A goods train leaves a station at a certain time and travels at a uniform…

Concept: In a catch-up (chase) problem, when a faster vehicle starts later from the same point and overtakes a slower one moving in the same direction, both have covered the SAME total distance from the start at the moment they meet. Setting (slower train's speed x its total travel time) equal to (faster train's speed x its travel time) gives the unknown speed; equivalently, the closing (relative) speed must clear the head-start gap in exactly the chase time.

Application:

1. Let the goods train's speed be x kmph.

2. The goods train has a 6-hour head start and keeps moving through the 4-hour chase, so by the moment it is caught it has been travelling for 6 + 4 = 10 hours.

3. Distance covered by the goods train when caught = 10x km.

4. Distance covered by the express train = 90 kmph x 4 h = 360 km.

5. Since both trains start from the same station and travel in the same direction, their distances are equal at the meeting point: 10x = 360.

6. Solving: x = 360 / 10 = 36.

Cross-check (relative-speed method):

During the 6-hour head start, the goods train opens a lead of 6x = 216 km. The express train closes this gap at a relative speed of (90 - x) = 54 kmph, taking 216 / 54 = 4 hours - exactly matching the given catch-up time, confirming x = 36 kmph.

So the goods train's speed is 36 kmph.