A car travels from Town A to Town B at 58 km/h, then returns from Town B to Town A at 52 km/h over the same distance. What is the approximate average speed for the entire round trip (equal distances)?

Introduction / Context:
When a vehicle goes and returns over equal distances at two different speeds, the “overall” or average speed is not the arithmetic mean. It is the harmonic-mean formula: average speed = 2ab / (a + b). This captures the fact that time spent at the lower speed weighs more in the total time.

Given Data / Assumptions:

• Outward speed a = 58 km/h.
• Return speed b = 52 km/h.
• Distances are equal in both legs.
• No halt time is mentioned; compute strictly on travel time.

Concept / Approach:
For equal one-way distance D, total time = D/58 + D/52. Total distance = 2D. Average speed = total distance / total time = 2D / (D/58 + D/52) = 2ab/(a + b).

Step-by-Step Solution:

Verification / Alternative check:
If you wrongly take the simple mean: (58 + 52)/2 = 55 km/h—here it coincidentally matches the rounded harmonic mean because the two speeds are close; but the correct derivation is harmonic mean for equal distances.

Why Other Options Are Wrong:
52 or 48 km/h underestimate; 60 km/h is impossible since both legs are at or below 58 km/h and 52 km/h.

Common Pitfalls:
Using arithmetic average for unequal times; forgetting that lower speed dominates time.

Final Answer:
55 km/h