The distance between A and B is 752 km. From A, Vishal goes to B with a speed…

Concept: When the same distance is covered at two different speeds, the average speed is total distance divided by total time, which reduces to the harmonic mean of the two speeds: 2v1v2/(v1 + v2). It is never the arithmetic mean of the two speeds, because more time is spent on the slower leg. The one-way distance cancels out of that expression, so its size never changes the answer.

Application: Here the two speeds are 96 km/hr from A to B and 94 km/hr from B to A, and each leg of the journey is 752 km.

1. Total distance for the round trip = 2 × 752 = 1504 km.

2. Time from A to B = 752 ÷ 96 = 47/6 hours ≈ 7.8333 hours.

3. Time from B to A = 752 ÷ 94 = 8 hours exactly.

4. Total time = 47/6 + 8 = 95/6 hours ≈ 15.8333 hours.

5. Average speed = total distance ÷ total time = 1504 ÷ (95/6) = (1504 × 6)/95 = 9024/95 = 94.98947… km/hr.

6. Rounded off to two decimal places, the average speed is 94.99 km/hr.

Cross-check: Substituting straight into the harmonic-mean formula gives 2 × 96 × 94/(96 + 94) = 18048/190 = 9024/95 = 94.99 km/hr — the same value, which confirms that the 752 km leg length never enters the result. The figure is also consistent with the concept: since the slower 94 km/hr leg takes longer, the average must sit between 94 and 96 km/hr and just below their arithmetic mean of 95 km/hr.