a train m eaves meerut at 5 a.m and reaches delhi at 9 am. another train delhi at 7 am and reaches meerut at 10.30 am at what time do the two trains cross each other

• suppose trains met t hrs after 7 am. and x is distance between Delhi and meerut.
  
  Then
  (x/3.5)*t + (x/4)*(t+2)=x
  or
  t/3.5 + (t+2)/4 =1
  
  Then
  t= 7/7.5 hrs or 56 mins.
  
  so trains will meet at 7.56 AM.
  
• 14 years agoHelpfull: Yes(35) No(6)
• crossing time ...7.56 AM.
  
  Ratio of speeds = 4/3.5 =8/7
• 14 years agoHelpfull: Yes(7) No(8)
• let t be the time, D be the distance.
  train m leaves meerut at 5.a.m.
  train n leaves delhi at 7.a.m.
  so,the two trains are differed by 2 hours.
  and let the first train travell for 4hours(from 5.a.m to 9.a.m),so the speed of the first train is:D/4kmph.
  and second train travell for 2.30hrs(from 7-10.30a.m)hence speed of second train is:D/2.30kmph=2D/5kmph.
  now the two trains meet after 7 am by
  t*(D/4)+(t-2)*(2D/5)=D
  by solving we get t=12hours.
  thus two trains are meet exactly 12 hours after 7.a.m
  i.e.at 7p.m
• 14 years agoHelpfull: Yes(2) No(24)
• let distance between station =x
  so speed of first train v1=x/4
  and speed of second train v2=x/3.5
  now let both train meet at time T then
  distance travelled by first equal to distance leftto travel for second train
  5:00 _____________________________(T)fisrt->_____________9:00
  10:30______________________________
• 14 years agoHelpfull: Yes(2) No(12)
• let distance between station =x
  so speed of first train v1=x/4
  and speed of second train v2=x/3.5
  now let both train meet at time T then
  distance travelled by first equal to distance leftto travel for second train
  5:00am _____________________________(T)fisrt->_____________9:00am
  10:30am_____________________________
• 14 years agoHelpfull: Yes(1) No(8)
• let distance between station =x
  so speed of first train v1=x/4
  and speed of second train v2=x/3.5
  now let both train meet at time T then
  distance travelled by first equal to distance leftto travel for second train
  5:00 ____T)fisrt->_____9:00
  10:30____
• 14 years agoHelpfull: Yes(1) No(6)
• First train takes 4 hours and the second train takes 3.5 hours.Time ratio is 8:7. Therefore, the speed ratio will be 7:8.Let the speeds be 7x and 8x, and distance be 28x ( 4×7 or 3.5×8).At 7 AM, the first train must have covered a distance of 14x.
  Therefore, at 7 A.M. the distance between the two trains is 28x-14x=14x.Time taken to meet = 14x/(7x+8x)=14/15 hour or 56 minutes.Hence, the two trains meet at 7.56 AM.
• 6 years agoHelpfull: Yes(1) No(0)
• if the distance between two stations is=x
  speed of fist tain= x/4
  and of second train=x/3.5
  given,n leaves 2 hours after m (7 a.m-5 a.m)
  let,in that 2 hours m travers a distance of y
  and remaining distance be x
  !.........!.........!
  < x km >< y km >
  when m traversed x km, 2 hours is over
  now for y distance in 2 hours the distance between n & m b [{(x/4)+{x/3.5)]
  or 7.5x/14 distance reduces in 2 hour between y km
  so,x distance traversing needs 56 minutes & including previous time for train m (7am-5am)=2 hours....so total time needed is 2 hours 56 minutes.
  so,they will meet on 5a.m+2 hour
• 14 years agoHelpfull: Yes(0) No(3)