Answer:
Explanation:
To find the number of child and adult tickets sold, we can use algebraic equations. Let's assume the number of child tickets sold is 'c' and the number of adult tickets sold is 'a'.
Given information:
Cost of child ticket: $6
Cost of adult ticket: $10
Total ticket sales: $600
1. Set up the equation for the total ticket sales:
6c + 10a = 600
2. We need to find a combination of 'c' and 'a' that satisfies this equation.
- Option 1: Let's assume all tickets sold were child tickets.
In this case, the equation becomes:
6c + 10(0) = 600
Simplifying, we get:
6c = 600
Solving for 'c':
c = 100
So, if all tickets were child tickets, 100 child tickets were sold.
- Option 2: Let's assume all tickets sold were adult tickets.
In this case, the equation becomes:
6(0) + 10a = 600
Simplifying, we get:
10a = 600
Solving for 'a':
a = 60
So, if all tickets were adult tickets, 60 adult tickets were sold.
- Option 3: Let's assume there were a combination of child and adult tickets sold.
We can try different values for 'c' and solve for 'a'.
For example, if we assume 'c' to be 50, the equation becomes:
6(50) + 10a = 600
Simplifying, we get:
300 + 10a = 600
Solving for 'a':
10a = 300
a = 30
So, if 50 child tickets and 30 adult tickets were sold, the equation is satisfied.
Therefore, there are multiple possible combinations of child and adult tickets sold that could result in $600 in ticket sales. The options are:
- 100 child tickets
- 60 adult tickets
- 50 child tickets and 30 adult tickets.