Alice walks down an upward escalator and counts 150 steps. Bob walks up the same escalator and counts 75 steps. Alice takes three times as many steps in a given time as Bob.

How many steps are visible on the escalator?

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    $\begingroup$ I hope Alice and Bob aren't using the escalators at the same time, other wise they're going to have a very awkward (or painful) meeting. $\endgroup$ – Nzall Jun 21 '16 at 14:44
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    $\begingroup$ Notice that nowhere it was said that escalator has constant speed, or even moves at all. $\endgroup$ – Revolver_Ocelot Jun 21 '16 at 16:15
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    $\begingroup$ @Revolver_Ocelot Assuming the escalator doesn't move -- I can't think of any reason not to just call them stairs then :-) $\endgroup$ – Insane Jun 22 '16 at 9:00

There are:

120 steps visible on the escalator (I think)

Since Alice takes 3x as many steps per time as Bob, and her trip was 150 steps while his was 75,

that means Alice was on the escalator for 2/3 as long as Bob was, because she took twice as many steps, but moves at 3x the speed.

So the equations we can set up are

$150 = n + 2/3s$
$75 = n - s$
$n$ is the number of steps showing, and $s$ is the number of steps that change per unit time.

Solving those gives us

$n=120$ and $s=15$

So just to double check

Let's say 15 steps cycle per minute, and that Alice takes 75 steps per minute, it takes her 2 minutes (150 steps) to cover the (120 + 2*15) 150 steps of going down the up escalator.
If Alice is taking 75 steps per minute, then Bob is taking 25, so that means his trip up the escalator takes 3 minutes, during which he walked 75 steps and it moved 3*15 or 45 steps.

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    $\begingroup$ Ha. I beat you by eleven whole seconds. $\endgroup$ – Gareth McCaughan Jun 21 '16 at 14:13
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    $\begingroup$ @GarethMcCaughan That was tight! At least we got the same thing. I think my extra 11 seconds were just me going "really?...." $\endgroup$ – Dan Russell Jun 21 '16 at 14:15
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    $\begingroup$ Both of you type(and probably solve) too fast :D $\endgroup$ – ABcDexter Jun 21 '16 at 14:29
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    $\begingroup$ @ABcDexter Just happened to see this one within 30 min of it being posted, and I immediately went into math-test mode. $\endgroup$ – Dan Russell Jun 21 '16 at 14:31
  • $\begingroup$ @DanRussell Haha, math-test mode. Similar case with me, I just saw a 1 in review, (A,B) was to be replaced by (Alice,Bob), the question was (or rather is) really interesting so jumped to get my pencil and paper :D $\endgroup$ – ABcDexter Jun 21 '16 at 14:35

We'll measure time in units of how long it takes for one step to emerge from the bottom or disappear at the top of the escalator and distance in units of steps.

Let the length of the escalator be $n$, Alice's speed be $a$, and Bob's speed be $b$. (So $a=3b$. These are speeds relative to the escalator steps.)

Alice goes down the escalator; her speed relative to the ground is $a-1$. So she takes time $n/(a-1)$ during which she passes $na/(a-1)$ steps.

Bob goes up; his speed relative to the ground is $b+1$. So he takes time $n/(b+1)$ during which he passes $nb/(b+1)$ steps.

Alice passes twice as many steps as Bob so (cancelling an $n$ on both sides) we have $a/(a-1)=2b/(b+1)$ or $a(b+1)=2b(a-1)$ or $ab-a-2b=0$. Now would be a good time to remember that $a=3b$ so this says $3b^2-5b=0$ so either $b=0$ (which indeed makes Alice count twice as many steps as Bob, namely none, but clearly isn't the solution here) or $b=5/3$ and accordingly $a=5$. We then have $150=na/(a-1)=n\cdot5/4$ so $n=120$. (Or, solving the other equation just to make sure we get the same answer, $75=nb/(b+1)=n\cdot(5/3)/(8/3)$ so $n=75\cdot8/5=120$: good.)

Super-simplistic sanity check

with every step, Alice moves 5 steps down the escalator while the escalator moves her up one step; net movement of 4 steps; so she takes 30 steps to reach the bottom. On each of those she counts 5 steps, yielding 150 as required.

