Every day a cyclist meets a train at a particular crossing. The road is straight before the crossing and both are traveling in the same direction. Cyclist travels with a speed of 10 Kmph. One day the cyclist comes late by 25 min. and meets the train 5km before the crossing. What is the speed of the train.

• Let the initial gap be x and speed of train be v.
  
  then
  
  meeting point from cyclist = 10x/(v - 10)
  
  On that particulate day.. initial gap = (x - 5v/12)
  
  so
  
  (10x/(v - 10)) - 5 = 10(x - 5v/12)/(v - 10)
  
  10x - 5v + 50 = 10x - 50v/12
  
  5v - 50v/12 = 50
  
  60v - 50v = 12*50
  
  10v = 600
  
  v = 60 km/hr
• 14 years agoHelpfull: Yes(21) No(11)
• hi dipin..i exactly didnt get the solution of this puzzle. will u xplain it once more?
• 12 years agoHelpfull: Yes(13) No(2)
• Cyclist and Train meet at 5 km point.
  
  If cyclist had left on time, they would be 25 min at 10km/h further
  10km/h * 25/60 h = 25/6 km
  
  This leaves the cyclist 5/6 km from the cross road, when the train is at 5 km
  
  Time cyclist takes to travel this 5/6 km = distance / speed
  
  (5/6) / 10 = 1 / 12 hours
  
  The train travels 5 km in 1/12 h
  
  speed = distance / time
  
  5 / (1/12) = 5 * 12 = 60 km / h
• 10 years agoHelpfull: Yes(11) No(4)
• 99090
• 14 years agoHelpfull: Yes(6) No(21)
• 60kmph
• 14 years agoHelpfull: Yes(4) No(4)
• yes. It is 60 kmph
• 14 years agoHelpfull: Yes(2) No(3)
• 12kmph
• 14 years agoHelpfull: Yes(2) No(12)
• 22 km/hr
• 14 years agoHelpfull: Yes(1) No(6)
• let x =speed of train. and it is less than speed of cycle
  5/10-x=25/60
• 14 years agoHelpfull: Yes(1) No(9)
• let us assume x=speed of train
  then by question:-
  (5/(10-x))=25/60
  x=2kmph
• 14 years agoHelpfull: Yes(1) No(21)
• time=distance/speed time=25min=25*60 hours, distance 5 km, speed of cycle is 10kmph and let trains is xkmpr.
  
  apply formula
  5/(X-10)=25*60
  x=60kmph
  
• 9 years agoHelpfull: Yes(1) No(5)