A train 150 m long runs at 72 km/h. Match List-I with List-II, where List-II gives the time in seconds.
| List-I | List-II | ||
| (A) | Time to pass a pole | (I) | 7.5 |
| (B) | Time to cross a platform 250 m long | (II) | 10 |
| (C) | Time to cross a train 100 m long running at 18 km/h in the opposite direction | (III) | 20 |
| (D) | Time to overtake a train 100 m long running at 54 km/h in the same direction | (IV) | 50 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
To change km/h into m/s, multiply by \(\frac{5}{18}\):
\(72 \times \frac{5}{18} = 20\) m/s, \(18 \times \frac{5}{18} = 5\) m/s, \(54 \times \frac{5}{18} = 15\) m/s
In each case, time = total length to be covered ÷ relative speed.
(A) Time to pass a pole
This matches (I) 7.5. The train covers its own length.
\(150 \div 20 = 7.5\)
(B) Time to cross a platform 250 m long
This matches (III) 20. The train covers its length plus the platform.
\((150 + 250) \div 20 = 20\)
(C) Time to cross a train 100 m long running at 18 km/h in the opposite direction
This matches (II) 10. In opposite directions the speeds add.
\((150 + 100) \div (20 + 5) = 10\)
(D) Time to overtake a train 100 m long running at 54 km/h in the same direction
This matches (IV) 50. In the same direction the speeds subtract.
\((150 + 100) \div (20 - 15) = 50\)