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Water mixing problem

  1. Feb 20, 2008 #1
    I've struggled with this problem for a bit and I ran out of ideas.

    "We have a tank of volume K liters of alcohol. We remove one liter from the tank and add one liter of water. From the mix, we remove one liter and add one liter of water. We do this one more time.

    At the end of the process, there should be 7 times more water than alcohol in the tank.

    What is the volume of the tank?"

    My attempt at a solution is attached below. I hope it's legible.

    The idea is that adding one liter of alcohol means multiplying the water and the alcohol components by (k-1)/k and adding 1 to the water component.

    The alcohol component has been written down as a_i, the water component is b_i, where:
    i={0,1,2,3} is the step
    {a0, b0} is the initial state (a0=k, b0=0)
    {a1,b1} is the state after adding one to b
    {a3,b3} is the final state.
     

    Attached Files:

    Last edited: Feb 20, 2008
  2. jcsd
  3. Feb 20, 2008 #2

    EnumaElish

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    Mixture(0) = alcohol.
    After the first iteration, mixture(1) = (K-1)/K alcohol + 1/K water
    After the second, mixture(2) = (K-1)/K mixture(1) + 1/K water
    After the third, mixture(3) = (K-1)/K mixture(2) + 1/K water, and alcohol(3)/water(3) = 1/7.
     
  4. Feb 20, 2008 #3
    Thanks. I mixed up stage one :)
     
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