document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1007208>Question 1007208</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A man rows a boat 910 feet upstream against a constant current in 14 minutes. He then rows 525 feet downstream (with the same current) in 7 minutes. Find the speed of the current and the equivalent rate at which he can row 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 #623186 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=623186\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the still water's rate, r.<BR>\n" );
document.write( "Call the current's rate, c.<BR>\n" );
document.write( "In general, speed is found by dividing distance by time...<BR>\n" );
document.write( "So we have two cases...<BR>\n" );
document.write( "r - c = 910/14 = 65<BR>\n" );
document.write( "r + c = 525/7 = 75<BR>\n" );
document.write( "Now add the equations and get<BR>\n" );
document.write( "2r = 140<BR>\n" );
document.write( "r = 70 feet per minute in still water<BR>\n" );
document.write( "which means the current must be <BR>\n" );
document.write( "c = 5 feet per minute </SPAN>\n" );
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