document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=793164>Question 793164</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Having travelled halfway to the end of a journey at an average speed of 15km/hr, how fast must you traverse the rest of the trip on order to average 20 km/hr?\r<BR>\n" );
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document.write( "How do I do this? I know that simply averaging the two speeds would be incorrect. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #480255 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=480255\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let y be the rate in second half and x the total distance\r<BR>\n" );
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document.write( "t=d/r\r<BR>\n" );
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document.write( "time first half + time second half = time for total journey\r<BR>\n" );
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document.write( "(x/2)/(15)+ (x/2) /y = x/20\r<BR>\n" );
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document.write( "x/30 -x/20 = - x/2y\r<BR>\n" );
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document.write( "(2x-3x)/60 = -x/2y\r<BR>\n" );
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document.write( "-x/60 = -x/2y\r<BR>\n" );
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document.write( "2y=60\r<BR>\n" );
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document.write( "y=30 km/h\r<BR>\n" );
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document.write( " </SPAN>\n" );
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