Math homework help??

Question

Rides at an amusement park require 3, 4, 5, or 6 tickets. A family purchases 59 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

The family has 61 tickets (59 + 2).

 

S = sum or 3 and 4 ticket rides

2S = sum of 5 and 6 ticket rides

 

2S + S = 61

3S = 61

S = 20 1/3           Under these constraints, somebody is selling fractions of a ticket.