Answer / singaravelan

Let the person be F
Let F takes 't' sec to take 50 steps down.
Then for reaching upstairs F takes'(t/5)' sec to take 50
steps up. So for taking 125 steps,
50step----- t/5sec
125step----- 0.5tsec
Let 'v' be the velocity with which the escalator moves.
So the no.of steps (distance) covered by F in moving
up and down will be the same.
Hence,
50+vt=125+0.5vt
0.5vt=75
vt=150
Therefore no.of steps in escalator will be
50+150=125+0.5*150=200
Thus there were 200 steps in the escalator....

Answer / krishna

112 and a half steps. Round it off to say 113 steps.

Answer / sabarish

169 approx.

let d be the steps in the esculator.

distance covered by man upwards = d+50
and downwards = d - 125

let s1 and s2 be the speed for upwards and downwards.

s1=s2 x 5

(d+50) / t = s1 ..........1
( d-125 ) / t = s2..........2

d= s1 x t - 50 d = s2 x t + 125

(s1 x t) - 50 = (s1 x t)/5 +125 solving (s1 x t) = 219.

sub in 1 we get d = 169.

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