Chapter 6.5, Problem 18PPS

z

To determine

To calculate the cost for 400 paintballs and lunch.

Answer to Problem 18PPS

$22

Explanation of Solution

Calculation:

Let x be the cost of lunch and y be the cost of the paintballs.

From the given figure,

The cost of 500 paintball with lunch is $25.

  $\Rightarrow x+500y=\mathrm{25......................}(1)$

The cost of 200 paintball with lunch is $16.

  $\Rightarrow x+200y=\mathrm{16......................}(2)$

Subtract equation (2) from equation (1),

  $\begin{array}{l}\Rightarrow (x+500y)-(x+200y)=25-16\\ \Rightarrow 300y=9\\ \Rightarrow y=0.03\end{array}$

Substitute the value of y in equation (1),

  $\begin{array}{l}\Rightarrow x+500(0.03)=25\\ \Rightarrow x+15=25\\ \Rightarrow x=25-15\\ \Rightarrow x=10\end{array}$

The cost of 400 paintballs with lunch is

  $\begin{array}{c}x+400y=10+400(0.03)\\ =10+12\\ =22\end{array}$

Conclusion:

Therefore, the cost of 400 paintballs with lunch is $22.