A person jogs at three times the speed of walking and runs at twice the speed…
2025
A person jogs at three times the speed of walking and runs at twice the speed of jogging. From his home to his office, he covers one-third of the distance by walking and the rest by jogging. On his way back home from office, he covers half the distance by jogging and the rest by running. The distance from his home to his office is 10 km, and he walks at a speed of 5 km/hr. Find his average speed (in km/hr) for the complete round trip (up to two decimal places).
Answer: D. 12.41 — Concept — Average speed is never the plain average of the separate speeds used on a journey. It is defined as total distance divided by total time. So…
- A.
11.50
- B.
10.57
- C.
10.65
- D.
12.41
Show answer & explanation
Correct answer: D
Concept — Average speed is never the plain average of the separate speeds used on a journey. It is defined as total distance divided by total time. So whenever a trip is broken into stretches covered at different speeds, each stretch's time must be found on its own from time = distance ÷ speed, all those times added up, and only then is the total distance divided by that total time.
Application — Fix the three speeds from the given ratios, then time each of the four stretches of the round trip.
Walking speed = 5 km/hr. Jogging is three times walking, so jogging = 3 × 5 = 15 km/hr. Running is twice jogging, so running = 2 × 15 = 30 km/hr.
Home to office, walking stretch: one-third of 10 km = 10/3 km at 5 km/hr, so time = (10/3) ÷ 5 = 2/3 hr.
Home to office, jogging stretch: the remaining 20/3 km at 15 km/hr, so time = (20/3) ÷ 15 = 4/9 hr.
Office to home, jogging stretch: half of 10 km = 5 km at 15 km/hr, so time = 5 ÷ 15 = 1/3 hr.
Office to home, running stretch: the remaining 5 km at 30 km/hr, so time = 5 ÷ 30 = 1/6 hr.
Total time = 2/3 + 4/9 + 1/3 + 1/6 = 12/18 + 8/18 + 6/18 + 3/18 = 29/18 hr.
Total distance = 10 + 10 = 20 km, so average speed = 20 ÷ (29/18) = 20 × 18/29 = 360/29 = 12.4137… ≈ 12.41 km/hr.
Cross-check — Collapse each 10 km leg into a single leg average speed and recombine the two legs independently.
Onward leg: 10 km in 2/3 + 4/9 = 10/9 hr, so the onward leg averages 10 ÷ (10/9) = 9 km/hr.
Return leg: 10 km in 1/3 + 1/6 = 1/2 hr, so the return leg averages 10 ÷ (1/2) = 20 km/hr.
The two legs are equal in distance, so the round-trip average is the harmonic mean of the leg speeds: (2 × 9 × 20) ÷ (9 + 20) = 360/29 ≈ 12.41 km/hr — the same value, reached by an independent route.
Note that the arithmetic mean of 9 and 20 is 14.5 km/hr, which is not the average speed of the trip; that gap is exactly what the concept above warns against.
Average speed for the complete round trip = 12.41 km/hr.