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More A in B or more B in A?

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  • 17-06-2012 4:52am
    #1
    charles_92688
    Registered Users Posts: 52 ✭✭


    You have two containers. Container A has one gallon of liquid A in it. Container B has one gallon of liquid B in it. You take a cup of A and pour it into the B container and mix it. Then you take a cup of the mix in the B container and pour it into the A container.

    Is there more liquid A in the B container or more liquid B in the A container?


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    #2
    scudzilla
    Registered Users Posts: 11,911 ✭✭✭✭scudzilla


    charles_92688 wrote: »
    You have two containers. Container A has one gallon of liquid A in it. Container B has one gallon of liquid B in it. You take a cup of A and pour it into the B container and mix it. Then you take a cup of the mix in the B container and pour it into the A container.

    Is there more liquid A in the B container or more liquid B in the A container?

    More liquid A in the B container


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    #3
    charles_92688
    Registered Users Posts: 52 ✭✭charles_92688


    scudzilla wrote: »
    More liquid A in the B container

    Nope.


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    #4
    FISMA
    Closed Accounts Posts: 2,616 ✭✭✭FISMA


    charles_92688 wrote: »
    You have two containers. Container A has one gallon of liquid A in it. Container B has one gallon of liquid B in it. You take a cup of A and pour it into the B container and mix it. Then you take a cup of the mix in the B container and pour it into the A container.

    Is there more liquid A in the B container or more liquid B in the A container?

    Are the liquids the same density?

    When you ask "more," do you mean mass-wise or volume-wise?


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    #5
    ray giraffe
    Registered Users Posts: 338 ✭✭ray giraffe


    They're the same! Nice puzzle :)


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    #6
    charles_92688
    Registered Users Posts: 52 ✭✭charles_92688


    ray giraffe wrote: »
    They're the same! Nice puzzle :)

    Yep. You are correct.


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    #7
    3DataModem
    Registered Users Posts: 3,957 ✭✭✭3DataModem


    Not if you use alcohol and water.


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    #8
    Gurgle
    Closed Accounts Posts: 5,064 ✭✭✭Gurgle


    3DataModem wrote: »
    Not if you use alcohol and water.
    Yes if you mix them together well and don't give them a chance to separate ;)


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    #9
    3DataModem
    Registered Users Posts: 3,957 ✭✭✭3DataModem


    Gurgle wrote: »
    3DataModem wrote: »
    Not if you use alcohol and water.
    Yes if you mix them together well and don't give them a chance to separate ;)

    Nope, the more you mix the more certain you can be they won't match.

    In fact, both containers will have less than a gallon.


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    #10
    locum-motion
    Registered Users Posts: 5,143 ✭✭✭locum-motion


    To avoid confusion, I'm going to refer to the containers and the liquids by different names, so A contains W and B contains X.

    According to this page, a cup is one sixteenth of a gallon.


    Therefore our starting point is that A contains 1 gal of W, and B contains 1 gal of X.

    Take 1/16 gal of W out of A, and pour it into B.

    At this point, A contains 15/16 gal of W, and B contains 17/16 gal of a mixture which we shall call Y.

    Y is a mixture of W and X in the proportion 1:16, or to put it another way: 1/17 of Y is W, and 16/17 of Y is X.

    Then take 1/16 gal of Y out of B, and pour it into A.

    Now, B contains 1 gal of Y, and A contains 1 gal of another mixture which we shall call Z.

    So, the question then becomes:

    Is the proportion of W in Y higher than the proportion of X in Z?

    We already know that the proportion of W in Y is 1/17.

    What is the proportion of X in Z?

    Z is made up of 15/16 gal of the original W, to which has been added 1/16 of a gal (or 1 cup) of liquid Y, of which 1/16 of a cup is W and 15/16 of a cup is X.15/16 of a cup = 15 / (16^2) of a gallon, or 15/256 of a gallon.

    So the proportion of X in Z is 15/256.

    Which is bigger: 1/17 or 15/256? Or are they the same?

    1/17 = 3/51 = 15/255, which is a bigger number than 15/266.

    So, there is a tiny little bit more W in container B than there is X in container A.

    This of course is only an answer for a "US customary cup" and a "US customary gallon".
    I tried to see if I could work it out algebraically for any vessel size*, but my head melted.

    *Or to be more accurate "for any proportion between vessel sizes"


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    #11
    locum-motion
    Registered Users Posts: 5,143 ✭✭✭locum-motion


    I'm wrong. I've spotted where I went wrong. I'll edit this post later when I've more time to explain why I'm wrong.

