A train met with an accident 3 hours after starting,which detains it for one hour,after which it proceeds at 75% of its original speed.It arrives at the destination 4 hours late.Had the accident taken plae 150km further along the railway line,the train would have arrived only 7/2 hours late.Find the length of the trip 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
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