document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=461149>Question 461149</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me!!<BR>\n" );
document.write( "Katie kayaks 20 km upstream in 3 hours and returns in 1.5 hours. How fast is the stream flowing and how fast does Katie paddle in still water?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #316304 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=316304\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let b = speed of the boat in still water, and s = speed of the stream.\r<BR>\n" );
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document.write( "Then 3(b-s) = 20 for upstream travel, and<BR>\n" );
document.write( "and 1.5(b+s) = 20 for downstream travel.\r<BR>\n" );
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document.write( "==> b - s = 20/3 and b+s = 40/3, or\r<BR>\n" );
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document.write( "b = 10 kph.\r<BR>\n" );
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document.write( "b + s = 40/3 ==> s = 10/3 kph. </SPAN>\n" );
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