document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1133860>Question 1133860</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A metallurgist has one alloy containing 21% copper and another containing 36%  copper. How many pounds of each alloy must he use to make 44 pounds of a third alloy containing 33% copper? (Round to two decimal places if necessary.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "_______ pounds of alloy containing 21% copper\r<BR>\n" );
document.write( "\n" );
document.write( "______ pounds of alloy containing 36% copper </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #751119 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=751119\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Here's an alternative to the traditional algebraic method for solving mixture problems like this involving two ingredients.  If you understand it, it will get you to the answer much faster and with far less work.<br><BR>\n" );
document.write( "(1) Think of starting with the 21% alloy and adding the 36% alloy until the mixture is 33% copper.<BR>\n" );
document.write( "(2) 33% is 4/5 of the way from 21% to 36%.  (33-21=12; 36-21=15; 12/15 = 4/5).<BR>\n" );
document.write( "(3) That means 4/5 of the mixture must be what you are adding -- the 36% alloy.<br><BR>\n" );
document.write( "So 4/5 of the total 44 pounds, or 35.2 pounds, is the 36% alloy and the other 8.8 pounds is the 21% alloy. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );