document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1076579>Question 1076579</A>: <I><SPAN STYLE=\"word-break: break-all;\">  In an alloy of 150 pounds of zinc and copper, there are 100 pounds of copper. How much copper must be added so that the alloy may be 10% zinc?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #691195 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=691195\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The fraction is Zn/Cu+Zn<BR>\n" );
document.write( "right now it is 50/150, if 150 pounds of both and 100 pounds of Cu, the rest has to be Zn. Let x=pounds of copper added<BR>\n" );
document.write( "50/(100+x+50)=0.1<BR>\n" );
document.write( "50/(150+x)=0.1<BR>\n" );
document.write( "50=15+0.1x<BR>\n" );
document.write( "35=0.1x<BR>\n" );
document.write( "x=350 pounds of copper<BR>\n" );
document.write( "At that point, the alloy will weigh 500 pounds, 50 of which will be zinc.<BR>\n" );
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