document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=431388>Question 431388</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A boat moves 7 km upstream in the same amount of time it moves 20 km downstream.<BR>\n" );
document.write( "if the rate of the current is 6 km per hour, find the rate of the boat 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 #299389 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=299389\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let s = the rate in still water<BR>\n" );
document.write( "Let c = the rate of the current = 6 km/hr<BR>\n" );
document.write( "Let t = time<BR>\n" );
document.write( "So the upstream rate = s - c<BR>\n" );
document.write( "And the downstream rate = s + c<BR>\n" );
document.write( "Since time = distance/rate, we can write<BR>\n" );
document.write( "t = 7/(s-c)<BR>\n" );
document.write( "t = 20/(s+c)<BR>\n" );
document.write( "Equating the RHS, and subsituting the value for c gives:<BR>\n" );
document.write( "7/(s-6) = 20/(s+6)<BR>\n" );
document.write( "Cross multiply:<BR>\n" );
document.write( "7s+42 = 20s-120 <BR>\n" );
document.write( "Solving for s gives:<BR>\n" );
document.write( "13s = 162 -> s = 12.462 km/hr </SPAN>\n" );
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