Walking at 6/7 of his usual speed, a man is 25 minutes too late. Find the…

Concept

When speed changes by a constant factor over the same fixed distance, time changes by the reciprocal of that factor — speed and time are inversely proportional for a fixed distance. If the new speed is a fraction k of the usual speed, the new time is 1/k of the usual time, and the extra time taken equals (1/k − 1) of the usual time.

Application

1. New speed = 6/7 of usual speed, so the time factor is the reciprocal: 7/6.

2. Extra time = (7/6 − 1) × (usual time) = 1/6 × (usual time).

3. This extra time is given as 25 minutes, so (1/6) × (usual time) = 25.

4. Usual time = 25 × 6 = 150 minutes = 2 hours 30 minutes.

Cross-check

Usual time = 150 min, so the new time = 150 × (7/6) = 175 min, meaning the man takes 175 − 150 = 25 min extra — exactly the delay stated. Hence the usual time is 2 hours 30 minutes.