document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1167838>Question 1167838</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Peter can row 11 miles downstream in the same time it takes him to row 7 miles upstream. He rows downstream for 3 hours, then turns and rows back for 4 hours, but finds he is still 5 miles from where he started his trip. What is the speed of the current?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #792444 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=792444\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> speed downstream is s+c and (s+c)*t=11<BR>\n" );
document.write( "speed upstream is s-c and (s-c)*t=7; so in 4 hours (s-c)*4+5=(s+c)*3<BR>\n" );
document.write( "we know that t=11/(s+c) and t=7(s-c)<BR>\n" );
document.write( "so that 11s-11c=7s+7c<BR>\n" );
document.write( "so that 4s=18c<BR>\n" );
document.write( "or 2s=9c<BR>\n" );
document.write( "or s=4.5c\r<BR>\n" );
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document.write( "Therefore, s-c=3.5 c<BR>\n" );
document.write( "and 14c+5=16.5c<BR>\n" );
document.write( "and 2.5c=5 and c=2 mph\r<BR>\n" );
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document.write( "s then is 9 mph, so that 3 hours downstream is 33 miles and 4 miles upstream is 28 miles. That checks.\r<BR>\n" );
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document.write( "11 miles downstream takes 1 hour, and that is the same time it takes him to row 7 miles upstream\r<BR>\n" );
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document.write( "The answer is 2 mph.<BR>\n" );
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