Question

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $20 and same-day tickets cost $40. For one performance, there were 45 tickets sold in all, and the total amount paid for them was $1500. How many tickets of each type were sold

Answer 1

They sold 15 advance tickets and 30 same day tickets.

Explanation:

(x means multiplication, our variables are a and s)

20a+40s=1500, and a+s=45

So, you just have to explore. The way I did it, first you pick a number. Say we do a=40, or 20x40=800, and s=5, or 40x5=200. 800+200=1000, which isn't enough money. So, figure out what 700/40 is (500 for what we didn't have in the last problem, 200 for what we got with 40), and it's 17.5. Doesn't work. So, try 800/40=20. Then, we subtract 1500-800, and we get 700. 700/20=35. This gets us the money we made, but not the right number of tickets. So, we try adding tickets to the one making us more money. 840/40=21, and 660/20=33. This gives us $1500, but 54 tickets. So, 880/40=22, and 620/20=31. $1500, 52 tickets. Then, we jump to 960/40=24, and 540/20=27. $1500, but 51 tickets. 1000/40=25, and 500/20=25. 1200/40=30, and 300/20=15. 1200+300=1500, and 30+15=45. So, they sold 15 advance tickets and 30 same day tickets.

Hope this doesn't confuse you.