document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1161874>Question 1161874</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How much of an alloy that is 10% copper should be mixed with 400 ounces of an alloy that is 60% copper in order to get an alloy that is 20% copper <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #785519 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=785519\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Here is a non-algebraic method for solving mixture problems like this.<br><BR>\n" );
document.write( "If you understand this method, it will get you to the answer much faster and with less work than the traditional algebraic method shown by the other tutor.<br><BR>\n" );
document.write( "Think of the problem this way:<br><BR>\n" );
document.write( "You are starting with a 60% copper alloy and you are adding 10% copper alloy, reducing the percentage of copper; you stop when the percentage reaches 20%.<br><BR>\n" );
document.write( "Now model that with numbers on a number line.  You are starting at 60 and heading towards 10, but you stop when you get to 20.<br><BR>\n" );
document.write( "What fraction of the distance have you gone?  From 60 to 10 is a difference of 50, the distance you went, from 60 to 20, was 40.  The fraction is 40/50 = 4/5.<br><BR>\n" );
document.write( "That means 4/5 of the final alloy has to be the 10% alloy that you are adding.<br><BR>\n" );
document.write( "So the 400 ounces you started with is 1/5 of the final alloy; that means the 4/5 of the final alloy that you added is 4*400 = 1600 ounces.<br><BR>\n" );
document.write( "The words of explanation make this sound like a lengthy process; but it is not.  Without the words of explanation, here is the complete solution:<br><BR>\n" );
document.write( "(1) 60 to 10 is 50; 60 to 20 is 40; 40/50 = 4/5<BR>\n" );
document.write( "(2) 4/5 of the final alloy is the 10% alloy you are adding; that's 4 times the 400 ounces you started with -- so 1600 ounces.<br><BR>\n" );
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