Question

At a metro station, a girl walks up a stationary escalator in 20 sec. If she remains stationary on the escalator, then the escalator take her up in 30 sec. The time taken by her to walk up on the moving escalator will be

• A. 25 sec
• B. 60 sec
• C. 12 sec
• D. 10 sec

Answer 1

The Correct Option is C

Let the length of the escalator be \( L \), the speed of the girl be \( v_g \), and the speed of the escalator be \( v_e \).
When the girl walks up the stationary escalator, she takes 20 sec. So, the speed of the girl is: \[ v_g = \frac{L}{20} \]
When the girl remains stationary on the escalator, the escalator takes her up in 30 sec. So, the speed of the escalator is: \[ v_e = \frac{L}{30} \] Now, when the girl walks on the moving escalator, the relative speed is \( v_g + v_e \). The time taken for her to reach the top is: \[ \text{Time} = \frac{L}{v_g + v_e} = \frac{L}{\frac{L}{20} + \frac{L}{30}} = \frac{L}{\frac{5L}{60}} = 12 \, \text{sec} \]
Thus, the time taken by her to walk up on the moving escalator is 12 sec.

The correct answer is (C) : 12 sec.

Answer 2

Let the length of the escalator be \( L \). 1. The girl's speed when she walks on the stationary escalator is: \[ \text{Speed of the girl} = \frac{L}{20} \, \text{m/s} \] 2. The speed of the escalator when she remains stationary on it is: \[ \text{Speed of the escalator} = \frac{L}{30} \, \text{m/s} \] Now, when she walks on the moving escalator, her effective speed is the sum of the speed of the girl and the speed of the escalator: \[ \text{Effective speed} = \text{Speed of the girl} + \text{Speed of the escalator} = \frac{L}{20} + \frac{L}{30} \] To simplify the above expression: \[ \text{Effective speed} = \frac{3L}{60} + \frac{2L}{60} = \frac{5L}{60} = \frac{L}{12} \, \text{m/s} \] The time taken to walk up the moving escalator is: \[ \text{Time taken} = \frac{L}{\text{Effective speed}} = \frac{L}{\frac{L}{12}} = 12 \, \text{sec} \] Thus, the time taken by her to walk up the moving escalator is 12 sec.