Average speed over equal distances: A traveler covers the first 160 km at 64 km/h and the next 160 km at 80 km/h. What is the average speed over the 320 km?

Introduction / Context: For two legs of equal distance, the average speed is the harmonic mean of the two speeds: 2ab/(a + b). This accounts correctly for time weighting when distances are the same but speeds differ on the segments.

Given Data / Assumptions:

• Leg 1: 160 km at 64 km/h.
• Leg 2: 160 km at 80 km/h.
• Total distance = 320 km.

Concept / Approach: Average speed = total distance / total time. Alternatively, use the harmonic mean for equal distances: V_avg = 2ab/(a + b), where a and b are the two speeds.

Step-by-Step Solution:

Verification / Alternative check: Time method: t1 = 160/64 = 2.5 h; t2 = 160/80 = 2 h; total time = 4.5 h; V_avg = 320/4.5 = 71.111… km/h, same result.

Why Other Options Are Wrong: 72 km/h is the simple average of speeds (incorrect for equal distances); 35.55 and 36 km/h are unrelated computations; 68 km/h understates the harmonic mean.

Common Pitfalls: Averaging speeds arithmetically when distances are equal; always use harmonic mean or compute via total time.

Final Answer: 71.11 km/h