Chapter 8, Problem 18PT

(a)

To determine

To find: The balance after 5 years.

Answer to Problem 18PT

The balance after 5 years is $8695

Explanation of Solution

Given:

Savings is $7500 at 3% compound interest.

Calculation:

Calculate the balance after 5 years.

  $\begin{array}{c}A=P{\left(1+\frac{r}{n}\right)}^{nt}\\ =7500{\left(1+\frac{3}{100}\right)}^5\\ =7500{\left(\frac{103}{100}\right)}^5\\ =\$8694.556\end{array}$

Thus, the balance after 5 years is $8695

(b)

To determine

To find: The time at which savings gets double.

Answer to Problem 18PT

The time at which savings gets double is 24 years.

Explanation of Solution

Given:

Savings is $7500 at 3% compound interest.

Calculation:

Calculate the time.

  $\begin{array}{c}A=P{\left(1+\frac{r}{n}\right)}^{nt}\\ 15000=7500{\left(1+\frac{3}{100}\right)}^t\\ 2={\left(\frac{103}{100}\right)}^t\\ \log 2=t\log 1.03\end{array}$

Solve further as,

  $\begin{array}{c}t=\frac{\log 2}{\log 1.03}\\ =23.449\\ \approx 24\text{ years}\end{array}$

Thus, the time at which savings gets double is 24 years.

(c)

To determine

To find: The years in which $10,000 is there in account.

Answer to Problem 18PT

There is $10,000 in account in about 10 years.

Explanation of Solution

Given:

Savings is $7500 at 3% compound interest.

Calculation:

Calculate the time.

  $\begin{array}{c}A=P{\left(1+\frac{r}{n}\right)}^{nt}\\ 10000=7500{\left(1+\frac{3}{100}\right)}^t\\ 1.33={\left(\frac{103}{100}\right)}^t\\ \log 1.33=t\log 1.03\end{array}$

Solve further as,

  $\begin{array}{c}t=\frac{\log 1.33}{\log 1.03}\\ =10\text{years}\end{array}$

Thus, there is $10,000 in account in about 10 years.