Chapter 11.1, Problem 79AYU

To determine

To calculate: The tabular representation of various possibilities of buying Hamburger, large fries and large colas.

Answer to Problem 79AYU

The tabular representation of various possibilities of buying Hamburger, large fries and large colas is provided below.

  $\begin{array}{ccc}\text{Hamburger}& \text{Large fries}& \text{Large colas}\\ 2.15& 0.83& 0.6\\ 2.1& 0.9& 0.65\\ 2.05& 0.916& 0.7\\ 2& 0.93& 0.75\\ 1.95& 0.95& 0.8\\ 1.9& 0.96& 0.85\\ 1.85& 0.983& 0.9\end{array}$

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.

Hamburgers costs between $\$1.75\text{ and }\$2.25$ , fries between $\$0.75\text{ and }\$1.00$ and colas between $\$0.60\text{ and }\$0.90$ .

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.

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,

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

Since, only two equations are there and three variables are there so the data is not sufficient to compute the price of each item.

It is provided that Hamburgers costs between $\$1.75\text{ and }\$2.25$ , fries between $\$0.75\text{ and }\$1.00$ and colas between $\$0.60\text{ and }\$0.90$ .

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,

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

Let $z=t$ that is for different values of z, other two variables will be evaluated.

Therefore, $x=\frac{11}{4}-t$ .

Now substitute $z=t$ and $x=\frac{11}{4}-t$ in the equation $8x+6y+6z=26.10$ .

Therefore, the value of y is $y=\frac{t}{3}+\frac{41}{60}$

Substitute $z=t$ in the equation

Hamburgers costs between $\$1.75\text{ and }\$2.25$ , fries between $\$0.75\text{ and }\$1.00$ and colas between $\$0.60\text{ and }\$0.90$ .

For the value of z that lie between $\$0.60\text{ and }\$0.90$ evaluate the value of x and y.

Therefore, the tabular representation of various possibilities of buying Hamburger, large fries and large colas is provided below.

  $\begin{array}{ccc}\text{Hamburger}& \text{Large fries}& \text{Large colas}\\ 2.15& 0.83& 0.6\\ 2.1& 0.9& 0.65\\ 2.05& 0.916& 0.7\\ 2& 0.93& 0.75\\ 1.95& 0.95& 0.8\\ 1.9& 0.96& 0.85\\ 1.85& 0.983& 0.9\end{array}$