Escalator Math

When Ted walked down the down-moving escalator, he reached the bottom after taking 50 steps. As an experiment, he then ran up the same escalator, one step at a time, reaching the top after taking 125 steps. Assuming that Ted went up five times as fast as he went down (that is, took five steps to every one step before), and that he made each trip at a constant speed, how many steps would be visible if the escalator stopped running?

Too freaking many. That's MY answer.

Ditto. That's a good one for Denis.

I'm going out on a limb here..

D = # of steps showing when the escalator is stopped
S = Ted's speed

The ratio of the number of steps that passes a given point
on the escalator when Ted is going up and down is
proportional to Ted's speed going up and down. So,




5D-250 = D+125
4D = 375
D = 93.75 steps

I get 100...

d = distance (or number of visible steps)
e = escalator speed (in steps per second)

Going down (assume a step takes 1 second):
took 50 steps, so 50 seconds. So:
d = 50 + 50e [1]

Going up (5 steps take 1 second (5 times faster)):
took 125 steps @ 5 per second, so 25 seconds. So:
d = 125 - 25e [2]

d = 50 + 50e [1]
2d=250 - 50e [2] * 2
================
3d = 300
d = 100

That gave me a headache...don't ask me to prove it!!

Denis is correct...

One hundred steps.

Let n be the number of steps visible when the escalator is not moving, and let a unit of time be the time it takes Ted to walk down one step. If he walks down the down-moving escalator in 50 steps, then n - 50 steps have gone out of sight in 50 units of time. It takes him 125 steps to run up the same escalator, taking five steps to every one step before. In this trip, 125 - n steps have gone out of sight in 125/5, or 25 units of time. Since the escalator runs at a constant speed, we have the following linear equation that readily yields a value for n of 100 steps:

n-50 125 - n
____ = _______
50 25