If a man walks 3 km/h, he is late to his office by 20 minutes. If he increases…

Concept

For a fixed distance, time = distance ÷ speed. When the same distance is covered at two different speeds, the difference between the two travel times equals the total gap between the two arrival outcomes. If one trip is late by L and the other is early by E (relative to the same scheduled time), the slower trip takes (L + E) more time than the faster trip.

Application

Let the distance be d km. The scheduled (on-time) duration is the same in both cases, so we compare the two actual travel times.

1. At 3 km/h the man is 20 min late and at 6 km/h he is 30 min early, so the slower trip takes 20 + 30 = 50 minutes longer than the faster trip.

2. Convert the gap to hours: 50 minutes = 50 ÷ 60 = 5/6 hour.

3. Travel time at 3 km/h is d/3 h; at 6 km/h it is d/6 h. Set their difference equal to the gap: d/3 − d/6 = 5/6.

4. Combine the left side: d/3 − d/6 = (2d − d)/6 = d/6. So d/6 = 5/6.

5. Multiply both sides by 6: d = 5. The distance is 5 km.

Cross-check

At 3 km/h, 5 km takes 5/3 h = 1 h 40 min; at 6 km/h, 5 km takes 5/6 h = 50 min. The difference is 1 h 40 min − 50 min = 50 min, matching the 20-min-late-plus-30-min-early gap. Hence 5 km is consistent.