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26 July, 23:05

Determine the amounts of 20% and 50% salt solutions that should be mixed to obtain 300 gallons of 41% salt solution.

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  1. 27 July, 02:44
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    Assume that solution with 20% and 50% salt is x and y gallons respectively.

    Therefore,

    x+y = 300 = > y=300-x

    Additionally

    20x+50y = 300*41 = 12300

    Substituting for y

    20x+50 (300-x) = 12300 = > 20x+15000-50x = 123000 = > - 30x = - 2700

    x = 90 gallons

    y = 300-x = 300 - 90 = 210 gallons
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