Chapter 3.1, Problem 2P

To determine

To find: All the possibilities for the number of each type of appliance he could have sold.

Answer to Problem 2P

The possibilities for the number of each kind of appliance he could have sold are,

1. No washing machines and five refrigerators.
2. Two washing machines and four refrigerators.
3. Four washing machines and three refrigerators.
4. Six washing machines and two refrigerators.
5. Eight washing machines and one refrigerator.
6. Ten washing machines and no refrigerator.

Explanation of Solution

Given information:

Bruce, an appliance salesman earns a commission of $\$50$ for each washing machine he sells and $\$100$ for each refrigerator. Last month he earned $\$500$ in commissions.

Calculation:

1. Choosing two variables to represent the numbers asked for,

Let’s represent the number of washing machines sold as $x$ and number of refrigerators sold as $y$ .

2. Writing an open sentence relating the variables,

The equation that represents the above situation is

$\begin{array}{c}50x+100y=500\\ x+2y=10\\ 2y=10-x\\ y=5-0.5x\end{array}$

3. Finding the possibilities for the number of each kind of appliance he could have sold,

Calculation:

For $x=0$ which means not selling a single washing machine,

  $\begin{array}{l}y=5-0.5\left(0\right)\\ y=5-0\\ y=5\end{array}$

Bruce could have sold $0$ washing machine and $5$ refrigerators.

For $x=2$ which means selling two washing machines,

  $\begin{array}{l}y=5-0.5\left(2\right)\\ y=5-1\\ y=4\end{array}$

Bruce could have sold $2$ washing machines and $4$ refrigerators.

For $x=4$ which means selling four washing machines,

  $\begin{array}{l}y=5-0.5\left(4\right)\\ y=5-2\\ y=3\end{array}$

Bruce could have sold $4$ washing machines and $3$ refrigerators.

For $x=6$ which means selling six washing machines,

  $\begin{array}{l}y=5-0.5\left(6\right)\\ y=5-3\\ y=2\end{array}$

Bruce could have sold $6$ washing machines and $2$ refrigerators.

For $x=8$ which means selling eight washing machines,

  $\begin{array}{l}y=5-0.5\left(8\right)\\ y=5-4\\ y=1\end{array}$

Bruce could have sold $8$ washing machines and $1$ refrigerator.

For $x=10$ which means selling eight washing machines,

  $\begin{array}{l}y=5-0.5\left(10\right)\\ y=5-5\\ y=0\end{array}$

Bruce could have sold $10$ washing machines and $0$ refrigerator.

The possibilities for the number of each kind of appliance he could have sold are,

1. No washing machines and five refrigerators.
2. Two washing machines and four refrigerators.
3. Four washing machines and three refrigerators.
4. Six washing machines and two refrigerators.
5. Eight washing machines and one refrigerator.
6. Ten washing machines and no refrigerator.

Thus we can find the possibilities for the number of each kind of appliance he could have sold.