Problems on Trains: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 3)

Let train length L = 285 m, train speed = 45 km/hr, and Arjun’s speed = 9 km/hr. They move towards each other, so relative speed = sum of speeds. Relative speed = 45 + 9 = 54 km/hr. Convert to m/sec: 54 x 5/18 = 15 m/sec. To fully pass Arjun, the train covers its own length L relative to Arjun. Time = L / relative speed = 285/15 = 19 seconds. Check: 15 m/sec x 19 sec = 285 m, which matches L.

Let train length L = 540 m, train speed = 99 km/hr, and Karan’s speed = 9 km/hr. They move towards each other, so relative speed = sum of speeds. Relative speed = 99 + 9 = 108 km/hr. Convert to m/sec: 108 x 5/18 = 30 m/sec. To fully pass Karan, the train covers its own length L relative to Karan. Time = L / relative speed = 540/30 = 18 seconds. Check: 30 m/sec x 18 sec = 540 m, which matches L.

Let train length L = 250 m, train speed = 99 km/hr, and Neha’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 99 – 9 = 90 km/hr. Convert to m/sec: 90 x 5/18 = 25 m/sec. To fully pass Neha, the train covers its own length L relative to Neha. Time = L / relative speed = 250/25 = 10 seconds. Check: 25 m/sec x 10 sec = 250 m, which matches L.

Let train length L = 225 m, train speed = 99 km/hr, and Sanjay’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 99 – 9 = 90 km/hr. Convert to m/sec: 90 x 5/18 = 25 m/sec. To fully pass Sanjay, the train covers its own length L relative to Sanjay. Time = L / relative speed = 225/25 = 9 seconds. Check: 25 m/sec x 9 sec = 225 m, which matches L.

Let train length L = 90 m, train speed = 117 km/hr, and Priya’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 117 – 9 = 108 km/hr. Convert to m/sec: 108 x 5/18 = 30 m/sec. To fully pass Priya, the train covers its own length L relative to Priya. Time = L / relative speed = 90/30 = 3 seconds. Check: 30 m/sec x 3 sec = 90 m, which matches L.

Let train length L = 60 m, train speed = 63 km/hr, and Priya’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 63 – 9 = 54 km/hr. Convert to m/sec: 54 x 5/18 = 15 m/sec. To fully pass Priya, the train covers its own length L relative to Priya. Time = L / relative speed = 60/15 = 4 seconds. Check: 15 m/sec x 4 sec = 60 m, which matches L.

Let train length L = 390 m, train speed = 117 km/hr, and Meera’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 117 – 9 = 108 km/hr. Convert to m/sec: 108 x 5/18 = 30 m/sec. To fully pass Meera, the train covers its own length L relative to Meera. Time = L / relative speed = 390/30 = 13 seconds. Check: 30 m/sec x 13 sec = 390 m, which matches L.

Let train length L = 60 m, train speed = 45 km/hr, and Sanjay’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 45 – 9 = 36 km/hr. Convert to m/sec: 36 x 5/18 = 10 m/sec. To fully pass Sanjay, the train covers its own length L relative to Sanjay. Time = L / relative speed = 60/10 = 6 seconds. Check: 10 m/sec x 6 sec = 60 m, which matches L.

Let train length L = 60 m, train speed = 81 km/hr, and Divya’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 81 – 9 = 72 km/hr. Convert to m/sec: 72 x 5/18 = 20 m/sec. To fully pass Divya, the train covers its own length L relative to Divya. Time = L / relative speed = 60/20 = 3 seconds. Check: 20 m/sec x 3 sec = 60 m, which matches L.

Let train length L = 160 m, train speed = 45 km/hr, and Meera’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 45 – 9 = 36 km/hr. Convert to m/sec: 36 x 5/18 = 10 m/sec. To fully pass Meera, the train covers its own length L relative to Meera. Time = L / relative speed = 160/10 = 16 seconds. Check: 10 m/sec x 16 sec = 160 m, which matches L.

Let train length L = 300 m, train speed = 63 km/hr, and Karan’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 63 – 9 = 54 km/hr. Convert to m/sec: 54 x 5/18 = 15 m/sec. To fully pass Karan, the train covers its own length L relative to Karan. Time = L / relative speed = 300/15 = 20 seconds. Check: 15 m/sec x 20 sec = 300 m, which matches L.

Let train length L = 120 m, train speed = 63 km/hr, and Rohan’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 63 – 9 = 54 km/hr. Convert to m/sec: 54 x 5/18 = 15 m/sec. To fully pass Rohan, the train covers its own length L relative to Rohan. Time = L / relative speed = 120/15 = 8 seconds. Check: 15 m/sec x 8 sec = 120 m, which matches L.

Let train length L = 350 m, train speed = 135 km/hr, and Sanjay’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 135 – 9 = 126 km/hr. Convert to m/sec: 126 x 5/18 = 35 m/sec. To fully pass Sanjay, the train covers its own length L relative to Sanjay. Time = L / relative speed = 350/35 = 10 seconds. Check: 35 m/sec x 10 sec = 350 m, which matches L.

Let train length L = 350 m, train speed = 135 km/hr, and Arjun’s speed = 9 km/hr. Both move the same way, so relative speed = difference of speeds. Relative speed = 135 – 9 = 126 km/hr. Convert to m/sec: 126 x 5/18 = 35 m/sec. To fully pass Arjun, the train covers its own length L relative to Arjun. Time = L / relative speed = 350/35 = 10 seconds. Check: 35 m/sec x 10 sec = 350 m, which matches L.

