Chapter 3.5, Problem 23PPS

To determine

To calculate: The amount of money invested in each account.

Answer to Problem 23PPS

The amount of money invested in account A is $\$55,000$ , account B is $\$20,000$ and account C is $\$25,000$ .

Explanation of Solution

Given information:

The amount money invested in 3 accounts is $\$100,000$ . The amount of money invested in account A is $\$30,000$ plus the amount invested in account C. The expected rate of interest to be earned is $\$6300$ .

  $\begin{array}{cc}\text{Account}& \text{Expected Interest}\\ A& 4\%\\ B& 8\%\\ C& 10\%\end{array}$

Calculation:

It is provided that the amount money invested in 3 accounts is $\$100,000$ . The amount of money invested in account A is $\$30,000$ plus the amount invested in account C. The expected rate of interest to be earned is $\$6300$ .

  $\begin{array}{cc}\text{Account}& \text{Expected Interest}\\ A& 4\%\\ B& 8\%\\ C& 10\%\end{array}$

Let amount of money invested in account A isx , amount of money invested in account Bbe y and amount of money invested in account Cbe z .

It is provided that the amount money invested in 3 accounts is $\$100,000$ .

  $x+y+z=100,000$

The amount of money invested in account A is $\$30,000$ plus the amount invested in account C.

  $x=30000+z$

The expected rate of interest to be earned is $\$6300$ .

  $\begin{array}{cc}\text{Account}& \text{Expected Interest}\\ A& 4\%\\ B& 8\%\\ C& 10\%\end{array}$

So,

  $\begin{array}{c}4x+8y+10z=630000\\ 2x+4y+5z=315000\end{array}$

The system of equations obtained is,

  $\begin{array}{c}x+y+z=100000\\ x-z=30000\\ 2x+4y+5z=315000\end{array}$

Since, $x=30000+z$

So,

  $\begin{array}{c}30000+z+y+z=100000\\ y+2z=70,000\end{array}$

And,

  $\begin{array}{c}2\left(30000+z\right)+4y+5z=315000\\ 4y+7z=255000\end{array}$

Therefore, equations are

  $\begin{array}{c}y+2z=70,000\\ 4y+7z=255000\end{array}$

Use the method of substitution to solve the new first equation and third equation of system above.

  $\begin{array}{c}4\left(70,000-2z\right)+7z=255000\\ -z=255000-280000\\ z=25000\end{array}$

Now substitute $z=25000$ in the equation $x-z=30000$ ,

  $\begin{array}{c}x-25000=30000\\ x=55000\end{array}$

Now substitute $x=55000$ and $z=25000$ in $x+y+z=100000$ .

  $\begin{array}{c}55000+y+25000=100000\\ y=20000\end{array}$

Therefore, solution is the coordinate pair, $\left(55000,20000,25000\right)$ .

Thus, the amount of money invested in account A is $\$55,000$ , account B is $\$20,000$ and account C is $\$25,000$ .