Question:
Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom's steps. Shyama gets to the top after taking 25 steps, while Vyom takes 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?
Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom's steps. Shyama gets to the top after taking 25 steps, while Vyom takes 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?
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Relative speed of person plus escalator determines total steps covered; set equal for both travellers.
Updated On: Aug 4, 2025
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The Correct Option is B
Solution and Explanation
Let escalator speed = $e$ steps/sec, Shyama speed = $s$, Vyom speed = $\frac23 s$.
Time for Shyama: $25/s$, distance covered = $(s+e)(25/s) = N$ steps (total length). Vyom: time $= 20/(\frac23 s) = 30/s$, distance = $(\frac23 s + e)(30/s) = N$. Equating: $(s+e)(25/s) = (\frac23 s + e)(30/s) \Rightarrow 25 + \frac{25e}{s} = 20 + \frac{30e}{s} \Rightarrow 5 = \frac{5e}{s} \Rightarrow e = s$. Then $N = 25 + 25 = 50$.
Time for Shyama: $25/s$, distance covered = $(s+e)(25/s) = N$ steps (total length). Vyom: time $= 20/(\frac23 s) = 30/s$, distance = $(\frac23 s + e)(30/s) = N$. Equating: $(s+e)(25/s) = (\frac23 s + e)(30/s) \Rightarrow 25 + \frac{25e}{s} = 20 + \frac{30e}{s} \Rightarrow 5 = \frac{5e}{s} \Rightarrow e = s$. Then $N = 25 + 25 = 50$.
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