nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/36 - More Bicycling

== by : $36$ ==
;More Bicycling
:Referring to the last puzzle, let us now consider the case where a third rider has to share the same bicycle.
:As a matter of fact, I understand that Anderson and Brown have taken a man named Carter into partnership, and the position today is this:
:Anderson, Brown and Carter walk respectively $4$, $5$ and $3$ miles per hour,
::and ride respectively $10$, $8$ and $12$ miles per hour.
:How are they to use that single bicycle so that all shall complete the $20$ miles journey at the same time?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $36$. -- More Bicycling
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $64$. More Bicycling
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/36 - More Bicycling
Henry Ernest Dudeney/Modern Puzzles/36 - More Bicycling/Solution

== by : $36$ ==
;More Bicycling
== Solution ==
Anderson:
:rides for $7 \tfrac {11} {27}$ miles, taking $44 \tfrac 4 9$ minutes
:walks the remaining $12 \tfrac {16} {27}$ miles in $3$ hours and $8 \tfrac 8 9$ minutes.
Brown:
:walks those $7 \tfrac {11} {27}$ miles in $1$ hour $28 \tfrac 8 9$ minutes
:rides for $1 \tfrac {13} {27}$ miles, taking $11 \tfrac 1 9$ minutes
:walks the remaining $11 \tfrac 1 9$ miles in $2$ hours $13 \tfrac 1 3$ minutes.
Carter:
:walks $8 \tfrac 8 9$ miles in $2$ hour $57 \tfrac {21} {27}$ minutes
:rides the remaining $11 \tfrac 1 9$ miles in $55 \tfrac 5 9$ minutes.
The total journey time is $3$ hours $53 \tfrac 1 3$ minutes.
== Proof ==
Let Anderson, Brown and Carter be denoted by $A$, $B$ and $C$ respectively.
To keep it simple, we will assume:
:$A$ starts by cycling, then leaves the bike for $B$
:$B$ takes over cycling from $A$, then leaves the bike for $C$
:$C$ finishes the journey on the cycle.
Let $d_1$ miles be the distance from the start to where $A$ dismounts to start walking and $B$ starts riding.
Let $d_2$ miles be the distance from $d_1$ to where $B$ dismounts to start walking.
Let $t$ hours be the time taken to do the total journey.
Let $t_{a_1}$ hours be the time taken by $A$ to travel $d_1$.
Let $t_{b_1}$ hours be the time taken by $B$ to travel $d_1$.
Let $t_{c_1}$ hours be the time taken by $C$ to travel $d_1$.
Let $t_{a_2}$ hours be the time taken by $A$ to travel $d_2$.
Let $t_{b_2}$ hours be the time taken by $B$ to travel $d_2$.
Let $t_{c_2}$ hours be the time taken by $C$ to travel $d_2$.
We have:
We set up this system of linear simultaneous equations in matrix form as:
:$\begin {pmatrix}
1 & 0 & -10 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & -5 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & -3 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & -4 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & -8 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & -3 & 0 \\
1 & 1 & -4 & 0 & 0 & -4 & 0 & 0 & 4 \\
1 & 1 & 0 & -5 & 0 & 0 & -5 & 0 & 5 \\
1 & 1 & 0 & 0 & -12 & 0 & 0 & -12 & 12 \\
\end {pmatrix}
\begin {pmatrix}
d_1 \\ d_2 \\ t_{a_1} \\ t_{b_1} \\ t_{c_1} \\ t_{a_2} \\ t_{b_2} \\ t_{c_2} \\ t \\
\end {pmatrix} = \begin {pmatrix}
0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 20 \\ 20 \\ 20 \\
\end {pmatrix}$
It remains to solve this matrix equation.
In reduced echelon form, this gives:
The result can be read directly and converted into the appropriate units.
== Sources ==
* : Solutions: $36$. -- More Bicycling
* : Answers: $64$. More Bicycling
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/36 - More Bicycling
Henry Ernest Dudeney/Modern Puzzles/36 - More Bicycling/Working

