document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=469583>Question 469583</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An artist has 30kg of an alloy that is 45% copper. How many kilograms of a second alloy that is 70% copper should be mixed with the first alloy to get a new alloy that is 60% copper </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #322159 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=322159\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> An artist has 30kg of an alloy that is 45% copper. How many kilograms of a second alloy that is 70% copper should be mixed with the first alloy to get a new alloy that is 60% copper<BR>\n" );
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document.write( "Equation:<BR>\n" );
document.write( "copper + copper = copper<BR>\n" );
document.write( "0.45*30 + 0.70x  = 0.60(x + 30)<BR>\n" );
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document.write( "Multiply thru by 100 to get:<BR>\n" );
document.write( "45*30 + 70x = 60x + 60*30<BR>\n" );
document.write( "15*30 = 10x<BR>\n" );
document.write( "x = 5 kg (amt of 70% alloy needed)<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( "============  </SPAN>\n" );
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