I believe the answer would be 1/9
The first thing we must do for this case is to define variables.
We have then:
x: ticket for seats in section a
y: ticket for seats in section b
z: ticket for seats in section c
We write the system of equations:
50x + 40y + 20z = 335000
x + y + z = 9500
y = x + z + 500
Solving the system we have:
x = 1500
y = 5000
z = 3000
Answer:
ticket for seats in section a = 1500
ticket for seats in section b = 5000
ticket for seats in section c = 3000
C is the correct answer as a dozen is twelve so the cost of one would be a dollar.