Answer: 8
Explanation:
Let's assume that the cost of an adult ticket is A dollars and the cost of a child ticket is C dollars.
From the information given on the first day, we know that 10 adult tickets were sold, so the total cost of adult tickets on the first day is 10A dollars. Similarly, 10 child tickets were sold, so the total cost of child tickets on the first day is 10C dollars. Together, these two amounts add up to $200, so we can write the equation:
10A + 10C = 200
On the second day, 13 adult tickets were sold, so the total cost of adult tickets on the second day is 13A dollars. Additionally, 2 child tickets were sold, so the total cost of child tickets on the second day is 2C dollars. Together, these two amounts add up to $172, so we can write the equation:
13A + 2C = 172
We now have a system of two equations with two unknowns:
10A + 10C = 200
13A + 2C = 172
To solve this system, we can use the method of substitution or elimination. I will use the method of elimination.
Multiplying the first equation by 13 and the second equation by 10, we get:
130A + 130C = 2600
130A + 20C = 1720
Now, let's subtract the second equation from the first equation:
(130A + 130C) - (130A + 20C) = 2600 - 1720
110C = 880
Dividing both sides of the equation by 110, we find:
C = 8
Therefore, each child ticket costs $8.