Heard on the Street question #1

[WushiX]

You are given two glass jugs. Each contains the same volume V of liquid. One jug contains pure alcohol and the other jug contains pure water. A modest quantity Q of water is poured from the water jug into the alcohol jug, which is then thoroughly mixed. The same modest quantity Q (now diluted) of alcohol is then poured back into the water jug to equalize the volumes of the jugs at their initial levels.
The initial concentration of alcohol in the alcohol jug equals the initial concentration of water in the water jug (at 100%). What is the relationship between the final concentrations of alcohol in the alcohol jug and water in the water jug?

According to the book, the answer is that both jugs have the same concentrations, specifically V/(V+Q).

Isn't that wrong? The book seems to be claiming that the concentration of water in the diluted alcohol is 0.

When you pour water into the alcohol, the concentration of the alcohol becomes V/(V+Q). So wouldn't this imply that the concentration of water in the diluted alcohol is c = Q/(V+Q)? Therefore when you pour this diluted alcohol into the water jug, the concentration of water in the water jug should become (V+Q*c)/(V+Q). Hence the concentration of water in the water jug is larger than the concentration of alcohol in the alcohol jug (which is V/(V+Q)).

Edit: Wow I'm stupid

 

[ThisGuy]

The easy way to see that the concentrations must be the same is that the volume in each jug begins and ends at the same value V. So however much alcohol was transferred into the water jug, the same amount of water must have been transferred into the alcohol jug.

 

[benzbubblecat]

the concentration of water in the water jug should become (V+Q*c)/(V+Q)

Remember that Q has been removed prior, so adding Q back takes the volume back to V only. The book is correct. IF we let Q = 10% of V, then the final concentrations of both are ~91%.

 

[Tom Maloney]

Step 0.
Jug1: volume V of alcohol, concentration 1
Jug2: volume V of water, concentration 1
Step 1. Pour Q of Jug2 into Jug1.
Jug1: volume V of alcohol, concentration V/(V+Q)
volume Q of water, concentration Q/(V+Q)
Jug2: volume V - Q of water, concentration 1
Step 2. Pour Q of Jug1 into Jug2.
Jug1: volume of alcohol = V * V/(V+Q), concentration V/(V+Q)
volume of water = V * Q/(V+Q) , concentration Q/(V+Q)
Jug2: volume of alcohol = Q * V/(V+Q), concentration Q/(V+Q)
volume of water = (V-Q) + Q * Q/(V+Q) = V * V/(V+Q), concentration V/(V+Q)

 

[WushiX]

Crap you're right thanks