document.write( "Question 421151: nurse has a solution that is 25% alcohol and another that is 50% alcohol. how much of each musst he use to make 20 liters of a solution that is 40% alcohol? \n" ); document.write( "
Algebra.Com's Answer #294139 by ptaylor(2198)\"\" \"About 
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Let x=amount of 25% alcohol needed
\n" ); document.write( "Then 20-x=amount of 50% alcohol needed
\n" ); document.write( "Now we know that the amount of pure alcohol that exists before the mixture takes place (0.25x+0.50(20-x)) has to equal the amount of pure alcohol in the final mixture (0.40*20). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.25x+0.50(20-x)=0.40*20 simplify
\n" ); document.write( "0.25x+10-0.50x=8 subtract 10 from each side and collect like terms
\n" ); document.write( "-0.25x=-2
\n" ); document.write( "x=8 liters------------------------amount of 25% alcohol needed
\n" ); document.write( "20-x=20-8=12 liters -----------------amount of 50% alcohol needed
\n" ); document.write( "CK
\n" ); document.write( "8*0.25 + 12*0.50=0.40*20
\n" ); document.write( "2+6=8
\n" ); document.write( "8=8\r
\n" ); document.write( "\n" ); document.write( "BTW problems such as this can also be solved using two equations and two unknowns, e.g.,
\n" ); document.write( "x+y=20--------------------eq1
\n" ); document.write( "0.25x+0.50y=0.40*20-------eq2\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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