document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=363358>Question 363358</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A metallurgist needs to make 12.4 lb of an alloy containing 50% gold. He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold. How much of each should he use? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #259078 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=259078\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 0.60x + 0.40(12.4 - x) = 0.50*12.4,<BR>\n" );
document.write( "0.60x + 4.96 - 0.40x = 6.2,<BR>\n" );
document.write( "0.20x = 1.24,<BR>\n" );
document.write( "x = 6.2. He has to mix 6.2 lb of the 60% gold alloy, and the same amount of the 40% gold alloy. </SPAN>\n" );
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