Problems on Trains: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

Let the train lengths be a metres and b metres. To fully overtake, the faster train must cover the sum of both lengths, relative to the slower one. Since they move the same way, relative speed = u – v m/sec (u greater than v). Time = distance / relative speed = (a + b) / (u – v).

Let the train lengths be a metres and b metres. To fully cross, together they must cover the sum of their lengths, a + b metres. Since they move towards each other, relative speed = u + v m/sec. Time = distance / relative speed = (a + b) / (u + v).

When two objects move towards each other, they approach from both sides at once. The gap between them closes at the combined rate of both speeds. So relative speed equals the sum of the two speeds. With speeds u and v, relative speed = u + v m/sec.

When two objects move the same way, the faster one slowly gains on the slower one. Relative speed measures how fast this gap closes. That rate equals the difference of the two speeds. With speeds u and v (u greater than v), relative speed = u – v m/sec.

The train starts crossing with its front at the platform’s near end. It finishes crossing once its rear clears the platform’s far end. So the front travels the platform length b, plus the train’s own length l. Total distance covered = (l + b) metres.

A pole or signal post is a single point, with no length of its own. The train has passed it only once the train’s rear also clears that point. So the front must travel a distance equal to the train’s own length. That distance is l metres.

Let speed of the train from A = Sa, and speed of the train from B = Sb. Let t = time in seconds for the two trains to meet. Before meeting, the A-train covers Sa x t and the B-train covers Sb x t. After meeting, each train still has to cover the other train’s pre-meeting distance. So Sb x t = Sa x 4, since the A-train covers that gap in 4 seconds. Also Sa x t = Sb x 25, since the B-train covers that gap in 25 seconds. Dividing these two equations gives (Sa/Sb)^2 = 25/4. So Sa : Sb = sqrt(25) : sqrt(4) = 5 : 2.

Let speed of the train from A = Sa, and speed of the train from B = Sb. Let t = time in seconds for the two trains to meet. Before meeting, the A-train covers Sa x t and the B-train covers Sb x t. After meeting, each train still has to cover the other train’s pre-meeting distance. So Sb x t = Sa x 9, since the A-train covers that gap in 9 seconds. Also Sa x t = Sb x 16, since the B-train covers that gap in 16 seconds. Dividing these two equations gives (Sa/Sb)^2 = 16/9. So Sa : Sb = sqrt(16) : sqrt(9) = 4 : 3.

Let speed of the train from A = Sa, and speed of the train from B = Sb. Let t = time in seconds for the two trains to meet. Before meeting, the A-train covers Sa x t and the B-train covers Sb x t. After meeting, each train still has to cover the other train’s pre-meeting distance. So Sb x t = Sa x 25, since the A-train covers that gap in 25 seconds. Also Sa x t = Sb x 49, since the B-train covers that gap in 49 seconds. Dividing these two equations gives (Sa/Sb)^2 = 49/25. So Sa : Sb = sqrt(49) : sqrt(25) = 7 : 5.

Let speed of the train from A = Sa, and speed of the train from B = Sb. Let t = time in seconds for the two trains to meet. Before meeting, the A-train covers Sa x t and the B-train covers Sb x t. After meeting, each train still has to cover the other train’s pre-meeting distance. So Sb x t = Sa x 9, since the A-train covers that gap in 9 seconds. Also Sa x t = Sb x 25, since the B-train covers that gap in 25 seconds. Dividing these two equations gives (Sa/Sb)^2 = 25/9. So Sa : Sb = sqrt(25) : sqrt(9) = 5 : 3.

Let speed of the train from A = Sa, and speed of the train from B = Sb. Let t = time in seconds for the two trains to meet. Before meeting, the A-train covers Sa x t and the B-train covers Sb x t. After meeting, each train still has to cover the other train’s pre-meeting distance. So Sb x t = Sa x 25, since the A-train covers that gap in 25 seconds. Also Sa x t = Sb x 36, since the B-train covers that gap in 36 seconds. Dividing these two equations gives (Sa/Sb)^2 = 36/25. So Sa : Sb = sqrt(36) : sqrt(25) = 6 : 5.

Let speed of the train from A = Sa, and speed of the train from B = Sb. Let t = time in seconds for the two trains to meet. Before meeting, the A-train covers Sa x t and the B-train covers Sb x t. After meeting, each train still has to cover the other train’s pre-meeting distance. So Sb x t = Sa x 16, since the A-train covers that gap in 16 seconds. Also Sa x t = Sb x 25, since the B-train covers that gap in 25 seconds. Dividing these two equations gives (Sa/Sb)^2 = 25/16. So Sa : Sb = sqrt(25) : sqrt(16) = 5 : 4.

