A train start running from station A,if made an accident after travelling 150 km.After accident the speed train reduces by 5/6 of the previous speed,so it reach 15 minute late at station B.Had the accident taken place 30 km further it have been 7 minutes late .find the speed of the train and distance between A & B station.

• 1125km/hr and 206.25km
  
  Let the distance be x and speed be s
  
  time t = x/s hours
  
  Since the total distance is x, distance traveled after accident = (x-150)
  and Speed at that time is reduced by 5/6, speed = s - 5/6*s = s/6
  time taken is increased by 15 min ,
  time = t + 15/60 hours
  = x/s + 15/60
  
  150/s + (x - 150)/s/6 = x/s + 15/60 ---(1)
  
  similarly the 2nd equation is
  
  180/s + (x-180)/s/6 = x/s + 7/60 ---(2)
  
  Solving (1) and (2)
  
  distance, x = 206.25km
  speed, s = 1125km/hr
• 10 years agoHelpfull: Yes(8) No(1)
• distance=1605km,speed=45km/hr
• 10 years agoHelpfull: Yes(0) No(0)
• Let the initial speed= x km/minute & distance= y km.
  original time taken= y/x min.
  After the accident speed reduces by 5/6 of previous speed,so speed after accident= x - (x(5/6)) = x/6.
  1st condition:
  (150/x) + ((6(y-150))/x) =(y/x) +15,
  or, 150 + 6y -900 = y + 15x,
  15x - 5y = -750 ----- (1)
  2nd condition,
  (180/x) + ((6(y-180))/x) = (y/x) + 7,
  or, 180 + 6y - 1080 = y + 7x,
  or, 7x - 5y = -900 ----(2)
  Solving (1) & (2), we get x= 75/4 & y= 206.25
  so, x = (75/4)km/minute = 1125 km/hr
  &, y = 206.25km
• 10 years agoHelpfull: Yes(0) No(1)