document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1173513>Question 1173513</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A metalworker has a metal alloy that is 20​% copper and another alloy that is 70​% copper. How many kilograms of each alloy should the metalworker combine to create  80kg of a 60​% 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 #798727 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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/cgi-bin/show-face.mpl?user=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=798727\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "If a formal algebraic solution is not required, try this quick and easy method for solving 2-part \"mixture\" problems like this:<br><BR>\n" );
document.write( "You are mixing 20% copper with 70% copper to get 60% copper.<BR>\n" );
document.write( "60% is 40/50 = 4/5 of the way from 20% to 70%.  (picture the three percentages on a number line, if it helps....)<BR>\n" );
document.write( "That means 4/5 of the mixture is the higher percentage alloy.<br><BR>\n" );
document.write( "ANSWER: 4/5 of 80kg, or 64kg, of 70% copper; the other 16kg of 20% copper.<br><BR>\n" );
document.write( "CHECK:<BR>\n" );
document.write( ".70(64)+.20(16) = 44.8+3.2 = 48<BR>\n" );
document.write( ".60(80) = 48<br><BR>\n" );
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