A train has been stopped for 6 minutes. Next station is 36 km away and to reach on scheduled time its speed is increased by 4 km/hr. Find its initial speed (in km/hr).

36

Let the initial speed of train = S km/h

ATQ,

⇒  \(\frac{36}{S}\) - \(\frac{36}{S\;+\;4}\) = \(\frac{6}{60}\)

⇒  36 [ \(\frac{S\;+\;4\;-\;S}{S\;(S\;+\;4)}\)] = \(\frac{1}{10}\)

⇒  \(\frac{36\;×\;4}{S\;(S\;+\;4)}\) = \(\frac{1}{10}\)

⇒  S (S + 4) = 1440

⇒  S (S+4) = 36 × (36 + 4)

⇒ Initial speed = 36 km/h