    Edit: I've highlightd above where I went wrong. I should have said:
    of which 1/17 of a cup is W and 16/17 of a cup is X.

    I'm not going to bother following through to the conclusion. I'm fairly sure now that it will show the answer "They're the same"!

    Thanks, Darjeeling.

    locum-motion wrote: »
    To avoid confusion, I'm going to refer to the containers and the liquids by different names, so A contains W and B contains X.

    According to this page, a cup is one sixteenth of a gallon.


    Therefore our starting point is that A contains 1 gal of W, and B contains 1 gal of X.

    Take 1/16 gal of W out of A, and pour it into B.

    At this point, A contains 15/16 gal of W, and B contains 17/16 gal of a mixture which we shall call Y.

    Y is a mixture of W and X in the proportion 1:16, or to put it another way: 1/17 of Y is W, and 16/17 of Y is X.

    Then take 1/16 gal of Y out of B, and pour it into A.

    Now, B contains 1 gal of Y, and A contains 1 gal of another mixture which we shall call Z.

    So, the question then becomes:

    Is the proportion of W in Y higher than the proportion of X in Z?

    We already know that the proportion of W in Y is 1/17.

    What is the proportion of X in Z?

    Z is made up of 15/16 gal of the original W, to which has been added 1/16 of a gal (or 1 cup) of liquid Y, of which 1/16 of a cup is W and 15/16 of a cup is X.15/16 of a cup = 15 / (16^2) of a gallon, or 15/256 of a gallon.

    So the proportion of X in Z is 15/256.

    Which is bigger: 1/17 or 15/256? Or are they the same?

    1/17 = 3/51 = 15/255, which is a bigger number than 15/266.

    So, there is a tiny little bit more W in container B than there is X in container A.

    This of course is only an answer for a "US customary cup" and a "US customary gallon".
    I tried to see if I could work it out algebraically for any vessel size*, but my head melted.

    *Or to be more accurate "for any proportion between vessel sizes"


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    #12
    darjeeling
    Registered Users Posts: 962 ✭✭✭darjeeling


    Bucket A contains volume v of liquid X
    Bucket B contains volume v of liquid Y
    1) Volume w is transferred from B to A.
    The liquid in A is mixed, and ...
    2) volume w is then transferred from A to B.

    After step 1:
    A contains v/(v + w) of X and w/(v + w) of Y by proportion. Multiply by (v + w) to get the volumes of X (=v) and Y (=w)
    B contains volume v-w of Y and 0 of X.
    Volume w is transferred from A to B in step 2. This contains vw/(v + w) of X and w^2/(v + w) of Y.

    After step 2, A contains:
    v - vw/(v + w) = (v^2 + vw - vw)/(v + w) = v^2/(v + w) of X
    and
    w - w^2/(v + w) = (vw + w^2 - w^2)/(v + w) = vw/(v + w) of Y.

    B contains:
    vw/(v + w) of X
    and
    v - w + w^2/(v + w) = (v^2 + vw - vw - w^2 + w^2)/(v + w) = v^2/(v + w) of Y

    Therefore the volume of X in B is equal to the volume of Y in A, and the volume of X in A is equal to the volume of Y in B.


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    #13
    darjeeling
    Registered Users Posts: 962 ✭✭✭darjeeling


    The real challenge is to think of a situation in which this would be useful to know. Anyone?


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    #14
    locum-motion
    Registered Users Posts: 5,143 ✭✭✭locum-motion


    darjeeling wrote: »
    The real challenge is to think of a situation in which this would be useful to know. Anyone?

    Imagine Ointment A contains ingredient a, and ointment B contains ingredient b.

    Now let's say a dermatologist has two patients with slightly different symptoms; maybe one is badly inflamed and a little bit infected, while the other is badly infected but only a little bit inflamed.

    If the doc thinks the first would benefit from 90% A mixed with 10% B, and the second would benefit from 90% B mixed with 10% A, then the pharmacist could use this fact to make up the two prescriptions.

    Although, admittedly, the likelihood of this happening is so slim as to be virtually non-existent!


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    #15
    darjeeling
    Registered Users Posts: 962 ✭✭✭darjeeling


    A much simpler answer says that, because no matter has been created or destroyed, so long as the volumes are equal after transferring liquid back and forth, the purities of the two liquids must be the same - i.e. there will be as much by volume of liquid X in bucket B as there is of liquid Y in A and of X in A as of Y in B.


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