Chapter 6.6, Problem 24IP

To determine

To find:Amount invested.

Answer to Problem 24IP

Amount invested $=p=\$1800$

Explanation of Solution

Given information:

Time $=6$ years

Simple interest $=\$567$ .

Interest percent $=5.25\%$

Concept:

  $I=prt$

Where

  $I=$ Simple interest.

  $p=$ Principal amount

  $t=$ time in years.

  $r=$ Annual interest rate.

Calculation:

From given information

Time $=t=6$ years.

Simple interest $=I=\$567$ .

Interest percent $=R=5.25\%$

We know

Annual interest rate $=r=\frac{R}{100}$

  $\begin{array}{l}\Rightarrow r=\frac{5.25}{100}\\ \Rightarrow r=0.0525\end{array}$

On substituting values in above mentioned concept

We get

  $\begin{array}{l}I=prt\\ \Rightarrow 567=\left(p\right)\left(0.0525\right)\left(6\right)\\ \Rightarrow 567=0.315p\\ \Rightarrow p=\frac{567}{0.315}\\ \Rightarrow p=\$1,800\end{array}$

Therefore,

Amount invested $=p=\$1800$ .