Train Problems – 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 - āĻĻ⧁āϟāĻž ⧰⧇āϞ⧰ āĻĻā§ˆā§°ā§āĻ˜ā§āϝ 400 āĻŽāĻŋ āφ⧰⧁ 600 āĻŽāĻŋāĨ¤ āĻ—āϤāĻŋ 110 āĻŽāĻŋ/āϛ⧇ āφ⧰⧁ 140 āĻŽāĻŋ/āϛ⧇āĨ¤
āĻāϕ⧇-āφāύāĻ• āĻ…āϤāĻŋāĻ•ā§ā§°āĻŽ āϕ⧰āĻŋāĻŦāϞ⧈ āĻ˛ā§‹ā§ąāĻž āϏāĻŽā§Ÿ āĻ•āĻŋāĻŽāĻžāύ ?


A) 2 s  B) 4 s  C) 5 s  D) 10 s


Ans / āωāĻ¤ā§āϤ⧰: B) 4 s


Explanation : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Total distance = 400 + 600 = 1000 m : āĻŽā§āĻ  āĻĻā§‚ā§°āĻ¤ā§āĻŦ = 400 + 600 = 1000 āĻŽāĻŋ
Relative speed = 110 + 140 = 250 m/s : āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ = 110 + 140 = 250 āĻŽāĻŋ/āϛ⧇
Time = 1000 ÷ 250 = 4 s : āϏāĻŽā§Ÿ = 1000 ÷ 250 = 4 āϛ⧇āϕ⧇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§ˆā§°ā§āĻ˜ā§āϝ⧰ (L) āĻĻ⧁āϟāĻž ⧰⧇āϞ āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļāϤ 60 āĻŽāĻŋ/āϛ⧇ āφ⧰⧁ 40 āĻŽāĻŋ/āϛ⧇ āĻ—āϤāĻŋāϤ āϚāϞ⧇āĨ¤
āϏāĻŽā§Ÿ āĻ•āĻŋāĻŽāĻžāύ ?


A) L/100  B) 2L/100  C) L/20  D) 2L/20


Ans / āωāĻ¤ā§āϤ⧰: B) 2L/100


Explanation : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž :
Total distance = L + L = 2L : āĻŽā§āĻ  āĻĻā§‚ā§°āĻ¤ā§āĻŦ = L + L = 2L
Relative speed = 60 + 40 = 100 m/s : āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ = 100 āĻŽāĻŋ/āϛ⧇
Time = 2L ÷ 100 : āϏāĻŽā§Ÿ = 2L ÷ 100


 Q3. When two trains move in opposite directions, relative speed is - āĻĻ⧁āϟāĻž ⧰⧇āϞ āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļāϤ āϚāϞāĻŋāϞ⧇ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ āĻš’āϞ -


A) Difference of speeds / āĻ—āϤāĻŋā§° āĻĒāĻžā§°ā§āĻĨāĻ•ā§āϝ
B) Sum of speeds / āĻ—āϤāĻŋā§° āϝ⧋āĻ—āĻĢāϞ
C) Average speed / āĻ—ā§œ āĻ—āϤāĻŋ
D) Product of speeds / āĻ—āϤāĻŋā§° āϗ⧁āĻŖāĻĢāϞ


Ans / āωāĻ¤ā§āϤ⧰: B) Sum of speeds / āĻ—āϤāĻŋā§° āϝ⧋āĻ—āĻĢāϞ


Explanation: Opposite direction → speeds are added. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļāϤ āϚāϞāĻŋāϞ⧇ → āĻ—āϤāĻŋ āϝ⧋āĻ— āϕ⧰āĻž āĻšā§ŸāĨ¤


Q4. Two trains cross each other in 5 seconds. Their combined length is 1250 m.
Relative speed is - āĻĻ⧁āϟāĻž ⧰⧇āϞ 5 āϛ⧇āϕ⧇āĻŖā§āĻĄāϤ āĻāϕ⧇-āφāύāĻ• āĻ…āϤāĻŋāĻ•ā§ā§°āĻŽ āϕ⧰⧇āĨ¤ āĻŽā§āĻ  āĻĻā§ˆā§°ā§āĻ˜ā§āϝ 1250 āĻŽāĻŋāĨ¤
āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ āĻ•āĻŋāĻŽāĻžāύ ?


