Chapter 3.7, Problem 47PPS

To determine

To calculate: The quantities of jelly-filled doughnuts were ordered.

Answer to Problem 47PPS

  $6$

Explanation of Solution

Given information:

The total cost of box containing 2 dozen doughnuts is $=$$\text{\$}\mathrm{12.60.}$

The plain doughnuts each cost $=$$\text{\$}0.50$ and the jelly-filled doughnuts each cost $=$$\text{\$}\mathrm{0.60.}$

Calculation:

Let the number of plain doughnuts be $x$

Let the number of jelly filled doughnuts be $y$

Total number of doughnuts is 2 dozens. Hence,

  $x+y=24$

Plain doughnuts costs $\text{\$}0.5$ cost of plain doughnuts is $0.5x$ .Jelly-filled doughnuts cost $\text{\$}0.60$ .Hence cost of jelly filed doughnuts is $0.6y$

Hence,

  $0.5x+0.6y=12.60$

Hence set of simultaneous equations are

  $\begin{array}{l}x+y=24\\ 0.5x+0.6y=12.60\end{array}$

Solve equations using Cramer's equation as follows:

The coefficient matrix is.

  $C=\left[\begin{array}{cc}1& 1\\ 0.5& 0.6\end{array}\right]$

The determinant of coefficient matrix is.

  $\begin{array}{l}\left|C\right|=\left|\begin{array}{cc}1& 1\\ 0.5& 0.6\end{array}\right|\\ =1\times 0.6-1\times 0.5\\ =0.1\end{array}$

Using Cramer's rule, the solution is:

  $\begin{array}{l}x=\frac{\left|\begin{array}{cc}m& b\\ n& g\end{array}\right|}{\left|C\right|}\\ x=\frac{\left|\begin{array}{cc}m& b\\ n& g\end{array}\right|}{\left|C\right|}\\ =\frac{24\times 0.6-1\times 12.6}{0.1}\\ =\frac{1.8}{0.1}\\ =18\end{array}$

Similarly,

  $\begin{array}{l}y=\frac{\left|\begin{array}{cc}a& m\\ f& n\end{array}\right|}{\left|C\right|}\\ =\frac{\left|\begin{array}{cc}1& 24\\ 0.5& 12.6\end{array}\right|}{0.1}\\ =\frac{1\times 12.6-24\times 0.5}{0.1}\\ =\frac{0.6}{0.1}\\ =6\end{array}$

Hence, the solution is,

  $\begin{array}{l}x=18\\ y=6\end{array}$

Hence, number of jelly filled doughnuts ordered was $6$ .