document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=384814>Question 384814</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A 30 L solution is 80% antifreeze. How much water must be added to produce a solution that is 60% antifreeze? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #272390 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=272390\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=amount of water needed<BR>\n" );
document.write( "Now we know that the amount of pure antifreeze before the mixture takes place (0.80*30) is equal to the amount of pure antifreeze after the mixture takes place (0.60(30+x)).  Sooo our equation to solve is:<BR>\n" );
document.write( "0.80*30=0.60(30+x) simplify<BR>\n" );
document.write( "24 =18+0.60x subtract 18 from each side<BR>\n" );
document.write( "0.60x=6<BR>\n" );
document.write( "x=10 L  amount of water needed\r<BR>\n" );
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document.write( "CK<BR>\n" );
document.write( "0.80*30=0.60*40<BR>\n" );
document.write( "24=24\r<BR>\n" );
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document.write( "Hope this helps---ptaylor </SPAN>\n" );
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