A) 200 m/s  B) 250 m/s  C) 300 m/s  D) 350 m/s


Ans / āωāĻ¤ā§āϤ⧰: B) 250 m/s


Explanation: Relative speed = Distance ÷ Time = 1250 ÷ 5 = 250 m/s


āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ = āĻĻā§‚ā§°āĻ¤ā§āĻŦ ÷ āϏāĻŽā§Ÿ = 1250 ÷ 5 = 250 āĻŽāĻŋ/āϛ⧇


Q5. Two trains move in the same direction at 140 m/s and 110 m/s.
Relative speed is - āĻĻ⧁āϟāĻž ⧰⧇āϞ āĻāϕ⧇āϟāĻž āĻĻāĻŋāĻļāϤ 140 āĻŽāĻŋ/āϛ⧇ āφ⧰⧁ 110 āĻŽāĻŋ/āϛ⧇ āĻ—āϤāĻŋāϤ āϚāϞ⧇āĨ¤
āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ āĻ•āĻŋāĻŽāĻžāύ ?


A) 250 m/s  B) 30 m/s  C) 110 m/s  D) 140 m/s


Ans / āωāĻ¤ā§āϤ⧰: B) 30 m/s


Explanation : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Same direction → subtract speeds : āĻāϕ⧇āϟāĻž āĻĻāĻŋāĻļāϤ āϚāϞāĻŋāϞ⧇ → āĻ—āϤāĻŋ āĻŦāĻŋā§Ÿā§‹āĻ—
140 − 110 = 30 m/s : 140 − 110 = 30 āĻŽāĻŋ/āϛ⧇


Very Important Shortcuts


• Crossing each other → Add lengths : āĻ…āϤāĻŋāĻ•ā§ā§°āĻŽ → āĻĻā§ˆā§°ā§āĻ˜ā§āϝ āϝ⧋āĻ—
• Opposite direction → Add speeds : āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļ → āĻ—āϤāĻŋ āϝ⧋āĻ—
• Same direction → Subtract speeds : āĻāϕ⧇āϟāĻž āĻĻāĻŋāĻļ → āĻ—āϤāĻŋ āĻŦāĻŋā§Ÿā§‹āĻ—
• Time = Distance ÷ Relative speed : āϏāĻŽā§Ÿ = āĻĻā§‚ā§°āĻ¤ā§āĻŦ ÷ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ


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 - āĻĻ⧁āϟāĻž ⧰⧇āϞ⧰ āĻĻā§ˆā§°ā§āĻ˜ā§āϝ 400 āĻŽāĻŋ āφ⧰⧁ 600 āĻŽāĻŋāĨ¤ āĻ—āϤāĻŋ 110 āĻŽāĻŋ/āϛ⧇ āφ⧰⧁ 140 āĻŽāĻŋ/āϛ⧇āĨ¤ āϏāĻŽā§Ÿ āĻ•āĻŋāĻŽāĻžāύ ?


A) 2 s  B) 4 s  C) 5 s  D) 10 s


Ans: B) 4 s


Explanation : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž
Total distance = 400 + 600 = 1000 m : āĻŽā§āĻ  āĻĻā§‚ā§°āĻ¤ā§āĻŦ = 1000 āĻŽāĻŋ
Relative speed = 110 + 140 = 250 m/s : āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ = 250 āĻŽāĻŋ/āϛ⧇
Time = 1000 ÷ 250 = 4s : āϏāĻŽā§Ÿ = 4 āϛ⧇āϕ⧇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§ˆā§°ā§āĻ˜ā§āϝ⧰ āĻĻ⧁āϟāĻž ⧰⧇āϞ āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļāϤ 60 āφ⧰⧁ 40 āĻŽāĻŋ/āϛ⧇ āĻ—āϤāĻŋāϤ āϚāϞ⧇āĨ¤ āϏāĻŽā§Ÿ āĻ•āĻŋāĻŽāĻžāύ ?


