nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles/Solution/Proof 1

== by : $62$ ==
;At Right Angles
== Solution ==
== Proof ==
Let $T$ be the time in question.
Let $m$ be the number of minutes past $5:00$ that $T$ is.
Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock at time $T$.
Let $\phi \degrees$ be the angle made by the hour hand with respect to twelve o'clock at time $T$.
Between $5:00$ and $6:00$ we have that $150 < \phi < 180$.
Hence from Condition for Valid Time Indication, we have that:
:$12 \paren {\phi - 150} = \theta$
Then we are told that:
:$\theta = \phi - 90$
as (from how soon) we are interested in the earliest instance of the hands being at right angles.
Hence:
Hence the result.
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles
Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles/Solution/Proof 2

== by : $62$ ==
;At Right Angles
== Solution ==
== Proof ==
From Time when Hour Hand and Minute Hand at Right Angle, the times when this is between $5$ and $6$ are:
:$05:10:55$
and:
:$05:43:38$
The first of these is the earlier of those two times.
== Sources ==
* : Solutions: $62$. -- At Right Angles
* : Answers: $54$. At Right Angles
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/62 - At Right Angles
Henry Ernest Dudeney/Puzzles and Curious Problems/63 - Westminster Clock

== by : $63$ ==
;Westminster Clock
:A man crossed over Westminster Bridge one morning between eight and nine o'clock by the tower clock
::(often mistakenly called Big Ben, which is the name of the large bell only. But this is by the way).
:On his return between four and five o'clock he noticed that the hands were exactly reversed.
:What were the exact times he made the crossings?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $63$. -- Westminster Clock
* : Arithmetical and Algebraical Problems: Clock Puzzles: $55$. Westminster Clock
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Clock Puzzles
Category:Ambiguous Times
Henry Ernest Dudeney/Puzzles and Curious Problems/63 - Westminster Clock/Solution

== by : $63$ ==
;Westminster Clock
== Solution ==
He travelled out at $08 : 23 \tfrac {71} {143}$ and returned at $04 : 41 \tfrac {137} {143}$.
:
== Proof ==
See Ambiguous Times for a full analysis.
== Sources ==
* : Solutions: $63$. -- Westminster Clock
* : Answers: $55$. Westminster Clock
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/64 - The Bath Chair

== by : $64$ ==
;The Bath Chair
:A correspondent informs us that a friend's house at $A$, where he was invited to lunch at $1$ p.m., is a mile from his own house at $B$.
:He is an invalid, and at $12$ noon started in his from $B$ towards $C$.
:His friend, who had arranged to join him and help push back, left $A$ at $12.15$ p.m., walking at $5$ miles per hour towards $C$.
:He joined him, and with his help they went back at $4$ miles per hour, and arrived at $A$ at exactly $1$ p.m.
:How far did our correspondent go towards $C$?
The illustration in the book shows $C$ as being in completely the opposite direction from $B$ to $A$.
That is, $B$ is directly between $A$ and $C$.
The real puzzle is why he set off in the direction of $C$ when his destination was in fact $A$, which is going the wrong way.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $64$. -- The Bath Chair
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $78$. The Bath Chair
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/64 - The Bath Chair/Solution

== by : $64$ ==
;The Bath Chair
== Solution ==
$\dfrac 2 3$ of a mile from $B$.
They met at $12:35$.
== Proof ==
Let $d$ miles be the distance towards $C$ at which our correspondent encountered the friend.
Let $t$ hours past $12$ be the time at which they met at that point.
We have:
== Sources ==
* : Solutions: $64$. -- The Bath Chair
* : Answers: $78$. The Bath Chair
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/65 - The Pedestrian Passenger

== by : $65$ ==
;The Pedestrian Passenger
:A train is travelling at the rate of $60$ miles per hour.
:A passenger at the back of the train wishes to walk to the front along the corridor,
::and in doing so walks at the rate of three miles per hour.
:At what rate is the man travelling over the permanent way?
:We will not involve ourselves here in quibbles and difficulties,
::similar to Zeno's paradox of the arrow, and Einstein's theory of relativity,
::but deal with the matter in the simple sense of motion in reference to the permanent way.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $65$. -- The Pedestrian Passenger
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $79$. The Pedestrian Passenger
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/65 - The Pedestrian Passenger/Solution

== by : $65$ ==
;The Pedestrian Passenger
== Solution ==
The puzzlingly obvious answer is: $63$ miles per hour.
 seems to believe that the result is somehow counterintuitive or paradoxical.
== Proof ==
Let $\mathbf v_T$ be the velocity of the train relative to the railway.
Let $\mathbf v_P$ be the velocity of the passenger relative to the train.
Then the velocity of the passenger relative to the railway is $\mathbf v_T + \mathbf v_P$.
The result follows when you plug in the numbers.
== Sources ==
* : Solutions: $65$. -- The Pedestrian Passenger
* : Answers: $79$. The Pedestrian Passenger
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/66 - Meeting Trains

== by : $66$ ==
;Meeting Trains
:At Wurzeltown Junction an old lady put her her head out of the window and shouted:
::"Guard! how long will the journey be from here to Mudville?"
:"All the trains take five hours ma'am, either way," replied the official.
:"And how many trains shall I meet on the way?"
:This absurd question tickled the guard, but he was ready with his reply:
:"A train leaves Wurzletown for Mudville, and also one from Mudville to Wurzletown, at five minutes past every hour. Right away!"
:The old lady induced one of her fellow passengers to work out the answer for her.
:What is the correct number of trains?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $66$. -- Meeting Trains
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $80$. Meeting Trains
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/66 - Meeting Trains/Solution

== by : $66$ ==
;Meeting Trains
== Solution ==
:$9$ trains.
== Proof ==
Let $W$ and $M$ denote Wurzeltown and Mudville respectively.
Each train takes $5$ hours for the journey.
At the point of departure from $W$, there will be:
:$1$ train just arriving at $W$ from $M$
:$1$ train just starting out from $M$
:Another $4$ trains already in transit from $M$ to $W$.
During the course of the journey from $W$ to $M$, $4$ more trains start off from $M$ to $W$.
At the arrival at $M$, another train is just about to leave $M$ for $W$.
So, not counting the two trains at either end (just arriving at $W$ and just about to leave $M$), there are $9$ trains met along the way.
== Sources ==
* : Solutions: $66$. -- Meeting Trains
* : Answers: $80$. Meeting Trains
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/67 - Carrying Bags

== by : $67$ ==
;Carrying Bags
:A gentleman had to walk to his railway station, four miles from his house,
::and was encumbered by two heavy bags of equal weight, but too heavy for him to carry alone.
:His gardener and the boy both insisted on carrying the luggage;
::but the gardener is an old man, and the boy not sufficiently strong,
::while the gentleman believes in a fair division of labour, and wished to take his own share.
:They started off with the gardener carrying one bag and the boy the other,
::while the gentleman worked out the best way of arranging that the three should share the burden equally among them.
:Now, how would you have managed it?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $67$. -- Carrying Bags
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $81$. Carrying Bags
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/67 - Carrying Bags/Solution

== by : $67$ ==
;Carrying Bags
== Solution ==
Let the boy carry his one bag for $1 \tfrac 1 3$ miles.
Then let him hand it to the gentleman, who will carry it to the station.
Let the gardener carry the other bag for $2 \tfrac 2 3$ miles.
Then let him hand it to the boy, who will then carry it to the station.
Each of the $3$ people have then carried one bag for $2 \tfrac 2 3$ miles.
Whether this is equal division of labour is debatable, because:
:the gardener is free to walk back home after having walked only $2 \tfrac 2 3$ miles, while the boy has to walk the entire journey
:the gentleman has to walk only one direction, while the others have to walk the $4$ miles home again.
However, while in modern times walking is considered a burdensome chore, it was just the way people got about a century ago, and not considered "labour".
== Sources ==
* : Solutions: $67$. -- Carrying Bags
* : Answers: $81$. Carrying Bags
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/68 - The Moving Staircase

== by : $68$ ==
;The Moving Staircase
:"I counted $50$ steps that I made in going down the moving staircase," said Walker.
:"I counted $75$ steps," said Trotman; "but I was walking down three times as quickly as you."
:If the staircase were stopped, how many steps would be visible?
:It is assumed that each man travelled at a uniform rate, and the speed of the staircase was also constant.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $68$. -- The Moving Staircase
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $82$. The Escalator
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/68 - The Moving Staircase/Solution

== by : $68$ ==
;The Moving Staircase
== Solution ==
:$100$ steps.
== Proof ==
Let $n$ be then number of visible steps when the escalator is stopped.
Let $t$ be the unit of time taken for one step to vanish at the bottom.
If you stand still on the escalator, it takes time $n t$ to reach the bottom.
If you take $x$ steps down the escalator, it takes time $\paren {n - x} t$ to reach the bottom.
Trotman takes $75$ steps in $\paren {n - 75} t$.
That is, he takes $3$ steps in $\dfrac {n - 75} {25} t$.
Walker takes $50$ steps in $\paren {n - 50} t$.
That is, he takes $1$ step in $\dfrac {n - 50} {50} t$.
But $3$ steps taken by Trotman take as much time as $1$ step taken by Walker.
Hence:
:$\dfrac {n - 75} {25} = \dfrac {n - 50} {50}$
which, after algebra, gives:
:$n = 100$
== Sources ==
* : Solutions: $68$. -- The Moving Staircase
* : Answers: $82$. The Escalator
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/69 - The Four Cyclists

== by : $69$ ==
;The Four Cyclists
:The four circles represent cinder paths.
:
:The four cyclists started at noon.
:Each person rode round a different circle,
::one at the rate of $6$ miles an hour,
::another at the rate of $9$ miles an hour,
::another at the rate of $12$ miles an hour,
::and the fourth at the rate of $15$ miles an hour.
:They agreed to ride until all met at the centre, from which they started, for the fourth time.
:The distances around each circle was exactly one-third of a mile.
:When did they finish their ride?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $69$. -- The Four Cyclists
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $83$. The Four Cyclists
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/69 - The Four Cyclists/Solution

== by : $69$ ==
;The Four Cyclists
== Solution ==
:$26$ minutes and $40$ seconds past noon.
== Proof ==
Let $t$ be the time in hours at which they all arrive together.
The times taken to complete one circuit are respectively:
:$\dfrac 1 3 \div 6 = \dfrac 1 {18}$ hours
:$\dfrac 1 3 \div 9 = \dfrac 1 {27}$ hours
:$\dfrac 1 3 \div 12 = \dfrac 1 {36}$ hours
:$\dfrac 1 3 \div 15 = \dfrac 1 {45}$ hours
We have that $\lcm {18, 27, 36, 45} = 540$
Hence let us clear the fractions by letting time $h$ be the time period that is $h = 1 {540}$ of an hour.
Let $t_1$, $t_2$, $t_3$ and $t_4$ be the times taken by each cyclist in units of $h$.
We have:
:$t_1 = 30$
:$t_2 = 20$
:$t_3 = 15$
:$t_4 = 12$
Now we have that:
:$\lcm {30, 20, 15, 12} = 60$
which is the smallest number of time units $h$ at which they are all at the middle together.
Thus:
:$t = \dfrac {60} {540} = \dfrac 1 9$
or $6 \tfrac 2 3$ minutes.
Four times this is $26 \tfrac 2 3$ minutes, or $26$ minutes and $40$ seconds.
== Sources ==
* : Solutions: $69$. -- The Four Cyclists
* : Answers: $83$. The Four Cyclists
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/6 - A New Partner

== by : $6$ ==
;A New Partner
:Two partners named Smugg and Williamson have decided to take a Mr. Rogers into partnership.
:Smugg has one and a half times as much capital invested in the business as Williamson
::and Rogers has to pay down $\pounds 2500$, which sum shall be divided between Smugg and Williamson,
::so that the three partners shall have an equal interest in the business.
:How shall that sum be divided?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $6$. -- A New Partner
* : Arithmetical and Algebraical Problems: Money Puzzles: $16$. A New Partner
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/6 - A New Partner/Solution

== by : $6$ ==
;A New Partner
== Solution ==
Rogers pays $\pounds 500$ to Smugg and $\pounds 2000$ to Williamson.
== Proof ==
The assumption that must be made is that the contribution of Rogers is one third of the total business value before he entered.
Let $T$ pounds be the total investment in the business.
Let $S$ and $W$ pounds be the amounts of capital invested by Smugg and Williamson respectively.
We have:
So Rogers pays $\pounds 500$ to Smugg and $\pounds 2000$ to Williamson.
== Sources ==
* : Solutions: $6$. -- A New Partner
* : Answers: $16$. A New Partner
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/70 - The Donkey Cart

== by : $70$ ==
;The Donkey Cart
:Atkins, Brown and Cranby had to go an journey of $40$ miles.
:Atkins could walk $1$ mile an hour,
::Brown could walk $2$ miles an hour,
::and Cranby could go in his donkey-cart at $8$ miles an hour.
:Cranby drove Atkins a certain distance, and, dropping him to walk the remainder,
::drove back to meet Brown on the way and carried him to their destination,
::where they all arrived at the same time.
:How long did the journey take?
:Of course, each went at a uniform rate throughout.
=== Click here for solution ===
== Also see ==
* Henry Ernest Dudeney/Modern Puzzles/37 - A Side-car Problem, exactly the same problem but with different numbers
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $70$. -- The Donkey Cart
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $84$. The Donkey Cart
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/70 - The Donkey Cart/Solution

== by : $70$ ==
;The Donkey Cart
:Atkins, Brown and Cranby had to go an journey of $40$ miles.
:Atkins could walk $1$ mile an hour,
::Brown could walk $2$ miles an hour,
::and Cranby could go in his donkey-cart at $8$ miles an hour.
:Cranby drove Atkins a certain distance, and, dropping him to walk the remainder,
::drove back to meet Brown on the way and carried him to their destination,
::where they all arrived at the same time.
:How long did the journey take?
:Of course, each went at a uniform rate throughout.
== Solution ==
The whole journey takes $10 \tfrac 5 {41}$ hours.
== Proof ==
Let Atkins, Brown and Cranby be denoted by $A$, $B$ and $C$ respectively.
We have that:
:$C$ sets off with $A$
:drops him off $d_1$ miles from the start
:then returns to pick up $B$ at a point $d_2$ miles from the start.
Let $t_1$ be the time $C$ and $A$ reach $d_1$.
Let $t_2$ be the time $C$ reaches $d_2$ to pick up $B$.
Let $t$ be the time they all arrive at their destination.
We have:
Then we have:
== Sources ==
* : Solutions: $70$. -- The Donkey Cart
* : Answers: $84$. The Donkey Cart
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/71 - The Three Motor-Cars

== by : $71$ ==
;The Three Motor-Cars
:Three motor-cars travelling along a road in the same direction are, at a certain moment, in the following positions in relation to one another.
:Andrews is a certain distance behind Brooks,
::and Carter is twice that distance in front of Brooks.
:Each car travels at its own uniform rate of speed,
::with the result that Andrews passes Brooks in seven minutes,
::and passes Carter five minutes later.
:Now, in how many minutes after Andrews would Brooks pass Carter?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $71$. -- The Three Motor-Cars
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $85$. The Three Cars
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/71 - The Three Motor-Cars/Solution

== by : $71$ ==
;The Three Motor-Cars
== Solution ==
:$6 \tfrac 2 3$ minutes.
== Proof ==
The trick of this puzzle is to consider the relative speeds instead of absolute speeds.
Let $A$, $B$ and $C$ denote the current positions of Andrews, Brooks and Carter respectively.
Let $d$ be the distance between where $A$ and $B$ are now.
Let $v_A$, $v_B$ and $v_C$ units per minute be the speeds of $A$, $B$ and $C$ respectively.
Let $t$ be the number of minutes since the start for Brooks to pass Carter.
We have:
 
Write $x = v_A - v_B$ and $y = v_A - v_C$. 
Then $v_B - v_C = y - x$.
So we have:
therefore Brooks passes Carter $6 \tfrac 2 3$ minutes after Andrew does.
== Sources ==
* : Solutions: $71$. -- The Three Motor-Cars
* : Answers: $85$. The Three Cars
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/72 - The Fly and the Motor-Cars

== by : $72$ ==
;The Fly and the Motor-Cars
:A road is $300$ miles long.
:A motor-car, $A$, starts at noon from one end and goes throughout at $50$ miles an hour,
::and at the same time another car, $B$, going uniformly at $100$ miles an hour, starts from the other end,
::together with a fly travelling $150$ miles an hour.
:When the fly meets the car $A$, it immediately turns and flies towards $B$.
:$(1)$ When does the fly meet $B$?
:The fly then turns towards $A$ and continues flying backwards and forwards between $A$ and $B$.
:$(2)$ When will the fly be crushed between the two cars if they collide and it does not get out of the way?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $72$. -- The Fly and the Motor-Cars
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $86$. The Fly and the Cars
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/72 - The Fly and the Motor-Cars/Solution

== by : $72$ ==
;The Fly and the Motor-Cars
== Solution ==
:$(1): \quad$ At $13:48$, at which point they are $120$ miles from $A$.
:$(2): \quad$ At $14:00$, at which point they are $100$ miles from $A$.
== Proof ==
Let $d_1$ miles from $A$ be the point at which the fly first meets $A$.
Let $d_2$ miles from $A$ be the point where $B$ is at that time.
Let $t_1$ hours after noon be the time at which this happens.
Let $d_3$ miles from $A$ be the point at which the fly then meets $B$.
Let $t_2$ hours after noon be the time at which this happens.
First we investigate where everybody is when the fly first meets $A$.
We have:
So when the fly first meets $A$, they are $75$ miles from $A$, while $B$ is $150$ miles from $A$ (which also happens to be $150$ miles from $B$).
This happens at $1 \tfrac 1 2$ hours after noon 
Now we investigate when the fly first meets $B$.
At this stage we are not interested in what happens to $A$, just $B$ and the fly.
We have:
So the fly meets $B$ at $120$ miles from $A$, which happens at $1 \tfrac 4 5$ hours after noon, or at $13:48$.
Let $d_4$ miles from $A$ be the point at which the cars collide.
Let this happen at $t_3$ hours after noon.
We have:
Thus $A$ and $B$ collide at $2$ hours past noon, $100$ miles from $A$.
We do not need to calculate the path of the fly, despite the fact that the naïve reader may be tempted into trying.
== Also see ==
* The Bulldozers and the Bee, of which this is a variant.
== Sources ==
* : Solutions: $72$. -- The Fly and the Motor-Cars
* : Answers: $86$. The Fly and the Cars
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Category:The Bulldozers and the Bee
Henry Ernest Dudeney/Puzzles and Curious Problems/73 - The Tube Stairs

== by : $73$ ==
;The Tube Stairs
:We ran up against Percy Longman, a young athlete, the other day when leaving Curley Street tube station.
:He stopped at the lift, saying, "I always go up by the stairs.
:A bit of exercise, you know.
:But this is the longest stairway on the line -- nearly $1000$ steps.
:I will tell you a queer thing about it that only applies to one other smaller stairway on the line.
:If I go up two steps at a time, there is one step left for the last bound;
::if I go up three at a time, there are two steps left;
::if I go up four at a time, there are three steps left;
::five at a time, four are left;
::six at a time, five are left;
::and if I went up seven at a time there would be six risers left over for the last bound.
:Now, why is that?"
:As he went flying up the stairs, three steps at a time, we laughed and said,
::He little suspects that if he went up twenty steps at a time there would be nineteen risers for his last bound!"
:How many risers are there in the Curley Street tube stairway?
:The platform does not count as a riser, and the top landing does.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $73$. -- The Tube Stairs
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $87$. The Subway Stairs
Category:Henry Ernest Dudeney/Puzzles and Curious Problems