document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1007274>Question 1007274</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a metal worker has a 30% copper and another alloy that is 60% copper. how many killograms of each alloy should the metal worker combine to create 80 kg of a 48% copper alloy </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #623273 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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=623273\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a metal worker has a 30% copper and another alloy that is 60% copper. how many killograms of each alloy should the metal worker combine to create 80 kg of a 48% copper alloy\r<BR>\n" );
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document.write( "let x=amt of 30% alloy to use<BR>\n" );
document.write( "80-x=amt of 60% alloy to use\r<BR>\n" );
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document.write( "30%x+60%(80-x)=48%*80<BR>\n" );
document.write( ".30x+48-.60x=38.4<BR>\n" );
document.write( ".30x=9.6<BR>\n" );
document.write( "x=32<BR>\n" );
document.write( "80-x=48<BR>\n" );
document.write( "amt of 30% alloy to use=32 kg<BR>\n" );
document.write( "amt of 60% alloy to use=48 kg </SPAN>\n" );
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