Question

3. The distance between Akola and Bhusawal is
168 km. An express train takes 1 hour less
than a passenger train to cover that distance.
Find the average speed of each train, if the
average speed of the express train is more by
14 km/h than that of the passenger train.
2)​

Answer 1

Answer:

The distance between Akola and Bhusawal is 168 km.

Suppose, average speed of passenger train is x km/hr.

∴ the average speed of express train is (x + 14) km/hr.

∴ the time required for passenger train = 168 / x hours

and the time required for express train = 168 / x +14 hours

∴ from the given condition,

168/x - 168/x+14 = 1

∴ (168x + 168×14 - 168x)/x(x +14) = 1

∴ x2 + 14x = 168 × 14

∴ x2 + 14x - 2352 = 0

∴ x2 + 56x - 42x -2352 = 0

∴ x(x + 56) - 42(x + 56) = 0

∴ x(x + 56)(x - 42) = 0

∴ x + 56 = 0 or x - 42 = 0

∴ x = - 56 or x = 42

But speed is not negative x = 42

∴ average speed of passenger train = 42 km/hr

and average speed of express train = (42 + 14) = 56 km/h

Hope! it will help you.