document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1007013>Question 1007013</A>: <I><SPAN STYLE=\"word-break: break-all;\">  A metal worker has a metal alloy that is 30% copper and another alloy that is 70% copper how many kilograms of each Alloway should the metalwork are combined to create 90 KG of a 62% copper alloy  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #623045 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=623045\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the amount of 30% alloy.<BR>\n" );
document.write( "Then 90-x will be the amount of 70% alloy.<BR>\n" );
document.write( "The set up looks like this then...<BR>\n" );
document.write( ".30x + .70(90-x) = .62(90)<BR>\n" );
document.write( ".30x + 63 - .70x = 55.8<BR>\n" );
document.write( "-.40x = -7.2<BR>\n" );
document.write( "x = 18 kg at 30% and so<BR>\n" );
document.write( "90-x = 72 kg at 70% </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );