document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1063662>Question 1063662</A>: <I><SPAN STYLE=\"word-break: break-all;\"> there were two different coper alloys of total weight 50 kg. The first contains 40% less copper than the second. Determine the percentage of copper in the first and  second alloys, if it is known that there were 6kg of copper in the first alloy and 12 kg in the second  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #678809 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=678809\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let n be the amount of the second alloy, and p be the percentage of copper. Then the amount of the first alloy is 50-n, and its' percentage is .6p. So:<BR>\n" );
document.write( "np=12<BR>\n" );
document.write( "(50-n)(.6p)=6<BR>\n" );
document.write( "Then:<BR>\n" );
document.write( "n=12/p<BR>\n" );
document.write( "(50-12/p).6p=6<BR>\n" );
document.write( "30p-7.2=6<BR>\n" );
document.write( "30p=13.2<BR>\n" );
document.write( "p=13.2/30=.44<BR>\n" );
document.write( ".6p=.264<BR>\n" );
document.write( "There was 26.4% copper in the first alloy, and 44% copper in the second alloy. &#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );