Question

How much should you deposit at the end of each month in an IRA that pays 11% compounded monthly to earn ​$50000 per year from interest​ alone, while leaving the principal​ untouched, to be withdrawn at the end of each year after you retire in 30 years?

Answer 1

Answer:$4545.45

Step-by-step explanation:

To determine how much you should deposit in an IRA each month to earn $50000 per year in interest alone, you can use the following formula:

Monthly deposit = (Annual interest / Interest rate) / Number of months per year

Plugging in the given values, you get:

Monthly deposit = ($50000 / 11%) / 12 months/year

= $4545.45

Therefore, you should deposit $4545.45 at the end of each month in your IRA to earn $50000 per year in interest alone. This assumes that the interest is compounded monthly, that the principal is left untouched, and that you will withdraw the full amount of interest earned at the end of each year after you retire in 30 years.

Answer 2

Answer:

$154.07

Step-by-step explanation:

$\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \\\\where:\\\\ \phantom{ww} \bullet A = final amount \\ \phantom{ww} \bullet P = principal amount \\ \phantom{ww} \bullet r = interest rate (in decimal form) \\ \phantom{ww} \bullet n = number of times interest is applied per year \\ \phantom{ww} \bullet t = time (in years) \\ \end{minipage}}$

First find the Future Account Value (target amount) you need at the beginning of your retirement in order to earn $50,000 of interest each year whilst leaving the principal untouched.

Given:

• A = P + $50,000
• P = P
• r = 11% = 0.11
• n = 12 (monthly)
• t = 1 year

Substitute the values in the compound interest formula to find the Future Account Value:

$\implies P+50000=P\left(1+\dfrac{0.11}{12}\right)^{12 \cdot 1}$

$\implies P+50000=P\left(\dfrac{1211}{1200}\right)^{12}$

$\implies 50000=P\left(\dfrac{1211}{1200}\right)^{12}-P$

$\implies 50000=P\left(\left(\dfrac{1211}{1200}\right)^{12}-1\right)$

$\implies P=\dfrac{50000}{\left(\left(\dfrac{1211}{1200}\right)^{12}-1\right)}$

$\implies P=432081.77$

Therefore, the Future Account Value needed at the beginning of your retirement is $432,081.77 to allow you to earn $50,000 per year from interest alone.

$\boxed{\begin{minipage}{8.5 cm}\underline{Savings Plan Formula}\\\\ FV=PMT\left[\dfrac{\left(1+\frac{r}{n}\right)^{nt}-1}{\frac{r}{n}} \right] \\\\where:\\\\ \phantom{ww} \bullet FV = future value\\ \phantom{ww} \bullet PMT = periodic payment \\ \phantom{ww} \bullet r = APR (in decimal form) \\ \phantom{ww} \bullet t = years \\ \phantom{ww} \bullet n = number of payments per year \\ \end{minipage}}$

Given:

• FV = $432,081.77
• r = 11% = 0.11
• t = 30 years
• n = 12 (monthly)

Substitute the values into the Savings Plan formula and solve for PMT to find the monthly payments:

$\implies 432081.77=PMT\left[\dfrac{\left(1+\frac{0.11}{12}\right)^{12 \cdot 30}-1}{\frac{0.11}{12}} \right]$

$\implies 432081.77=PMT\left[\dfrac{\left(\frac{1211}{1200}\right)^{360}-1}{\frac{11}{1200}} \right]$

$\implies 432081.77=PMT\left[2804.519736 \right]$

$\implies PMT=\dfrac{432081.77}{2804.519736}$

$\implies PMT=154.0662255$

Therefore, you should deposit $154.07 at the end of each month to be able to withdraw $50,000 per year from interest alone at the end of each year after you retire in 30 years.