Question

A train goes 4692 km from Station A to station B and completes the journey in time. In return journey, due to some engine problem train starts 10 hrs late and to reach on time trains average speed is increased by 5 km/h. What was the original speed of train? 

1. 46 km/h
2. 44 km/h
3. 49 km/h
4. 51 km/h

Answer 1

Correct Answer - Option 1 : 46 km/h

Let V be the average speed of the train and T the time required for the train to complete the one-way journey.

As we know, the average speed is inversely proportional to the time taken if the distance covered is constant.

According to the question, 4692 = VT = (V + 5) × (T - 10)

⇒ VT = (V + 5) × (T - 10) = VT + 5T – 10V – 50 ⇒ T = 2V + 10

⇒ VT = 2V² + 10V ⇒ 4692 = 2V² + 10V ⇒ V² + 5V – 2346 = 0

⇒ V² + 51V – 46V – 2346 = 0 ⇒ (V + 51) × (V - 46) = 0

⇒ V = -51 or 46 ⇒ V = 46 km/h