SOLUTION: Can you help me answer this: A car radiator contains 10 liters of 30% solution. How many liters will have to be replaced with pur antifreeze if the resulting solution is to be 50%
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Question 484542: Can you help me answer this: A car radiator contains 10 liters of 30% solution. How many liters will have to be replaced with pur antifreeze if the resulting solution is to be 50% antifreeze?
Is 4 liters correct? Answer by chessace(471) (Show Source):
You can put this solution on YOUR website! Nope.
As usual x = amount replaced.
You are going to take out x which consists of .3 AF (and .7 not).
Then put in x of all AF.
Required: 50% AF after this operation.
I.e., new + old AF = 50 % of 10 liters = 5 liters.
The old AF is the amount in middle of operation * its % AF
Mid amount: 10 - x, AF still 30% or .3
x + (10 - x) * .3 = 5
x + 10*0.3 - .3 x = 5
(1-0.3)x + 3 = 5
.7 x = 5 - 3
7 x = 20
x = 20/7 liters [< 3]
