document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=195507>Question 195507</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Respected sir,\r<BR>\n" );
document.write( "\n" );
document.write( "1] A train travelling 25km an hour leaves Delhi at 9 a.m. & another train travelling 35 km an hour starts at 2 p.m. in the same direction . how many km from Delhi are they together ?\r<BR>\n" );
document.write( "\n" );
document.write( "2] Walking 3/4 of his usual rate,a man is 1 1/2 (one an half hour) hours too late.find his usual time.\r<BR>\n" );
document.write( "\n" );
document.write( "plz sir solve my difficulty as soon as possible.\r<BR>\n" );
document.write( "\n" );
document.write( "thank u. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #146638 by </B><A HREF=/tutors/aboutme.mpl?userid=ejazali_syed><B>ejazali_syed(7)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ejazali_syed><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=146638\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1)A train travelling 25km an hour leaves Delhi at 9 a.m. & another train travelling 35 km an hour starts at 2 p.m. in the same direction . how many km from Delhi are they together ? \r<BR>\n" );
document.write( "\n" );
document.write( "speed  = distance/time  => distance = speed *time\r<BR>\n" );
document.write( "\n" );
document.write( "d = 25*5 = 125 kms.     i.e. by the time the second train starts travelling which is 5 hrs after the first one the first train has covered a distance of 125kms.\r<BR>\n" );
document.write( "\n" );
document.write( "for the first train => d1 = 25*t1;  <BR>\n" );
document.write( "for the second train => d2 = 35*t2\r<BR>\n" );
document.write( "\n" );
document.write( "so the second train will catch the first train in t2 hrs.\r<BR>\n" );
document.write( "\n" );
document.write( " relation of t1 in terms of t2 => t1 = t2+5 (since train 1 has started moving 5 hrs before t2)\r<BR>\n" );
document.write( "\n" );
document.write( "we need the condition d1 = d2   =>   25(t2+5) = 35(t2)  <BR>\n" );
document.write( "                                                                    35t2-25t2 = 125 => t2 = 125/10 \r<BR>\n" );
document.write( "\n" );
document.write( "   hence they will meet eachother after 12.5 hrs of train 2 leaving the station.\r<BR>\n" );
document.write( "\n" );
document.write( "i.e. 2.30 am<BR>\n" );
document.write( "------------------------------------------------<BR>\n" );
document.write( "2)Walking 3/4 of his usual rate,a man is 1 1/2 (one an half hour) hours too late.find his usual time. <BR>\n" );
document.write( "plz sir solve my difficulty as soon as possible. \r<BR>\n" );
document.write( "\n" );
document.write( "same formula  s = d/t   --------- 1\r<BR>\n" );
document.write( "\n" );
document.write( "let s be the usual rate.   d be the distance travelled in t hrs\r<BR>\n" );
document.write( "\n" );
document.write( "today he waliking at 3/4 of usual rate = 3s/4. and to cover the distance d it took 1.5 hrs more i.e. t+1.5\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "so    3s/4 = d/(t+1.5)  ---------------2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "usual time t = ?\r<BR>\n" );
document.write( "\n" );
document.write( "from 1    d = st<BR>\n" );
document.write( "from 2   d = 3s(t+1.5)/4\r<BR>\n" );
document.write( "\n" );
document.write( "therefore     st = s(3t+4.5)/4   ==>   4t = 3t +4.5   => t = 4.5 hrs is the usual time. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );