Train Problems â MCQS
Q1. Two trains of lengths 400 m and 600 m run towards each other at speeds 110 m/s and 140 m/s.
Time taken to cross each other is - āĻĻā§āĻāĻž ā§°ā§āϞ⧰ āĻĻā§ā§°ā§āĻā§āϝ 400 āĻŽāĻŋ āĻā§°ā§ 600 āĻŽāĻŋāĨ¤ āĻāϤāĻŋ 110 āĻŽāĻŋ/āĻā§ āĻā§°ā§ 140 āĻŽāĻŋ/āĻā§āĨ¤
āĻāĻā§-āĻāύāĻ āĻ
āϤāĻŋāĻā§ā§°āĻŽ āĻā§°āĻŋāĻŦāϞ⧠āϞā§ā§ąāĻž āϏāĻŽā§ āĻāĻŋāĻŽāĻžāύ ?
A) 2 s B) 4 s C) 5 s D) 10 s
Ans / āĻāϤā§āϤ⧰: B) 4 s
Explanation : āĻŦā§āϝāĻžāĻā§āϝāĻž:
Total distance = 400 + 600 = 1000 m : āĻŽā§āĻ āĻĻā§ā§°āϤā§āĻŦ = 400 + 600 = 1000 āĻŽāĻŋ
Relative speed = 110 + 140 = 250 m/s : āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ = 110 + 140 = 250 āĻŽāĻŋ/āĻā§
Time = 1000 ÷ 250 = 4 s : āϏāĻŽā§ = 1000 ÷ 250 = 4 āĻā§āĻā§āĻŖā§āĻĄ
Q2. Two trains of equal length L cross each other in opposite directions at 60 m/s and 40 m/s.
Time taken is - āϏāĻŽāĻžāύ āĻĻā§ā§°ā§āĻā§āϝ⧰ (L) āĻĻā§āĻāĻž ā§°ā§āϞ āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļāϤ 60 āĻŽāĻŋ/āĻā§ āĻā§°ā§ 40 āĻŽāĻŋ/āĻā§ āĻāϤāĻŋāϤ āĻāϞā§āĨ¤
āϏāĻŽā§ āĻāĻŋāĻŽāĻžāύ ?
A) L/100 B) 2L/100 C) L/20 D) 2L/20
Ans / āĻāϤā§āϤ⧰: B) 2L/100
Explanation : āĻŦā§āϝāĻžāĻā§āϝāĻž :
Total distance = L + L = 2L : āĻŽā§āĻ āĻĻā§ā§°āϤā§āĻŦ = L + L = 2L
Relative speed = 60 + 40 = 100 m/s : āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ = 100 āĻŽāĻŋ/āĻā§
Time = 2L ÷ 100 : āϏāĻŽā§ = 2L ÷ 100
Q3. When two trains move in opposite directions, relative speed is - āĻĻā§āĻāĻž ā§°ā§āϞ āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļāϤ āĻāϞāĻŋāϞ⧠āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ āĻš’āϞ -
A) Difference of speeds / āĻāϤāĻŋā§° āĻĒāĻžā§°ā§āĻĨāĻā§āϝ
B) Sum of speeds / āĻāϤāĻŋā§° āϝā§āĻāĻĢāϞ
C) Average speed / āĻā§ āĻāϤāĻŋ
D) Product of speeds / āĻāϤāĻŋā§° āĻā§āĻŖāĻĢāϞ
Ans / āĻāϤā§āϤ⧰: B) Sum of speeds / āĻāϤāĻŋā§° āϝā§āĻāĻĢāϞ
Explanation: Opposite direction → speeds are added. āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļāϤ āĻāϞāĻŋāϞ⧠→ āĻāϤāĻŋ āϝā§āĻ āĻā§°āĻž āĻšā§āĨ¤
Q4. Two trains cross each other in 5 seconds. Their combined length is 1250 m.
Relative speed is - āĻĻā§āĻāĻž ā§°ā§āϞ 5 āĻā§āĻā§āĻŖā§āĻĄāϤ āĻāĻā§-āĻāύāĻ āĻ
āϤāĻŋāĻā§ā§°āĻŽ āĻā§°ā§āĨ¤ āĻŽā§āĻ āĻĻā§ā§°ā§āĻā§āϝ 1250 āĻŽāĻŋāĨ¤
āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ āĻāĻŋāĻŽāĻžāύ ?
