A train left station A for station B at a certain speed. After travelling for 100 km, the train meets with an accident and could travel at 4/5th of the original speed and reaches 45 minutes late at station B. Had the accident taken place 50 km further on , it would have reached 30 minute late at station B. what is the distance between station A and B ?

Let original speed = 5x km/h. After 100 km, speed = 4x km/h. If accident at 100 km: delay 45 min. If accident at 150 km (50 km further): delay 30 min. Extra 50 km at 4x instead of 5x saves 15 min. Time difference: 50/4x - 50/5x = 15/60 hr. 12.5x - 10x = 0.25 hr. 2.5x = 0.25, x = 0.1. Speed = 0.5 km/h. Let remaining distance after 100 km be D. D/0.4 - D/0.5 = 45/60. 2.5D - 2D = 0.75. 0.5D = 0.75, D = 150. Total = 100 + 150 = 250 km. Option E is correct.