Fatty and Skinny are two thieves, who stole gold coins and now divide them between themselves. Skinny says, “That’s unfair, Fatty, in your bag there are three times more coins than in mine.” Fatty answers, “OK Skinny, I am giving you one more coin for each year of your age.” Skinny is still not happy, “But you still have double my coins.” “This is only fair, as I am twice your age,” Fatty concludes. Later, Fatty leaves for a short time, and Skinny takes one coin from Fatty’s bag for each year of Fatty’s age, and puts these coins in his own bag. Now, which bag has more coins, Fatty’s or Skinny’s?
The answer is:
Now they each have an equal number of coins.
At the start
Let's say Skinny is y years old. So Fatty is double that: 2y.
Let's say Skinny has x coins at the start. Fatty has triple that: 3x.
Fatty then gives Skinny y coins, so Skinny has x+y coins, and Fatty has 3x-y. We are told Fatty still has double.
So, 2(x+y) = 3x-y ... x=3y
Then Skinny takes 2y coins. Now Skinny has x+y+2y = x+3y = 6y, and Fatty has 3x-y-2y = 9y-3y = 6y.
We can represent the relationships with three variables:
f - the starting coins for fatty
s - the starting coins for skinny
a - the age of skinny (the age of fatty is 2a)
The formula at the start:
Fatty has 3 times the coins of Skinny
f = 3s
After fatty gives skinny a number of coins equaling skinny's age:
Fatty (f) minus skinny's age (a) is still twice Skinny (s) + skinny's age (a)
f - a = 2(s + a)
In terms of Fatty's starting coins:
Simplify f - a = 2(s + a)
f - a = 2s + 2a
f = 2s + 3a
in terms of s since f is 3s
3s = 2s + 3a
s = 3a
s = 3aand
f = 3sso at at the start Skinny has
3acoins and Fatty has
After Fatty giving away
acoins, Fatty has
8aand Skinny with
4abecause Fatty then has double the coins.
After Skinny takes another
2acoins, Fatty has
6aand Skinny has
Therefore after all is said and done, they have the same number of coins.