== Working for by : $36$ -- More Bicycling ==
The simultaneous equations in matrix form:
:$\begin {pmatrix}
1 & 0 & -10 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & -5 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & -3 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & -4 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & -8 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & -3 & 0 \\
1 & 1 & -4 & 0 & 0 & -4 & 0 & 0 & 4 \\
1 & 1 & 0 & -5 & 0 & 0 & -5 & 0 & 5 \\
1 & 1 & 0 & 0 & -12 & 0 & 0 & -12 & 12 \\
\end {pmatrix}
\begin {pmatrix}
d_1 \\ d_2 \\ t_{a_1} \\ t_{b_1} \\ t_{c_1} \\ t_{a_2} \\ t_{b_2} \\ t_{c_2} \\ t \\
\end {pmatrix} = \begin {pmatrix}
0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 20 \\ 20 \\ 20 \\
\end {pmatrix}$
when converted to reduced echelon form, gives:
:$\begin {pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end {pmatrix}
\begin {pmatrix}
d_1 \\ d_2 \\ t_{a_1} \\ t_{b_1} \\ t_{c_1} \\ t_{a_2} \\ t_{b_2} \\ t_{c_2} \\ t \\
\end {pmatrix} = \begin {pmatrix}
200/27 \\ 40/27 \\ 20/27 \\ 40/27 \\ 200/81 \\ 10/27 \\ 5/27 \\ 40/81 \\ 35/9 \\
\end {pmatrix}$
== Proof ==
Category:Henry Ernest Dudeney/Modern Puzzles/36 - More Bicycling
Henry Ernest Dudeney/Modern Puzzles/37 - A Side-car Problem

== by : $37$ ==
;A Side-car Problem
:Atkins, Baldwin and Clarke had to go a journey of $52$ miles across country.
:Atkins had a motor-bicycle with sidecar for one passenger.
:How was he to take one of his companions a certain distance,
::drop him on the road to walk the remainder of the way,
::and return to pick up the second friend,
::so that they should all arrive at their destination at exactly the same time?
:The motor-bicycle could do $20$ miles per hour,
::Baldwin could walk $5$ miles per hour,
::and Clarke could walk $4$ miles per hour.
:Of course, each went at his proper speed throughout and there was no waiting.
:I might have complicated the problem by giving more passengers,
::but I have purposely made it easy,
::and all the distances are an exact number of miles -- without fractions.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $37$. -- A Side-car Problem
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $65$. A Sidecar Problem
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/37 - A Side-car Problem/Solution

== by : $37$ ==
;A Side-car Problem
== Solution ==
There are $2$ solutions:
:$(1): \quad$ Atkins sets off with Clarke and drives $40$ miles, leaving Clarke to walk the remaining $12$ miles
:::Atkins picks up Baldwin $16$ miles from the start and continues to the destination.
:$(2): \quad$ Atkins sets off with Baldwin and drives $36$ miles, leaving Baldwin to walk the remaining $16$ miles
:::Atkins picks up Clarke $12$ miles from the start and continues to the destination.
The whole journey takes $5$ hours.
== Proof ==
Let Atkins, Baldwin and Clarke be denoted by $A$, $B$ and $C$ respectively.
First let us assume that:
:$A$ sets off with $C$
: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 $A$ and $C$ reach $d_1$.
Let $t_2$ be the time $A$ reaches $d_2$ to pick up $B$.
Let $t$ be the time they all arrive at their destination.
We have:
Then we have:
Running the scenario in reverse gives another solution.
== Sources ==
* : Solutions: $37$. -- A Side-car Problem
* : Answers: $65$. A Sidecar Problem
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/38 - The Despatch-Rider

== by : $38$ ==
;The Despatch-Rider
:If an army $40$ miles long advances $40$ miles
::while a despatch-rider gallops from the rear to the front,
::delivers a despatch to the commanding general,
::and returns to the rear,
::how far has he to travel?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $38$. -- The Despatch-Rider
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $66$. The Dispatch Rider
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/38 - The Despatch-Rider/Solution

