You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator.
How many steps do you need if the escalator stands still?
For something somewhat related please see "Walking down an escalator"
Let the speed of the escalator be P steps per second, and let the total number of steps in the escalator be T.
Also,let the respective speed of the individual moving upward and downward (in steps per second), together with the escalator, be u and d.
Then, by the given conditions:
(i) P = u-1 = 5-d
(ii) (50/1)*u = (125/5)*d = T
From (i) and (ii), we have:
50(P+1) = 25(5-P)
or, 50P + 50 = 125- 25P
or, 75P = 75
or, P = 1, so that from (i), we have: u = 2
Substituting this value in (ii), we have: T = 50*u = 100
Consequently, the total number steps that will be covered in the upward and the downward journey, if the escalator stands still will be equal to 2*T = 200
Edited on September 17, 2008, 4:08 pm