document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1184584>Question 1184584</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Two metal alloys, containing 6% zinc and 16% zinc are to be mixed together. How much of each is needed to get to 150 kg of an 14% zinc alloy? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #815234 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13241)</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=815234\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "You should know and understand the standard formal algebraic method for solving 2-part mixture problems, as shown by the other tutor.<br><BR>\n" );
document.write( "If a formal algebraic solution is not required, here is a simple non-algebraic way for solving 2-part mixture problems like this.  When the numbers are \"nice\", as they are in this problem, the answer can be obtained in a matter of seconds.<br><BR>\n" );
document.write( "(1) Observe/calculate that 14% is 4/5 of the way from 6% to 16%<br><BR>\n" );
document.write( "(2) That means 4/5 of the mixture is the higher percentage ingredient.<br><BR>\n" );
document.write( "ANSWER: 4/5 of 150kg, or 120kg, of 16% zinc; 1/5 of 150kg, or 30kg, of 6% zinc.<br><BR>\n" );
document.write( "CHECK:<BR>\n" );
document.write( ".16(120)+.06(30)=19.2+1.8=21<BR>\n" );
document.write( ".14(150)=21<br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );