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 #136097 by Mathtut(3670) $\"\"$ $\"About$

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let x be the amount drained and added back to the radiator
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\n" ); document.write( "we know that the original mixture is 35% water so there is 1.05 liters of water
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\n" ); document.write( "we also know that as we drain this, whatever we drain has 35% water. so the drained portion of water(or water lost while draining), would be .35x.
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\n" ); document.write( "we also know what ever we do drain we have to add back, that is x
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\n" ); document.write( "our end result needs to be 50% of capacity( which is 3 liters)
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\n" ); document.write( "so .35(3)-.35x + x =.5(3)
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\n" ); document.write( "1.05-.35x + x=1.5
\n" ); document.write( ":
\n" ); document.write( ".65x=.45
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\n" ); document.write( " $\"highlight%28x=.69%29\"$ liters \n" ); document.write( "

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