This has to be harder than it seems, but all I know is that it is too hard for me.
There are two glasses of soda, both containing equal amounts, but the amount is unknown.
Let’s say one is Pepsi, and one is Sprite. You pour some unknown amount of the Sprite into the Pepsi glass first. You then stir the mixture. Next perform the same step in the opposite direction, using an equal amount from the Pepsi glass. Stir the mixture now in the Sprite glass.
Which glass will contain the most?
They’d be equal. It’s easier to visualize if you assign numbers to “unknown”
So let’s say we’ve got 20ml of Pepsi and 20ml of Sprite.
Take 5ml from the Sprite, and add it to the Pepsi, making 25ml Pepsi, 15ml Sprite.
Then take 5ml of Pepsi, and add it back to the sprite, making 20ml of Pepsi and 20ml of Sprite.
I’ve heard a difference variant of the puzzle but it has the same answer.
Say you have a glass of Sprite and an equal glass of Pepsi. Take a measure, say teaspoon, from the Sprite glass and put it into the Pepsi glass. Stir. Now take the same teaspoon and take a measure from the Sprite/Pepsi mixture and put it in the glass of Sprite.
What are the proportions in the two glasses?
The answer is that both have a majority/minority ratio that is exactly the same. Same puzzle, just expressed in a more complicated format.
Or you can visualize it without numbers, if that helps:
You have transferred an equal amount of liquid from one glass to the other and then back. So now the Pepsi glass has some Sprite in it and the Sprite glass has some Pepsi in it. If you started and ended with two equal amounts, then whatever Pepsi was removed from the Pepsi glass has been replaced with an equal amount of Sprite. The removed amount of Pepsi is now in the Sprite glass to fill the void from the Sprite that was removed from the Sprite glass.
So there is an equal amount of “contamination” in each glass.
I thought it had to be equal, but as I said in the OP, I thought this was to obvious. I was given this “puzzle” as a brain teaser, so so much for the teasing.
Thanks for the confirmation.
I suspect there is a subtlety to the original question which was missed, and which makes it a more interesting question:
Start with a glass of sprite and a glass of pepsi, equal amounts (not necessarily a known amount).
Take a measured amount (e.g. a tablespoon) from the sprite glass and put it in the pepsi glass. Stir the pepsi glass.
Now take the same amount of pepsi/sprite mixture from the pepsi glass, and put it in the sprite glass. Stir the sprite glass.
Now, clearly the amount of liquid in each glass is the same, but is the pepsi/sprite ratio in the pepsi glass the same as the sprite/pepsi ratio in the sprite glass? If not, which is greater?(Answer in spoiler box)
There is more sprite in the sprite glass than there is pepsi in the pepsi glass.
Um, galt? If the amounts (of “contamination”) are the same, aren’t the ratios the same?
But the amounts of contaminant aren’t the same. you put say 20 ml of 100% sprite into the pepsi glass, then put 20 ml of 99.1% (or whatever) pepsi into the sprite glass.
Okay, you’re adding less pepsi to the sprite glass than you added sprite to the pepsi glass, but there’s less sprite there to add it to (than there was pepsi when you added the sprite to the pepsi). The proportions are the same.
Look at it this way: at the very end, when the levels in both glasses are equal, however much sprite is now in the pepsi glass must be missing from the sprite glass, and to keep the levels equal it must have been replaced by the same amount of pepsi.
Or read Scarlett67’s post again.
They’d be contaminated in the same ratio. Try it this way:
Start: 10 Units Pepsi 10 units Sprite
-1 Unit Pepsi +1 Unit Pepsi
Subtot. 9 Units Pepsi 11 units, 10/11 sprite/pepsi ratio)
+1 unit 10/11 s/p mix -1 unit 10/11 s/p mix
Final 10 units ? s/p mix 10 units 10/11 s/p mix
Solve for ?: 9(11/11) + 1(1/11) = (99 +1)/110 = 10/11 ratio of Pepsi to Sprite
Ha, you’re all wrong!
Read the OP again. When some fizzy water (why fizzy water?) is poured into the first glass and STIRRED, some of the carbon dioxide will come out of solution and bubble out. That will reduce the volume in that glass. When an equal amount is poured back into the second glass, it contains less carbon dioxide, so when the glass is stirred, less come out of solution. Ergo, the second glass has more liquid by volume (but not by much).
What do I win?
Urf. Try to imagine that in two columns, one for each glass.
I’d bet some lazy chemist with only 2 clean beakers worked this out originally. 
For what it’s worth, here’s yet another explanation I managed to Google up. This one refers to the problem in its more traditional “water and wine” formulation, which would probably avoid sishoch’s (smart-alecky yet valid) objection.
Yeah I think it is easier to understand using simple numbers.
20 ml in each glass
If you take 5 ml of Sprite and add it to the Pepsi, you now have 25ml of solution in the Pepsi glass and that solution is 20% Sprite. (5 divided by 25 x 100).
Now you take 5 ml of that solution and put it back in the Sprite. But what you are adding back is really 4ml of Pepsi and 1ml of Sprite (20% Sprite). So now the Sprite glass has 4ml of Pepsi in a 20ml glass and we have our same 20% solution (4 divided by 20 x 100).
Let P represent the pepsi glass and S represent the Sprite glass.
P = y; S = y; (both have y units).
P = y + x; S = y - x; (pour x units from S into P).
P = y + x - x = y; S = y - x + x = y. (pour x units from P back into S). Same result as 1).
This was my first response to the person who gave me the problem, but they just glared and said “no smartass”.
No, he’s actually right. I had calculated and re-calculated, and every time, I said, “aha, one’s a n% solution and the other’s an (100-n)% solution”. Turns out I’m a dumbass who forgot the parameters of my original question (that the ratios be compared as inverses of one another).