document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1076102>Question 1076102</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A goldsmith has two gold alloys. The first alloy is 30% gold; the second alloy is 80% gold. How many grams of each should be mixed to produce 50 grams of an alloy that is 68% gold? <BR>\n" );
document.write( "amount of 30% gold _______grams<BR>\n" );
document.write( "amount of 80% gold _______grams  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #690786 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=690786\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let n be the amount of 80% alloy. Then the 30% alloy is 50-n. So:<BR>\n" );
document.write( ".8n+.3(50-n)=.68(50)<BR>\n" );
document.write( ".5n+15=34<BR>\n" );
document.write( ".5n=19<BR>\n" );
document.write( "n=38<BR>\n" );
document.write( "He needs 38 gms of 80% alloy, and 12 gms 30% alloy. &#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );