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Math/drinking teaser

  1. Jan 13, 2005 #1


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    For your hot date, you have a decanter with 1 litre of red wine and a decanter with 1 litre of white wine.

    While you're off stoking the fire, your conniving little brother takes one teaspoon of red wine from its decanter, pours it into the decanter with the white wine, and mixes it thoroughly.

    He then takes one teaspoon of the now-slightly-diluted white wine mixture and mixes it back into the red wine decanter.

    You are furious and send him to his room. You're now going to serve only one of them, the least tainted.

    Which decanter (red or white) is less diluted?
  2. jcsd
  3. Jan 13, 2005 #2


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    Both are equally messed up.
  4. Jan 14, 2005 #3


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    ->The red is less diluted.
    Bet you would still drink both bottles. :rofl:
  5. Jan 14, 2005 #4
    i think red will be less diluted, though the difference will be very very less. almost negligible.
  6. Jan 14, 2005 #5
    I think the red wine is less diluted!
    (because if I use two teaspoons simultaneously the two liquids will be equal diluted)

    DaveC426913, do you have the right answer?
  7. Jan 14, 2005 #6


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    ::Bottle I has 1 unit of A, and 0 units of B (A : red wine, B : white...whatever)
    Bottle II has 0 units of A, 1 unit of B

    First transfer x (x<1) units from I to II

    Bottle I has (1-x) unit of A, and 0 units of B
    Bottle II has (0+x) units of A, 1 unit of B

    Bottle II has a total (1+x) units of wine, of which the fraction of A is x/(1+x) and the fraction of B is 1/(1+x).

    Now x units are transfered from II back to I.

    These x units will have A and B in the above proportions. So, it actually contains x*x/(1+x) {or x^2/(1+x)} units of A and x/(1+x) units of B.

    Since this is removed from Bottle II, Bottle II must now have x - (x^2)/(1+x) {=x/(1+x)} units of A and 1 - (x)/(1+x) {=1/(1+x)} units of B.

    And since this is transfered to Bottle I, Bottle I must now have (1-x) + (x^2)/(1+x) {=1/(1+x)} units of A and 0 + (x)/(1+x) {=x/(1+x)} units of B.

    Compare the last two statements. The dilution fractions are identical (ie : volume of A in Bottle II = volume of B in Bottle I, and vice versa)

  8. Jan 14, 2005 #7


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    Gokul got the right answer, but *whew* I'll bet you worked up a sweat!

    Here's the really short way:

    They are both equally impure.

    If the kid removed one teaspoon of wine, and then put one teaspoon of wine back, how much does each decanter now contain? They both contain exactly 1 litre.

    Since we started with an identical amount of each wine, the two decanters must have identical amounts of impurities in them.
  9. Jan 14, 2005 #8


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    Aargh ! I've been making myself feel a lot like a jackass, lately. This was just in keeping with that spirit. :yuck:

    PS : When you write out my monstrous looking thing on paper, it just takes up a couple of lines...and it's more convincing, as a proof. I'll bet there will be folks that will still argue your reasoning ! :wink:
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