A train covered a certain distance at a uniform speed. If the train had been 6…

For a journey covered in time t at a constant speed x, the distance D = x × t remains the same however the speed and time are described. If changing the speed by a fixed amount and adjusting the time accordingly still describes the same journey, equating the two expressions for D gives a linear equation relating speed and time. Two independent changes of this kind give two simultaneous linear equations in speed and time, which can be solved together.

1. Let the usual speed be x km/hr and the scheduled time be t hours, so the distance is D = x × t.

2. Speed 6 km/hr faster and 4 hours less than scheduled: (x + 6)(t − 4) = xt. Expanding gives xt − 4x + 6t − 24 = xt, so 6t − 4x = 24, i.e. 3t − 2x = 12 — Equation (1).

3. Speed 6 km/hr slower and 6 hours more than scheduled: (x − 6)(t + 6) = xt. Expanding gives xt + 6x − 6t − 36 = xt, so 6x − 6t = 36, i.e. x − t = 6, so x = t + 6 — Equation (2).

4. Substituting Equation (2) into Equation (1): 3t − 2(t + 6) = 12, which gives 3t − 2t − 12 = 12, so t = 24 hours.

5. From Equation (2): x = t + 6 = 24 + 6 = 30 km/hr.

6. Distance D = x × t = 30 × 24 = 720 km.

Check: at 36 km/hr (30 + 6), the time taken would be 720 ÷ 36 = 20 hours, which is 24 − 4 hours — matching the first condition. At 24 km/hr (30 − 6), the time taken would be 720 ÷ 24 = 30 hours, which is 24 + 6 hours — matching the second condition. Both conditions hold, confirming the distance is 720 km.