A) 200 m/s B) 250 m/s C) 300 m/s D) 350 m/s
Ans / āĻāϤā§āϤ⧰: B) 250 m/s
Explanation: Relative speed = Distance ÷ Time = 1250 ÷ 5 = 250 m/s
āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ = āĻĻā§ā§°āϤā§āĻŦ ÷ āϏāĻŽā§ = 1250 ÷ 5 = 250 āĻŽāĻŋ/āĻā§
Q5. Two trains move in the same direction at 140 m/s and 110 m/s.
Relative speed is - āĻĻā§āĻāĻž ā§°ā§āϞ āĻāĻā§āĻāĻž āĻĻāĻŋāĻļāϤ 140 āĻŽāĻŋ/āĻā§ āĻā§°ā§ 110 āĻŽāĻŋ/āĻā§ āĻāϤāĻŋāϤ āĻāϞā§āĨ¤
āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ āĻāĻŋāĻŽāĻžāύ ?
A) 250 m/s B) 30 m/s C) 110 m/s D) 140 m/s
Ans / āĻāϤā§āϤ⧰: B) 30 m/s
Explanation : āĻŦā§āϝāĻžāĻā§āϝāĻž:
Same direction → subtract speeds : āĻāĻā§āĻāĻž āĻĻāĻŋāĻļāϤ āĻāϞāĻŋāϞ⧠→ āĻāϤāĻŋ āĻŦāĻŋā§ā§āĻ
140 − 110 = 30 m/s : 140 − 110 = 30 āĻŽāĻŋ/āĻā§
Very Important Shortcuts
• Crossing each other → Add lengths : āĻ
āϤāĻŋāĻā§ā§°āĻŽ → āĻĻā§ā§°ā§āĻā§āϝ āϝā§āĻ
• Opposite direction → Add speeds : āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļ → āĻāϤāĻŋ āϝā§āĻ
• Same direction → Subtract speeds : āĻāĻā§āĻāĻž āĻĻāĻŋāĻļ → āĻāϤāĻŋ āĻŦāĻŋā§ā§āĻ
• Time = Distance ÷ Relative speed : āϏāĻŽā§ = āĻĻā§ā§°āϤā§āĻŦ ÷ āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ
Q1. Two trains of lengths 400 m and 600 m run towards each other at speeds 110 m/s and 140 m/s.
Time taken to cross each other is - āĻĻā§āĻāĻž ā§°ā§āϞ⧰ āĻĻā§ā§°ā§āĻā§āϝ 400 āĻŽāĻŋ āĻā§°ā§ 600 āĻŽāĻŋāĨ¤ āĻāϤāĻŋ 110 āĻŽāĻŋ/āĻā§ āĻā§°ā§ 140 āĻŽāĻŋ/āĻā§āĨ¤ āϏāĻŽā§ āĻāĻŋāĻŽāĻžāύ ?
A) 2 s B) 4 s C) 5 s D) 10 s
Ans: B) 4 s
Explanation : āĻŦā§āϝāĻžāĻā§āϝāĻž
Total distance = 400 + 600 = 1000 m : āĻŽā§āĻ āĻĻā§ā§°āϤā§āĻŦ = 1000 āĻŽāĻŋ
Relative speed = 110 + 140 = 250 m/s : āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ = 250 āĻŽāĻŋ/āĻā§
Time = 1000 ÷ 250 = 4s : āϏāĻŽā§ = 4 āĻā§āĻā§āĻŖā§āĻĄ
Q2. Two trains of equal length L cross each other in opposite directions at 60 m/s and 40 m/s. Time taken is - āϏāĻŽāĻžāύ āĻĻā§ā§°ā§āĻā§āϝ⧰ āĻĻā§āĻāĻž ā§°ā§āϞ āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļāϤ 60 āĻā§°ā§ 40 āĻŽāĻŋ/āĻā§ āĻāϤāĻŋāϤ āĻāϞā§āĨ¤ āϏāĻŽā§ āĻāĻŋāĻŽāĻžāύ ?
