document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=987996>Question 987996</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Gold and copper are heavier than water by 19 and 9 times respectively. In what proportion should they be mixed to form an alloy so that it will be heavier than water by 12 times? <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #609768 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=609768\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x = the fraction of the gold alloy.  Then 1-x will be the fraction of the copper alloy.<BR>\n" );
document.write( "So we can write the following equation:<BR>\n" );
document.write( "19x + 9(1-x) = 12<BR>\n" );
document.write( "Solve for x:<BR>\n" );
document.write( "10x = 3<BR>\n" );
document.write( "x = 3/10 = 0.3<BR>\n" );
document.write( "1-x = 0.7<BR>\n" );
document.write( "So the proportion is 7:3 copper to gold </SPAN>\n" );
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