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Just one more h.w. problem

  1. Jun 25, 2010 #1
    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

    ?

    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. jcsd
  3. Jun 25, 2010 #2
    Your labels need a little fixing.

    Glass 2: [ x / (x+1) ] * 100% wine , [ 1 / (x+1) ] * 100% water
     
  4. Jun 25, 2010 #3
    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 [tex]\frac{x}{x+1}[/tex] and the fraction of wine is [tex]\frac{1}{x+1}[/tex]

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

    So, water left in glass 2 = [tex]x - \frac{x}{x+1} = \frac{{x}^{2}}{x+1} [/tex]ounces

    Wine in glass 1 = [tex]x - 1 + \frac{1}{x+1} = \frac{{x}^{2}}{x+1} [/tex]ounces
     
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