Chapter 4.6, Problem 12CYU

To determine

To find: the number of quarters and dimes Kevin have.

Answer to Problem 12CYU

The number of quarters and dimes are 10 and 15 respectively..

Explanation of Solution

Given information:

Kevin had 25 quarters and dimes.

The total value of all the coins is $\$4$ .

Formula used:

Inverse of a $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ is, where $ad-bc\ne 0$ .

The value of a second order determinant is the difference of the products of the products of the two diagonals.

Steps to solve the equations are:-

Step1. Find the inverse of the coefficient matrix.

Step 2. Multiple each side of the matrix equation by the inverse matrix.

Calculation:

According to the question ,

Let x be the number of quarters .

And y be the number of dimes.

Thus, the system of equation formed is

  $\begin{array}{l}x+y=25\\ 0.25x+0.10y=4\end{array}$

The matrix equation is

  $\left[\begin{array}{cc}1& 1\\ 0.25& 0.10\end{array}\right].\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}25\\ 4\end{array}\right]$ .

Step 1.

  $=A^{-1}=\frac{1}{(1)(0.10)-(1)(0.25)}\left[\begin{array}{cc}0.10& -1\\ -0.25& 1\end{array}\right]$

  $\begin{array}{l}=\frac{1}{0.10-0.25}\left[\begin{array}{cc}0.10& -1\\ -0.25& 1\end{array}\right]\\ =-\frac{1}{0.15}\left[\begin{array}{cc}0.10& -1\\ -0.25& 1\end{array}\right]\end{array}$

Step2.

  $\begin{array}{l}=-\frac{1}{0.15}\left[\begin{array}{cc}0.10& -1\\ -0.25& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0.25& 0.10\end{array}\right].\left[\begin{array}{c}x\\ y\end{array}\right]=-\frac{1}{0.15}\left[\begin{array}{cc}0.10& -1\\ -0.25& 1\end{array}\right]\left[\begin{array}{c}25\\ 4\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=-\frac{1}{0.15}\left[\begin{array}{c}(0.10\times 25)+(-1\times 4)\\ (-0.25\times 25)+(1\times 4)\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=-\frac{1}{0.15}\left[\begin{array}{c}2.5-4\\ -6.25+4\end{array}\right]\\ =\left[\begin{array}{c}x\\ y\end{array}\right]=-\frac{1}{0.15}\left[\begin{array}{c}-1.5\\ -2.25\end{array}\right]\end{array}$

Or

  $=\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}10\\ 15\end{array}\right]$

Thus, the number of quarters are 10 and number of dimes are 15.