document.write( "Question 181478: The radiator in a car is filled with a solution of 65 per cent antifreeze and 35 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent? \n" ); document.write( "

Algebra.Com's Answer #136099 by Alan3354(69443) $\"\"$ $\"About$

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The radiator in a car is filled with a solution of 65 per cent antifreeze and 35 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
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\n" ); document.write( "At the start, it has 3 liters of 65% antifreeze, so that's 3*0.65 = 1.95 l of AF.
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\n" ); document.write( "At the end, we want 3 liters of 50% AF, which will be 1.5 l of AF in the radiator. So 0.45 liters of AF must be removed.
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\n" ); document.write( "To drain 0.45 l of a solution that's 65% AF, we must drain 0.45/0.65 liters. See that? Each liter drained removes 0.65 liters of AF. We need to remover 0.45 liters, so we drain 0.45/0.65 liters, = 9/13 liters, apx 692 milliliters.
\n" ); document.write( "Then add the water. \n" ); document.write( "

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