Question

A person walks up a stalled escalator in 80 sec. When standing on the same escalator, now moving, he is carried up in 20 s. The time taken by him to walk up the moving escalator is

• A. \( 4 \) s
• B. \( 8 \) s
• C. \( 12 \) s
• D. \( 16 \) s

Hint:

Consider the speeds of the person and the escalator as rates of covering the length of the escalator. When moving together in the same direction, their speeds add up. Use the relationship time = distance/speed.

Answer 1

The Correct Option is D

Let the length of the escalator be \( L \).
The speed of the person walking on the stalled escalator is \( v_p = \frac{L}{80} \) m/s.
The speed of the moving escalator is \( v_e = \frac{L}{20} \) m/s.
When the person walks up the moving escalator, his effective speed is the sum of his walking speed and the escalator's speed, since both are in the same direction.
Effective speed \( v_{eff} = v_p + v_e = \frac{L}{80} + \frac{L}{20} \).
To add these fractions, find a common denominator, which is 80: \( v_{eff} = \frac{L}{80} + \frac{4L}{80} = \frac{5L}{80} = \frac{L}{16} \) m/s.
The time taken by the person to walk up the moving escalator is the length of the escalator divided by the effective speed: Time \( t = \frac{L}{v_{eff}} = \frac{L}{\frac{L}{16}} = L \cdot \frac{16}{L} = 16 \) s.