document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=334722>Question 334722</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A copper alloy that is 40% copper is to be combined with a copper alloy that is 80% copper to produce 120 kilograms of an alloy that is 70% copper. How many kilograms of each alloy must be used? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #239844 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=239844\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A copper alloy that is 40% copper is to be combined with a copper alloy that is 80% copper to produce 120 kilograms of an alloy that is 70% copper. How many kilograms of each alloy must be used?<BR>\n" );
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document.write( "Equation:<BR>\n" );
document.write( "copper + copper = copper<BR>\n" );
document.write( "0.40x + 0.80(120-x) = 0.70*120<BR>\n" );
document.write( "---<BR>\n" );
document.write( "Multiply thru by 100 to get:<BR>\n" );
document.write( "40x + 80*120 - 80x = 70*120<BR>\n" );
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document.write( "-40x = -10*120<BR>\n" );
document.write( "x = (1/4)120<BR>\n" );
document.write( "x = 30 kg (amt. of 40% alloy needed in the mixture.<BR>\n" );
document.write( "=================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );