Chapter 11.2, Problem 31E

a.

To determine

To find the interest that would be earned if it is compounded once, twice, four times, twelve times, or $365$ times in the year.

Explanation of Solution

Given:

The investment is $\$1000$ and rate of interest is $5\%$ .

Calculation:

Using the compound interest formula as follows-

For once in a year-

  $\begin{array}{l}A=P{\left(1+\frac{r}{n}\right)}^{nt}\\ A=1000{\left(1+\frac{5}{365}\right)}^{365\times \frac{1}{365}}\\ A=1000{\left(1+\frac{1}{73}\right)}^1\\ A\approx 1014\end{array}$

For twice in a year-

  $\begin{array}{l}A=P{\left(1+\frac{r}{n}\right)}^{nt}\\ A=1000{\left(1+\frac{5}{365}\right)}^{365\times \frac{2}{365}}\\ A=1000{\left(1+\frac{1}{73}\right)}^2\\ A\approx 1028\end{array}$

For four times in a year-

  $\begin{array}{l}A=P{\left(1+\frac{r}{n}\right)}^{nt}\\ A=1000{\left(1+\frac{5}{365}\right)}^{365\times \frac{4}{365}}\\ A=1000{\left(1+\frac{1}{73}\right)}^4\\ A\approx 1056\end{array}$

For twelve times in a year-

  $\begin{array}{l}A=P{\left(1+\frac{r}{n}\right)}^{nt}\\ A=1000{\left(1+\frac{5}{365}\right)}^{365\times \frac{12}{365}}\\ A=1000{\left(1+\frac{1}{73}\right)}^{12}\\ A\approx 1177\end{array}$

For $365$ times in a year-

  $\begin{array}{l}A=P{\left(1+\frac{r}{n}\right)}^{nt}\\ A=1000{\left(1+\frac{5}{365}\right)}^{365\times 1}\\ A=1000{\left(1+\frac{1}{73}\right)}^{365}\\ A\approx 143461\end{array}$

b.

To determine

To find out the account to choose for the highest return.

Answer to Problem 31E

Super Saver Account.

Explanation of Solution

Given:

The given table is −

  

Calculation:

On observing each account very closely, it can be estimated that the interest rate is almost same for each account but on super saver the amount is compounded daily that will result in the highest return. So the Super Saver account will return the highest return.