FV\): \[ P = \frac{2,200,000 \cdot 0.0041667}{(1 + 0.0041667)^{600} - 1} \]
Next, we compute \((1 + 0.0041667)^{600}\): \[ (1 + 0.0041667)^{600} \approx 6.022575 \]
Now plugging this back into the equation for \(P\): \[ P = \frac{2,200,000 \cdot 0.0041667}{6.022575 - 1} \] \[ P = \frac{9,166.67}{5.022575} \approx 1,826.45 \]
Therefore, you should deposit approximately $1,826.45 at the end of each month into your IRA to achieve your goal of earning $110,000 per year from interest after 50 years.
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