There are two beakers, one containing water, the other wine. A certain amount of water is transferred to the wine, then the same amount of the mixture is transferred back to the water. Is there now more water in the wine than there is wine in the water? (from M.Gardner)
Please explain when you answer.
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6 comments:
There are equal amounts of water in wine as there is wine in the water.
Rachel, you are absolutely right. However we all would like to see the detailed explanation.
Also, please email me your address to maria@marialando.com to get a prize!
w = water, y = wine
x = the fraction of water
First w - wx = y + wx
then (w - wx) + (y + wx)x = (y+wx) - (y+wx)x
w - wx + yx + wx^2 = y + wx - yx - wx^2
w = y
This was a difficult puzzle and you still solved it in hours!
Here is my interpretation of the solution:
Assume both beakers contain 100 units of liquid. We take X units of water and transfer from Beaker1 to Beaker2.
Beaker1 contains now 100-X units of water.
Beaker2 contains 100 units of wine and X units of water.
Now transfer X units of the mixture from Beaker2 to Beaker1. Transferred amount contain X*(100/(100+X)) wine and X*(X/(100+X)) water.
So, now in Beaker1 we have X*(100/(100+X)) wine and the rest is water. We do not actually care how much. In Beaker2 we have X-X*(X/(100+X)) water and we do not care how much wine.
The amount of wine in Beaker1 is equal to the amount of water in Beaker2 as:
X*(100/(100+X)) = X-X*(X/(100+X))
100*X/(100+X) = 100*X/(100+X)
I would need to find a way to split the prize between Rachel and Peter.
I am joining this group of Archimedeses late in the game so no prize for me this time. Still want to share with you how I intuitively guessed the right answer: since it’s a closed system and the same amount of liquid is being transferred back and forth the balance should stay the same. Then, of course, I confirmed my guess by drawing step-by-step schematics.
And don’t you try to mix wine and water at our table as we only serve exceptional wines! :-)
Sasha S.
Here's how I verified my intuition - with a simple hypothetical. Beaker A contains 1 cup of water. Beaker B, 1 cup wine. Take 1/2 cup water from beaker A and pour into Beaker B. Beaker B now contains 1/3 water and 2/3 wine. Now take 1/2 cup liquid from Beaker B (which is, of course, 1/3 water and 2/3 wine) and pour back into Beaker A. Beaker B is of course still 1/3 water and 2/3 wine. Half of beaker A is now water, and 1/2 is 2/3 wine. Therefore Beaker A is now 1/3 wine and 2/3 water. There is no more water in the wine than wine in the water. -- Heather
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