A) L/100  B) 2L/100  C) L/20  D) 2L/20


Ans: B) 2L/100


Explanation : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž
Total distance = L + L = 2L : āĻŽā§āĻ  āĻĻā§‚ā§°āĻ¤ā§āĻŦ = 2L
Relative speed = 60 + 40 = 100 m/s : āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ = 100 āĻŽāĻŋ/āϛ⧇


Q3. When two trains move in opposite directions, relative speed is - āĻĻ⧁āϟāĻž ⧰⧇āϞ āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļāϤ āϚāϞāĻŋāϞ⧇ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ āĻš’āϞ -


A) Difference  B) Sum  C) Average  D) Product


Ans: B) Sum


Explanation: Opposite direction → speeds are added. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļāϤ āϚāϞāĻŋāϞ⧇ → āĻ—āϤāĻŋ āϝ⧋āĻ— āϕ⧰āĻž āĻšā§ŸāĨ¤


Q4. Two trains cross each other in 5 s. Combined length is 1250 m. Relative speed is - ā§Ģ āϛ⧇āϕ⧇āĻŖā§āĻĄāϤ āĻ…āϤāĻŋāĻ•ā§ā§°āĻŽ āϕ⧰⧇, āĻŽā§āĻ  āĻĻā§ˆā§°ā§āĻ˜ā§āϝ 1250 āĻŽāĻŋāĨ¤ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ āĻ•āĻŋāĻŽāĻžāύ ?


A) 200  B) 250  C) 300  D) 350 m/s


Ans: B) 250 m/s


Explanation : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Speed = Distance ÷ Time = 1250 ÷ 5 = 250 m/s : āĻ—āϤāĻŋ = āĻĻā§‚ā§°āĻ¤ā§āĻŦ ÷ āϏāĻŽā§Ÿ = 1250 ÷ 5 = 250 āĻŽāĻŋ/āϛ⧇


Q5. Two trains move in the same direction at 140 m/s and 110 m/s. Relative speed is - āĻāϕ⧇āϟāĻž āĻĻāĻŋāĻļāϤ 140 āφ⧰⧁ 110 āĻŽāĻŋ/āϛ⧇āĨ¤ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ āĻ•āĻŋāĻŽāĻžāύ ?


A) 250  B) 30  C) 110  D) 140 m/s


Ans: B) 30 m/s


Explanation: Same direction → subtract speeds = 140 − 110. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻāϕ⧇āϟāĻž āĻĻāĻŋāĻļāϤ → āĻ—āϤāĻŋ āĻŦāĻŋā§Ÿā§‹āĻ— = 140 − 110


Q6. A train 200 m long crosses a pole in 20 s. Speed is - ⧍ā§Ļā§Ļ āĻŽāĻŋ āĻĻā§ˆā§°ā§āĻ˜ā§āϝ⧰ ⧰⧇āϞ⧇ ⧍ā§Ļ āϛ⧇āϕ⧇āĻŖā§āĻĄāϤ āϖ⧁āρāϟāĻŋ āĻĒāĻžā§° āϕ⧰⧇āĨ¤ āĻ—āϤāĻŋ āĻ•āĻŋāĻŽāĻžāύ ?


A) 5  B) 8  C) 10  D) 12 m/s


Ans: C) 10 m/s


Explanation : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:
Distance = length of train = 200 m : āĻĻā§‚ā§°āĻ¤ā§āĻŦ = ⧰⧇āϞ⧰ āĻĻā§ˆā§°ā§āĻ˜ā§āϝ = 200 āĻŽāĻŋ
Speed = 200 ÷ 20 = 10 m/s : āĻ—āϤāĻŋ = 10 āĻŽāĻŋ/āϛ⧇


Q7. Correct formula for time taken by two trains to cross is - āĻĻ⧁āϟāĻž ⧰⧇āϞ āĻ…āϤāĻŋāĻ•ā§ā§°āĻŽ āϕ⧰āĻžā§° āϏāĻ āĻŋāĻ• āϏ⧂āĻ¤ā§ā§° -


A) (S₁+S₂)/(L₁+L₂)  B) (L₁+L₂)/(S₁+S₂)  C) (L₁−L₂)/(S₁−S₂)  D) (S₁−S₂)/(L₁+L₂)


Ans: B


Explanation : Time = Total distance ÷ Relative speed : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āϏāĻŽā§Ÿ = āĻŽā§āĻ  āĻĻā§‚ā§°āĻ¤ā§āĻŦ ÷ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ


Q8. A train of length 150 m crosses a man in 15 s. Speed is - ā§§ā§Ģā§Ļ āĻŽāĻŋ āĻĻā§ˆā§°ā§āĻ˜ā§āϝ⧰ ⧰⧇āϞ⧇ ā§§ā§Ģ āϛ⧇āϕ⧇āĻŖā§āĻĄāϤ āĻŽāĻžāύ⧁āĻš āĻāϜāύ āĻĒāĻžā§° āϕ⧰⧇āĨ¤ āĻ—āϤāĻŋ āĻ•āĻŋāĻŽāĻžāύ ?


