document.write( "Question 121689This question is from textbook College Algebra
\n" ); document.write( ": It is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 70 liters of 20% solution. How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength? \n" ); document.write( "

Algebra.Com's Answer #89345 by josmiceli(19441) $\"\"$ $\"About$

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Let $\"x\"$ = the amount of 20% solution that I need to
\n" ); document.write( "drain off.
\n" ); document.write( "Then $\"x\"$ will also be the amount of pure antifreeze
\n" ); document.write( "that I add afterwards.
\n" ); document.write( "My basic equation in words is:
\n" ); document.write( "(% antifreeze solution) = (amount of antifreeze) / (antifreeze + water)
\n" ); document.write( "My (antifreeze + water) does not change. I started with $\"70\"$ l
\n" ); document.write( "and I end up with $\"70\"$ l.
\n" ); document.write( "I'm looking for a 40% solution
\n" ); document.write( ".4 = (amount of antifreeze)/70
\n" ); document.write( "I start off with $\"70\"$ l of 20% solution.
\n" ); document.write( "That means that it contains
\n" ); document.write( " $\".2%2A70+=+14\"$ l of antifreeze
\n" ); document.write( "I first drained off $\"x\"$ amount of 20% solution.
\n" ); document.write( "That means I drained off $\".2x\"$ amount of antifreeze
\n" ); document.write( "The I add $\"x\"$ amount of antifreeze
\n" ); document.write( " $\".4+=+%2814+-+.2x+%2B+x%29+%2F+70\"$
\n" ); document.write( " $\".4+=+%2814+%2B+.8x%29+%2F+70\"$
\n" ); document.write( " $\"28+=+14+%2B+.8x\"$
\n" ); document.write( " $\".8x+=+14\"$
\n" ); document.write( " $\"x+=+17.5\"$ l answer
\n" ); document.write( "I'd like to check this
\n" ); document.write( "What did I start with?
\n" ); document.write( " $\"14\"$ l antifreeze and $\"56\"$ l water for a $\"14+%2B+56+=+70\"$ l solution
\n" ); document.write( "I drain off $\"17.5\"$ l of this 20% solution
\n" ); document.write( "I drain off $\".2%2A17.5+=+3.5\"$ l antifreeze
\n" ); document.write( "and $\".8%2A17.5+=+14\"$ l water
\n" ); document.write( "Now I've got $\"14+-+3.5+=+10.5\"$ l antifreeze
\n" ); document.write( "and $\"56+-+14+=+42\"$ l water
\n" ); document.write( "The last step: I want to add $\"17.5\"$ l of antifreeze
\n" ); document.write( "Now I've got $\"10.5+%2B+17.5+=+28\"$ l of antifreeze
\n" ); document.write( "and still $\"42\"$ l of water
\n" ); document.write( "going back to:
\n" ); document.write( "(% antifreeze solution) = (amount of antifreeze) / (antifreeze + water)
\n" ); document.write( "(% antifreeze solution) = $\"28+%2F+%2842+%2B+28%29\"$
\n" ); document.write( "(% antifreeze solution) = $\"28+%2F+70\"$
\n" ); document.write( "(% antifreeze solution) = 40%
\n" ); document.write( "OK \n" ); document.write( "

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