Chapter 4, Problem 90RE

(a)

To determine

To Find: The amount of the investment after $1\frac{1}{2}years$

Answer to Problem 90RE

  $\$5664.98$

Explanation of Solution

Given: A sum of $\$5000$ is invested at an interest rate of $8\frac{1}{2}\%$ per year, compounded semiannually

FORMULA USED:

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

Given:

  $\begin{array}{l}P=\$5000\\ r=0.085\\ n=2\end{array}$

Then

  $\begin{array}{l}A(t)=P{\left(1+\frac{r}{n}\right)}^{nt}\\ A(t)=5000{\left(1+\frac{0.085}{2}\right)}^{2(1.5)}\\ A(t)=\$5664.98\end{array}$

Hence, the amount of the investment after $1\frac{1}{2}years$ is $\$5664.98$ .

(b)

To determine

To Find: The period of time such that amount of the investment becomes $\$7000$ after that time.

Answer to Problem 90RE

  $4.04\text{ years}$

Explanation of Solution

Given: A sum of $\$5000$ is invested at an interest rate of $8\frac{1}{2}\%$ per year, compounded semiannually

FORMULA USED:

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

Given:

  $\begin{array}{l}P=\$5000\\ r=0.085\\ n=2\\ A(t)=7000\end{array}$

Then

  $\begin{array}{l}A(t)=P{\left(1+\frac{r}{n}\right)}^{nt}\\ 7000=5000{\left(1+\frac{0.085}{2}\right)}^{2t}\\ 2t\log \left(1+\frac{0.085}{2}\right)=\log \frac{7000}{5000}\\ t=4.04\text{ years}\end{array}$

(c)

To determine

To Find: Time for the amount to grow $\$7000$ if interest were compounded continuously instead of semiannually.

Answer to Problem 90RE

  $3.96\text{ years}$

Explanation of Solution

Given: A sum of $\$5000$ is invested at an interest rate of $8\frac{1}{2}\%$ per year, compounded semiannually

FORMULA USED:

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

Given:

  $\begin{array}{l}P=\$5000\\ r=0.085\\ A(t)=7000\end{array}$

Then

  $\begin{array}{l}7000=5000e^{0.085t}\\ \ln 7000=\ln 5000e^{0.085t}\\ t=\frac{\ln \left(\frac{7000}{5000}\right)}{0.085}\\ t=3.96\text{ years}\\ \end{array}$