Chapter 1.7, Problem 40E

Solution

To Find: The invested amount at each rate if $1$ year interest of $\$1571$ was received.

The invested amounts are $\$18,000$ and $\$7,000.$

Given:

Principal amount $p=\$25000$

Initial Annual interest $r_1=5.5\%$

Final Annual interest $r_2=8.3\%$

Total interest payment $I=\$1571$

Calculation:

Let $p$ is the principle, $t$ is the time and $r$ is the rate of interest per annum.

Find the total interest:

  $I=\frac{p\cdot t\cdot r}{100}$

Substitute $I=I_1,p=x,t=1,r=5.5$ in the above formula:

  $I_1=\frac{x\cdot 1\cdot \left(5.5\right)}{100}$

Now, again substitute $I=I_2,p=25000-x,t=1,r=8.3$ in $I=\frac{p\cdot t\cdot r}{100}$ :

  $I_2=\frac{\left(25000-x\right)\cdot 1\cdot \left(8.3\right)}{100}$

Total interest can be calculated by,

  $I=I_1+I_2$

Substitute $I=\$1571$ , then the values of $I_1=\frac{x\cdot 1\cdot \left(5.5\right)}{100}$ and $I_2=\frac{\left(25000-x\right)\cdot 1\cdot \left(8.3\right)}{100}$ in the above formula:

  $\begin{array}{c}1571=\frac{x\cdot 1\cdot \left(5.5\right)}{100}+\frac{\left(25000-x\right)\cdot 1\cdot \left(8.3\right)}{100}\\ 1571=\frac{5.5x+\left(25000\right)\left(8.3\right)-x\left(8.3\right)}{100}\\ 157100=5.5x+207500-8.3x\\ 2.8x=50400\\ x=\frac{50400}{2.8}\\ x=18000\end{array}$

The initial investment is $\$18,000$ .

To find the final investment subtract $\$18,000$ from the principal amount:

  $\$25,000-\$18,000=\$7,000$

Thus, the invested amounts are $\$18,000$ and $\$7,000.$