document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1102919>Question 1102919</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many pounds of a chicken feed that is 35% corn must be mixed with 320 lb of a feed that is 80% corn to make a chicken feed that is 75% corn?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #717655 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13203)</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=717655\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Here is another, very different, way to solve this kind of \"mixture\" problems.<br><BR>\n" );
document.write( "For me, this other method gets me to the answer much faster, and with far less work, than the standard algebraic method shown by the other tutor.<br><BR>\n" );
document.write( "(1) Determine how far the percentage of the mixture is from the percentages of each ingredient: 80%-75% = 5%; 75%-35% = 40%.<BR>\n" );
document.write( "(2) The ratio of the percentage differences is 5:40, or 1:8.<BR>\n" );
document.write( "(3) That means the two ingredients must be mixed in the ratio 1:8; and since the mixture percentage is much closer to 80% than to 35%, the larger portion must be the 80% feed.<BR>\n" );
document.write( "(4) There are 320 lb of the 80% feed; the 8:1 ratio means there must be 40 lb of the 35% feed. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );