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Can you solve this century old puzzle? Watch

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    We'll say 10% gets poured first and that A & B have the same volume.

    At the first pour A to B, B_Glass is (1/10A + 10/10B)
    Inorder to obtain 10% we'll take 1/11A
    as it's percent in comparison with the rest of B_Glass.

    When the second exchange takes place
    A_Glass = (9/10A) + 1/10 times (1/11A + 10/11B)
    which is (9/10A + 1/110A + 10/110B)

    therefore when they are both half again
    A_Glass = ((9x11)/(10x11)A + 1/110A + 10/110B) = (100/110A + 10/110B)
    B_Glass = (10/110A + 100/110B)

    so we've distributed just as much A in B as B in A.

    Another way, if you poured all of A_Glass into B_Glass
    and then poured half of that back into the A_Glass there would be an equal amount of A and B in both Glasses.
 
 
 
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