document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1173324>Question 1173324</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many grams of chocolate worth P600 per kg must be mixed with 10 jg of chocolate worth P450 per kilogram to produce a moxture worth P500 per kilogram? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #798547 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=798547\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Here is a quick and easy way to solve 2-part \"mixture\" problems like this without formal algebra (if, of course, a formal algebraic solution is not required).<br><BR>\n" );
document.write( "The price of the mixture per kilogram, P500, is 1/3 of the way from P450 to P600, the prices per kg of the two ingredients.  (look at it as 500 begin 1/3 of the way from 450 to 600 on a number line....)<br><BR>\n" );
document.write( "That means 1/3 of the mixture is the more expensive ingredient.<br><BR>\n" );
document.write( "That means the 10kg of the less expensive chocolate is 2/3 of the mixture; and that mean 1/3 of the mixture is 5kg.<br><BR>\n" );
document.write( "ANSWER: 5kg<br><BR>\n" );
document.write( "CHECK:<BR>\n" );
document.write( "10(450)+5(600) = 4500+3000 = 7500<BR>\n" );
document.write( "15(500) = 7500<br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );