If a man walks at the rate of 4 kmph, he misses a train by only 6 min. However, if he walks at the rate of 5 kmph he reaches the station 6 minutes before the arrival of the train. The distance covered by him to reach the station is

• 4 km
  
  Let the distance covered be d and time taken to reach the station in time t
  
  d/4 = t+6/60---(1)
  d/5 = t-6/60---(2)
  
  (1)-(2) gives
  
  d/4 - d/5 = (6+6)/60
  d/20 = 12/60
  d = 12*20/60
  =4 km
  
• 10 years agoHelpfull: Yes(3) No(0)
• Let d be the distance ,
  s be the speed ,
  t be the time
  
  6 mins => 6/60 = 1/10hrs
  
  speed = 4, late by 1/10hrs
  d/4 = t + 1/10
  => d = 4t + 2/5 --> (1)
  speed = 5, early by 1/10hrs
  d/5 = t - 1/10
  => d = 5t - 1/2 --> (2)
  
  Equate "d"
  
  4t + 2/5 = 5t - 1/2
  t = 2/5 + 1/2 => t = 9/10
  t = 9/10 hrs
  
  d = 5 * (9/10) - 1/2
  => 18 - 1/2
  => 35/2
  d= 17.5 km
  
  Ans : 17.5 km
• 10 years agoHelpfull: Yes(0) No(1)