document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=825839>Question 825839</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A certain metal is 40% tin. How many kilograms of this metal must be mixed with 80 kilograms of a metal that is 70% tin to get a metal that is 50% tin? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #497622 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=497622\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A certain metal is 40% tin. How many kilograms of this metal must be mixed with 80 kilograms of a metal that is 70% tin to get a metal that is 50% tin?<BR>\n" );
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document.write( "Equation:<BR>\n" );
document.write( "tin + tin = tin<BR>\n" );
document.write( "0.40x + 0.70*80 = 0.50(x+80)<BR>\n" );
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document.write( "40x + 70*80 = 70x + 50*80<BR>\n" );
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document.write( "-30x = -20*80<BR>\n" );
document.write( "x = (2/3)80<BR>\n" );
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document.write( "x = 53 1/3 kg (amt. of 40% metal needed)<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );