SOLUTION: A movie theater has a seating capacity of 315. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults. There are half as many adults as there are children

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Question 1157006: A movie theater has a seating capacity of 315. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults. There are half as many adults as there are children. If the total ticket sales was $ 2294, How many children, students, and adults attended?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13215)   (Show Source): You can put this solution on YOUR website!


There is more than one way to get total ticket sales of $2294 with the prices shown. With some of those ways, not all the seats are filled.

The statement of the problem is deficient; we have to know how many of the 315 seats were filled.


Answer by ikleyn(52906)   (Show Source): You can put this solution on YOUR website!
.

            Since the condition says nothing about the attendance,  it is possible,  by default,  to assume that all  315  tickets were sold.

            Actually,  we do not need to know about the theater' capacity - we only need to know the number of the ticket sold.

            With this assumption,  see my solution below.


Let x be the number of adults.

Then the number of children is 2x, according to the condition,  and

the students are the rest (315 - x - 2x) = 315 - 3x persons.


Now you can write the money equation (revenue equation)


    5*(2x) + 7*(315-3x) + 12*x = 2294.


Simplify and solve for x


    10x + 7*315 - 21x + 12x = 2294

    x = 2294 - 7*315 = 89.


The problem is just solved.


ANSWER.  89 adults;  2*89 = 178 children, and the rest  315 - 89 - 178 = 48 are students.


CHECK.  89*12 + 178*5 + 48*7 = 2294 dollars.    ! Precisely correct !

Solved.

---------------

The lesson to learn from this solution:

    For the first glance, 3 unknowns are required to solve the problem.

    But, actually, it can be solved using only one unknown.


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