Escalator -- interesting Puzzle.

An escalator is descending at constant speed. A walks down and takes 50 steps to reach the bottom. B runs down and takes 90 steps in the same time as A takes 10 steps.

How many steps are visible when the escalator is not operating?

-- James.

Really a very interesting puzzle. BUt I could not get it. I am stuck with the approach of finding the solution. Could someone give the answer and the method for solving the problem.

Actually the question was incomplete. I apologise for this.

Here is the corrected one.


An escalator is descending at constant speed. A walks down and takes 50 steps to reach the bottom. B runs down faster and takes 90 steps to reach the bottom.If B takes 90 steps in the same time as A takes 10 steps then how many steps are visible when the escalator is not operating?


-- James.

James can you suggest the answer and the approach for getting the solution. Really a very interesting puzzle. But I am struck with the approach for solving it.

There are 100 steps in the esclator.

Case : When A & B walk on unmoving surface
--------------------------------------------
10 Steps of A = 90 Steps of B .................... [Eq.1]
=> Speed of A : Speed of B :: 1 : 9
=> Time of A : Time of B :: 9 : 1

Case : When A & B walk on esclator
-----------------------------------
Steps taken by B : 90
TIme taken by B : T
Steps taken by A : 50
Time taken by A : 5T [From [Eq.1]]

Time of A : Time of B :: 5 : 1

let e be the number of steps moved by the esclator when A takes 1 step

So

(1+ e) / (9 + e) = 1/5
=> e = 1 step for every step of A

A takes 50 steps
=> Esclator has moved by 50 steps
Total number of steps = 50 + 50 = 100

Cheers
hsundar...

I think some data are missing,Here is not clear that B reaches the bottom or not..