SOLUTION: A Radiator has 10L of 30% Fluid (30% Anti Freeze and 70% Water). How many liters need to be replaced to have a 50% solution?

Question 156664: A Radiator has 10L of 30% Fluid (30% Anti Freeze and 70% Water). How many liters need to be replaced to have a 50% solution?
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
for there to be 30% antifreeze, there must be 3 liters of antifreeze and 70 liters of water.
to make 50% antifreeze, there must be 5 liters of antifreeze and 5 liters of water.
so you need to add 2 liters of antifreeze and take out 2 liters of water.
how to do?
to take out 2 liters of water from a 70% solution of water, you need to take out 2/.7 = 2.857142857 liters of mixture.
by taking out that much of the mixture, you have taken out 2 liters of water and .857142857 liters of anti-freeze.
you now need to add 2.857142857 liters of anti-freeze.
this is the 2 that you wanted to add in the first place plus the .857142857 that you took out.
you are left with 5 liters of anti-freeze and 5 liters of water which is where you want to be for a 50% mixture.

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