Problems on Trains – Average speed for equal distances (find return speed): A train goes from A to B at 64 km/h and returns from B to A at a slower speed. If the average speed for the round trip is 56 km/h, what is the return speed?

Introduction / Context:For equal distances each way, the average speed is the harmonic mean: v_avg = 2uv/(u + v), where u and v are the onward and return speeds. With average and one leg’s speed known, solve for the other speed.

Given Data / Assumptions:

• Onward speed u = 64 km/h.
• Average speed v_avg = 56 km/h.
• Distances A→B and B→A are equal.

Concept / Approach:Use v_avg = 2uv/(u + v). Rearranging for v gives: 56(64 + v) = 128v ⇒ solve linearly.

Step-by-Step Solution:56(64 + v) = 128v.3584 + 56v = 128v ⇒ 3584 = 72v ⇒ v = 3584/72 = 49.777… km/h.

Verification / Alternative check:Harmonic mean with u = 64 and v ≈ 49.777 gives v_avg ≈ 56, confirming correctness.

Why Other Options Are Wrong:48, 52, 47.46, 50 km/h do not satisfy the harmonic-mean equation with 64 km/h producing 56 km/h average.

Common Pitfalls:Using arithmetic mean instead of harmonic mean; assuming average = (64 + v)/2, which is incorrect for equal distances at different speeds.

Final Answer:49.77 km/hr.