document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=383994>Question 383994</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Going from point A to point B a driver averages 60mph. On the return trip along the same route he averages 40mph. What is the average rate for the entire trip? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #271814 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=271814\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It's not 50 mph.  More time is spent moving at 40 mph than moving at 60 mph.  It's always less than the average of the 2 speeds, unless they're equal.<BR>\n" );
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document.write( "d = rt<BR>\n" );
document.write( "t = d/r<BR>\n" );
document.write( "d = the distance, so t = 2d/r for the round trip<BR>\n" );
document.write( "2t = d/60 + d/40 = 100d/2400 = d/24 (total time)<BR>\n" );
document.write( "t = d/48<BR>\n" );
document.write( "r = 48 mph<BR>\n" );
document.write( " </SPAN>\n" );
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