Question

On weekdays, a movie theater charges different rates for adults and children. If 3 adults and 2 children go for a movie on a weekday, the total cost of the tickets is $31. If 2 adults and 3 children go on a weekday, the total cost of the tickets is $29. If a group of adults and 6 children go to the movie theater on a weekday and pay $58 for tickets, how many adults are in this group?

4

5

7

8

Answer 1

Let the cost of an adult ticket be \(x\) and the cost of a child ticket be \(y\). We can set up the following equations based on the information provided:

1. From the first scenario (3 adults and 2 children):
\[
3x + 2y = 31
\]

2. From the second scenario (2 adults and 3 children):
\[
2x + 3y = 29
\]

We can solve this system of equations to find the values of \(x\) and \(y\).

First, we will multiply the first equation by 3 and the second equation by 2 to eliminate \(y\):

- Multiply the first equation by 3:
\[
9x + 6y = 93
\]

- Multiply the second equation by 2:
\[
4x + 6y = 58
\]

Now, subtract the second equation from the first:
\[
(9x + 6y) - (4x + 6y) = 93 - 58
\]
This simplifies to:
\[
5x = 35
\]
Thus,
\[
x = 7
\]

Now substitute \(x = 7\) back into one of the original equations to find \(y\). We can use the first equation:
\[
3(7) + 2y = 31
\]
This simplifies to:
\[
21 + 2y = 31
\]
Subtract 21 from both sides:
\[
2y = 10
\]
Thus,
\[
y = 5
\]

Now we know the ticket prices are:
- Adult ticket: \(x = 7\)
- Child ticket: \(y = 5\)

Now we need to determine how many adults are in the group of 6 children that paid $58 for tickets. Let the number of adults be \(a\). The equation for the total cost becomes:
\[
7a + 6(5) = 58
\]
This simplifies to:
\[
7a + 30 = 58
\]
Subtract 30 from both sides:
\[
7a = 28
\]
Dividing both sides by 7 gives:
\[
a = 4
\]

Thus, the number of adults in the group is \( \boxed{4} \).