- By The Anonymous BEACON Puzzler (who wishes to remain so)
Wine to Water
You have a glass containing 100 spoonfuls of pure wine and a second glass containing 100 spoonfuls of pure water.
You transfer one spoonful of of the pure wine into the glass of water. You then stir that mixture fully mixing the two and transfer one spoonful of that mixture back into the wine glass. Both glasses once again contain 100 spoonfuls liquid, but is there now more volume of wine in the water or more water in the wine?
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First of all, an even bigger puzzle is why you would be messing up a nice glass of wine?
At any rate the first puzzle answer is "No." There is exactly the same amount of wine in the water as water in the wine.
Not buying it?
OK. But first we will give you a couple of hints and suggestions. They may help or possibly just frustrate you (infuriate maybe?) even more before looking at the real explanation:
It makes no difference at all that you thoroughly mix the wine into the water before transferring a spoonful from the water glass back into the wine glass.
It makes no difference that the initial amounts in the two glasses were the same as long they are returned to their original volumes.
You can mix the contents in any manner you wish as long as you return both glasses to their original amounts of liquid.
You can conduct enough actual experiments.
In all cases but the last the final total amount of wine in the water will be identical to the amount of water that is in the wine. In the last case the test materials may be missing prior to a conclusion.
OK- The Real Solution Solution
Finally! A puzzle that lets me use that line.
When the mixing of the two glasses is complete and they are each at their initial levels:
1) Whatever amount of wine missing from the wine glass has to be in the water glass.
2) Since the water glass has to have that wine in it, yet it is at its original level, for that to be true that same amount of water has to now be in the wine glass.
Still not buying it? Drink enough test results and who gives a hoot!
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Which Weigh to Go?
There are ten stacks of ten coins each. Nine of the stacks contain fresh from the mint silver dollars. One of the stacks contains 10 seemingly perfect counterfeits. You know the weight of a real silver dollar and are told that a counterfeit dollar weighs one gram more than a real silver dollar. You have an accurate digital scale.
What is the fewest number of weighings to guarantee you find the counterfeits?
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The counterfeit stack can be found in a single weighing. Not your answer? Think about it, or you can whimp out and view the explanation now.
Explanation
You select one coin from stack #1, 2 coins from stack #2, and so on up to all 10 coins from the tenth stack.
Place all those 55 coins on the scale.
The number of grams more than the weight of 55 real coins is the number of the counterfeit stack.
For example, if the scale reading is six grams more that the weight of 55 real coins, then stack #6 is the counterfeit stack.
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