Average speed for three equal-distance parts at different speeds: A journey is divided into three equal parts of distance. The traveler covers them at 15 km/h, 10 km/h, and 5 km/h respectively. What is the average speed for the whole trip?

Introduction / Context:When equal distances are traveled at different speeds, the average speed is not the arithmetic mean. It equals the harmonic-mean-like combination: v_avg = 3 / (1/v1 + 1/v2 + 1/v3).

Given Data / Assumptions:

• Three equal parts by distance
• Speeds v1 = 15 km/h, v2 = 10 km/h, v3 = 5 km/h

Concept / Approach:Let each distance be d. Total distance = 3d. Total time = d/15 + d/10 + d/5. Then v_avg = (3d) / (d(1/15 + 1/10 + 1/5)) = 3 / (1/15 + 1/10 + 1/5).

Step-by-Step Solution:1/15 + 1/10 + 1/5 = 0.0666667 + 0.1 + 0.2 = 0.3666667v_avg = 3 / 0.3666667 ≈ 8.1818 km/h

Verification / Alternative check:Because the slowest speed (5 km/h) applies to one-third of the distance, the overall average must be closer to 5 than to 15, which 8.18 satisfies.

Why Other Options Are Wrong:2.28 is far too low; 9 and 14 are arithmetic-mean style distractors; 7.5 underestimates the harmonic combination.

Common Pitfalls:Taking the average of speeds rather than the reciprocal-time weighted form for equal distances.

Final Answer:8.18 km /h