document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=809938>Question 809938</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A woman on foot leaves a restaurant heading west at noon. Three hours later, a man on a bicycle leaves the resturant pedaling west at a rate of 12 miles per hour. at six o clock that evening, a woman on a scooter leaves the resturant driving west at a speed of 39 miles per hour. if the first woman was walking at a rate of 3 miles per hour, calculate the distances below.\r<BR>\n" );
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document.write( "A. how far from the resturant will the woman on foot be when the man on the bicycle reaches her?<BR>\n" );
document.write( "B. How long will it take for the woman on the scooter to reach the woman on foot?<BR>\n" );
document.write( "C. How far from the resturant will the man on the bicycle be when the woman on the scooter reaches him </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #487862 by </B><A HREF=/tutors/aboutme.mpl?userid=devipriya><B>devipriya(2)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=devipriya><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=487862\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Part 1: <BR>\n" );
document.write( "Let the time be measured as 't'. <BR>\n" );
document.write( "The distance traveled when the woman meets the cyclist will be the same (d=d). The time for the cyclist, however, will be three hours lass than the time for the woman. <BR>\n" );
document.write( "Distance = Rate * Time <BR>\n" );
document.write( "For the woman, d = 3 * t. <BR>\n" );
document.write( "For the cyclist, d = 12 * (t - 3). <BR>\n" );
document.write( "The distance is equal, so: <BR>\n" );
document.write( "3t = 12(t-3), <BR>\n" );
document.write( "3t = 12t - 36, <BR>\n" );
document.write( "36 = 9t, <BR>\n" );
document.write( "t = 4. <BR>\n" );
document.write( "That's twelve miles. \r<BR>\n" );
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document.write( "Part 2: <BR>\n" );
document.write( "Similar to part 1: <BR>\n" );
document.write( "d = 3t, <BR>\n" );
document.write( "d = 39(t - 6); <BR>\n" );
document.write( "so <BR>\n" );
document.write( "3t = 39(t - 6), <BR>\n" );
document.write( "3t = 39t - 234, <BR>\n" );
document.write( "234 = 36t, <BR>\n" );
document.write( "t = 6.5 (hours), that's 19.5 miles away when they meet! \r<BR>\n" );
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document.write( "Part 3: <BR>\n" );
document.write( "The woman on foot no longer needs considered- the offset of the man-on-bicycle and woman-on-scooter is three hours: <BR>\n" );
document.write( "d = 12t, <BR>\n" );
document.write( "d = 39(t - 3); so, <BR>\n" );
document.write( "12t = 39(t - 3), <BR>\n" );
document.write( "12t = 39t - 117, <BR>\n" );
document.write( "117 = 27t, <BR>\n" );
document.write( "so the time for the man-on-bicycle is 4.333 hours (the woman-on-scooter takes 1.333 hours). \r<BR>\n" );
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document.write( "Either way, the distance is 52 miles.  </SPAN>\n" );
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