CAT 2022 Slot 3 · Quant · Question 12

Let the escalator's length be L, Moody's normal walking speed w and the escalator's speed e.

Walking normally, the two speeds add, so L=30\left( w+e \right).

Walking at double speed, L=20\left( 2w+e \right).

The length is the same either way, so 30\left( w+e \right)=20\left( 2w+e \right).

Expanding, 30w+30e=40w+20e.

Collecting terms, 10e=10w, so e=w — the escalator moves at exactly his walking pace.

Substituting back, L=30\left( w+w \right)=60w.

Standing still, only the escalator moves him, at speed e=w.

The time is \frac{L}{e}=\frac{60w}{w}=60 seconds.

Shortcut. Doubling his walking speed cuts the time from 30 to 20 seconds, so the walking contribution and the escalator contribution must be equal. Standing still therefore removes exactly half the combined speed, doubling the 30 seconds to 60.