Question

A radiator contains 25 quarts of a water and anti-freeze, of which 60% is antifreeze. How much if this solution should be drained and replaced with water for the new solution to be 40% antifreeze?
Consider just the amount of anti-freeze present. If x quarts of 60% solution are drained and replaced with water (0% solution), then you have

0.60(25-x) + 0.00(x) = 0.40(25)

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