SOLUTION: An automobile radiator contains 16L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that

Question 84494: An automobile radiator contains 16L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that there will be 50% antifreeze? (Not from a textbook.) I'm ok to solve is just adding antifreeze, but can't figure out how to set it up so that the solution is drained and replaced with pure antifreeze.
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
you're going to drain xL of 30% antifreeze leaving (16-x)L ... to this you'll add xL of 100% antifreeze to get 16L of 50%

.3(16-x)+x=.5(16) ... 4.8-.3x+x=8 ... .7x=3.2 ... x=32/7 or about 4 1/2 L
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