Chapter 7.5, Problem 26PPS

a.

To determine

To find: the amount that D should add to ensure that he has $1500 in 5 years.

Answer to Problem 26PPS

  $\$\text{304}\text{.99}$

Explanation of Solution

Given:

D deposited $500 that earns interest monthly and after 8 years the amount in his account is $807.07.

Calculation:

Let the rate of interest be $r$ per annum.

For the investment of $500, equation can be set up as:

  $\begin{array}{c}807.07=500{\left(1+\frac{r}{12}\right)}^{8\times 12}\\ 807.07=500{\left(1+\frac{r}{12}\right)}^{96}\\ {\left(1+\frac{r}{12}\right)}^{96}=\frac{807.07}{500}\\ {\left(1+\frac{r}{12}\right)}^{96}=\text{1}\text{.61414}\\ 1+\frac{r}{12}=\text{1}\text{.0049999}\\ \frac{r}{12}=0.\text{0049999}\\ r=\text{0}\text{.05999976}\end{array}$

Now, he wants to have $1500 in 5 years in his account. Let P be amount needed to be in his account in present.

Then,

  $\begin{array}{c}A=P{\left(1+\frac{r}{12}\right)}^{12n}\\ 1500=P{\left(1+\frac{\text{0}\text{.05999976}}{12}\right)}^{12\times 5}\\ 1500=P{\left(\text{1}\text{.0049999}\right)}^{60}\\ P=\frac{1500}{{\left(\text{1}\text{.0049999}\right)}^{60}}\\ P=\text{1112}\text{.06}\end{array}$

That means, he should add $\text{\$1112}\text{.06}-\$807.07=\$\text{304}\text{.99}$ .

b.

To determine

To explain: The solution process used in part (a).

Explanation of Solution

First find the rate of interest that gives the amount $807.07 after 8 years on investment of $500.

Then, find the amount that needs to be invested that would grow to $1500 in years at that rate of interest.

Finally, subtract the amount $807.07 from the amount found to get the money that needs to be added to the account.

c.

To determine

To write: The assumption made in calculating the amount.

Explanation of Solution

The assumption made in calculating the amount that needs to be added in the account to have $1500 in years is that the rate of interest remains the same for coming years as that of the last 8 years.