SOLUTION: A radiator contains 8 quartz of mixture of water and antifreezer of 40% of mixture in antifreezer, how percent of mixture should be drained and replaced by pure antifreezer so that
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Question 1049640: A radiator contains 8 quartz of mixture of water and antifreezer of 40% of mixture in antifreezer, how percent of mixture should be drained and replaced by pure antifreezer so that the resultant mixture contain 60% antifreezer? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A radiator contains 8 quarts of a mixture of water and antifreeze that is 40% antifreeze,
how much of the mixture should be drained and replaced by pure antifreeze so that the resultant mixture contain 60% antifreeze?
:
let x = of pure antifreeze and also = the amt of mixture to be removed
:
.4(8-x) + x = .60(8)
3.2 - .4x + x = 4.8
.6x = 4.8 - 3.2
.4x = 1.6
x = 1.6/.4
x = 4 qts removed and 4 qts of pure antifreeze added
:
In terms of percent, 50% removed and 4 qts added
