Math 2000 - Solutions 7
Fall 2001
Unit 4B 14. To create a retirement fund worth $2,000,000 by making monthly deposits for 30 years, assuming an APR of 9%, you must deposit

\begin{displaymath} \frac{\$2,000,000 \times \frac{0.09}{12}}{(1+\frac{0.09}{12})^{12 \times 30}-1} = \$1092.45 \end{displaymath}

each month. 21. First, we need to figure out how much you must save by the time you reach 60. In order to draw an annual income of $50,000 from then on, without end, if we assume an APR of 8%, then the amount saved must satisfy $A \times 0.08 = \$50,000$. Hence, $A= \frac{\$50,000}{0.08}= \$625,000$. Next, in order to save $625,000 after 30 years of monthly deposits at 8%, you must deposit

\begin{displaymath} \frac{\$625,000 \times \frac{0.08}{12}}{(1+\frac{0.08}{12})^{12 \times 30}-1} = \$419.36 \end{displaymath}

each month. Unit 4C 16. A 2-year loan of $4000 at an APR of 8% requires monthly payments of

\begin{displaymath} \frac{\$4,000 \times \frac{0.08}{12}}{1-\left(1+\frac{0.08}{12}\right)^{(-12\times 2)}}=\$180.91, \end{displaymath}

which is much more than you can afford. Borrowing the same amount of money for 3 years at 9% requires monthly payments of

\begin{displaymath} \frac{\$4,000 \times \frac{0.09}{12}}{1-\left(1+\frac{0.09}{12}\right)^{(-12\times 3)}}=\$127.20, \end{displaymath}

which is affordable. A 4-year loan of $4000 at 10% requires monthly payments of

\begin{displaymath} \frac{\$4,000 \times \frac{0.10}{12}}{1-\left(1+\frac{0.10}{12}\right)^{(-12\times 4)}}=\$101.45, \end{displaymath}

which is even more affordable. The second and third options both meet your needs. You can check that the total payments are greater with the third loan, because of the extra year of payments, so you will probably want to go with the second loan option. 22. Suppose now we make monthly payments of $300 towards a balance of $1200 at an APR of 18%, so that the monthly interest rate is 1.5%. Then, at the end of each month, the balance is reduced by $300 but also increased by $75, plus the interest on the previous month's balance. At the end of the first month your balance becomes: $\$1200.00-\$300+\$75 +(1.5\% \times \$1200.00) = \$993.00$. At the end of the second month it's: $\$993.00-\$300+\$75 +(1.5\% \times \$993.00) = \$782.90$. Continuing in this way, we get the entries in last column of the following table.

$ \begin{tabular}{\vert r\vert c\vert c\vert c\vert c\vert} \hline \emph{M} &\... ...$353.18 \\ 5 & \$300 & \$75 & \$5.30 & \$133.48 \\ \hline \end{tabular} $

At the end of the fifth month the balance becomes: $\$353.18-\$200+\$75 +(1.5\% \times \$353.18) = \$133.48$. The sixth payment of $300 is more than we need to pay this off; a partial payment suffices. 34a. A 25-year loan of $60,000 at an APR of 8% requires monthly payments of

\begin{displaymath} \frac{\$60,000 \times \frac{0.08}{12}}{1-\left(1+\frac{0.08}{12}\right)^{(-12\times 25)}}=\$463.09. \end{displaymath}

b. If we pay off the loan in 15 years instead of 25 years, the payments increase to:

\begin{displaymath} \frac{\$60,000 \times \frac{0.08}{12}}{1-\left(1+\frac{0.08}{12}\right)^{(-12\times 15)}}=\$573.39. \end{displaymath}

c. For the 25-year loan, the total payments are $12 \times 25 \times \$463.09 = \$138,927.00$, and for the 15-year loan they are $12 \times 15 \times \$573.39 = \$103,210.20$.