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I was reading the following puzzle:

There is a glass of juice and a glass of water with identical volumes. We take 1 full spoon of the juice and put it in the glass of water.
After briefly stirring the mixture, we put a full spoon of the mixture back into the glass of juice.
Question: which amount is greater? The amount of juice in the water or the amount of water in the juice?

I worked it out that the amounts are equal by calculating the ratio of juice in the water etc and assuming a prefect mixture.

But although the solution reported the same result, I don't understand really understand its logic and seems faulty to me.
The solution states:

After the exchange the amount of juice in the water must be the same as the amount of water in the juice, since the volumes of the two glasses after the exchange are the same. The juice missing from the one glass was moved to the other having the missing volume replaced by water.

So I don't have a problem proving it algebraically. I am trying to understand the intuition behind ths stated answer.

Isn't it based on the assumption that e.g. the juice poured in the water glass might be in lower parts and we just remove using the spoon, water from the surface?
Otherwise why is the equal volume used here? It seems to me that we remove a volume of juice from the juice glass (e.g. x which is the volume of the spoon) and we put back in a volume of a mixture (not water).

epimorphic
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Jim
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2 Answers2

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It doesn’t matter just what combination of water and juice is returned to the juice glass: since the glasses end up equally full, whatever amount of juice is missing from the juice glass must be exactly compensated for by the amount of water brought in from the water glass. This can be verified algebraically as follows.

At the end the juice glass contains some amount $J$ of juice and some amount $w$ of water, and the water glass contains some amount $W$ of water and $j$ of juice. We know that

$$J+j=W+w$$

and

$$J+w=W+j\,.$$

Subtracting the second equation from the first, we find that

$$j-w=w-j\,,$$

so $w=j$ and hence $W=J$. Note that the argument does not assume any specific kind of mixing (e.g., uniform) in the water glass after the first transfer.

Brian M. Scott
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  • I could work it out algebraically. So the problem is not finding a way to prove it. I just wanted to understand the intuitive solution – Jim Dec 24 '20 at 23:30
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    @Jim: But the algebraic work that you described in the question is not sufficient to justify the conclusion, since it assumes a specific mixing in the water glass after the first transfer. The calculation that I gave is independent of the amount of mixing at that stage. If you think about why it works no matter what the mixing is, you may start to get a handle on the intuitive argument. Or perhaps this will help: since the glasses end up equally full, whatever amount of juice is missing from the juice glass must be exactly compensated for by the amount of water brought in from the water glass. – Brian M. Scott Dec 24 '20 at 23:34
  • since the glasses end up equally full, whatever amount of juice is missing from the juice glass must be exactly compensated for by the amount of water brought in from the water glass. Yes this helps! – Jim Dec 24 '20 at 23:41
  • @Jim: Okay; I’ll add that at the top of the answer. – Brian M. Scott Dec 24 '20 at 23:42
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Alternative perspective.

Adopt the syntax that $J$ is the juice that remains in the juice glass, $j$ is the amount of juice that was transferred, and $w$ is the amount of water that was transferred.

The challenge is to prove that $j = w.$

It is totally irrelevant whether the initial volume of the juice and water glasses are equal. It is also totally irrelevant whether there is one transfer or multiple transfers. The only critical point is that the volume of the mixture in the Juice glass after the transfer(s) is identical to the initial volume in the Juice glass.

Therefore,

$$J + j ~=~ \text{starting volume in Juice glass} ~=~ \text{ending volume in Juice glass} ~=~ J + w.$$

Therefore,

$$j = w.$$

user2661923
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