A train met with an accident 50 km away from station A. It completed the remaining journey at three-fourth of its original speed and reached station B 35 minutes late. Had the accident occurred 24 km further down the line, it would have been only 25 minutes late. What is the original speed of the train?
48 km/hr
Let the delay formula be \(\dfrac{D-d}{3s}\) where d is the accident distance, D is total distance, s is original speed.
Case 1 (d=50, delay 35/60 hr): \(D-50 = 3s\times\tfrac{35}{60} = 1.75s\).
Case 2 (d=74, delay 25/60 hr): \(D-74 = 3s\times\tfrac{25}{60} = 1.25s\).
Subtracting: \((D-50)-(D-74) = 1.75s-1.25s \Rightarrow 24 = 0.5s \Rightarrow s = 48\).
Hence, the original speed of the train is 48 km/hr.
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