Question

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $30 and same-day tickets cost $15. For one performance, there were 50 tickets sold in all, and the total amount paid for them was 975 . How many tickets of each type were sold?

Answer 1

15 advance tickets and 35 same-day tickets were sold for the show.

Step-by-step explanation:

Let's assume that the number of advance tickets sold is x.

Since there were 50 tickets sold in total, the number of same-day tickets sold would be 50 - x.

The total amount paid for advance tickets would be 30x and the total amount paid for same-day tickets would be 15(50 - x).

According to the information given in the question, the total amount paid for all the tickets was $975.

So we can write the equation:

30x + 15(50 - x) = 975

Now we can solve this equation to find the value of x, which represents the number of advance tickets sold.

Expanding the equation, we have:

30x + 750 - 15x = 975

Combining like terms, we get:

15x + 750 = 975

Subtracting 750 from both sides, we get:

15x = 225

Dividing both sides by 15, we get:

x = 15

Therefore, 15 advance tickets were sold and 35 same-day tickets were sold.