A train’s average speed on the onward (outbound) journey is 25% higher than on the return journey. After reaching the destination, the train halts for 2 hours. The total time for the complete to-and-fro trip is 32 hours, and the total distance covered (to and fro) is 1600 km (i.e., 800 km each way). Find the train’s average speed on the onward journey (in km/h).

Introduction / Context:This problem tests weighted time calculations with different speeds on two legs of a round trip and a fixed halt. Translating words into equations and keeping careful units is the key.

Given Data / Assumptions:

• Total distance (to and fro) = 1600 km ⇒ 800 km each way.
• Onward speed = 1.25 * return speed.
• Halt at destination = 2 h.
• Total elapsed time (travel + halt) = 32 h.

Concept / Approach:If return speed is v, onward speed is 1.25v. Travel time = 800/(1.25v) + 800/v. Add the 2 h halt and set equal to 32 h.

Step-by-Step Solution:

Verification / Alternative check:At v = 48, times are 640/48 = 13.333… h and 800/48 = 16.666… h. Sum = 30 h; plus 2 h halt = 32 h as required.

Why Other Options Are Wrong:Values like 56.25, 66.50, or 67 km/h violate the time equation after including the halt; they do not total 32 h.

Common Pitfalls:Forgetting the halt, or assuming 1600 km is one-way. Also mixing up “25% more speed” with “25% less time” can lead to algebra errors.

Final Answer:60 km/h