Let train length = x metres and train speed = s m/sec. Crossing the 60-m platform takes 23 seconds: x + 60 = s x 23. Crossing the 100-m platform takes 27 seconds: x + 100 = s x 27. Subtracting the two equations cancels x: 60 – 100 = s x (23 – 27). So s = (60-100)/(23-27) = 10 m/sec. Substitute back: x = s x 23 – 60 = 10×23 – 60 = 170 m. Check: x + 100 = 270 m, and s x 27 = 270 m, which match.

Let train length = x metres and train speed = s m/sec. Crossing the 100-m platform takes 28 seconds: x + 100 = s x 28. Crossing the 170-m platform takes 35 seconds: x + 170 = s x 35. Subtracting the two equations cancels x: 100 – 170 = s x (28 – 35). So s = (100-170)/(28-35) = 10 m/sec. Substitute back: x = s x 28 – 100 = 10×28 – 100 = 180 m. Check: x + 170 = 350 m, and s x 35 = 350 m, which match.

Let train length = x metres and train speed = s m/sec. Crossing the 150-m platform takes 20 seconds: x + 150 = s x 20. Crossing the 140-m platform takes 19 seconds: x + 140 = s x 19. Subtracting the two equations cancels x: 150 – 140 = s x (20 – 19). So s = (150-140)/(20-19) = 10 m/sec. Substitute back: x = s x 20 – 150 = 10×20 – 150 = 50 m. Check: x + 140 = 190 m, and s x 19 = 190 m, which match.

Let train length = x metres and train speed = s m/sec. Crossing the 110-m platform takes 12 seconds: x + 110 = s x 12. Crossing the 130-m platform takes 13 seconds: x + 130 = s x 13. Subtracting the two equations cancels x: 110 – 130 = s x (12 – 13). So s = (110-130)/(12-13) = 20 m/sec. Substitute back: x = s x 12 – 110 = 20×12 – 110 = 130 m. Check: x + 130 = 260 m, and s x 13 = 260 m, which match.

Let train length = x metres and train speed = s m/sec. Crossing the 70-m platform takes 7 seconds: x + 70 = s x 7. Crossing the 50-m platform takes 6 seconds: x + 50 = s x 6. Subtracting the two equations cancels x: 70 – 50 = s x (7 – 6). So s = (70-50)/(7-6) = 20 m/sec. Substitute back: x = s x 7 – 70 = 20×7 – 70 = 70 m. Check: x + 50 = 120 m, and s x 6 = 120 m, which match.

Let train length = x metres and train speed = s m/sec. Crossing the 200-m platform takes 30 seconds: x + 200 = s x 30. Crossing the 50-m platform takes 15 seconds: x + 50 = s x 15. Subtracting the two equations cancels x: 200 – 50 = s x (30 – 15). So s = (200-50)/(30-15) = 10 m/sec. Substitute back: x = s x 30 – 200 = 10×30 – 200 = 100 m. Check: x + 50 = 150 m, and s x 15 = 150 m, which match.

Let train length = x metres and train speed = s m/sec. Crossing the 190-m platform takes 18 seconds: x + 190 = s x 18. Crossing the 50-m platform takes 11 seconds: x + 50 = s x 11. Subtracting the two equations cancels x: 190 – 50 = s x (18 – 11). So s = (190-50)/(18-11) = 20 m/sec. Substitute back: x = s x 18 – 190 = 20×18 – 190 = 170 m. Check: x + 50 = 220 m, and s x 11 = 220 m, which match.

Let train length = x metres and train speed = s m/sec. Crossing the 80-m platform takes 12 seconds: x + 80 = s x 12. Crossing the 110-m platform takes 14 seconds: x + 110 = s x 14. Subtracting the two equations cancels x: 80 – 110 = s x (12 – 14). So s = (80-110)/(12-14) = 15 m/sec. Substitute back: x = s x 12 – 80 = 15×12 – 80 = 100 m. Check: x + 110 = 210 m, and s x 14 = 210 m, which match.

Let train length = x metres and train speed = s m/sec. Crossing the 130-m platform takes 24 seconds: x + 130 = s x 24. Crossing the 170-m platform takes 28 seconds: x + 170 = s x 28. Subtracting the two equations cancels x: 130 – 170 = s x (24 – 28). So s = (130-170)/(24-28) = 10 m/sec. Substitute back: x = s x 24 – 130 = 10×24 – 130 = 110 m. Check: x + 170 = 280 m, and s x 28 = 280 m, which match.

Let train length = x metres and train speed = s m/sec. Crossing the 50-m platform takes 23 seconds: x + 50 = s x 23. Crossing the 100-m platform takes 28 seconds: x + 100 = s x 28. Subtracting the two equations cancels x: 50 – 100 = s x (23 – 28). So s = (50-100)/(23-28) = 10 m/sec. Substitute back: x = s x 23 – 50 = 10×23 – 50 = 180 m. Check: x + 100 = 280 m, and s x 28 = 280 m, which match.

Let train length L = 150 m and speed = 90 km/hr. Convert speed to m/sec: 90 x 5/18 = 25 m/sec. In 39 seconds, the train covers speed x time = 25 x 39 = 975 m. This distance equals train length plus platform length. So platform length = 975 – 150 = 825 metres. Check: 150 + 825 = 975 m, the total distance covered.