Chapter 11.1, Problem 80AYU

To determine

To calculate: The price of the food items Hamburger, large fries and large colas.

Answer to Problem 80AYU

The price of the food items are hamburger $\$1.95$ , large fries $\$0.95$ and large colas $\$0.80$ .

Explanation of Solution

Given information:

One group of customers paid $\$26.10$ for 8 deluxe hamburgers, 6 orders of large fries and 6 large colas.

Second group of customers paid $\$31.60$ for 10 deluxe hamburgers, 6 orders of large fries and 8 large colas.

Third group of customers paid $\$10.95$ for 3 deluxe hamburgers, 2 orders of large fries and 4 large colas.

Formula used:

Calculation:

Consider the provided information that one group of customers paid $\$26.10$ for 8 deluxe hamburgers, 6 orders of large fries and 6 large colas.

Second group of customers paid $\$31.60$ for 10 deluxe hamburgers, 6 orders of large fries and 8 large colas.

Third group of customers paid $\$10.95$ for 3 deluxe hamburgers, 2 orders of large fries and 4 large colas.

Let x denote the price of hamburger, y denote the price of large fries and z denote the price of large colas.

From the above data, the equations are formulated as,

  $\begin{array}{c}8x+6y+6z=26.10\\ 10x+6y+8z=31.6\\ 3x+2y+4z=10.95\end{array}$

Since, there are three equations are there and three variables are there so the data is sufficient to compute the price of each item.

Eliminate the variable y from the above two equations,

Subtract the equations,

  $\begin{array}{c}8x+6y+6z=26.10\\ 10x+6y+8z=31.6\end{array}$

To obtain,

  $\begin{array}{c}8x+6y+6z=26.10\\ 10x+6y+8z=31.6\\ 2x+2z=5.5\\ x+z=2.75\end{array}$

Now multiply the equation $3x+2y+4z=10.95$ by 3 and subtract is from $10x+6y+8z=31.6$ .

  $\begin{array}{c}10x+6y+8z=31.6\\ 9x+6y+12z=32.85\end{array}$

To obtain,

  $x-4z=1.25$ .

Now, solve the equations $x-4z=1.25$ and $x+z=2.75$ . Subtract both the equation, to obtain,

  $z=0.8$ . Now substitute the obtained value in $x+z=2.75$ , therefore, $x=1.95$ .

Now substitute $z=0.8$ and $x=1.95$ in the equation $8x+6y+6z=26.10$ .

Therefore, the value of y is $y=0.95$

Thus, the price of the food items are hamburger $\$1.95$ , large fries $\$0.95$ and large colas $\$0.80$ .