What is the largest possible number of 6-ticket rides that the family can go on subject to these constraints?

Question

Rides at an amusement park require 3, 4, 5, or 6 tickets. A family purchases 107 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 1

S represents sum of 3-4 ticket rides

2S represents sum of 5-6 ticket rides

 

2s + 2 = 109

3S = 109

S = 36 1/3                    doubt if one can have 1/3 of a ticket!

 

2S = 72 2/3                  

 

The family has 72 5-6 tickets.

The most 6 ticket rides would be 72.

 

Unless I am missing something.