document.write( "Question 682458: The radiator in a car is filled with a solution of 75 percent antifreeze and 25 percent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 percent antifreeze. If the capacity of the radiator is 3.3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 percent?
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Algebra.Com's Answer #423205 by josmiceli(19441)\"\" \"About 
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The capacity of the radiator is 3.3 liters.
\n" ); document.write( "The radiator is filled with a solution of 75% antifreeze
\n" ); document.write( "which is \"+.75%2A3.3+=+2.475+\" liters of antifreeze
\n" ); document.write( "Let \"+x+\" = liters,of coolant to be drained and
\n" ); document.write( "replaced with water
\n" ); document.write( "\"+.75x+\" liters of coolant is in the liters of the
\n" ); document.write( "\"+x+\" liters drained
\n" ); document.write( "It started with \"+3.3+\" liters and ended with \"+3.3+\" liters
\n" ); document.write( "--------------------
\n" ); document.write( "\"+%28+2.475+-+.75x+%29+%2F+3.3+=+.5+\"
\n" ); document.write( "\"+2.475+-+.75x+=+.5%2A3.3+\"
\n" ); document.write( "\"+2.475+-+.75x+=+1.65+\"
\n" ); document.write( "\"+.75x+=+2.475+-+1.65+\"
\n" ); document.write( "\"+.75x+=+.825+\"
\n" ); document.write( "\"+x+=+1.1+\"
\n" ); document.write( "1.1 liters of the 75% solution must be drained and
\n" ); document.write( "replaced with pure water
\n" ); document.write( "check:
\n" ); document.write( "\"+3.3+-+1.1+=+2.2+\" liters left after draining
\n" ); document.write( "\"+.75%2A2.2+=+1.65+\" liters of antifreeze left
\n" ); document.write( "Now add \"+1.1+\" liters of water
\n" ); document.write( "The concentration is
\n" ); document.write( "\"+1.65+%2F+3.3+=+.5+\" or 50%
\n" ); document.write( "OK
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