document.write( "Question 88593: It is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 60 liters of 20% solution. How many liter of this shoud be drained and replaced with 100% antifreeze to get the desired strength? Please help me with this problem. Thanks very much. \n" ); document.write( "
Algebra.Com's Answer #64426 by ankor@dixie-net.com(22740)\"\" \"About 
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It is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 60 liters of 20% solution. How many liter of this should be drained and replaced with 100% antifreeze to get the desired strength? Please help me with this problem.
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\n" ); document.write( "Let x = amt drained and amt replaced
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\n" ); document.write( "The remaining amt of 20% solution will = (60-x)
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\n" ); document.write( ".20(60-x) + 1,00(x) = .40(60)
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\n" ); document.write( "12 - .2x + 1x = 24
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\n" ); document.write( "1x - .2x = 24 - 12
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\n" ); document.write( ".8x = 12
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\n" ); document.write( "x = 12/.8
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\n" ); document.write( "x = 15 liters drained and replace by 15 liters of 100% solution
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\n" ); document.write( "Check solution, amt of 20% solution = 60-15 = 45 liters
\n" ); document.write( ".20(45) + 1.00(15) = .40(60)
\n" ); document.write( "9 + 15 = 24; proves the solution
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\n" ); document.write( "Not really that hard, right?? \n" ); document.write( "
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