A plane travels along the three sides of an equilateral triangle at speeds of…
2025
A plane travels along the three sides of an equilateral triangle at speeds of 20 km/h, 40 km/h, and 60 km/h respectively. Find the average speed of the plane for the entire journey.
- A.
30.55 km/h
- B.
32.73 km/h
- C.
36.36 km/h
- D.
31.96 km/h
Show answer & explanation
Correct answer: B
Concept: When equal distances are covered at different speeds, the average speed for the whole trip is NOT the arithmetic mean of the speeds — it is total distance divided by total time. For n equal segments of length d at speeds v1, v2, …, vn, average speed = nd / (d/v1 + d/v2 + … + d/vn) = n / (1/v1 + 1/v2 + … + 1/vn).
Application: Let each side of the equilateral triangle be d km.
Time taken on each side: d/20 hours, d/40 hours, and d/60 hours respectively.
Total distance covered = 3 equal sides = 3d km.
Total time = d/20 + d/40 + d/60. Using the LCM of 20, 40, and 60 (which is 120): total time = (6d + 3d + 2d)/120 = 11d/120 hours.
Average speed = total distance ÷ total time = 3d ÷ (11d/120) = (3 × 120)/11 = 360/11 ≈ 32.73 km/h.
Cross-check: Applying the harmonic-style shortcut directly, average speed = 3 / (1/20 + 1/40 + 1/60) = 3 / (11/120) = 360/11 ≈ 32.73 km/h, confirming the step-by-step result. Notice this is well below the plain arithmetic mean of the three speeds (40 km/h) — expected, since the plane spends the most time travelling at its slowest speed (20 km/h), which pulls the average down.