I suppose we might as well do the general case, just in case other variants of this problem get asked in future. So: let's say Bob goes up the escalator at speed $b$ and Alice goes up at speed $a=kb$ ($b$ unknown, $k$ known). Alice counts $p$ upward steps, Bob counts $q$ upward steps. In the example above we have $k=-3$, $p=-150$, and $q=75$. (It's important to get the signs right.) Then

$p=na/(a+1)=nkb/(kb+1)$ and $q=nb/(b+1)$ so $p/q=(kb+k)/(kb+1)=1+(k-1)/(kb+1)$, whence $b=((k-1)q/(p-q)-1)/k=(kq-p)/[k(p-q)]$ and hence $a=(kq-p)/(p-q)$. Then $n=q(1+1/b)=q(1+[k(p-q)]/(kq-p))=pq(k-1)/(kq-p)$.

Sanity check: for the particular case asked here we get


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    $\begingroup$ Every step you take... every move you make.... $\endgroup$ – Lightness Races in Orbit Jun 23 '16 at 10:59
  • $\begingroup$ I have the strangest feeling I'm being watched. $\endgroup$ – Gareth McCaughan Jun 23 '16 at 11:14
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    $\begingroup$ Can't you see? You belong to me. $\endgroup$ – Lightness Races in Orbit Jun 23 '16 at 11:41
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    $\begingroup$ So in this case it's five steps forward, one step back. $\endgroup$ – Peregrine Rook Jun 25 '16 at 18:21

I'm assuming that "steps are visible" refers to steps beeing seen, i.e "at the surface" ?

If so, - and taking any of the computed N steps answers from the others

N = 120, see above

I believe the corect answer should be

N + 1 = 121 ... at least most of the time.


Most of the time you will see a fraction of the first step and a fraction of the last step during transition on an escalator, but you count them as one each.

enter image description here

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  • $\begingroup$ Tell me, do you count the base of the stairs(on the same level as the ground itself) as a step of the stairs ? ( I do get your point though) $\endgroup$ – ABcDexter Jun 21 '16 at 14:40
  • $\begingroup$ @ABcDexter It really depends on what is meant by "How many steps are visible on the escalator?". If you do interpret "step" as "vertical jump" it still works, because most of the time the top and the bottom "steps" will be a fraction of their "full step" only. $\endgroup$ – BmyGuest Jun 21 '16 at 14:48
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    $\begingroup$ In any case: If my answer is correct, then it is a trick-question and if it isn't it is a boring math task and not a puzzle (i.e. VTC!) $\endgroup$ – BmyGuest Jun 21 '16 at 14:49
  • $\begingroup$ You posted a picture, cool :) Yes, you are right, there must have been an option d) 121, which was meant to be overlooked once the 120 is found. $\endgroup$ – ABcDexter Jun 21 '16 at 14:53
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    $\begingroup$ I like the lateral thinking angle. Also, it makes me wonder how many total steps (including non-visible) are in this or any other escalator. Yours has 9-10 visible of 20 total, but now I'm imaging a giant subterranean circle with thousands of steps, only a tiny fraction of which are visible. $\endgroup$ – Dan Russell Jun 21 '16 at 19:17

"How many steps are visible on the escalator? "

Well, that depends on many mathmatical things, what counts as a step, etc. So, I'm going to round off and say:

Half of them, of course!

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    $\begingroup$ What an abstract of an answer it is. Bazzinga! $\endgroup$ – ABcDexter Jun 21 '16 at 16:57
  • $\begingroup$ Well, it just seemed like the obvious and self-evident answer to me... :) $\endgroup$ – Self Evident Jun 21 '16 at 19:00

As already said by (almost) everyone else,


But all the other answers are just working through the numbers. Let's algebra this!

The system can be expressed as:

$$ v_A T_A = l + T_Ae = S_A \\ v_B T_B = l - T_Be = S_B \\ v_A = \alpha v_B $$

Where $v_A$ = speed of Alice against direction of escalator (steps per time period - say seconds), $v_B$ = speed of Bob in direction of escalator, $T_A$ = time for Alice to complete journey, $T_B$ = time for Bob to complete journey, $l$ = length of escalator in steps, $e$ = speed of escalator in steps per time period, $S_A$ = steps seen by Alice, $S_B$ = steps seen by Bob. We also introduce $\alpha$ = ratio of Alice's speed compared to Bob's.

We can re-write those equations as:

$${\alpha v T_A \over S_A} = {l + T_Ae \over S_A} = {vT_B \over S_B} = {l - T_Be \over S_B} = 1$$

Where $v$ = $v_B$ from above and we have substituted $v_A = \alpha v_B$.

