Question 190731
    Adjusting Antifreeze: Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of a 40% antifreeze solution. How many quarts of each should she use?
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Let x = quarts of pure antifreeze
20-x = quarts of 40% antifreeze
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"amt of antifreeze from pure" + "amt from 40%" = "50% of total mixture"
x + .40(20-x) = .50(20)
x + 8 - .40x = 10
x - .40x = 2
.60x = 2
x = 2/.60
x = 20/6
x = 10/3 quarts (3.333)
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Quarts of 40% antifreeze:
20-x = 20-10/3 = 60/3 - 10/3 = 50/3 quarts (16.667)