== by : $38$ ==
;The Despatch-Rider
== Solution ==
:$40 + \sqrt {2 \times 40^2} \approx 96 \cdotp 568$ miles.
== Proof ==
We have that the length of the army is $40$ miles.
Let $d_1$ miles be the distance travelled by the despatch-rider from the rear to the front.
Let $d_2$ miles be the distance travelled by the despatch-rider returning from the front to the rear.
Let $t_1$ be the time taken to achieve the first leg of the journey.
Let $t_2$ be the time taken to achieve the second leg of the journey.
Let $v_d$ be the speed of the despatch-rider.
Let $v_a$ be the speed of the army.
We have that:
:$40 = v_a \paren {t_1 + t_1}$
During the first leg, the speed relative to the army of the despatch-rider is $v_d - v_a$.
During the second leg, the speed relative to the army of the despatch-rider is $v_d + v_a$.
:$40 = t_1 \paren {v_d - v_a} = t_2 \paren {v_d + v_a}$
Thus:
:$40 = v_a \paren {t_1 + t_1} = v_a \paren {\dfrac {40} {v_d - v_a} + \dfrac {40} {v_d + v_a} }$
Let $r = \dfrac {v_d} {v_a}$.
Hence as $v_d > v_a$ we have that:
:$r > 1$
Hence:
:$1 = \dfrac 1 {r - 1} + \dfrac 1 {r + 1} = \dfrac {2 r} {r^2 - 1}$
This leads to:
:$r^2 - 2 r - 1 = 0$
which, from the Quadratic Formula, gives us:
:$r = 1 \pm \sqrt 2$
As $r > 1$:
:$r = 1 + \sqrt 2$
which leads to:
:$d_1 + d_2 = 40 + 40 \sqrt 2 = 40 + \sqrt {2 \times 40^2}$
== Sources ==
* : Solutions: $38$. -- The Despatch-Rider
* : Answers: $66$. The Dispatch Rider
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/39 - The Two Trains

== by : $39$ ==
;The Two Trains
:Two railway trains, one $400$ feet long and the other $200$ feet long, ran on parallel rails.
:It was found that when they went in opposite directions they passed each other in $5$ seconds,
::but when they ran in the same direction the faster train would pass the other in $15$ seconds.
:Now, a curious passenger worked out from these facts the rate per hour at which each train ran.
:Can the reader discover the correct answer?
:Of course, each train ran with a uniform velocity.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $39$. -- The Two Trains
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $67$. The Two Trains
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/39 - The Two Trains/Solution

== by : $39$ ==
;The Two Trains
== Solution ==
The faster train went at $54 \tfrac 6 {11}$ miles per hour.
The slower train went at $27 \tfrac 3 {11}$ miles per hour.
== Proof ==
Let $v_1$ feet per second denote the speed of the faster train.
Let $v_2$ feet per second denote the speed of the slower train.
We have:
The result follows on converting $80$ feet per second and $40$ feet per second into miles per hour.
== Sources ==
* : Solutions: $39$. -- The Two Trains
* : Answers: $67$. The Two Trains
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/3 - Dollars and Cents

== by : $3$ ==
;Dollars and Cents
:An American correspondent tells me that a man went into a store and spent one-half of the money that was in his pocket.
:When he came out he found that he had just as many cents as he had dollars when he went in
:and half as many dollars as he had cents when he went in.
:How much money did he have on him when he entered?
=== Click here for solution ===
== Also see ==
* Pocket-money, which follows exactly the same pattern.
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $3$. -- Dollars and Cents
* : Arithmetical and Algebraical Problems: Money Puzzles: $2$. Dollars and Cents
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/3 - Dollars and Cents/Solution

== by : $3$ ==
;Dollars and Cents
== Solution ==
The total amount he had on him was $\$ 99.98$.
== Proof ==
Let $S$ be the amount he started with: $S_d$ dollars and $S_c$ cents.
Let $F$ be the amount he finished with: $F_d$ dollars and $F_c$ cents.
We recall the conversion factors:
:$100$ cents make one dollar.
Hence any cent quantities in either $S$ or $F$ cannot be greater than $99$.
That is:
:$S_c < 99$
:$F_c < 99$
We are given that:
We have that:
The smallest values of $S_d$ and $S_c$ that satisfy the above equation are:
As $S_c \le 99$ it follows that there can be no other solution.
Hence:
:$D = \$ 99.98$
== Sources ==
* : Solutions: $3$. -- Dollars and Cents
* : Answers: $2$. Dollars and Cents
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/40 - Pickleminster to Quickville

== by : $40$ ==
;Pickleminster to Quickville
:Two trains, $A$ and $B$, leave Pickleminster for Quickville at the same time as two trains, $C$ and $D$, leave Quickville for Pickleminster.
:$A$ passes $C$ $120$ miles from Pickleminster and $D$ $140$ miles from Pickleminster.
:$B$ passes $C$ $126$ miles from Quickville and $D$ half-way between Pickleminster and Quickville.
:Now, what is the distance from Pickleminster to Quickville?
:Every train runs uniformly at an ordinary rate.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $40$. -- Pickleminster to Quickville
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $68$. Pickleminster to Quickville
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/40 - Pickleminster to Quickville/Solution

