A train covered a distance at a uniform speed. If the train had been 6 km/hr…

Concept: For a journey covered at a uniform speed, distance = speed × time (d = s × t). When the speed changes by a fixed amount, comparing the resulting distance (at the new speed and new time) to the same actual distance d gives one linear equation relating speed and time. Two such altered-speed conditions give two equations that can be solved simultaneously for the original speed and time.

Application: Let the usual speed be s km/hr and the usual time be t hr, so the distance is d = s × t.

1. Faster by 6 km/hr takes 4 hours less: (s + 6)(t − 4) = d. Expanding gives st − 4s + 6t − 24 = st, which simplifies to 6t − 4s = 24 … (i)

2. Slower by 6 km/hr takes 6 hours more: (s − 6)(t + 6) = d. Expanding gives st + 6s − 6t − 36 = st, which simplifies to 6s − 6t = 36, i.e., s − t = 6 … (ii)

3. From (ii), s = t + 6. Substituting into (i): 6t − 4(t + 6) = 24 ⟹ 6t − 4t − 24 = 24 ⟹ 2t = 48 ⟹ t = 24 hr.

4. Then s = t + 6 = 30 km/hr.

5. Distance d = s × t = 30 × 24 = 720 km.

Cross-check: At 36 km/hr (30 + 6), time = 720/36 = 20 hr, which is 4 hr less than 24 hr — matches the first condition. At 24 km/hr (30 − 6), time = 720/24 = 30 hr, which is 6 hr more than 24 hr — matches the second condition. Both conditions hold, confirming the distance is 720 km.