Question 1180809: A movie theater has a seating capacity of 129. 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 sales was $930, how many children, students and adults attended?
Found 2 solutions by ikleyn, josgarithmetic:
Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website! .
A movie theater has a seating capacity of 129.
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 sales was $930, 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, according to the condition,
and the number of students is the rest (129-x-2x) = (129-3x).
The "money" equation (the revenue equation) is
12x + 5*(2x) + 7*(129-3x) = 930 dollars.
12x + 10x + 7*129 - 21x = 930
x = 930 - 7*129 = 27 is the number od adults.
ANSWER. 27 adults; 2*27 = 54 children and the rest (129-27-54) = 48 are students.
Solved.
The lesson to learn from the solution is THIS :
This problem is to be solved using ONE unknown and one equation - not three.
Answer by josgarithmetic(39630)  (Show Source):
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