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Math Help - percent mixture problem

  1. #1
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    percent mixture problem

    "How much water should be added to 30 gallons of a solution that is 70% antifreeze in order to get a mixture that is 60% antifreeze?"

    I am having trouble figuring out how to set up the problem.
    If I do .6(30+x) = .7(30) + 1x the answer is -7.5 gallons.
    How do I set up this equation?
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  2. #2
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    Re: percent mixture problem

    Hello, rainbowx!

    How much water should be added to 30 gallons of a solution that is 70% antifreeze
    in order to get a mixture that is 60% antifreeze?

    Since water is added to the solution, we must think in terms of water.

    The original solution is 30 gallons which is 30% water.
    . . It contains: 0.30 \times 30 \,=\,9 gallons of water.

    Then x gallons of water is added.

    The final mixture contains: x + 9 gallons of water. .[1]


    But we know that the final mixture will be x+30 gallons which is 40% water.
    . . It contains: 0.40(x+30) gallons of water. .[2]


    We just described the final amount of water in two ways.

    There is our equation! . . . x+9 \;=\;0.4(x+30)

    Got it?

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  3. #3
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    Re: percent mixture problem

    another way to solve this.
    The conc line starts at 0% AF goes to 60 and then to 70%. Therefor 60 =parts 70% and 10 =parts water so answer for 30 gals 70% is 5 gals water
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