Chapter 7.4, Problem 90E

To determine

To calculate:amount moneyborrowed at $3\%$ , $4\%$ and $6\%$ rate from corporation

Answer to Problem 90E

the money borrowed at $3\%$ rate: $\$200000$

the money borrowed at $4\%$ rate: $\$500000$

the money borrowed at $6\%$ rate: $\$800000$

Explanation of Solution

Given information:

A corporation borrowed $\$1,500,000$ to expand its line of shoes.Some of the money was borrowed at $3\%$ ,some at $4\%$ and some at $6\%$ .Given that annual interest was $\$74,000$ and the amount borrowed at $6\%$ was four times the amount borrowed at $3\%$ .

Formula used:

Calculation:

Let $x$ , $y$ and $z$ be the money borrowed at $3\%$ , $4\%$ and $6\%$ rate

Total money borrowed is $\$1,500,000$

  $x+y+z=1500000$

Now, the interest at $3\%$ rate: $\frac{3x}{100}$

the interest at $4\%$ rate: $\frac{4y}{100}$

the interest at $6\%$ rate: $\begin{array}{l}\frac{6z}{100}=4\left(\frac{3x}{100}\right)\\ 12x-6z=0\end{array}$

the annual interest was $\$74,000$

  $\begin{array}{l}\frac{3x}{100}+\frac{4y}{100}+\frac{6z}{100}=74000\\ 0.03x+0.04y+0.06z=74000\end{array}$

the amount borrowed at $6\%$ was four times the amount borrowed at $3\%$

  $z=4x$

Equation 1: $x+y+z=1500000$

Equation 2: $0.03x+0.04y+0.06z=74000$

Equation 3: $z=4x$

Now solving these equation

  $\left|A\right|=\left|\begin{array}{ccc}1& 1& 1\\ 0.03& 0.04& 0.06\\ 4& 0& -1\end{array}\right|=0.07$

Since $D\ne 0$ so given system of equation has a solution

  $\begin{array}{l}AX=B\\ X=A^{-1}B\end{array}$

Where $A=\left(\begin{array}{ccc}1& 1& 1\\ 0.03& 0.04& 0.06\\ 4& 0& -1\end{array}\right)$ , $X=\left(\begin{array}{l}x\\ y\\ z\end{array}\right)$ and $B=\left(\begin{array}{l}1500000\\ 74000\\ 0\end{array}\right)$

  $\begin{array}{l}{A}^{-1}=\frac{1}{\left|A\right|}adjA\\ \left|A\right|=0.07\\ adjA=\left(\begin{array}{ccc}-0.04& 1& 0.02\\ 0.27& -5& -0.03\\ -0.16& 4& 0.01\end{array}\right)\\ AX=B\\ \left(\begin{array}{ccc}1& 1& 1\\ 0.03& 0.04& 0.06\\ 4& 0& -1\end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}1500000\\ 74000\\ 0\end{array}\right)\\ X=A^{-1}B\\ \left(\begin{array}{l}x\\ y\\ z\end{array}\right)={\left(\begin{array}{ccc}1& 1& 1\\ 0.03& 0.04& 0.06\\ 4& 0& -1\end{array}\right)}^{-1}\left(\begin{array}{l}1500000\\ 74000\\ 0\end{array}\right)\\ \left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\frac{1}{0.07}\left(\begin{array}{ccc}-0.04& 1& 0.02\\ 0.27& -5& -0.03\\ -0.16& 4& 0.01\end{array}\right)\left(\begin{array}{l}1500000\\ 74000\\ 0\end{array}\right)\\ \left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}200000\\ 500000\\ 800000\end{array}\right)\end{array}$

Therefore,

the money borrowed at $3\%$ rate: $\$200000$

the money borrowed at $4\%$ rate: $\$500000$

the money borrowed at $6\%$ rate: $\$800000$