A) L/100 B) 2L/100 C) L/20 D) 2L/20
Ans: B) 2L/100
Explanation : āĻŦā§āϝāĻžāĻā§āϝāĻž
Total distance = L + L = 2L : āĻŽā§āĻ āĻĻā§ā§°āϤā§āĻŦ = 2L
Relative speed = 60 + 40 = 100 m/s : āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ = 100 āĻŽāĻŋ/āĻā§
Q3. When two trains move in opposite directions, relative speed is - āĻĻā§āĻāĻž ā§°ā§āϞ āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļāϤ āĻāϞāĻŋāϞ⧠āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ āĻš’āϞ -
A) Difference B) Sum C) Average D) Product
Ans: B) Sum
Explanation: Opposite direction → speeds are added. āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļāϤ āĻāϞāĻŋāϞ⧠→ āĻāϤāĻŋ āϝā§āĻ āĻā§°āĻž āĻšā§āĨ¤
Q4. Two trains cross each other in 5 s. Combined length is 1250 m. Relative speed is - ā§Ģ āĻā§āĻā§āĻŖā§āĻĄāϤ āĻ āϤāĻŋāĻā§ā§°āĻŽ āĻā§°ā§, āĻŽā§āĻ āĻĻā§ā§°ā§āĻā§āϝ 1250 āĻŽāĻŋāĨ¤ āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ āĻāĻŋāĻŽāĻžāύ ?
A) 200 B) 250 C) 300 D) 350 m/s
Ans: B) 250 m/s
Explanation : āĻŦā§āϝāĻžāĻā§āϝāĻž:
Speed = Distance ÷ Time = 1250 ÷ 5 = 250 m/s : āĻāϤāĻŋ = āĻĻā§ā§°āϤā§āĻŦ ÷ āϏāĻŽā§ = 1250 ÷ 5 = 250 āĻŽāĻŋ/āĻā§
Q5. Two trains move in the same direction at 140 m/s and 110 m/s. Relative speed is - āĻāĻā§āĻāĻž āĻĻāĻŋāĻļāϤ 140 āĻā§°ā§ 110 āĻŽāĻŋ/āĻā§āĨ¤ āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ āĻāĻŋāĻŽāĻžāύ ?
A) 250 B) 30 C) 110 D) 140 m/s
Ans: B) 30 m/s
Explanation: Same direction → subtract speeds = 140 − 110. āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻāĻā§āĻāĻž āĻĻāĻŋāĻļāϤ → āĻāϤāĻŋ āĻŦāĻŋā§ā§āĻ = 140 − 110
Q6. A train 200 m long crosses a pole in 20 s. Speed is - ⧍ā§Ļā§Ļ āĻŽāĻŋ āĻĻā§ā§°ā§āĻā§āϝ⧰ ā§°ā§āϞ⧠⧍ā§Ļ āĻā§āĻā§āĻŖā§āĻĄāϤ āĻā§āĻāĻāĻŋ āĻĒāĻžā§° āĻā§°ā§āĨ¤ āĻāϤāĻŋ āĻāĻŋāĻŽāĻžāύ ?
A) 5 B) 8 C) 10 D) 12 m/s
Ans: C) 10 m/s
Explanation : āĻŦā§āϝāĻžāĻā§āϝāĻž:
Distance = length of train = 200 m : āĻĻā§ā§°āϤā§āĻŦ = ā§°ā§āϞ⧰ āĻĻā§ā§°ā§āĻā§āϝ = 200 āĻŽāĻŋ
Speed = 200 ÷ 20 = 10 m/s : āĻāϤāĻŋ = 10 āĻŽāĻŋ/āĻā§
Q7. Correct formula for time taken by two trains to cross is - āĻĻā§āĻāĻž ā§°ā§āϞ āĻ āϤāĻŋāĻā§ā§°āĻŽ āĻā§°āĻžā§° āϏāĻ āĻŋāĻ āϏā§āϤā§ā§° -
A) (Sâ+Sâ)/(Lâ+Lâ) B) (Lâ+Lâ)/(Sâ+Sâ) C) (Lâ−Lâ)/(Sâ−Sâ) D) (Sâ−Sâ)/(Lâ+Lâ)
Ans: B
Explanation : Time = Total distance ÷ Relative speed : āĻŦā§āϝāĻžāĻā§āϝāĻž: āϏāĻŽā§ = āĻŽā§āĻ āĻĻā§ā§°āϤā§āĻŦ ÷ āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ
Q8. A train of length 150 m crosses a man in 15 s. Speed is - ā§§ā§Ģā§Ļ āĻŽāĻŋ āĻĻā§ā§°ā§āĻā§āϝ⧰ ā§°ā§āϞ⧠⧧ā§Ģ āĻā§āĻā§āĻŖā§āĻĄāϤ āĻŽāĻžāύā§āĻš āĻāĻāύ āĻĒāĻžā§° āĻā§°ā§āĨ¤ āĻāϤāĻŋ āĻāĻŋāĻŽāĻžāύ ?
