SOLUTION: A movie theater has a seating capacity of 291. 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

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: A movie theater has a seating capacity of 291. 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      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1176309: A movie theater has a seating capacity of 291. 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 $2116, How many children, students, and adults attended?
Answer by ikleyn(52771) About Me  (Show Source):
You can put this solution on YOUR website!
.

We assume that all 291 tickets are sold.


Let "a" be the number of adults;  then the number of children is  (2a).


The number of students is the rest, (291-a-2a) = 291-3a.


The "money" equation is


    12a + 5*(2a) + 7*(291-3a) = 2116  dollars.


Simplify and solve for "a".


    12a + 10a + 7*207 - 21a = 2116 - 7*291,

    a = 79.


ANSWER.  79 adults;  2*79 = 158 children  and  (291-3*79) = 54 students.


CHECK.  79*12 + 5*158 + 54*7 = 2116  dollars, total revenue.   ! Correct !

Solved.

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

This problem is to be solved using one equation in one unknown.

If you want to see many other similar solved problems, look into the lessons
    - Advanced word problems to solve using a single linear equation
    - HOW TO algebreze and solve these problems using one equation in one unknown
in this site.