document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1092605>Question 1092605</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How much of an alloy that is 20% copper should be mixed with 300 ounces of an alloy that is 80% copper in order to get an alloy that is 30% &#8203;copper? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #707247 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=707247\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let n be the amount of 20% alloy. Then:<BR>\n" );
document.write( "0.2n+0.8(300)=0.3(300+n)<BR>\n" );
document.write( "0.2n+240=0.3n+90<BR>\n" );
document.write( "0.1n=150<BR>\n" );
document.write( "n=1500 oz.<BR>\n" );
document.write( "&#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
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