Question

The distance between Alkola and Bhusawal is 168 km. As express train takes 1 hour less than a passenger train to cover the distance. Find the average speed of each train if the average speed of the express train is more by 14 km/hr than the speed of the passenger train.

Answer 1

speed of passenger train = 42 km /hr

speed of express train = 56km / hr

Answer 2

Answer:

42 km/hr, 56 km/hr

Step-by-step explanation:

Let the speed of the passenger train be x km/hr.

Given that speed of the express train is more by 14 km/hr.

So, speed of the express train = (x + 14) km/hr.

Given, Distance = 168 km.

∴ Time = Distance/Speed.

(i) Time taken by the passenger train = 168/x

(ii) Time taken by the express train = 168/(x + 14)

Given that express train takes 1 hour less than a passenger train.

⇒ (168/x) - (168/x + 14) = 1

⇒ 168(x + 14) - 168x = x(x + 14)

⇒ 168x + 2352 - 168x = x² + 14x

⇒ 2352 = x² + 14x

⇒ x² + 14x - 2352 = 0

⇒ x² + 56x - 42x - 2352 = 0

⇒ x(x + 56) - 42(x + 56) = 0

⇒ (x - 42)(x + 56) = 0

⇒ x = 42, -56 {∵ Speed cannot be negative}

⇒ x = 42

When x = 42:

⇒ x + 14 = 56

∴ Speed of express train is 56 km/hr

∴ Speed of Passenger train is 42 km/hr.

Hope it helps!