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    sburlington6's Avatar
    sburlington6 Posts: 2, Reputation: 1
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    #1

    Oct 31, 2006, 02:54 PM
    Mixture Problems
    How many ounces of pure water must be added to 80oz of a 20% acid solution to make a solution that is 12% acid? (hint: Pure water is 0% Acid.)
    s_cianci's Avatar
    s_cianci Posts: 5,472, Reputation: 760
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    #2

    Oct 31, 2006, 08:01 PM
    let x = # of ounces of pure water
    You have 80 oz. of 20% acid (before adding the water)
    You have x + 80 oz. of 12% acid after adding the water
    The amount of acid is the same before the water is added and after the water is added
    So: .2(80) = .12(x + 80)
    Solve the above equation for x
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