document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1133341>Question 1133341</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How much of an alloy that is 20% copper should be mixed with 600 ounces of an alloy that is 60% copper in order to get an alloy that is 30% ​copper? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #750527 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=750527\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The traditional algebraic solution method....<br><BR>\n" );
document.write( "20% of x ounces, plus 60% of 600 ounces, equals 30% of (x+600) ounces.<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=.20%28x%29%2B.60%28600%29+=+.30%28x%2B600%29 ALIGN=MIDDLE  ALT=\".20%28x%29%2B.60%28600%29+=+.30%28x%2B600%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "The equation is easily solved with basic algebra.<br><BR>\n" );
document.write( "I leave that to you....<br><BR>\n" );
document.write( "A faster path to the solution of any mixture problem like this involving two ingredients....<br><BR>\n" );
document.write( "30% is 1/4 of the way from 20% to 60%.  (The distance from 20% to 60% is 40%; the distance from 20% to 30% is 10%; 10% is 1/4 of 40%.)<br><BR>\n" );
document.write( "That means 1/4 of the alloy must be the higher percentage.<br><BR>\n" );
document.write( "The 600 ounces of 60% copper are 1/4 of the total, so the total is 600*4 = 2400 ounces; that means 1800 ounces of the 20% alloy. </SPAN>\n" );
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