document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=8643>Question 8643</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a chemist has one solution that is 60% chlorinated and another that is 40% chlorinated. How much of each solution is needed to make a 100 L solution that is 50% chlorine?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #4798 by </B><A HREF=/tutors/aboutme.mpl?userid=rapaljer><B>rapaljer(4671)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rapaljer><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=rapaljer><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=4798\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Normally this is a word problem that must be worked out with an equation (see my Lesson Plan on Word Problems--Coin and Mixture Problems--Part I).  However, there is an intuitive solution to this one if you think about it.  You are mixing some 60% and some 40% solutions, trying to obtain a 50% solution, which is exactly halfway between the 60% and the 40%.  You will obviously need to mix them half and half.  The solution is 50 L of each.\r<BR>\n" );
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document.write( "R^2 at SCC </SPAN>\n" );
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