document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1173675>Question 1173675</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A cabin cruiser leaves a harbour and travels south at 15km/h. One hour later a speed boat leaves the same harbour and travels on the same course at 25 km/h. How far from the harbour will the speed boat pass the cruiser? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #798947 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=798947\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> One way is the cabin cruiser has a 15 km head start and is being caught at 10 km/h, so it is 1.5 hours chase and 2.5 hours after leaving the harbor, or 37.5 km.<BR>\n" );
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document.write( "another way is cruiser has gone 15km/h*x h km.<BR>\n" );
document.write( "the speed boat has gone 25 km/h (x-1) h km.<BR>\n" );
document.write( "Set those equal to each other'<BR>\n" );
document.write( "15x=25x-25<BR>\n" );
document.write( "10x=25<BR>\n" );
document.write( "x=2.5 hours, time the cabin cruiser has gone 15*2.5 or 37.5 km and the speed boat 25*1.5 or 37.5 km. </SPAN>\n" );
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