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# Can you solve this century old puzzle?

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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016

2. (Original post by ServantOfMorgoth)

Are the glasses the same size? In theory they should be equal

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3. It's equal
4. There should be a higher chance of there being more A in B, because in the first transfer, it is 100% definite that only liquid A is being poured into liquid B's flask, but in the second transfer, since there is already some liquid A in liquid B's flask, some of liquid A might be poured back into it's own flask.

So statistically, I would say more A in B.
(prob wrong though :I )
5. I (originally) agreed with the above poster since

a) that was my conclusion
b ) the fact that someone has concluded this and felt confident enough to post it slightly increases the chance it's the right answer

However it probably isn't the right answer since I arrived at the conclusion in 20 seconds and the puzzle is apparently really old.

Cheers
6. More A in B due to concentration of B becoming less when it was transferred back to A
7. (Original post by ServantOfMorgoth)

ok

so if we are assuming that there's no loss of liquid during transfer and that the "small amount" poured is the same and the glasses are thoroughly mixed after liquids are poured

they're still not equal

so when you pour liquid A into B and mix you have A+B
when you pour liquid A+B back into A you have more A

lets put some numbers into this, pour 1/2 of A into B mix and you have 1/2 of B and 1/4 A in one glass
let's say that we pour half the mixture of A+B back into A
so 1/4 A added to 1/4 of B and 1/8 of A so in total 3/8 of A and 1/4 B
in the other glass we have 1/4 B and 1/8 of A right?

conclusion: more of A in B?
8. (Original post by moggis)
I (originally) agreed with the above poster since

a) that was my conclusion
b ) the fact that someone has concluded this and felt confident enough to post it slightly increases the chance it's the right answer

However it probably isn't the right answer since I arrived at the conclusion in 20 seconds and the puzzle is apparently really old.

Cheers
I didn't say it was unsolved lol

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9. (Original post by ServantOfMorgoth)

It is a simple two step logic process.

Assuming that the two vessels have identical sizes and that no liquid is lost in the transfers, we can see that both vessels end up with the same amount of liquid after the double transfer that they had before it.

However much of liquid A has been transferred from A to B must therefore be matched by the amount of liquid B that has been transferred to vessel A.

This is true no matter what degree of mixing has taken place, and no matter what size the vessels are.

Edit: on reflection, I don't think my first assumption is necessary anyway.
10. (Original post by ServantOfMorgoth)
I didn't say it was unsolved lol

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I didn't think it was unsolved.

Merely that I probably wouldn't solve it in 20 seconds but if I did it's likely not much of a puzzle.
11. It doesn't actually give exact amounts of how much is transferred each time, so 10% could have been moved in the first transaction and 20% could have been moved in the second transaction.

So, who knows?
12. (Original post by Fractite)
It doesn't actually give exact amounts
They aren't necessary as it says each glass holds the same amount (being half-full) at both the beginning and end.
13. Oh nevermind, I just re-read the puzzle again.

Yeah, they're both equal.
14. Surely if they're both full, the liquid would spill out when poured into the other ?????

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15. Never mind , I didn't read the 'half' bit

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16. Here's my thinking.
I'm interpreting that the two glasses are of the exact same volume and that 'A' and 'B' refer to the type of liquids, not the amount of liquid.

I'm gonna let 'gA' be the glass with A, and 'gB' be the glass with B.

When you pour A into B, the amount in Glass B is more than half full by x amount; x is the volume of A transferred.
Glass A on the other hand is still going to have only A, but is less than half full by x amount.

You pour it back into A such that the volume in each glass is equal, and back to half of the volume of the glass.

But you don't just pour back the amount of A you just transferred; what you pour into Glass A is going to be A and B. This is because in Glass B, the proportion of the liquid is made up by B so there's a high probability that you're going to pour some B into A.
Also the text says mixture so it's suggesting A and B have mixed to some degree.
Of course, it's possible that A and B mix in such a way that the particles of A lay near the bottom and never reach the top so that they're poured back. But if we're just saying that you pour it back after the first transfer, without any additional stirring or shaking or swirling, then it's likely there are still particles of A to be poured back (alongside with B).

In both transfers you are transferring x amount.
In the first transfer, x was composed of only A.
In the second transfer, x was composed of A and B.
We could let y be the amount of B that was transferred in this transfer and it must be true that y is less than x

The question is asking if there's more A in B, or more B in A.
There's x amount of A in B, but there's only a lesser amount of B in A.

So there's more A in B.
17. A in B.
18. (Original post by Dodgypirate)
A in B.
No explanation?

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19. (Original post by ServantOfMorgoth)
No explanation?
(Original post by ServantOfMorgoth)

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Actually forget that, there's the same amount.

I cba to write the whole explanation down because it's doing my head in... but it's simple algebra.
20. I'll post the answer by tomorrow

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