Question 159495
    A vat contains 450 gallons of 30% alcohol solution, but the manufacturer wants a 42.6% alcohol solution. How much solution must be drained and replaced with pure alcohol to obtain the desired results?
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Let x = amount drained and replaced with pure alcohol
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"amount of alcohol added" + "amt of alcohol in original" = "desired mixture"
x + .30(450-x) = .426(450)
x + 135 - .30x = 191.7
.70x + 135 = 191.7
.70x = 56.7
x = 81 gallons