SOLUTION: A movie theater has a seating capacity of 383. 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 1132189: A movie theater has a seating capacity of 383. 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 $ 2778, How many children, students, and adults attended?
Found 3 solutions by rothauserc, MathTherapy, greenestamps:
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
let c be the number of children, a the number of adults, s be the number of students
:
a = c/2
:
1) c +c/2 +s = 383
:
2) 5c +6c +7s = 2778
:
Note 12c/2 = 6c
:
solve equation 1 for s
:
3c +2s = 766
:
2s = 766 -3c
:
s = (766 -3c)/2
:
substitute for s in equation 2
:
11c +7(766 -3c)/2 = 2778
:
22c +5362 -21c = 5556
:
c = 194
:
a = 194/2 = 97
:
s = 383 -(194+97) = 92
:
******************************************************
there are 194 children, 97 adults, 92 students
:
check answer using equation 2
:
194*5 +97*12 +92*7 = 2778
:
970 +1164 +644 =2778
:
2778 = 2778
:
answer is ok
***************************************************
:

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
A movie theater has a seating capacity of 383. 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 $ 2778, How many children, students, and adults attended?
Let number of students and adults be S, and A, respectively
Then number of children = 2A
We then get: S + A + 2A = 383 =====> S + 3A = 383 =====> S = 383 - 3A ------- eq (i)
Also, 5(2A) + 7S + 12A = 2,778 ------- eq (ii)
5(2A) + 7(383 - 3A) + 12A = 2,778 ------- Substituting 383 - 3A for S in eq (ii)
10A + 2,681 - 21A + 12A = 2,778
10A - 21A + 12A = 2,778 - 2,681
A, or number of adults =
Number of children:
Number of students:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Two other tutors have provided solutions, by very different algebraic paths, of what was PROBABLY the INTENDED problem -- that the theater was filled to capacity.

But the problem only says the capacity of the theater is 383 -- it does NOT say all the theater was filled to capacity.

Take away the constraint that all the seats are filled, and the problem has multiple solutions.

The number of adults is half the number of children, so let x be the number of adults and y be the number of students; then the number of children is 2x. The only other information we have is that the total ticket sales, at $5 per child, $7 per student, and $12 per adult, is $2778.

So we have a single equation with two unknowns; and the unknowns have positive integer values. Under those circumstances, we MIGHT get a single solution; but more likely we will get a family of solutions.





To find the solution(s) to an equation like this, we can solve the equation for one variable in terms of the other, then use the fact that the solutions are positive integers to find the solutions.




A general technique for finding solutions is to perform the division as a whole number quotient plus a remainder:




On the left, y has to be a whole number; and x has to be a whole number, so (397-3x) is a whole number. That means (x+1)/7 has to be a whole number.

The whole number values that make (x+1)/7 a whole number are

6, 13, 20, 27, ...; any whole number of the form 6+7t.

So

Then



At this point, we have shown that every solution in integers of the equation



must be of the form (x,y) = (6+7t, 378-22t)

The total number of people is children plus students plus adults = 2x+y+x = 3x+y:



The seating capacity of the theater is 383. Since we have found that the total number of people is 396-t, we can see that the theater is full to capacity when t=13 and there are empty seats when t is greater than 13.

This leads to a family of solutions to the problem in which the theater does not need to be filled to capacity.
       number of   number of    number of    total number in
  t    adults, x  students, y  children, 2x    the theater
        (6+7t)     (378-22t)    (12+14t)        (396-t)
 -----------------------------------------------------------
  13      97          92          194             383
  14     104          70          208             382
  15     111          48          222             381
  16     118          26          236             380
  17     125           4          250             379


So, without the requirement that the theater be filled to capacity, there are 5 combinations of children, students, and adults in which the number of adults is half the number of children and the total ticket sales is $2778.

Of course, the first of the solutions in the table above is the solution the other two tutors found, by including the requirement that the theater be filled to capacity.

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