Question

A movie theater has a seating capacity of 319. The theater charges $5.00 for children, $7.00 for students, and $12.00 for adults. There are half as many adults as there are children. If the total ticket sales was $ 2326, How many children, students, and adults attended?

Answer 1

There were 186 children, 93 students and 40 adults attended the movie theater.

Explanation:

To solve this problem, create a system of equations.

First, define the variables:

• Let x be the number of children.
• Let y be the number of students.
• Let z be the number of adults.

From the information given, we know that the total number of people in the theater must equal the seating capacity:

`\implies x+y+z=319`

The number of adults is half the number of children:

`\implies z=(1)/(2)x`

The total ticket sales is the sum of the cost of each ticket type multiplied by the number of people who bought that type of ticket:

`\implies 5x+7y+12z=2326`

Now we have a system of equations:

`\begin{cases}\begin{aligned}x+y+z&=319\\z&=(1)/(2)x\\5x+7y+12z&=2326\end{aligned}\end{cases}`

Substitute the second equation into the first equation and rearrange to isolate y:

`\implies x+y+(1)/(2)x=319`

`\implies y+(3)/(2)x=319`

`\implies y=319-(3)/(2)x`

Substitute the equation for z and the equation for y into the third equation and solve for x:

`\implies 5x+7\left(319-(3)/(2)x\right)+12\left((1)/(2)x\right)=2326`

`\implies 5x+2233-(21)/(2)x+6x=2326`

`\implies (1)/(2)x+2233=2326`

`\implies (1)/(2)x=93`

`\implies x=186`

Now we have found the value of x, substitute this into the second equation and solve for z:

`\implies z=(1)/(2) (186)`

`\implies z=93`

Finally, substitute the found value of x into the equation for y and solve for y:

`\implies y=319-(3)/(2)(186)`

`\implies y=319-279`

`\implies y=40`

Therefore, the number of children, students, and adults who attended the theater was:

• 186 children
• 93 students
• 40 adults