Let speed of the train from A = Sa, and speed of the train from B = Sb. Let t = time in seconds for the two trains to meet. Before meeting, the A-train covers Sa x t and the B-train covers Sb x t. After meeting, each train still has to cover the other train’s pre-meeting distance. So Sb x t = Sa x 4, since the A-train covers that gap in 4 seconds. Also Sa x t = Sb x 9, since the B-train covers that gap in 9 seconds. Dividing these two equations gives (Sa/Sb)^2 = 9/4. So Sa : Sb = sqrt(9) : sqrt(4) = 3 : 2.

Let speed of the train from A = Sa, and speed of the train from B = Sb. Let t = time in seconds for the two trains to meet. Before meeting, the A-train covers Sa x t and the B-train covers Sb x t. After meeting, each train still has to cover the other train’s pre-meeting distance. So Sb x t = Sa x 9, since the A-train covers that gap in 9 seconds. Also Sa x t = Sb x 36, since the B-train covers that gap in 36 seconds. Dividing these two equations gives (Sa/Sb)^2 = 36/9. So Sa : Sb = sqrt(36) : sqrt(9) = 6 : 3.

Let the faster train be 120 m long at 126 km/hr. Let the slower train be 220 m long at 90 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 126 – 90 = 36 km/hr. Convert to m/sec: 36 x 5/18 = 10 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 120 + 220 = 340 m. Time = distance / relative speed = 340/10 = 34 seconds. Check: 10 m/sec x 34 sec = 340 m, which matches.

Let the faster train be 240 m long at 108 km/hr. Let the slower train be 220 m long at 36 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 108 – 36 = 72 km/hr. Convert to m/sec: 72 x 5/18 = 20 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 240 + 220 = 460 m. Time = distance / relative speed = 460/20 = 23 seconds. Check: 20 m/sec x 23 sec = 460 m, which matches.

Let the faster train be 160 m long at 90 km/hr. Let the slower train be 260 m long at 54 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 90 – 54 = 36 km/hr. Convert to m/sec: 36 x 5/18 = 10 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 160 + 260 = 420 m. Time = distance / relative speed = 420/10 = 42 seconds. Check: 10 m/sec x 42 sec = 420 m, which matches.

Let the faster train be 160 m long at 117 km/hr. Let the slower train be 200 m long at 99 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 117 – 99 = 18 km/hr. Convert to m/sec: 18 x 5/18 = 5 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 160 + 200 = 360 m. Time = distance / relative speed = 360/5 = 72 seconds. Check: 5 m/sec x 72 sec = 360 m, which matches.

Let the faster train be 130 m long at 99 km/hr. Let the slower train be 260 m long at 63 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 99 – 63 = 36 km/hr. Convert to m/sec: 36 x 5/18 = 10 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 130 + 260 = 390 m. Time = distance / relative speed = 390/10 = 39 seconds. Check: 10 m/sec x 39 sec = 390 m, which matches.

Let the faster train be 260 m long at 126 km/hr. Let the slower train be 230 m long at 108 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 126 – 108 = 18 km/hr. Convert to m/sec: 18 x 5/18 = 5 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 260 + 230 = 490 m. Time = distance / relative speed = 490/5 = 98 seconds. Check: 5 m/sec x 98 sec = 490 m, which matches.

Let the faster train be 190 m long at 126 km/hr. Let the slower train be 140 m long at 90 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 126 – 90 = 36 km/hr. Convert to m/sec: 36 x 5/18 = 10 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 190 + 140 = 330 m. Time = distance / relative speed = 330/10 = 33 seconds. Check: 10 m/sec x 33 sec = 330 m, which matches.

Let the faster train be 190 m long at 144 km/hr. Let the slower train be 130 m long at 72 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 144 – 72 = 72 km/hr. Convert to m/sec: 72 x 5/18 = 20 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 190 + 130 = 320 m. Time = distance / relative speed = 320/20 = 16 seconds. Check: 20 m/sec x 16 sec = 320 m, which matches.

Let the faster train be 290 m long at 81 km/hr. Let the slower train be 240 m long at 63 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 81 – 63 = 18 km/hr. Convert to m/sec: 18 x 5/18 = 5 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 290 + 240 = 530 m. Time = distance / relative speed = 530/5 = 106 seconds. Check: 5 m/sec x 106 sec = 530 m, which matches.

Let the faster train be 150 m long at 81 km/hr. Let the slower train be 190 m long at 45 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 81 – 45 = 36 km/hr. Convert to m/sec: 36 x 5/18 = 10 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 150 + 190 = 340 m. Time = distance / relative speed = 340/10 = 34 seconds. Check: 10 m/sec x 34 sec = 340 m, which matches.

Let the faster train be 260 m long at 90 km/hr. Let the slower train be 170 m long at 72 km/hr. They move the same way, so relative speed = difference of speeds. Relative speed = 90 – 72 = 18 km/hr. Convert to m/sec: 18 x 5/18 = 5 m/sec. To fully overtake, the faster train covers the sum of both lengths relative to the slower one. Distance = 260 + 170 = 430 m. Time = distance / relative speed = 430/5 = 86 seconds. Check: 5 m/sec x 86 sec = 430 m, which matches.