A) 5  B) 8  C) 10  D) 12 m/s


Ans: C) 10


Explanation: Speed = 150 ÷ 15 āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻ—āϤāĻŋ = 150 ÷ 15


Q9. Speed of 72 km/h in m/s is - 72 āĻ•āĻŋāĻŽāĻŋ/āϘāĻŖā§āϟāĻž = ?


A) 18  B) 20  C) 22  D) 25


Ans: B) 20


Explanation: km/h → m/s = × 5/18 āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: km/h → m/s = × 5/18


Q10. When a train crosses a platform, distance covered is - ⧰⧇āϞ āĻāϟāĻžāχ āĻĒā§āϞ⧇āϟāĻĢā§°ā§āĻŽ āĻĒāĻžā§° āϕ⧰āĻŋāϞ⧇ āĻĻā§‚ā§°āĻ¤ā§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§āϝāĻžāĻ–ā§āϝāĻž: ⧰⧇āϞ⧇ āύāĻŋāϜ⧰ āφ⧰⧁ āĻĒā§āϞ⧇āϟāĻĢā§°ā§āĻŽā§° āĻĻā§ˆā§°ā§āĻ˜ā§āϝ āĻĻā§ā§Ÿā§‹āϟāĻž āĻĒāĻžā§° āϕ⧰⧇āĨ¤


Q11. A train crosses a pole in 10 s at 15 m/s. Length is - ā§Ģ āĻŽāĻŋ/āϛ⧇ āĻ—āϤāĻŋāϤ ā§§ā§Ļ āϛ⧇āϕ⧇āĻŖā§āĻĄāϤ āϖ⧁āρāϟāĻŋ āĻĒāĻžā§° āϕ⧰⧇āĨ¤ āĻĻā§ˆā§°ā§āĻ˜ā§āϝ āĻ•āĻŋāĻŽāĻžāύ ?


A) 100  B) 120  C) 150  D) 180 m


Ans: C) 150


Explanation : Length = Speed × Time = 15 × 10 āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻĻā§ˆā§°ā§āĻ˜ā§āϝ = āĻ—āϤāĻŋ × āϏāĻŽā§Ÿ


Q12. Two trains of lengths 300 m and 500 m cross each other in 8 s. Relative speed is - ā§Šā§Ļā§Ļ āφ⧰⧁ ā§Ģā§Ļā§Ļ āĻŽāĻŋ ⧰⧇āϞ ā§Ž āϛ⧇āϕ⧇āĻŖā§āĻĄāϤ āĻ…āϤāĻŋāĻ•ā§ā§°āĻŽ āϕ⧰⧇āĨ¤ āĻ—āϤāĻŋ āĻ•āĻŋāĻŽāĻžāύ ?


A) 80  B) 90  C) 100  D) 120 m/s


Ans: C) 100


Explanation: Total distance = 800 m → 800 ÷ 8 āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻŽā§āĻ  āĻĻā§‚ā§°āĻ¤ā§āĻŦ = 800 āĻŽāĻŋ → 800 ÷ 8


Q13. Relative speed in same direction is found by - āĻāϕ⧇āϟāĻž āĻĻāĻŋāĻļāϤ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ āϕ⧇āύ⧇āĻ•ā§ˆ āĻĒā§‹ā§ąāĻž āϝāĻžā§Ÿ ?


A) Add  B) Multiply  C) Divide  D) Subtract


Ans: D


Explanation: Same direction → subtract speeds. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻāϕ⧇āϟāĻž āĻĻāĻŋāĻļāϤ → āĻ—āϤāĻŋ āĻŦāĻŋā§Ÿā§‹āĻ—āĨ¤


Q14. A train crosses a 200 m platform in 25 s. Train length is 300 m. Speed is - ⧍ā§Ļā§Ļ āĻŽāĻŋ āĻĒā§āϞ⧇āϟāĻĢā§°ā§āĻŽ ⧍ā§Ģ āϛ⧇āϕ⧇āĻŖā§āĻĄāϤ āĻĒāĻžā§° āϕ⧰⧇āĨ¤ ⧰⧇āϞ⧰ āĻĻā§ˆā§°ā§āĻ˜ā§āϝ ā§Šā§Ļā§Ļ āĻŽāĻŋāĨ¤ āĻ—āϤāĻŋ ?


