A person travels from home to office at 4 km/h and returns at 6 km/h over the same route. What is the average speed for the entire trip?

Introduction / Context:Average speed over equal distances at two different speeds is not the arithmetic mean; it is the harmonic mean. This scenario appears whenever you go and return over the same path but at different speeds.

Given Data / Assumptions:

• Outbound speed = 4 km/h.
• Return speed = 6 km/h.
• Distances out and back are equal.

Concept / Approach:For equal distances with speeds u and v, average speed v_avg = 2uv / (u + v). This comes from total distance divided by total time, with time = distance/speed for each leg.

Step-by-Step Solution:Let each leg be distance d. Total distance = 2d.Total time = d/4 + d/6 = (3d + 2d)/12 = 5d/12 hours.Average speed = total distance / total time = 2d / (5d/12) = 24/5 = 4.8 km/h.

Verification / Alternative check:Harmonic mean formula gives v_avg = 2*4*6/(4+6) = 48/10 = 4.8 km/h, confirming the calculation.

Why Other Options Are Wrong:4.5 and 5 km/h are near but incorrect; 10 km/h is impossible as it exceeds both segment speeds.

Common Pitfalls:Using the arithmetic mean (which would be 5 km/h) instead of the harmonic mean for equal distances.

Final Answer:4.8 km/h