document.write( "Question 467328: I'm really stuck on this word problem.word problem are my weakness: It is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 40 liters of 20% solution. How many liters of this should be drained and replaced with 100% anitfreeze to get the desired strength? \n" ); document.write( "
Algebra.Com's Answer #320627 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.
\n" ); document.write( " The radiator now has 40 liters of 20% solution.
\n" ); document.write( " How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?
\n" ); document.write( ":
\n" ); document.write( "Let x = amt of original anti-freeze to be drained, also the amt of pure antifreeze to be added
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\n" ); document.write( "Write a decimal amt of antifreeze equation
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\n" ); document.write( ".20(40-x) + 1x = .40(40)
\n" ); document.write( "8 - .2x + 1x = 16
\n" ); document.write( "-.2x + 1x = 16 - 8
\n" ); document.write( ".8x = 8
\n" ); document.write( "x = \"8%2F.8\"
\n" ); document.write( "x = 10 liters removed and 10 liters of pure antifreeze to be added
\n" ); document.write( ":
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\n" ); document.write( "Check this in the original equation
\n" ); document.write( ".20(40-10) + 10 = .4(40)
\n" ); document.write( ".2(30) + 10 = 16
\n" ); document.write( "6 + 10 = 16; confirms our solution of x = 10 \n" ); document.write( "
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