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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!

Thank you for any help you can provide,
-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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