document.write( "Question 392688: Gerry mixes different solutions with concentrations of 25%, 40%, and 50% to get 200 liters of a 32% solution. If he uses twice as much of the 25% solution as the 40% solution, find out how may liters of each kind he uses. \n" ); document.write( "
Algebra.Com's Answer #278774 by ptaylor(2198)\"\" \"About 
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Let x=amount of the 40% solution
\n" ); document.write( "Then 2x=amount of the 25% solution
\n" ); document.write( "And (200-3x)=amount of 50% solution
\n" ); document.write( "Now we know that the amount of pure stuff that exists before the mixture takes place has to equal the amount of pure stuff that exists after the mixture takes place. So our equation to solve is:
\n" ); document.write( "0.40x+0.25*2x+0.50(200-3x)=0.32*200 simplify
\n" ); document.write( "0.40x+0.50x+100-1.50x=64
\n" ); document.write( "-0.60x=-100+64
\n" ); document.write( "-0.60x=-36
\n" ); document.write( "x=60 liters-----amount of 40% solution
\n" ); document.write( "2x=120 liters--amount of 25% solution
\n" ); document.write( "200-3x=200-180=20 liters----amount of 50% solution
\n" ); document.write( "CK
\n" ); document.write( "60*0.40+120*0.25+0.50*20=0.32*200
\n" ); document.write( "24+30+10=64
\n" ); document.write( "64=64\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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