document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=84494>Question 84494</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An automobile radiator contains 16L of antifreeze and water.  This mixture is 30% antifreeze.  How much of this mixture should be drained and replaced with pure antifreeze so that there will be 50% antifreeze?  (Not from a textbook.)  I'm ok to solve is just adding antifreeze, but can't figure out how to set it up so that the solution is drained and replaced with pure antifreeze. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #60893 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=60893\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you're going to drain xL of 30% antifreeze leaving (16-x)L ... to this you'll add xL of 100% antifreeze to get 16L of 50%\r<BR>\n" );
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document.write( ".3(16-x)+x=.5(16) ... 4.8-.3x+x=8 ... .7x=3.2 ... x=32/7 or about 4 1/2 L </SPAN>\n" );
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