document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1195987>Question 1195987</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Kimani and mwaura live 40km apart.one day kimani left his house at 10:30am and cycled towards mwaura's house at an average speed of 15km/hr.Mwaura left his house at 12:00noon on the same day and cycled towards kimani's houseat an average speed of 25km/hr.determine the distance from kimani's house when the two met<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #828639 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=828639\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The problem asks for the distance from kimani's house when the two meet, so what we need to determine is the distance kimani travels before they meet.<br><BR>\n" );
document.write( "He leaves 1.5 hours before mwaura; traveling 15km/h, he travels 1.5(15) = 22.5km before mwaura starts; that leaves 40-22.5 = 17.5km.<br><BR>\n" );
document.write( "When they are both riding, the ratio of their speeds is 15:25 = 3:5.  That means kimani rides 3/8 of the remaining 17.5km before they meet.<br><BR>\n" );
document.write( "(3/8)(17.5) = (3/8)(35/2) = 105/16 = 6.5625 km<br><BR>\n" );
document.write( "So the total distance kimani rides before they meet is<br><BR>\n" );
document.write( "22.5 + 6.5625 = 29.0625km<br><BR>\n" );
document.write( "ANSWER: 29.0625km<br><BR>\n" );
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