Question

Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is 60 . For one performance, 40 advance tickets and 15 same-day tickets were sold. The total amount paid for the tickets was 1400. What was the price of each kind of ticket?

Answer 1

Answer: Let's call the price of an advance ticket "a" and the price of a same-day ticket "s". We know that the combined cost of one advance ticket and one same-day ticket is 60, so we can set up the equation: a + s = 60.

We also know that 40 advance tickets and 15 same-day tickets were sold for a total of 1400 dollars. So we can set up the equation: 40a + 15s = 1400

We have a system of two equations with two variables. We can use substitution or elimination method to solve for the unknowns.

Using substitution:

Solve the first equation for s, s=60-a

substitute it into the second equation, 40a + 15(60-a) = 1400

Expanding the second equation: 40a + 900 - 15a = 1400

25a = 500

a = 20

substitute a back into the first equation: s = 60 - 20 = 40

So the price of an advance ticket is $20, and the price of a same-day ticket is $40.

Explanation: