Question

A train met with an accident 60km away from station a. It completed the remaining journey at 5/6th of the original speed and reached station b 1hr 12mins late. Had the accident taken place 60km further, it would have been only 1hr late. What was the original speed of the train?

Answer 1

Let say original speed = S km/h

Let say distance between a & B = D km

ORIGINAL TIME = D/S

Time taken for 60 km = 60/S

Remaining Distance = D - 60

Speed after accident = 5S/6

Time taken after accident = (D-60)/(5S/6) = (6*(D-60)/5S

60/S + (6*(D-60)/5S = D/S + 1.2    (as train reached  1hr 12 minute Late = 1.2 Hr)

300 + 6D-360 = 5D +6S

D - 6S = 60  - Eq A

Time taken for 120 km = 120/S

Remaining Distance = D - 120

Time taken after accident = (D-120)/(5S/6) = (6*(D-120)/5S

120/S + (6*(D-120)/5S = D/S + 1    (as train would reached  1hr Late )

600 + 6D - 720 = 5D + 5S

D - 5S = 120  Eq B

Eq B - Eq A

S = 60 km/h

Original Speed of train = 60 km/Hr

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