Ryan and Andy have some carnival tickets. If Ryan sells 10 tickets per day and Andy sells 5 tickets per day, Ryan will have 40 tickets left when Andy has sold all his tickets. If Ryan sells 5 tickets per day and Andy sells 10 tickets per day, Ryan will have 70 tickets left when Andy has sold all his tickets. a) How many tickets does Ryan have? b) Each tickets costs $12. If Ryan and Andy manage to sell all their tickets, how much more does Ryan collect than Andy?
Let the number of tickets Ryan and Andy have be R and A.
Case 1: How long would it take Andy to sell all his tickets? It would take A/5 days, because, in this case, he sells 5 each day. In that time Ryan would sell (A/5)×10=2A tickets—twice as many as Andy. Ryan would have 2A fewer tickets than he had at the beginning, so, since Ryan would have 40 tickets left, R-2A=40. We can write R=40+2A.
Case 2: If Andy sells 10 tickets a day it would take A/10 days to sell all his tickets, and in this time, Ryan would sell 5(A/10)=½A half as many as Andy. He will have R-½A tickets left, so R-½A=70 tickets. We can write R=70+½A.
We have two equations to express R so the two expressions must be the same:
40+2A=70+½A. We can rearrange these terms:
2A-½A=70-40. That is 3A/2=30 because 1½ is the same as 3/2.
If 3A/2=30, 3A=60 and A=20. Andy has 20 tickets. R=40+2A=40+40=80, so (a) Ryan has 80 tickets.
(b) Each ticket costs $12, so Andy’s tickets are with 20×12=$240 and Ryan’s are worth 80×12=$960. They each sell all their tickets, so R-A=80-20=60, Ryan sold 60 more tickets than Andy, and so he collects 60×12=$720 more than Andy (also 960-240=$720).
CHECK.
Case 1: It would take Andy 20/5=4 days to sell all his tickets and Ryan would have sold 40 tickets in this time, 80-40=40 tickets left.
Case 2: It would take Andy 20/10=2 days to sell all his tickets and Ryan would have sold 10 tickets in this time, 80-10=70 tickets left.
So the solution fits the facts.
