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CAT 2022 Slot 3 — QA Question 5

EscalatorEasy

Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same direction. He takes 20 seconds to finish riding the escalator if he walks at twice his normal speed in the same direction. If Moody decides to stand still on the escalator, then the time, in seconds, needed to finish riding the escalator is

Answer & solution

Answer: 60

Solution

Easy

Classic escalator problem. Let Moody's normal speed be mm and the escalator ee (steps/sec), with NN steps total. Each scenario gives an equation N=(combined speed)×(time)N=(\text{combined speed})\times(\text{time}). Standing still means time =N/e=N/e.

1

Set up both walking scenarios (Moody walks with the escalator, so speeds add):

normal speed, 30 s: N=(m+e)⋅30...(1)double speed, 20 s: N=(2m+e)⋅20...(2)\begin{aligned} &\text{normal speed, 30 s:}\ &N&=(m+e)\cdot30\quad\text{...(1)}\\ &\text{double speed, 20 s:}\ &N&=(2m+e)\cdot20\quad\text{...(2)} \end{aligned}
2

Eliminate mm. Take 4×(1)−3×(2)4\times(1)-3\times(2) so the mm terms cancel (120m−120m120m-120m):

4N−3N=120(m+e)−60(2m+e)N=120m+120e−120m−60eN=60e\begin{aligned} &4N-3N=120(m+e)-60(2m+e)\\ &N=120m+120e-120m-60e\\ &N=60e \end{aligned}
3

Standing still, only the escalator moves him, so the time is:

Ne=60ee=60 seconds\frac{N}{e}=\frac{60e}{e}=60\text{ seconds}
Time standing still=60 seconds\text{Time standing still}=\mathbf{60}\text{ seconds}
CAT 2022 Slot 3 QA Q5: Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same directi — Solution | TheCATExam