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Mary has a solution that is 60% alcohol and another that is 20% alcohol. How much of each should she use to make 100 milliliters of a solution that is 52% alcohol?

  • math -

    let the volume of 60% solution used be x
    then the volume of the 20% has to be 100-x

    solve:

    .6x +.2(100-x) = .52(100)

  • math -

    where did you get 100-x???

  • math -

    If the both of us together have 100 cents, and I have x, then you have 100-x

    Mary wants to end up with 100 ml
    if x ml are of the 60% type, then
    100-x ml must be the 40% kind.

    notice (x) + (100-x) = 100

  • math -

    and that doesnt answer the question, the question asks how much of each solution she should use

  • math -

    its 20% not 40% percent

  • math -

    if the 60% one is x, then 100-x = 100-60 which equals 40, not 20

  • math -

    Emmet, look how I defined x.
    Once you find x you have the amount of 60% solution used, plug x into 100-x and you have the 20% solution.

    Do you know how to solve the equation?

    (the 40% is a typo, look at my equation, I used 20%)

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