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 \( a \) and the cost of a child ticket be \( c \). We can set up the following equations based on the information provided:

1. For 3 adults and 2 children:
\[
3a + 2c = 31 \quad \text{(1)}
\]

2. For 2 adults and 3 children:
\[
2a + 3c = 29 \quad \text{(2)}
\]

Now, we will solve these two equations simultaneously. First, we can multiply equation (1) by 3 and equation (2) by 2 to make the coefficients of \( c \) the same:

\[
9a + 6c = 93 \quad \text{(3)}
\]
\[
4a + 6c = 58 \quad \text{(4)}
\]

Next, we can subtract equation (4) from equation (3):

\[
(9a + 6c) - (4a + 6c) = 93 - 58
\]
\[
5a = 35
\]
\[
a = 7
\]

Now that we have \( a \), we can substitute it back into either equation (1) or equation (2) to find \( c \). Let's substitute into equation (1):

\[
3(7) + 2c = 31
\]
\[
21 + 2c = 31
\]
\[
2c = 31 - 21
\]
\[
2c = 10
\]
\[
c = 5
\]

So, the cost of an adult ticket is \( a = 7 \) and the cost of a child ticket is \( c = 5 \).

Now we know that the problem states a group consisting of some adults and 6 children paid $58 for tickets. We can set up the equation:

\[
xa + 6c = 58
\]

Substituting \( a \) and \( c \):

\[
7x + 6(5) = 58
\]
\[
7x + 30 = 58
\]
\[
7x = 58 - 30
\]
\[
7x = 28
\]
\[
x = 4
\]

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