A) 5 B) 8 C) 10 D) 12 m/s
Ans: C) 10
Explanation: Speed = 150 ÷ 15 āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻāϤāĻŋ = 150 ÷ 15
Q9. Speed of 72 km/h in m/s is - 72 āĻāĻŋāĻŽāĻŋ/āĻāĻŖā§āĻāĻž = ?
A) 18 B) 20 C) 22 D) 25
Ans: B) 20
Explanation: km/h → m/s = × 5/18 āĻŦā§āϝāĻžāĻā§āϝāĻž: km/h → m/s = × 5/18
Q10. When a train crosses a platform, distance covered is - ā§°ā§āϞ āĻāĻāĻžāĻ āĻĒā§āϞā§āĻāĻĢā§°ā§āĻŽ āĻĒāĻžā§° āĻā§°āĻŋāϞ⧠āĻĻā§ā§°āϤā§āĻŦ āĻš’āϞ -
A) Train length B) Platform length C) Train + Platform D) Speed × Time
Ans: C
Explanation: Train must cover its own length plus platform length. āĻŦā§āϝāĻžāĻā§āϝāĻž: ā§°ā§āϞ⧠āύāĻŋāĻā§° āĻā§°ā§ āĻĒā§āϞā§āĻāĻĢā§°ā§āĻŽā§° āĻĻā§ā§°ā§āĻā§āϝ āĻĻā§ā§ā§āĻāĻž āĻĒāĻžā§° āĻā§°ā§āĨ¤
Q11. A train crosses a pole in 10 s at 15 m/s. Length is - ā§Ģ āĻŽāĻŋ/āĻā§ āĻāϤāĻŋāϤ ā§§ā§Ļ āĻā§āĻā§āĻŖā§āĻĄāϤ āĻā§āĻāĻāĻŋ āĻĒāĻžā§° āĻā§°ā§āĨ¤ āĻĻā§ā§°ā§āĻā§āϝ āĻāĻŋāĻŽāĻžāύ ?
A) 100 B) 120 C) 150 D) 180 m
Ans: C) 150
Explanation : Length = Speed × Time = 15 × 10 āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻĻā§ā§°ā§āĻā§āϝ = āĻāϤāĻŋ × āϏāĻŽā§
Q12. Two trains of lengths 300 m and 500 m cross each other in 8 s. Relative speed is - ā§Šā§Ļā§Ļ āĻā§°ā§ ā§Ģā§Ļā§Ļ āĻŽāĻŋ ā§°ā§āϞ ā§Ž āĻā§āĻā§āĻŖā§āĻĄāϤ āĻ āϤāĻŋāĻā§ā§°āĻŽ āĻā§°ā§āĨ¤ āĻāϤāĻŋ āĻāĻŋāĻŽāĻžāύ ?
A) 80 B) 90 C) 100 D) 120 m/s
Ans: C) 100
Explanation: Total distance = 800 m → 800 ÷ 8 āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻŽā§āĻ āĻĻā§ā§°āϤā§āĻŦ = 800 āĻŽāĻŋ → 800 ÷ 8
Q13. Relative speed in same direction is found by - āĻāĻā§āĻāĻž āĻĻāĻŋāĻļāϤ āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ āĻā§āύā§āĻā§ āĻĒā§ā§ąāĻž āϝāĻžā§ ?
A) Add B) Multiply C) Divide D) Subtract
Ans: D
Explanation: Same direction → subtract speeds. āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻāĻā§āĻāĻž āĻĻāĻŋāĻļāϤ → āĻāϤāĻŋ āĻŦāĻŋā§ā§āĻāĨ¤
Q14. A train crosses a 200 m platform in 25 s. Train length is 300 m. Speed is - ⧍ā§Ļā§Ļ āĻŽāĻŋ āĻĒā§āϞā§āĻāĻĢā§°ā§āĻŽ ⧍ā§Ģ āĻā§āĻā§āĻŖā§āĻĄāϤ āĻĒāĻžā§° āĻā§°ā§āĨ¤ ā§°ā§āϞ⧰ āĻĻā§ā§°ā§āĻā§āϝ ā§Šā§Ļā§Ļ āĻŽāĻŋāĨ¤ āĻāϤāĻŋ ?
