Chapter 4.2, Problem 36E

(a)

To determine

The compound interest semi-annually in the rate $5\frac{1}{8}\%$.

Answer to Problem 36E

The compound interest semi-annually in the rate $5\frac{1}{8}\%$ is $\$1006.25$.

Explanation of Solution

Given: The interest rate is $5\frac{1}{8}\%$ per year and compounded semi-annually.

Calculation:

The formula for calculate the compound interest is,

$A\left(r\right)=P{\left(1+\frac{r}{n}\right)}^{nt}$

Let,

$\begin{array}{c}t=1\text{year}\\ P=\$1000\end{array}$

Substitute 1 for t, 0.00625 for r, $\$1000$ for P, 2 for n to find compound interest semi-annually,

$\begin{array}{c}A\left(0.00625\right)=1000{\left(1+\frac{0.00625}{2}\right)}^{2\times 1}\\ =1000{\left(1+0.0005\right)}^2\\ =1000\times 1.006\\ =\$1006.25\end{array}$

Hence, the amount after $2\frac{1}{2}\%$ rate per year is $\$1006.25$.

(b)

To determine

The compound interest continuously in the rate $2\%$.

Answer to Problem 36E

The amount after $5\%$ rate per year is $\$1051.26$.

Explanation of Solution

Given: The interest rate is $5\%$ per year and compounded continuously.

Calculation:

The formula for calculate the compound interest continuously is,

$A\left(r\right)=P{e}^{rt}$

Let,

$\begin{array}{c}t=1\text{year}\\ P=\$1000\end{array}$

Substitute 1 for t, 0.05 for r, $\$1000$ for P to find compound interest continuously,

$\begin{array}{c}A\left(0.02\right)=1000\times e^{\left(0.05\times 1\right)}\left[\because e=2.718\right]\\ =1000\times {\left(2.718\right)}^{0.05}\\ =1000\times 1.051\\ =\$1051.26\end{array}$

Hence, the amount after $5\%$ rate per year is $\$1051.26$.

On compare the all above parts, part (a) is the best because its interest rate is lowest.

So, part (a) compound interest semi annually in the rate $5\frac{1}{8}\%$ is the best.