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Math Help - Mixture Type 2 - Problem

  1. #1
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    Angry Mixture Type 2 - Problem

    The question is:
    A solution of 27 gallons of acid contains 9 gallons of pure acid.How much water in gallons should be added to produce a 25% solution of this acid ?
    The answer is 9. However here is how i approached it.

    Quantity %age Cal
    water x 100% x
    acid solution 27 9/27 or (33.33%) 0.33 x 27 = 9
    Required Solution x+27 25% Equation
    Equation is :
    x+9 = 0.25(x+27)
    x+9 = 0.25x + 6.75

    Could anyone tell me what am i doing wrong i am getting a negative value ??
    Last edited by MikeNoob; May 28th 2012 at 08:57 PM.
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  2. #2
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    Re: Mixture Type 2 - Problem

    Wouldn't it just be:
    you are adding water which has a 0% acid concentration
    since you have 9gallons of acid, you want the total solution to be 36 gallons.
    Add 36 - 27 gallons of water = 9 gallons.

    Doing it in your table you have conc of water is 100%, but actually it's 0% as you're talking about the acid concentration.
    so it would be
    Quantity %age Cal
    Initially 27 1/3 9
    addWater x 0 0
    Required Solution x+27 25% 9
    so (27+x)/4 = 9
    x + 27 = 36
    x= 9
    Last edited by linalg123; May 28th 2012 at 11:35 PM.
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  3. #3
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    Re: Mixture Type 2 - Problem

    @Linalg123 You are absolutely correct. Thanks for pointing this out!!!.
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