Puzzle: mischievous child

Kids can sometimes get out of control at parties, and today’s puzzle is about a child that finds it amusing to mess with the drink setup.

Here is the puzzle:

At a dinner party, there are two large bowls filled with juice. One bowl holds exactly 1 gallon of apple juice and another has 1 gallon of fruit punch.

A mischievous child notices the bowls and decides to have a little fun. The child fills up a ladle of apple juice and mixes it into the bowl with fruit punch. Not content to stop here, he decides to do the reverse. He fills up a ladle of the fruit punch/apple juice mixture and returns it to the apple juice bowl.

The child would proceed further, but his mother notices what he is doing and makes him stop. The child apologizes to the hosts, who decide to shrug off the matter as little harm was done.

But an interesting question does arise about the two mixtures of juice.

In the end, the two bowls ended up with some of the other juice. The question is: which bowl has more of the other juice? That is, does the fruit punch bowl have more apple juice or does the apple juice bowl have more fruit punch?

Assume the ladel holds a volume of 1 cup and the juices were mixed thoroughly when the child transferred the juices.

Can you solve it?

The solution to the story is in the comments section.

Open unsolved question: In the story, the mother catches the child after 1 round of transferring juice back and forth. But if she didn’t catch him, we know both bowls would eventually end up as a 50/50 mixture of both juices. The question is this: how many rounds does it take to get to the 50/50 mixture? Remember the ladle contains 1 cup of juice, and each bowl starts with 1 gallon (16 cups) of juice.

(I suspect the problem can be generalized for bowl size B and ladle size L, but I have not worked on that problem yet)



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  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Answer to the puzzle

    The hard way to solve this problem is doing the math. You can calculate that both juice bowls end up with an equal concentration of the other juice, and thus the transferred volumes must be equal.

    The easier way is to think logically. Notice that both bowls begin and end up with exactly 1 gallon of liquid. This means that whatever apple juice ended up in the fruit punch bowl must have been replaced by the same volume of fruit punch that went into the apple juice bowl. Therefore, the two volumes are equal.

    The problem is known as the wine/water puzzle. If you’d like a more detailed solution, I found a nice explation here: wine/water problem solution

  • Mark R.

    regarding the open question, the mixture will asymptotically tend to 50/50 but never reach it.

  • ChrisZ

    My intuition was that the answer to the open problem is that they approach a 50/50 solution but never become exactly equal – at least under the false assumption that punch is continuous. My initial simplified run through seems to bear this out. I tried it with bowls of punch that were each 2 cups, moving one cup each round, and got the following pattern:

    Looking just at Bowl A (since it’s symmetric):

    Round 0 – 2 cups A
    Round 1 – 5/3 cups A, 1/3 cups B
    Round 2 – 11/9 cups A, 7/9 cups B
    Round 3 – 29/27 cups A, 25/27 cups B

    So just extrapolating from the pattern (i.e. not bothering to prove it), you get the formula that after Round n Bowl A will have (3^n + 2)/3^n cups A and (3^n – 2)/3^n cups B.

    If I wanted to extrapolate again just based on that pattern and some very rough mental math, I would guess that the general formula for the contents of Bowl A after n rounds with a bowl of size B and ladle of size L is ((B-L)(B+L)^n + BL)/(B+L)^n cups A and ((L)(B+L)^n – BL)/(B+L)^n, but I’ve really only done some very simple pattern seeking and a couple test cases so I’m sure I messed up some terms in there. Maybe I’ll come back when it’s not 3am and work on it a bit more.

  • john

    For the problem, no assumption about sizes is needed other than the same amount is transferred in each direction (you don’t have half a ladle in one direction and a full one in the other.

    for the “open” question, it seems obvious that the ratio would approach 50-50 asymptotically (never reaching it), as long as the extra assumption is made: that the bowls are thoroughly mixed between each transfer. But here is a calculation anyway.

    To start I define C(n) to be the concentration of apple juice in bowl 1 (the one that started as apple juice) after n transfers (back and forth so that both bowls have a volume B). From the first problem, we know that C is also the concentration of Punch in bowl 2 and 1-C is the concentration of punch in bowl 1 and of apple juice in bowl 2. Now I define d(n) such that C(n)=1/2+d(n).

    When transferring a ladle (volume L) from bowl 1 to bowl 2, only the concentration of bowl 2 changes. On transferring back, only the concentration of bowl 1 changes. So we only need to look at the change in concentration in bowl 2 on the first transfer, then the first problem will give all the concentrations for the completed transfer.

    So after n+1 transfers, the total volume of apple juice in the second bowl is L*C(n)+B*(1-C(n)) and the total volume is L+B. Rewriting the volume of apple juice in terms of d(n) gives L*(1/2+d(n))+B*(1/2-d(n))= (L+B)/2+(L-B)*d(n). Dividing by L+B gives the concentration as 1/2+(L-B)*d(n)/(L+B). That is the concentration of apple juice in bowl 2, so that is equal to 1-C(n+1), and C(n+1)=1/2+(B-L)*d(n)/(L+B).

    So, as long as L is smaller than B (it certainly can’t be larger – if it’s equal, then the 50-50 mix will be achieved after the first transfer), d(n+1) will be smaller than d(n) and will have the same sign, and will not be 0. In other words, the concentration will approach 1/2 asymptotically.

    In fact, the concentration will be C(n)=1/2+[(B-L)/(B+L)]^n.

  • john

    This is a correction to my last comment. in the last equation I forgot to put d(0) which is 1/2, so:

    C(n)=1/2+d(0)X[(B-L)/(B+L)]^n=(1+[(B-L)/(B+L)]^n)/2

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