So we want to solve for $l$. Let's rearrange:

$${l + T_Ae \over S_A} = {l - T_Be \over S_B}$$ $$\rightarrow l = {{T_AeS_B + T_BeS_A} \over {S_A-S_B}} \tag{1}$$


$$\alpha v{T_A \over S_A} = v{T_B \over S_B}$$ $$\rightarrow T_A = T_B{S_A \over \alpha S_B} \tag{2}$$


$${l - T_Be \over S_B} = 1$$ $$\rightarrow T_Be = l - S_B \tag{3}$$

Combining (1) and (2) gives:

$$l = {T_BeS_A(1+1/\alpha) \over S_A-S_B} \tag{4}$$

Then combining (3) and (4) gives:

$$l = {(l-S_B)S_A(1+1/\alpha) \over S_A-S_B}$$

Which (eventually) rearranges to:

$$l = S_A{\alpha+1 \over \alpha+{S_A \over S_B}} \tag{*}$$

All we have to do now is plug in some values. We're told that $\alpha = 3$, $S_A = 150$ and $S_B = 75$:

$$l = S_A{3+1 \over 3+{150 \over 75}} = S_A{4 \over 5} = 150 \times {4 \over 5} = 120$$

And we can find out the relative speeds too (absolute speeds can't be found since we have no time values given in the question):

$${e \over v} = \alpha {S_A-S_B \over S_A+\alpha S_B} = 3 {150-75 \over 150+3 \times 75} = {3 \over 5}$$

Which means:

Bob is travelling at 5/3 the speed of the escalator and Alice is traveling at 5 times the speed of the escalator. Wikipedia says escalators move at 0.3–0.6m/s, so Alice would be moving at 1.5-3.0m/s (~3–7mph). Quite fast for walking down stairs and an accident waiting to happen, but plausible.

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    $\begingroup$ Welcome to Puzzling, and great answer! $\endgroup$ – Deusovi Jun 22 '16 at 22:28

Knowing only that Alice covers 3 steps each time that Bob covers 1 step, nothing further needs to be considered about speeds.   In fact, the escalator may move erratically, or even not at all.

Picture five identical escalators in sequence, effectively forming a single escalator 5 times as long. Start Alice coming down from the very top at the same time as Bob starts up from the bottom.

   They will collide after Alice descends 3 escalator lengths and Bob ascends 2 lengths.

This works out so simply because each step that disappears at the top, behind Alice, is simultaneously compensated by a new step that appears at the bottom, behind Bob.

                                           (always visible)
     Still-visible Alice's steps             Bob's steps
     |                         |             |         |
                                |           |           |              |
                                Steps between           As many steps as
                                Alice and Bob          those of Alice that
                                                      disappeared at the top

The total number of visible steps is always the sum of Alice's and Bob's steps so far plus the remaining steps between the two.

When Alice and Bob collide, that total is just their combined steps.   So why 5 escalators?

While Alice takes 150 steps to cover an escalator, Bob takes takes 75 steps but each of those steps takes 3 times as long as Alice's.
Thus the ratio of times for a single escalator is $~$ Alice$\bf\kern 2mu : \kern 2mu$Bob $~$=$~$ 150$\bf\kern 3mu : \kern 2mu$75$\scriptsize\raise1mu\times$3 $~$=$~$ 2$\bf\kern2mu : \kern2mu$3.
For Alice and Bob to meet after the same amount of time, the ratio of their distances must be 3$\bf\kern2mu : \kern2mu$2.
The simplest number that divides into portions with that ratio is 5. Could also have used 1 escalator divided into 3/5 and 2/5, or 100 escalators divided into 60 and 40, and so on.

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  • $\begingroup$ “… nothing further needs to be considered about speeds.” I guess I agree. “In fact, the escalator may move erratically, or even not at all.” I disagree. The escalator moves in such a way that it scrolls 30 steps while Alice is traversing it and 45 steps while Bob is. If the escalator were not moving at all, how could the two users get different numbers of steps? … That note aside, this is a neat, innovative approach, so I’m going to usher you into Club 5000. $\endgroup$ – Peregrine Rook Jun 25 '16 at 18:30

As per my calculations, answer is:



Let us assume that the number of steps is $x$
The upward speed of the escalator is $y$ $\frac{steps}{second}$.

