SOLUTION: A radiator contains 8 liters of mixture of water and antifreeze. If 40% of the mixture is antifreeze, how much of the mixture shall be drained and replaced with pure antifreeze so

Question 472759: A radiator contains 8 liters of mixture of water and antifreeze. If 40% of the mixture is antifreeze, how much of the mixture shall be drained and replaced with pure antifreeze so that the resultant mixture will contain 60% antifreeze?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A radiator contains 8 liters of mixture of water and antifreeze.
If 40% of the mixture is antifreeze, how much of the mixture shall be drained
and replaced with pure antifreeze so that the resultant mixture will contain 60% antifreeze?
:
Let x = the amt to be drained, and the amt of pure antifreeze to be added
:
.40(8-x) + 1x = .60(8)
3.2 - .4x + 1x = 4.8
-.4x + 1x = 4.8 - 3.2
.6x = 1.6
x =
x = 2 liters drained and 2 liters of pure antifreeze is added
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