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### Stairs on an Escalator

```Date: 05/21/2001 at 04:15:17
From: Dario
Subject: Problem Solving

Hello.

I have a problem that I cannot get my head around. It goes like this.

A woman is walking down a downward-moving escalator and steps down 10
steps to reach the bottom. Just as she reaches the bottom of the
escalator, a sale commences on the floor above. She runs back up the
downward moving escalator at a speed five times that which she walked
down. She covers 25 steps in reaching the top. How many steps are
visible on the escalator when it is switched off?

Could you please explain to me how this is done? I find it puzzling!

-Dario
```

```
Date: 05/22/2001 at 13:39:39
From: Doctor Greenie
Subject: Re: Problem Solving

Hello, Dario -

Thanks for sending your question to us here at Dr. Math.

I had some fun (and a few mild headaches) searching for the solution
to your problem. I agree - it is puzzling!

If you can see through the wording to set up your equations correctly,
there are probably easier ways to solve this problem than the way I
did it, but here is my solution method.

Let x be the number of steps showing on the escalator when it is
stopped. Also, suppose that, when the escalator is running, the steps
move at n steps per second.

On the trip down the escalator, the woman steps down 10 of the steps,
so the number of steps the escalator moves in that time is x-10. With
the escalator running at n steps per second, the time required for the
woman to walk down the moving escalator is (x-10)/n seconds.

When the sale starts and she turns around and runs up the downward-
moving escalator, she moves 5 times as fast as on the way down.  But
she only covers half of 5 times as many steps; so the time required
to reach the top is one-half of the time required for the trip down,
and the trip up the escalator takes (x-10)/(2n) seconds.

With the escalator moving at n steps per second, it will move (x-10)/2
steps in the (x-10)/(2n) seconds it takes the woman to run up. So the
number of steps the woman must run up is the number of steps showing
when the escalator is still, plus the number of steps the escalator
moves while the woman is running up; and we are told that in running
up the escalator she goes up 25 steps. So we have an equation which we
can solve for x to get the answer to the problem:

x + (x-10)/2 = 25

I hope this helps. Write back if you have any further questions on
this problem.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/23/2001 at 15:05:00
From: Doctor Greenie
Subject: Re: Problem Solving

Hello again, Dario -

Here is another explanation of this problem that is perhaps a bit
easier to understand than the first one I gave.

Let x be the number of stairs visible on the escalator when it is
stopped. Let r be the rate (number of stairs per second) at which the
escalator moves when it is running.

The woman walks down the escalator at one rate and runs back up the
escalator at a rate 5 times as fast; as we showed before, the time she
takes walking down is twice as much as the time she takes running back up.
Let t be the time (seconds) she takes to run up; then 2t is the time she
takes to walk down.

In walking down the escalator, the number of stairs she walks down is
equal to the number of stairs on the stopped escalator, minus the
number of stairs that the escalator moves in the time 2t; we are told
she walks down 10 steps. So we have:

x - (r)(2t) = 10   ..............................[1]

In running up the escalator, the number of stairs she runs up is equal
to the number of stairs on the stopped escalator, plus the number of
stairs that the escalator moves in time t; we are told that she runs
up 25 stairs. So we have:

x + (r)(t) = 25   ...............................[2]

We can solve equations [1] and [2] simultaneously by getting the "rt"
terms to drop out:

x - 2rt = 10
2x + 2rt = 50
--------------
3x       = 60
x       = 20

Thanks again for sending us your unusual problem.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/08/2003 at 05:59:49
From: Doctor Anthony
Subject: Re: Problem Solving

Here is another way to solve the problem.  Let

s = the number of steps on the stationary escalator.

u = the speed of escalator

v = the speed of woman walking down, relative to escalator

The time while the woman moves down 10 steps equals the time that the
escalator moves s-10 steps.  Also the time that woman moves 25 steps
equals the time that the escalator moves 25-s steps.

s-10    10            25-s    25      5
----- = ----    and    ---- = ----  = ---
u       v              u     5v      v

Dividing the equations we get

s-10    10
---- =  ---  = 2
25-s     5

s - 10 = 50 - 2s

3s = 60

s = 20

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```Date: 06/09/2003 at 10:55:37
From: Doctor Ian
Subject: Re: Problem Solving

For those who prefer the straightforward, plodding approach, suppose N is
the number of steps visible when the escalator isn't moving.  On the way
down, the woman moves N steps in t seconds, and the escalator moves
(N-10) steps in the same time:

distance = rate * time

= [10/t + (N-10)/t]*t

On the way up, the woman moves at 5 times her former rate,
while the escalator continues to move as before.  This time,
they're working against each other:

distance = rate * time

= [50/t - (N-10)/t]*T

So have two equations, and three unknowns (N, t, and T).  Holy
underconstrained system, Batman!  What now?

Suppose she spent the same amount of time running up as she spent
running down.  She would cover 50 steps, right?  But she only covers
25 steps, so she must spend half as much time running up as she spends
running down, and the time required to reach the top is one-half of
the time required for the trip down.  So we really only have two
unknowns, and if we set the distance up equal to the distance down
we have

[10/t + (N-10)/t]*t = [50/t - (N-10)/t]*(t/2)

The t's cancel,

10 + (N-10) = [50 - (N-10)]*(1/2)

and now we just have to solve for N.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Puzzles

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