document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=352062>Question 352062</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 40 oz of a 35% gold alloy are mixed with 52 oz of a 26% alloy. Find the concentration of the resulting gold alloy. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #251524 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=251524\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 40 oz of a 35% gold alloy<BR>\n" );
document.write( "52 oz of a 26% alloy. <BR>\n" );
document.write( "Total mix = 40+52= 92 oz.<BR>\n" );
document.write( "35*40+52*26=92*x<BR>\n" );
document.write( "1400+1352 =92x<BR>\n" );
document.write( "x=29.9 % = 30% </SPAN>\n" );
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