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| Question 1170750:  A movie theater has a seating capacity of 299. 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 $ 2164, How many children, students, and adults attended?
 
 Answer by ikleyn(52879)
      (Show Source): 
You can put this solution on YOUR website! . A movie theater has a seating capacity of 299.
 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 $ 2164, How many children, students, and adults attended?
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Let x be the number of adults.
Then the number of children is  2x
and the number of students is  (299-x-2x) = 299-3x  (assuming full capacity is occupied).
Next you write the money equation (= the revenue equation)
    12x + 5*(2x) + 7*(299-3x) = 2164   dollars
It is your basic (governing) equation for this problem.
Now simplify and solve it
    12x + 10x +7*299 - 21x = 2164
           x               = 2164 - 7*299 = 71.
ANSWER.  71 adults,  2*71 = 142 children  and the rest  299-71-142 = 86 are students.
CHECK.   71*12 + 142*5 + 86*7 = 2164 dollars is the total revenue.    ! Correct !
Solved.
 
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 The amazing unexpected fact is that the problem is solved using one single unknown and one equation.
 
 
 
 
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