A) 16 B) 18 C) 20 D) 24 m/s
Ans: C) 20
Explanation : āĻŦā§āϝāĻžāĻā§āϝāĻž :
Total distance = 200 + 300 = 500 m : āĻŽā§āĻ āĻĻā§ā§°āϤā§āĻŦ = 500 āĻŽāĻŋ
Speed = 500 ÷ 25 : āĻāϤāĻŋ = 500 ÷ 25
Q15. A train passes a man in 6 s. Length = 120 m. Speed is - ā§Ŧ āĻā§āĻā§āĻŖā§āĻĄāϤ āĻŽāĻžāύā§āĻš āĻāĻāύ āĻĒāĻžā§° āĻā§°ā§āĨ¤ āĻĻā§ā§°ā§āĻā§āϝ ⧧⧍ā§Ļ āĻŽāĻŋāĨ¤ āĻāϤāĻŋ ?
A) 15 B) 18 C) 20 D) 25
Ans: C) 20
Explanation : āĻŦā§āϝāĻžāĻā§āϝāĻž
Speed = 120 ÷ 6 : āĻāϤāĻŋ = 120 ÷ 6
Q16. Which quantities are added in crossing problems ? āĻ āϤāĻŋāĻā§ā§°āĻŽ āϏāĻŽāϏā§āϝāĻžāϤ āĻā§āύāĻŦā§ā§° āϝā§āĻ āĻā§°āĻž āĻšā§ ?
A) Speed only B) Length only C) Length & speed D) Time
Ans: C
Explanation: Crossing → add lengths, opposite → add speeds. āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻ āϤāĻŋāĻā§ā§°āĻŽ → āĻĻā§ā§°ā§āĻā§āϝ āϝā§āĻ, āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļ → āĻāϤāĻŋ āϝā§āĻāĨ¤
Q17. If distance is doubled and speed same, time will - āĻĻā§ā§°āϤā§āĻŦ āĻĻā§āĻŦāĻŋāĻā§āĻŖ āĻšāϞ⧠āϏāĻŽā§ -
A) Half B) Double C) Same D) Four times
Ans: B
Explanation: Time ∝ Distance āĻŦā§āϝāĻžāĻā§āϝāĻž: āϏāĻŽā§ ∝ āĻĻā§ā§°āϤā§āĻŦ
Q18. Relative speed of 90 km/h and 54 km/h in opposite direction is - ⧝ā§Ļ āĻā§°ā§ ā§Ģā§Ē āĻāĻŋāĻŽāĻŋ/āĻāĻŖā§āĻāĻž āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļāϤāĨ¤ āĻāĻĒā§āĻā§āώāĻŋāĻ āĻāϤāĻŋ ?
A) 36 B) 90 C) 144 D) 150
Ans: C) 144
Explanation: Opposite direction → add speeds = 90 + 54 āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļāϤ → āĻāϤāĻŋ āϝā§āĻ
Q19. Conversion from km/h to m/s is - km/h ā§° āĻĒā§°āĻž m/s āϞ⧠-
A) ×18/5 B) ×5/18 C) ÷5 D) ÷18
Ans: B
Explanation: Standard conversion factor = ×5/18 āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻŽāĻžāύāĻ ā§°ā§āĻĒāĻžāύā§āϤ⧰ = ×5/18
Q20. Time is calculated using - āϏāĻŽā§ āĻā§āύā§āĻā§ āĻāĻŖāύāĻž āĻā§°āĻž āĻšā§ ?
A) Speed ÷ Distance B) Distance × Speed C) Distance ÷ Speed D) Speed − Distance
Ans: C
Explanation: Time = Distance ÷ Speed āĻŦā§āϝāĻžāĻā§āϝāĻž: āϏāĻŽā§ = āĻĻā§ā§°āϤā§āĻŦ ÷ āĻāϤāĻŋ
Final Exam Shortcut
Crossing → add lengths : āĻ
āϤāĻŋāĻā§ā§°āĻŽ → āĻĻā§ā§°ā§āĻā§āϝ āϝā§āĻ
Opposite direction → add speeds : āĻŦāĻŋāĻĒā§°ā§āϤ āĻĻāĻŋāĻļ → āĻāϤāĻŋ āϝā§āĻ
Same direction → subtract speeds : āĻāĻā§āĻāĻž āĻĻāĻŋāĻļ → āĻāϤāĻŋ āĻŦāĻŋā§ā§āĻ