A Theatrical Play
On the first night of the school play, the Drama Club sells 44 student tickets and 55 adult tickets and they take in a total of $416.02. On the second night, they sell 75 student tickets and 37 adult tickets and take in a total of $306.20.
good for them!
306.20
To solve this problem, we need to set up a system of equations.
Let's say the cost of a student ticket is 's' dollars and the cost of an adult ticket is 'a' dollars.
From the given information, we can set up the following equations:
Equation 1: 44s + 55a = 416.02
This equation represents the total revenue from the first night, where 44 student tickets and 55 adult tickets were sold, totaling $416.02.
Equation 2: 75s + 37a = 306.20
This equation represents the total revenue from the second night, where 75 student tickets and 37 adult tickets were sold, totaling $306.20.
Now, we need to solve these equations to find the values of 's' and 'a':
To solve this system of equations, we can use either the substitution method or the elimination method.
Let's use the elimination method:
Multiply Equation 1 by 75 and Equation 2 by 44 to eliminate 's':
(44 * 75) * s + (55 * 75) * a = 416.02 * 75
(75 * 44) * s + (37 * 44) * a = 306.20 * 44
Simplifying these equations gives us:
3300s + 4125a = 31201.5
3300s + 1628a = 13488.8
Now, subtract the second equation from the first equation:
(3300s + 4125a) - (3300s + 1628a) = (31201.5) - (13488.8)
Simplifying this equation gives us:
2497a = 17712.7
Dividing both sides of the equation by 2497, we get:
a = 7.09
Now, substitute the value of 'a' back into either Equation 1 or Equation 2 to solve for 's'. Let's use Equation 1:
44s + 55(7.09) = 416.02
Simplifying this equation gives us:
44s + 389.95 = 416.02
Subtracting 389.95 from both sides, we get:
44s = 26.07
Dividing both sides by 44, we get:
s ≈ 0.592
So, the cost of a student ticket is approximately $0.592 and the cost of an adult ticket is approximately $7.09.