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?
2 Answers
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
So
s = 3a
andf = 3s
so at at the start Skinny has3a
coins and Fatty has9a
coins.
After Fatty giving awaya
coins, Fatty has8a
and Skinny with4a
because Fatty then has double the coins.
After Skinny takes another2a
coins, Fatty has6a
and Skinny has6a
coins.
Therefore after all is said and done, they have the same number of coins.