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    Topics || Problems

    How much of a 40% solution of alcohol and how much of an 80% solution should be mixed to give 40 gallons of 50% solution? Solution:
    A(Alcohol), W(water), (M)Mixture
    A40%80%50%
    W60%20%50%
    xy40 Gallons

    \(x+y = 40\)

    \(x = 40-y\)

    \(0.40x +0.80y = 0.50(40) \)

    \(0.40(40-y) +0.80y = 0.50(40) \)

    \(16 +0.4 y = 20 \)

    \(y = 10\)

    \(x = 40-10\)

    \(x =30\)

    Thus, to create a solution a 30 gallons is needed from the 40% solution and 10 gallons from the other.