document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=765318>Question 765318</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In order for coffee to be labeled \"kona blend\", it must contain at least 65% kona beans.  Barking Dog Roasters has 50 pounds of 100% kona coffee beans.  How much Sumatran coffee must they add to the kona beans if they wish to market the Kona coffee as kona blend? <BR>\n" );
document.write( "has to be set up like <BR>\n" );
document.write( "let x =<BR>\n" );
document.write( "let y =<BR>\n" );
document.write( "and equations thank you! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #466079 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39617)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><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=466079\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The problem needs just one variable.<BR>\n" );
document.write( "Let p = the number of pounds of Sumatran coffee to add to the Kona beans.\r<BR>\n" );
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document.write( "You want the resulting blend to be 65% Kona.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=50%2A100%2F%2850%2Bp%29=65 ALIGN=MIDDLE  ALT=\"50%2A100%2F%2850%2Bp%29=65\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Adding p pounds of Sumatra coffee will reduce the concentration of Kona in the blend from 100% down to whatever concentration is desired.<BR>\n" );
document.write( "SOLVE FOR p.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=5000=65%28p%2B50%29 ALIGN=MIDDLE  ALT=\"5000=65%28p%2B50%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=5000=65p%2B3250 ALIGN=MIDDLE  ALT=\"5000=65p%2B3250\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=65p=5000-3250 ALIGN=MIDDLE  ALT=\"65p=5000-3250\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=65p=1750 ALIGN=MIDDLE  ALT=\"65p=1750\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=p=26%2612%2F13 ALIGN=MIDDLE  ALT=\"p=26%2612%2F13\"   DISABLED_style= \"max-width:90%; height:auto;\" > pounds, but some judgement should be made here about what is most practical.  Best would be to bring that fraction down just a little bit, say make it to about <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=7%2F8 ALIGN=MIDDLE  ALT=\"7%2F8\"   DISABLED_style= \"max-width:90%; height:auto;\" > of a pound to ensure at least 65% Kona.  Say <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28p=26%267%2F8%29 ALIGN=MIDDLE  ALT=\"highlight%28p=26%267%2F8%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > pounds. </SPAN>\n" );
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