document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1002373>Question 1002373</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A car radiator needs a 40&#8203;% antifreeze solution. The radiator now holds 21 liters of a 10&#8203;% solution. How many liters of this should be drained and replaced with&#8203; 100% antifreeze to get the desired&#8203; strength? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #619321 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=619321\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 21 liters of 0.10=2.1 liters of pure antifreeze.<BR>\n" );
document.write( "Want 21 liters *0.4=8.4 liters of pure antifreeze. Let x be the amount that is new.<BR>\n" );
document.write( "Then in the radiator, there will be (21-x) liters of 10% and x liters of 100%<BR>\n" );
document.write( "(21-x)*(0.1)+1(x)=8.4<BR>\n" );
document.write( "2.1-0.1x+x=8.4<BR>\n" );
document.write( "0.9x=6.3l<BR>\n" );
document.write( "x=7 liters.<BR>\n" );
document.write( "7 liters of new<BR>\n" );
document.write( "14 liters of old<BR>\n" );
document.write( "Total is 7+(14)(0.1)=7+1.4=8.4 liters<BR>\n" );
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