a train met with an accident 60 km away from anantpur station.It completed the remaining journey at 5/6 th of the previous speed and reached barmula station 1 hour 12 in late.had acciden takn place 60 km furthr ,it woud have been only 1 hour late a What is th normalspeed of trainb wha isthe distance between anantpur and barmula...pls explain ur answers.its getting difficult fr me to understand it.

1) 60 km/hr, 420km
2) 80km/hr.800km
3) 8km/hr.80km
4) none of these

• LET THE NORMAL SPEED OF TRAIN BE "x km/hr"
  LET THE TOTAL DISTANCE BE "y km"
  
  -Accident occur after 60km from anantpur i.e; train has travelled 60km with normal speed. hence time taken at normal speed till accident= (60/x) hr.
  -Rest distance from accident to barmula= (y-60) km.
  -Speed for rest of distance= (5/6)x = (5x/6).
  -time taken to travel rest distance= (y-60)/(5x/6) = ((6(y-60))/5x)
  -if accident would not have occurred den time taken = y/x
  -train late by 1hr 12min= 1+(12/60)= 6/5 hr
  - Hence 1st equation is =
  (60/x)+ ((6(y-60))/5x) = y/x + 6/5 ------eqn 1
  - similarly eqn 2 can be written as
  (120/x) + ((6(y-120))/5x) = y/x + 1 ------eqn 2
  
  simplify and solve both the equations...
  ans will be x = 60 km/hr and y = 420 km
  
  
• 9 years agoHelpfull: Yes(0) No(0)