document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=522519>Question 522519</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A local hospital is holding a two day marathon walk to raise funds for a new research facility. The total distance of the marathon is 26.2 miles. On the first day, Martha starts walking at 10:00 am. She walks 4 miles per hour. Carol starts two hours later than Martha but decides to run to catch up to Martha. Carol runs at a speed of 6 miles per hour. Carol wants to reduce the time she takes to catch up to Martha by 1 hour. How can she do this by changing her starting time? How can she do this by changing her speed? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #346898 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=346898\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Martha --speed = 4mph---10.00 am\r<BR>\n" );
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document.write( "Carol----speed ---6mph---12.00 noon\r<BR>\n" );
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document.write( "x/4-2=x/6\r<BR>\n" );
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document.write( "x/4-x/6=2<BR>\n" );
document.write( "(6x-4x)/24=2<BR>\n" );
document.write( "2x=48<BR>\n" );
document.write( "x=24 distance where she catches up\r<BR>\n" );
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document.write( "Carol distance =24<BR>\n" );
document.write( "carol speed = 6\r<BR>\n" );
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document.write( "carol time = 4 hours\r<BR>\n" );
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document.write( "reduce by 1 hour\r<BR>\n" );
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document.write( "carol time will be 3 hours<BR>\n" );
document.write( "distance = 24miles<BR>\n" );
document.write( "speed = 24/3 = 8 mph </SPAN>\n" );
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