document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=114633>Question 114633</A>: <I><SPAN STYLE=\"word-break: break-all;\"> \r<BR>\n" );
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document.write( " How much of an alloy that is 10%  copper should be mixed with 300 ounces of an <BR>\n" );
document.write( "alloy that is 90%  copper in order to get an alloy that is 50% copper?\r<BR>\n" );
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document.write( "___ ounces </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #83393 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=83393\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .1X+.90*300=.5(300+X)<BR>\n" );
document.write( ".1X+270=150+.5X<BR>\n" );
document.write( ".1X-.5X=150-270<BR>\n" );
document.write( "-.4X=-120<BR>\n" );
document.write( "X=-120/-.4<BR>\n" );
document.write( "X=300 OUNCES OF 10% COPPER IS REQUIRED.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( ".10*300+.9*300=.5*600<BR>\n" );
document.write( "30+270=300<BR>\n" );
document.write( "300=300<BR>\n" );
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