Stairs on an EscalatorDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/