Chapter 7, Problem 54CR

To determine

To Calculate: The amount placed in each investment where an inheritance $\$40000$ is to be divided among the three investments which yield $\$3500$ per year, the interest rate for the three investments are $7\%,9\%and11\%$ at simple interest respectively, and the second and thirds amounts are $\$3000$ and $\$5000$ lesser than the first amount.

Answer to Problem 54CR

  $\text{16000 be the amount invested at 7\%, 13000 be the amount invested at 9\% and 11000 be the amount invested at 11\%}\text{.}$

Explanation of Solution

Calculation: Let x be the amount invested at $7\%$ , y be the amount invested at $9\%$ , and z be the amount invested at $11\%$ .

It is given that the second and third amounts are $\$3000$ and $\$5000$ lesser than the first amount.

Thus,

  $y=x-\mathrm{3000........}(1)$

And,

  $z=x-\mathrm{5000........}(2)$

  $\$40000$ is to be divided among the three investments $7\%$ , $9\%$ and $11\%$ at simple interest

  $\begin{array}{l}x+y+z=40,\mathrm{000................}(3)\\ 0.07x+0.09y+0.11z=\mathrm{3500.....................}(4)\end{array}$

Use equation ( $1$ )and ( $2$ ) in ( $3$ )

  $\begin{array}{l}x+x-3000+x-5000=40,000\\ 3x=40,000+8000\\ 3x=48000\\ x=16000\\ \end{array}$

Use the value of x in ( $1$ )

  $\begin{array}{l}y=16000-3000\\ y=13000\end{array}$

Use the value of x in ( $2$ )

  $\begin{array}{l}z=16000-5000\\ z=11000\end{array}$

Put all the values 0f x ,y, z in equation( $4$ )

  $\begin{array}{l}0.07\times 16000+0.09\times 13000+0.11\times 11000\\ =1120+1170+1210\\ =3500\end{array}$

  $\text{16000 be the amount invested at 7\%, 13000 be the amount invested at 9\% and 11000 be the amount invested at 11\%}\text{.}$