Question 1015397: Rides at an amusement park require 3, 4, 5, or 6 tickets. A family purchases 58 tickets and receives 2 tickets from a park guest who had leftover tickets. The family wants to use all their tickets and wants the sum of 5- and 6-ticket rides to be twice as much as the sum of 3- and 4-ticket rides. What is the largest possible number of 6-ticket rides that the family can go on subject to these constraints?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the problem doesn't state that you have to have all of them.
in other words, you can have only two types out of the four and you would satisfy the requirements.
looking at the maximum possible of 6 ticket rides, you can get 10 rides that require 6 tickets apiece.
that's not good because there is no 2 to 1 ratio between 5 or 6 ticket rides and 3 or 4 ticket rides, since you have no 3 or 4 ticket rides.
you have to have more than zero of 3 or 4 ticket rides.
so drop down to 9 rides that require 6 tickets each.
9 * 6 = 54 tickets, which leave 6 tickets for the 3 or 4 ticket rides.
you can have 2 rides that require 3 tickets apiece.
you have 9 rides that take 6 tickets each.
you have 2 rides that take 3 tickets each.
the total number of tickets used is 60 so that's ok.
however, 9 rides that can be 5 or 6 ticket rides divided by 2 rides that can be 3 or 4 ticket rides is a ratio of 9/2 which is not the same as 2/1, so the constraint that 5 or 6 ticket rides have to be twice the amount of 3 or 4 ticket rides is not satisfied.
drop down to 8 rides that require 6 tickets apiece.
8 * 6 = 48 tickets.
you have 60 - 48 = 12 tickets spare.
this will allow you to have 4 rides that require 3 tickets apiece or 3 rides that require 4 tickets apiece.
4 rides that require 3 tickets apiece is your choice because then you wind up with 8 rides that are in the 5 or 6 ticket per ride category and you have 4 rides that are in the 3 or 4 ticket per ride category.
the ratio of number of 5 or 6 ticket rides to the number of 3 or 4 ticket rides is 8/4 which is the same as 2/1.
you have the required ratio of twice the number of 5 or 6 tickets rides to 3 or 4 ticket rides and you have used a total of 60 tickets.
4 * 3 ticket rides = 12 tickets and 8 * 6 ticket rides = 48 tickets and 12 + 48 = 60 tickets so you're good.
i believe the maximum 6 ticket rides the family can go on with these constraints is 8.
the combination that allows that and keeps faith with the constraints is:
4 rides that require 3 tickets apiece.
0 rides that require 4 tickets apiece.
0 rides that require 5 tickets apiece.
8 rides that require 6 tickets apiece.
the maximum of 8 rides that require 6 tickets apiece does not allow for any rides that require 4 tickets apiece nor any rides that require 5 tickets apiece.
since the problem didn't state that you had to have any 4 ticket or 5 ticket rides, the constaints are still satisfied.
8 rides that require 6 tickets apiece and 4 rides that require 3 tickets apiece is you solution as far as i can tell.
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