document.write( "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? \n" ); document.write( "

Algebra.Com's Answer #115485 by gonzo(654) $\"\"$ $\"About$

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for there to be 30% antifreeze, there must be 3 liters of antifreeze and 70 liters of water.
\n" ); document.write( "to make 50% antifreeze, there must be 5 liters of antifreeze and 5 liters of water.
\n" ); document.write( "so you need to add 2 liters of antifreeze and take out 2 liters of water.
\n" ); document.write( "how to do?
\n" ); document.write( "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.
\n" ); document.write( "by taking out that much of the mixture, you have taken out 2 liters of water and .857142857 liters of anti-freeze.
\n" ); document.write( "you now need to add 2.857142857 liters of anti-freeze.
\n" ); document.write( "this is the 2 that you wanted to add in the first place plus the .857142857 that you took out.
\n" ); document.write( "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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