document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=892529>Question 892529</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A radiator contains 8 liters of a 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 66% antifreeze solution? (Round your answer to two decimal places.) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #540572 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=540572\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x = the # of liters to be drained and replaced.<BR>\n" );
document.write( "After drainage, there is 8 - x liters left, with 25% antifreeze.  The amount x liters is 100% antifreeze. The resulting 8 liters of solution has 66% antifreeze.<BR>\n" );
document.write( "Hence 0.25*(8 - x) +1.00x = 0.66*8<BR>\n" );
document.write( "==> 2- 0.25x + x = 5.28<BR>\n" );
document.write( "==> 0.75x = 3.28<BR>\n" );
document.write( "==> x = 3.28/0.75<BR>\n" );
document.write( "+==> x = 4.37 liters of pure antrifreeze </SPAN>\n" );
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