document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=170153>Question 170153</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A GROCER MIXES 6 POUNDS OF COFFEE THAT IS 60% COLUMBIAN WITH ONE THAT 80% COLUMBIAN HOW MANY POUNDS OF THE SECOND TYPE SHOULD BE USED TO OBTAIN  A MIXTURE THAT IS 78% COLUMBIAN? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #125601 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=125601\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A GROCER MIXES 6 POUNDS OF COFFEE THAT IS 60% COLUMBIAN WITH ONE THAT 80% COLUMBIAN HOW MANY POUNDS OF THE SECOND TYPE SHOULD BE USED TO OBTAIN  A MIXTURE THAT IS 78% COLUMBIAN?<BR>\n" );
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document.write( "Columbian Equation: 0.60*6 + 0.80x = 0.78(6+x)<BR>\n" );
document.write( "Multiply thru by 100 to get:<BR>\n" );
document.write( "60*6 + 80(x) = 78*6 + 78x<BR>\n" );
document.write( "2x = 18*6<BR>\n" );
document.write( "x = 9*6 = 54 lbs.  (amt. of 80% coffee in the mixture)<BR>\n" );
document.write( "===========================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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