Chapter 11.1, Problem 76AYU

To determine

To calculate: The number of adults, children and senior citizens that brought the tickets with condition at twice as many children’s tickets were sold as adult tickets. It is given that A Movie Theater charges $8.00 for adults, $4.50 for children, and $6.00 for senior citizen. One day theater sold 450 tickets and collected $2320.

Answer to Problem 76AYU

There are 110 adults, 220 children and 75 senior citizens that brought the tickets with condition at twice as many children’s tickets were sold as adult tickets.

Explanation of Solution

Given information:

It is given that A Movie Theater charges $8.00 for adults, $4.50 for children, and $6.00 for senior citizen. One day theater sold 450 tickets and collected $2320 with condition at twice as many children’s tickets were sold as adult tickets.

Calculation:

Consider the given information that A Movie Theater charges $8.00 for adults, $4.50 for children, and $6.00 for senior citizen.

Now, let

  $\begin{array}{c}x=\text{ the number of adult tickets}\\ y=\text{the number of child tickets}\\ z=\text{ the number of senior citizen tickets}\end{array}$

Since, the total numbers of tickets are 405. So,

  $x+y+z=405$

Since, the total revenue is $2320.So,

  $8x+4.50y+6z=2320$

Since, twice as many children’s tickets were sold as adult tickets. So,

  $y=2x$

Now, system of equation is,

  $\{\begin{array}{c}x+y+z=405\\ 8x+4.50y+6z=2320\\ y=2x\end{array}$

Substitute y in the first equation as,

  $\begin{array}{c}x+y+z=405\\ x+2x+z=405\\ 3x+z=405\end{array}$

Substitute y in the second equation as,

  $\begin{array}{c}8x+4.50y+6z=2320\\ 8x+4.50\left(2x\right)+6z=2320\\ 8x+9x+6z=2320\\ 17x+6z=2320\end{array}$

Now, the new system of equation is,

  $\{\begin{array}{c}3x+z=405\\ 17x+6z=2320\end{array}$

Solve the above system of equation as,

  $\begin{array}{l}\underset{\_}{\begin{array}{c}\left(-6\right)3x+\left(-6\right)z=\left(-6\right)405\\ 17x+6z=2320\end{array}}\\ x=110\\ \end{array}$

Now,

  $\begin{array}{l}y=2x\\ y=2\left(110\right)\\ y=220\\ \end{array}$

Next is,

  $\begin{array}{c}x+y+z=405\\ 110+220+z=405\\ z=405-110-220\\ z=75\end{array}$

Therefore, there are 110 adults, 220 children and 75 senior citizens that brought the tickets with condition at twice as many children’s tickets were sold as adult tickets.