Question 42783: Two cans x and y both contain some water. From x Jim pours as much water into y as y already contains. Then from y he pours as much water into x as y presently has. Finally he pours from x to y as much water as y presently has. Each can now contains 24 ounces of water. how many ounces of water were in each can at the start?
Answer by psbhowmick(878)   (Show Source):
You can put this solution on YOUR website! Let the cans initially contain 'u' and 'v' amounts of water respectively.
As Jim pours ,from X, as much water into Y as Y already contains so after the first operation, X contains (u-v) and Y contains (v + v =) '2v' amount of water.
Next, he pours 'v' amount of water from Y into X (as presently Y has 'v' amount of water).
So volume of water in Y becomes 'v' and amount of water in X becomes (u-v) + v = u.
Again, Jim pours 'v' amount of water into Y (as presently Y has 'v' amount of water).
So volume of water in Y becomes '2v' and amount of water in X becomes (u-v).
These are the final amounts of water present in X and Y.
Final content in each vessel is 24 ounces.
Thus we have two equations:
u - v = 24 and 2v = 24.
The equation: 2v =24 gives v = 24/2 = 12.
Then from u - v = 24 we get u = 24 + v = 24 + 12 = 36.
As the initial contents of the vessels were 'u' and 'v' so they contained 36 and 12 ounces of water respectively.
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