If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.

• Let the required distane be x km.
  Differene in the times taken at two speeds = 12 min = 1/5 hr.
  => x/5 - x/6 = 1/5
  => 6x - 5x = 6
  => x = 6
  Hence, the required distane is 6 km.
• 12 years agoHelpfull: Yes(10) No(2)
• Answer is 6km.
  
  Explanation:
  distance = speed * time, we have:
  (1) s = 5 * (t + 7/60)
  (2) s = 6 * (t - 5/60)
  solving these equations gives:
  (1) s = 5t + 7/12
  (2) s = 6t - 1/2
  Rearranging gives the following simultaneous equations:
  (1) s - 5t = 7/12
  (2) s - 6t = -1/2
  We need to find s so multiply (1) by 6 and (2) by 5 to get:
  (1) 6s - 30t = 42/12 = 7/2
  (2) 5s - 30t = -5/2
  Subtract (1) from (2) to get
  s = 12/2 = 6
  Hence the distance to the station is 6 km
  
• 12 years agoHelpfull: Yes(9) No(1)
• Let d be the distance travelled to the train in kilometres.
  
  Let s be the speed travelled to reach the train on time in km/h.
  
  Let t be the time taken to reach the train on time, in hours.
  
  d = st
  
  d = 5(t+7/60) ;- converting 7min to hours = 7/60
  
  d = 6(t-5/60)
  
  5(t+7/60) = 6(t-5/60)
  
  5t + 35/60 = 6t - 30/60
  
  t = 65/60
  
  d = 5(72/60) = 360/60 = 6 kilometres.
• 8 years agoHelpfull: Yes(2) No(0)
• walking 6 kmph
  (...distance(D)........) (....T=5 min before....) . train starts at this point
  
  (......................x..........................)
  
  walking 5 kmph (....t=7 min after...)
  (..............distance (d)............................................)
  (...........time between the two length............)
  T + t =12 min
  i.e .,difference in Time of two distance=12 min= 0.2 hr
  
  time = distance / speed ,so x/5 - x/6 = 0.2 => x=6 ,
  Henc , he want to cover 6 km to reach the train.
  
• 10 years agoHelpfull: Yes(0) No(0)