SOLUTION: A movie theater has a seating capacity of 207. 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 1141992: A movie theater has a seating capacity of 207. 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 $ 1498, How many children, students, and adults attended?
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39792) About Me  (Show Source):
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a adults
x students
c children

system%28a%2Bc%2Bx=207%2Cc%2Fa=2%2C5c%2B7x%2B12a=1498%29
Simplify and solve the system.
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a%2B2a%2Bx=207
3a%2Bx=207
x=207-3aandc=2a

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The revenue equation can be written in terms of just one variable, a.
10a%2B7%28207-3a%29%2B12a=1498
Solve for a;
use it to find x and c.



FURTHER STEPS
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10a%2B1449-21a%2B12a=1498
a%2B1449=1498
a=98-49
highlight%28a=49%29--------adults
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c=2a
c=2%2A49
highlight%28c=98%29-------children
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x=207-a-c
x=207-49-98
highlight%28x=60%29---------students

Answer by ikleyn(53751) About Me  (Show Source):
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.

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


The number of students is the rest, (207-a-2a) = 207-3a.


The "money" equation is


    12a + 5*(2a) + 7*(207-3a) = 1498  dollars.


Simplify and solve for "a".


    12a + 10a + 7*207 - 21a = 1498,

    a = 1498 - 7*207

    a = 49.


ANSWER.  49 adults;  2*49 = 98 children  and  (207-3*49) = 60 students.

Solved.

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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.


Answer by greenestamps(13330) About Me  (Show Source):
You can put this solution on YOUR website!


Just a comment about the problem; you have already received two responses....

It should be noted that, EXACTLY AS PRESENTED, the problem is flawed. A unique answer can't be found; there are many different combinations of children, students, and adults that can produce total ticket sales of $1498.

The trouble with the presentation of the problem is that it does not tell the number of seats that were filled. With only the statement that the SEATING CAPACITY is 207, there is insufficient information to solve the problem. In order to find a unique solution (as shown in the other responses), one has to assume that all the seats were filled.

And, in general, it is never safe to assume any information in a problem that is not given.