document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=304959>Question 304959</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?\r<BR>\n" );
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document.write( "please give a complete step by step answer to this question - Thank You\r<BR>\n" );
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document.write( "I last took algebra in college back in the early 90s, so I know how to tackle the questions but I don't remember the correct way to solve many of the algebra questions I am facing.  I am reviewing for the TEAS nursing aptitude entrance exam so, it very important that I review and re-master my weak subject areas  like higher-math and Algebra specifically.\r<BR>\n" );
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document.write( "Thanks for you help in advance and I look forward to hearing back from you soon.\r<BR>\n" );
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document.write( "Matt<BR>\n" );
document.write( "mcpcot@hotmail.com </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #218396 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=218396\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?\r<BR>\n" );
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document.write( "The first cyclist rides at 6 mph.<BR>\n" );
document.write( "So in 3 hours he rides 18 miles\r<BR>\n" );
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document.write( "Let the distance where they meet be x+18\r<BR>\n" );
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document.write( "By the time second cyclist travels x+18 miles, first cyclist rides x miles\r<BR>\n" );
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document.write( "Time taken by fist cyclist = distance / rate = x/6 hours\r<BR>\n" );
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document.write( "time taken by second cyclist = x+18 / 10\r<BR>\n" );
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document.write( "The times are same in both the case\r<BR>\n" );
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document.write( "x/6 = x+18/10\r<BR>\n" );
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document.write( "6(x+18)=10x\r<BR>\n" );
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document.write( "6x+108= 10x\r<BR>\n" );
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document.write( "4x=108\r<BR>\n" );
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document.write( "x= 27 miles\r<BR>\n" );
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document.write( "Distance / rate = time\r<BR>\n" );
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document.write( "x+18 /10 is the time taken by second cyclist to cross the first cyclist\r<BR>\n" );
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document.write( "27+18 /10 = 45/10 = 4.5 hours \r<BR>\n" );
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