After travelling 3 hours a train meets with an accident due to this it stops for an hour. After this the train moves at 75% speed of its original speed and reaches to destination 4 hours late. If the accident would occur at 150 km ahead in the same line then the train reaches only 3.5 hours late. Then find the distance ofjourney and the original speed of the train?
Answers
Answer:
Original Speed = 100 km/Hr
Distance of Journey = 1200 km
Step-by-step explanation:
Let say original speed of train = S km/Hr
Original time to reach station = T hr
Total Distance = ST km
Distance covered in 3 Hours = 3S km
reduced Speed = (75/100)S = 3S/4 km/Hr
Time run with reduced speed = ( T + 4 -3 - 1) = T hr ( total time taken increased by 4 hour , 3 hrs already run and one hour stop time)
Distance covered with reduced speed = 3ST/4
ST = 3S + 3ST/4
4ST = 12S + 3ST
ST = 12S
T = 12
Original time to journey = 12 hr
if accident was 150 km ahead of the point
then time taken with Original Speed = 150/S
Time taken with Reduced Speed = 150/(3S/4) = 200/S
200/S - 150/S = 1/2 (4 - 3.5 = 0.5 = 1/2)
=> 200 - 150 = S/2
=> S = 100
Original Speed = 100 km/Hr
Distance of Journey = ST = 100 * 12 = 1200 km
Answer:
Step-by-step explanation:
150(1/5) /(3/4) (1/2) =100kmph
100*75*4/25= 1200