Rides at an amusement park require 3 4 5 or 6 tickets A fami

Rides at an amusement park require 3, 4, 5, or 6 tickets. A family purchases 90 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?

Solution

totally family has 92 tickets with them

to enter into park it require 3,4,5 or 6 tickets

given that 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.

to find  the largest possible number of 6-ticket rides is 11

that is

11(6) =66 tickets

1(5) = 5

adding the no of passes = 11+1 =12

now we have already gave 71 tickets

we have 92 - 71 = 21 tickets

we can have three 3 ticket passes = 9

and three 4 ticket passes =21

so this sum upto 3+3 =6 passes

now checking  the sum of 5- and 6-ticket rides to be twice as much as the sum of 3- and 4-ticket rides.

1 + 11 = 2(3+3)

12=12

true

so the largest possible number of 6-ticket rides that the family can go on subject to these constraints is 11 rides