Question

A movie theater has a seating capacity of 235. 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 $ 1704, How many children, students, and adults attended?

Answer 1

adults: 59, students:58 and children 118

Explanation:

let A for adults, and C = children and S for students

There are half as many adults as there are children=

A=C/2 , C=2A

A+C+S=235 or

A+2A+S=235 first equation

3A+S=235

12A+5C+7S =1704 or

12A+10A+7S=1704

22A + 7S=1704 second equation

3A+S=235 first

solve by addition and elimination

22A+7S=1704

21 A+7S=1645 subtract two equations

A=59 adults

C=2A=2(59)=118

substitute in :A+S+C=235

S=235-(118+59)=58

check: 5C+7S+12A=1704

5(118)+7(58)+12(59)=1704

Answer 2

Hey there! :)

Answer:

118 children

58 students

59 adults

Explanation:

We can solve this problem by setting up a system of equations:

Let a = adults

2a = children (since double the # of adults were children), and

s = students

Set up the equations:

1704 = 5(2a) + 7s + 12(a)

1704 = 10a + 7s + 12a

235 = 2a + a + s

Simplify the equations:

1704 = 22a + 7s

235 = 3a + s

Subtract the bottom equation from the top by multiplying the bottom equation by 7 to eliminate the 's' variable:

1704 = 22a + 7s

7(235 = 3a + s)

1704 = 22a + 7s

1645 = 21a + 7s

---------------------- (Subtract)

59 = a

This is the number of adults. Substitute this number into an equation to solve for the number of students:

235 = 3(59) + s

235 = 177 + s

s = 58.

Since the number of children is equivalent to 2a, solve:

2(59) = 118 children.

Therefore, the values for each group are:

118 children

59 adults

58 students.