The only way to get to Ship Island is to take the ferry which is free for babies (2 and under), $17 for children (3-10 years old), $25 for seniors (62 and older), and $27 for everyone else. Debbie paid $115 for 6 people in her family to go to the island for the day. Is it possible that Great-grandma Schowalter (91 years old) when on the trip?
Assume there were no babies in the group
number of children --- x
number of adults ----- y
number of seniors ---- 6-x-y
17x + 27y + 25(6-x-y) = 115
17x + 27y + 150 - 25x - 25y = 115
-8x + 2y = -35
8x - 2y = 35
The left side of this equation is even, the right side is odd,
this equation has no integer solution.
My assumption of no baby is false, ....
So there had to be at least one baby
Assume there was 1 baby
number of children --- x
number of adults ----- y
number of seniors ----- 5-x-y
17x + 27y + 25(5-x-y) = 115
17x + 27y + 125 - 25x - 25y = 115
...
8x - 2y = 10
4x - y = 5 , both x and y less than 5 , and integers
We get a solution when x=2 and y = 3
1 babies, 2 children, 3 adults, no seniors
Assume 2 babies
....
17x + 27y + 25(4-x-y) = 115
-8x + 2y = 15
Again, the left side is even, right side is odd, no solution,
no 2 babies
We could create a chart showing all possibilities
headings: b c a s cost for (b)abie, (c)hildren , etc
b c a s cost
0 0 6 0 xx
0 0 5 1 xx ($160)
0 0 4 2 xx
0 0 3 3 xx
...
1 2 3 0 √√ ($115)
..
I found no other combination , so grandma could did not go
alternate argument:
notice that the left side of the equation does not change from
8x - 2y = .....
and depending on the number of babies, we get the following set of equations
8x - 2y = 35 for 0 babies
8x - 2y = 10 for 1 baby, which had a solution , x=2 , y = 3
8x - 2y = -15 for 2 babies, no positive integer solution
8x - 2y = -40 for 3 babies, no positive integer solution
...
The only combination that worked is:
1 babies
2 children
3 adults
0 seniors -----> 1($0) + 2($17) + 3($27) + 0($25) = $115
GRANDMA DID NOT GO