Two trains are traveling from point A to point B such that the speed of first train is 65 kmps and the speed of 2 train is 29 kmph. Where is the distance b/w A&B such that the slower train reached 5 hrs late compared to the faster?

• if x is the distance, then
  x/29 - x/65 = 5
  
  Then x = 5*65*29/(65-29) = 5*65*29/36 = 261.8055 kms
• 13 years agoHelpfull: Yes(34) No(2)
• please see 65 kmps
  and 29 kmph
• 13 years agoHelpfull: Yes(7) No(3)
• DIPIN and his method, both are right. Its 261.8055 kms from A
• 13 years agoHelpfull: Yes(5) No(3)
• suppose distance is x km.and time taken by faster abd slower trains are t1 and t2 respectivly.
  t2-t1=5;
  (x/29)-(x/65)=36x/29*65
  5=36x/29*65
  x=261.805.
• 13 years agoHelpfull: Yes(5) No(0)
• 65-29=36*5=180
• 13 years agoHelpfull: Yes(0) No(18)
• for every 1 hr the second train is late by 36kms from the first train.so why can't the answer be 36*5=180kms
  
• 11 years agoHelpfull: Yes(0) No(4)
• Distance is same for both trains so
  let distance = D
  speed of first train s1 = 65kmph
  speed of second train s1 = 29kmph
  (D/s1)-(D/s2) = t1 - t2
  (D/65)-(D/29) = 5
  just solve the equation
• 7 years agoHelpfull: Yes(0) No(0)
• use this formula
  time=distance/velocity
  given, v1-v2=5, d1=d2
  put the data and get the result
  Ans is(261.805)
• 7 years agoHelpfull: Yes(0) No(0)