Just one more h.w. problem

1. Jun 25, 2010

Jamin2112

1. The problem statement, all variables and given/known data

We have two identical glasses. Glass 1 contains x ounces of wine; glass 2 contains x ounces of water (x≥1). We remove 1 ounce of wine from glass 1 and add it to glass 2. The wine and water in glass 2 mix uniformly. We now remove 1 ounce of liquid from glass 2 and add it to glass 1. Prove that the amount of water in glass 1 is now the same as the amount of wine in glass 2.

2. Relevant equations

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3. The attempt at a solution

First we have

Glass 1: x ounces of wine
Glass 2: x ounces of water

After the first transaction we have

Glass 1: (x-1) ounces of wine
Glass 2: x ounces of water + 1 ounce of wine

Since we're mixing uniformly, the makeup of each glass will be

Glass 1: 100% wine
Glass 2: [ x / (x+1) ] * 100% ounces of wine , [ 1 / (x+1) ] * 100% ounces of water

Am I right so far?

2. Jun 25, 2010

Phrak

Your labels need a little fixing.

Glass 2: [ x / (x+1) ] * 100% wine , [ 1 / (x+1) ] * 100% water

3. Jun 25, 2010

graphene

First we have

Glass 1: x ounces of wine
Glass 2: x ounces of water

After the first transaction we have

Glass 1: (x-1) ounces of wine
Glass 2: x ounces of water + 1 ounce of wine

Since we're mixing uniformly, the makeup of each glass will be
1: (x-1) ounces of wine (still)
2: x ounces of water + 1 ounce of wine, hence the fraction of water is $$\frac{x}{x+1}$$ and the fraction of wine is $$\frac{1}{x+1}$$

Removing 1 ounce of liquid from glass 2 would mean removing $$\frac{x}{x+1}$$ * 1 ounces of water and $$\frac{1}{x+1}$$ * 1 ounces of wine and adding them to glass 1.

So, water left in glass 2 = $$x - \frac{x}{x+1} = \frac{{x}^{2}}{x+1}$$ounces

Wine in glass 1 = $$x - 1 + \frac{1}{x+1} = \frac{{x}^{2}}{x+1}$$ounces