document.write( "Question 454937: Six hundred liters of 50% alcohol solution was diluted to make a 40% alcohol solution. How many liters of the 50% solution needed to be replaced by pure water in order to bring the alcohol concentration down to 40%? \n" ); document.write( "

Algebra.Com's Answer #312404 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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Six hundred liters of 50% alcohol solution was diluted to make a 40% alcohol solution.
\n" ); document.write( " How many liters of the 50% solution needed to be replaced by pure water in
\n" ); document.write( " order to bring the alcohol concentration down to 40%?
\n" ); document.write( ":
\n" ); document.write( "Let x = amt or solution removed and amt of pure water added to replace it
\n" ); document.write( ":
\n" ); document.write( "Remember pure water is a 0% solution
\n" ); document.write( ":
\n" ); document.write( ".50(600-x) = .40(600)
\n" ); document.write( "300 - .50x = 240
\n" ); document.write( "-.50x = 240 - 300
\n" ); document.write( "-.50x = -60
\n" ); document.write( "x = $\"%28-60%29%2F%28-.50%29\"$
\n" ); document.write( "x = +120 liters of 50% solution removed and 120 liters of pure water added. \n" ); document.write( "

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