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Question 1175370: A movie theater has a seating capacity of 361. 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 $ 2614, How many children, students, and adults attended?
Found 4 solutions by Boreal, ikleyn, greenestamps, MathTherapy:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! C=2A
S=361-3A, since C=2A and we are subtracting everything else from 361 to get the students
5(2A)+7(361-3A)+12A=2614
10A+2527-21A+12A=2614
A=87 ($1044)
C=174 ($870)
S=100 ($700)
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
A movie theater has a seating capacity of 361. 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 $ 2614, How many children, students, and adults attended?
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x adults
2x children
and (361 - x - 2x) = (361-3x) students (the rest . . . )
Total money equation
12x + 5*(2x) + 7*(361-3x) = 2614 dollars
Simplify and solve
12x + 10x - 21x = 2614 - 7*361
x = 87
ANSWER. 87 adults; 2*87 = 174 children and the rest 361-87 - 174 = 100 are students.
CHECK. 87*12 + 5*174 + 7*100 = 2614 dollars, in total. ! Correct !
Solved.
Notice that I solved the problem using only one equation and only one single unknown.
It is how this problem SHOULD be solved (!)
Also, notice that the capacity of the theater is IRRELEVANT to this problem.
We only should know (instead of the total capacity) the total number of SOLD TICKETS.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
The correct response to this problem is that there is not enough information to solve the problem.
The capacity of the theater is, as one of the tutors states, irrelevant; what matters is the total number of tickets sold.
Since the problem only gives the capacity of the theater and not the total number of tickets sold, the problem can't be solved. More precisely, there are probably many different combinations of tickets sold which will give a total ticket sales of $2614 with 361 OR FEWER total tickets sold.
Both responses you have received previously assumed that all the tickets were sold, so that the number of tickets sold was equal to the capacity of the theater.
But assuming information that is not given in the statement of a problem is BAD mathematics....
Bottom line: The statement of the problem is deficient.
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Dear imbecile tutor @MathTherapy....
I might be older than you, but I at least have some intelligence.
WHEN ARE YOU GOING TO STOP MAKING ABSURD PRONOUNCEMENTS THAT OTHER TUTORS' CORRECT AND ACCURATE RESPONSES ARE WRONG AND NEED TO BE IGNORED?
The statement of this problem does NOT NOT NOT say that tickets for all the seats were sold. There ARE other combinations of tickets sold that would bring in the same total amount of money.
By posting a response that says my accurate response needs to be ignored, you are demonstrating your ignorance and your pitiful arrogance.
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website!
A movie theater has a seating capacity of 361. 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 $ 2614, How many children, students, and adults attended?
IGNORE the response from the OLD person who claims that enough info wasn't presented. If the capacity is 361, then it's obvious
that that's the number of occupants that attended. Again, IGNORE that response!
That's a COMMON response from a PERSON who believes he knows everything!!
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