document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=281844>Question 281844</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you solve a mixture problem like this?:<BR>\n" );
document.write( "One solution is 80% acid and another one is 30% acid. How many of each solution is needed to make a 200L solution that is 62% acid? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #204724 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=204724\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let 80% acid required be x\r<BR>\n" );
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document.write( "30% acid will be 200 -x\r<BR>\n" );
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document.write( "80%-----------------------30%---------------------62%\r<BR>\n" );
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document.write( "x-------------------------200-x-------------------200   quantity\r<BR>\n" );
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document.write( "0.8x+0.3(200-x)=0.62*200\r<BR>\n" );
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document.write( "0.8x+60-0.3x=124\r<BR>\n" );
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document.write( "0.5x=124-60\r<BR>\n" );
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document.write( "0.5x=64\r<BR>\n" );
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document.write( "x=64/0.5\r<BR>\n" );
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document.write( "=128 liters  ---- 80% acid\r<BR>\n" );
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document.write( "The balance will be 30 % acid solution=72 liters\r<BR>\n" );
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document.write( " </SPAN>\n" );
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