Also, let the speed of Bob(normal speed of walking) be $b$ $\frac{steps}{second}$,
then the speed of Alice will be $3b$ $\frac{steps}{second}$

One thing that is same for both of them is: $x$, as they both travelled through the same escalator.
As we have distance = speed x time,
Thus, we have two equations:
$x= (b+y)*t_1$ $ \ldots \space$ eqn i
$x= (3*b-y)*t_2$ $ \ldots \space$ eqn. ii

Using $t_1 = \frac{75}{b}$ and $t_2 = \frac{150}{3b}$ (as the actual time taken will be actual distance, which is the count of foot-steps, divided by their speed),

we can solve the equation,
$(b+y)*\frac{75}{b} = (3*b-y)*\frac{150}{3b}$

which gives, $3b = 5y$


putting this in either of the two equations(i or ii),
in eqn. ii
$x= (5y-y) \frac{150}{5y}$
$x= \frac{4*y*150}{5y}$
the final answer is $x = 120$

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This is another derivation of the same result, but shows that the speed of the escalator can not be determined and could be any suitable constant value (provided that it does not outrun Alice).

Alice walks $a$ steps per unit of time. Bob walks at $b$ steps per unit of time. The escalator moves up $u$ steps per unit of time.

Alice is slowed down by the escalator, but in a time $T_a$ she will have walked enough to compensate the loss of speed, and will have covered $n$ steps, the length of the escalator.

\begin{equation} (a - u) T_a = n ~~~~~~~~~~ (1) \end{equation}

Note that it must be $a > u$.

Alice physically takes more than $n$ steps, precisely she takes $150$ steps during that time $T_a$.

\begin{equation} a T_a = 150 ~~~~~~~~~~ (2) \end{equation}

Bob is sped up by the escalator, and in a time $T_b$ he will have walked enough to cover $n$ steps, the length of the escalator.

\begin{equation} (b + u) T_b = n ~~~~~~~~~~ (3) \end{equation}

Bob physically takes less than $n$ steps, precisely he takes $75$ steps during that time $Tb$.

\begin{equation} b T_b = 75 ~~~~~~~~~~ (4) \end{equation}

Alice walks faster than Bob by a factor of $3$.

\begin{equation} a = 3 b ~~~~~~~~~~ (5) \end{equation}

From (2), (4) and (5) it follows that $T_a / T_b = 2/3$. Then we rewrite (1) and (3) like this:

\begin{equation} a T_a - n = u T_a ~~~~~~~~~~ (6) \end{equation}

\begin{equation} b T_b - n = -u T_b ~~~~~~~~~~ (7) \end{equation}

And we divide (6) by (7). On the right side the $u$ factors cancel out and $T_a/T_b=2/3$, hence:

\begin{equation} \frac{a T_a - n}{b T_b - n} = -2/3 ~~~~~~~~~~ (8) \end{equation}

Note that (8) is well defined for any non-zero $u$.

Replacing (2) and (4) in (8) we obtain:

\begin{equation} \frac{150 - n}{75 - n} = -2/3 ~~~~~~~~~~ (9) \end{equation}

And from (9) it is easy to derive $n = 120$.

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The answer is

215 steps


Bob starts going up the stairs. When he is 25 stairs up, Alice starts coming down. They meet roughly in the middle and have a very painful tumble!

Sanity check

This and none of the other answers could be correct. What escalator has more than 20 steps?!?

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  • $\begingroup$ Regarding your sanity check. I have been on an escalator in Prague (longest in the EU) at Náměstí Míru station. It is 87 m long and approx 43 m high. Going up it at rush hour would be suicidal (and a good check of sanity)! Wikipedia suggests there is an escalator almost 50% longer in St Petersburg. $\endgroup$ – Penguino Jun 22 '16 at 4:28
  • $\begingroup$ @Penguino wow, okay. I know this is a riddlt anyway, so it need not make sense, but that it is intriging that those exist irl. $\endgroup$ – user64742 Jun 22 '16 at 4:37
  • $\begingroup$ Clearly you have never been on the London underground; tfl.gov.uk/corporate/about-tfl/what-we-do/london-underground/… (27.4m = 154 steps assuming a standard 7 inch rise) $\endgroup$ – Dave Jun 22 '16 at 21:16
  • $\begingroup$ The escalator at the Wheaton station of the Washington D.C. Metro (underground/subway) features the longest set of single-span escalators in the Western Hemisphere, each featuring a length of 230 feet (70 m) with a vertical rise of 115 feet (35 m) — deeper (higher) than the London one.  Even at the more conservative 8 inch step estimate, that's about 172.5 steps. $\endgroup$ – Peregrine Rook Jun 25 '16 at 18:06

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