document.write( "Question 491716: how many liters of 60% alcohol solution and 40% alcohol solution should be mixed to obtain twenty liters of a 45% alcohol solution? \n" ); document.write( "
Algebra.Com's Answer #334746 by ptaylor(2198)\"\" \"About 
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Let x=amount of 60% solution needed
\n" ); document.write( "Then 20-x=amount of 40% solution needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure alcohol that exists before the mixture takes place is equal to the amount of pure alcohol after the mixture takes place, so:\r
\n" ); document.write( "\n" ); document.write( "0.60x+0.40(20-x)=amount of pure alcohol before the mixture takes place, and
\n" ); document.write( "0.45*20=amount of pure alcohol after the mixture takes place,so\r
\n" ); document.write( "\n" ); document.write( "0.60x+0.40(20-x)=0.45*20 simplify
\n" ); document.write( "0.60x+8-0.40x=9 subtract 8 from each side
\n" ); document.write( "0.20x=1
\n" ); document.write( "x=5 liters-------------------------amount of 60% solution needed
\n" ); document.write( "20-x=20-5=15 liters----------------amount of 40% solution needed\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "0.60*5+0.40*15=0.45*20
\n" ); document.write( "3+6=9
\n" ); document.write( "9=9\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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