A) 16  B) 18  C) 20  D) 24 m/s


Ans: C) 20


Explanation : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž :


Total distance = 200 + 300 = 500 m : āĻŽā§āĻ  āĻĻā§‚ā§°āĻ¤ā§āĻŦ = 500 āĻŽāĻŋ
Speed = 500 ÷ 25 : āĻ—āϤāĻŋ = 500 ÷ 25


Q15. A train passes a man in 6 s. Length = 120 m. Speed is - ā§Ŧ āϛ⧇āϕ⧇āĻŖā§āĻĄāϤ āĻŽāĻžāύ⧁āĻš āĻāϜāύ āĻĒāĻžā§° āϕ⧰⧇āĨ¤ āĻĻā§ˆā§°ā§āĻ˜ā§āϝ ⧧⧍ā§Ļ āĻŽāĻŋāĨ¤ āĻ—āϤāĻŋ ?


A) 15  B) 18  C) 20  D) 25


Ans: C) 20


Explanation : āĻŦā§āϝāĻžāĻ–ā§āϝāĻž
Speed = 120 ÷ 6 : āĻ—āϤāĻŋ = 120 ÷ 6


Q16. Which quantities are added in crossing problems ? āĻ…āϤāĻŋāĻ•ā§ā§°āĻŽ āϏāĻŽāĻ¸ā§āϝāĻžāϤ āϕ⧋āύāĻŦā§‹ā§° āϝ⧋āĻ— āϕ⧰āĻž āĻšā§Ÿ ?


A) Speed only  B) Length only  C) Length & speed  D) Time


Ans: C


Explanation: Crossing → add lengths, opposite → add speeds. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻ…āϤāĻŋāĻ•ā§ā§°āĻŽ → āĻĻā§ˆā§°ā§āĻ˜ā§āϝ āϝ⧋āĻ—, āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļ → āĻ—āϤāĻŋ āϝ⧋āĻ—āĨ¤


Q17. If distance is doubled and speed same, time will - āĻĻā§‚ā§°āĻ¤ā§āĻŦ āĻĻā§āĻŦāĻŋāϗ⧁āĻŖ āĻšāϞ⧇ āϏāĻŽā§Ÿ -


A) Half  B) Double  C) Same  D) Four times


Ans: B


Explanation: Time ∝ Distance āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āϏāĻŽā§Ÿ ∝ āĻĻā§‚ā§°āĻ¤ā§āĻŦ


Q18. Relative speed of 90 km/h and 54 km/h in opposite direction is - ⧝ā§Ļ āφ⧰⧁ ā§Ģā§Ē āĻ•āĻŋāĻŽāĻŋ/āϘāĻŖā§āϟāĻž āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļāϤāĨ¤ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āĻ—āϤāĻŋ ?


A) 36  B) 90  C) 144  D) 150


Ans: C) 144


Explanation: Opposite direction → add speeds = 90 + 54 āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļāϤ → āĻ—āϤāĻŋ āϝ⧋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§āϝāĻžāĻ–ā§āϝāĻž: āĻŽāĻžāύāĻ• ā§°ā§‚āĻĒāĻžāĻ¨ā§āϤ⧰ = ×5/18


Q20. Time is calculated using - āϏāĻŽā§Ÿ āϕ⧇āύ⧇āĻ•ā§ˆ āĻ—āĻŖāύāĻž āϕ⧰āĻž āĻšā§Ÿ ?


A) Speed ÷ Distance  B) Distance × Speed  C) Distance ÷ Speed  D) Speed − Distance


Ans: C


Explanation: Time = Distance ÷ Speed āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āϏāĻŽā§Ÿ = āĻĻā§‚ā§°āĻ¤ā§āĻŦ ÷ āĻ—āϤāĻŋ


 Final Exam Shortcut


Crossing → add lengths : āĻ…āϤāĻŋāĻ•ā§ā§°āĻŽ → āĻĻā§ˆā§°ā§āĻ˜ā§āϝ āϝ⧋āĻ—
Opposite direction → add speeds : āĻŦāĻŋāĻĒā§°ā§€āϤ āĻĻāĻŋāĻļ → āĻ—āϤāĻŋ āϝ⧋āĻ—
Same direction → subtract speeds : āĻāϕ⧇āϟāĻž āĻĻāĻŋāĻļ → āĻ—āϤāĻŋ āĻŦāĻŋā§Ÿā§‹āĻ—