== by : $40$ ==
;Pickleminster to Quickville
== Solution ==
:$210$ miles.
== Proof ==
Let $V_A$, $V_B$, $V_C$ and $V_D$ miles per hour be the speeds of $A$, $B$, $C$ and $D$ respectively.
Let $d$ be the distance between Pickleminster and Quickville.
From the given conditions:
This leads to the $2$ solutions:
:$d = 144$
:$d = 210$
The first of these suggests that $A$ travels $140$ miles while $D$ travels $4$ miles, which would mean $A$ goes $35$ times as fast as $D$.
The second answer is more sensible: it has $A$ going twice as fast as $B$ and $D$, and $C$ going at a speed between these.
== Sources ==
* : Solutions: $40$. -- Pickleminster to Quickville
* : Answers: $68$. Pickleminster to Quickville
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/41 - The Damaged Engine

== by : $41$ ==
;The Damaged Engine
:We were going by train from Anglechester to Clinkerton, and an hour after starting some accident happened to the engine.
:We had to continue the journey at $\tfrac 3 5$ of the former speed, and it made us $2$ hours late at Clinkerton,
::and the driver said that if only the accident had happened $50$ miles farther on the train would have arrived $40$ minutes sooner.
:Can you tell from that statement just how far it is from Anglechester to Clinkerton?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $41$. -- The Damaged Engine
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $69$. The Damaged Engine
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/41 - The Damaged Engine/Solution

== by : $41$ ==
;The Damaged Engine
== Solution ==
:$200$ miles, of which the first $50$ miles were at $50$ miles per hour, and the rest at $30$ miles per hour.
== Proof ==
Let $d$ miles be the distance from Anglechester to Clinkerton.
Let $v$ miles per hour be the speed of the train when working properly.
Let $d_1$ miles be the distance from Anglechester to where the train malfunctioned.
Let $t$ hours be the time the train would normally take to travel the journey.
We have:
== Sources ==
* : Solutions: $41$. -- The Damaged Engine
* : Answers: $69$. The Damaged Engine
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/42 - The Puzzle of the Runners

== by : $42$ ==
;The Puzzle of the Runners
:Two men ran a race round a circular course, going in opposite directions.
:Brown was the best runner and gave Tompkins a start of $\tfrac 1 8$ of the distance.
:But Brown, with a contempt for his opponent, took things too easily at the beginning,
::and when he had run $\tfrac 1 6$ of his distance he met Tompkins,
::and saw that his chance of winning the race was very small.
:How much faster than he went before must Brown now run in order to tie with his competitor?
:The puzzle is quite easy when once you have grasped its initial conditions.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $42$. -- The Puzzle of the Runners
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $70$. The Puzzle of the Runners
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/42 - The Puzzle of the Runners/Solution

== by : $42$ ==
;The Puzzle of the Runners
== Solution ==
$20 \tfrac 1 4$ times faster.
Probably not technically achievable.
== Proof ==
While Brown has run $\dfrac 1 6$ of the course, or $\dfrac 4 {24}$ of it, Tomkins has already run $\dfrac 5 6 - \dfrac 1 8$, that is $\dfrac {17} {24}$ of it.
Thus Tomkins's pace has up till now been $\dfrac {11} 4$ of Brown's.
Brown now has $\dfrac 5 6$ of the course to run, while Tomkins has $\dfrac 1 6$ left.
Thus Brown must speed up to $5$ times as fast as Tomkins in order to pass the finish line at the same time as Tomkins.
Thus he must go at $5 \times \dfrac {17} 4$, which is $\dfrac {85} 4$ times as fast as he went at first.
Then we note that $\dfrac {85} 4$ times as fast is the same thing as $\dfrac {85} 4 - 1$ times faster.
Hence the correct answer is $\dfrac {81} 4 = 20 \tfrac 1 4$ times faster.
== Sources ==
* : Solutions: $42$. -- The Puzzle of the Runners
* : Answers: $70$. The Puzzle of the Runners
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/43 - The Two Ships

== by : $43$ ==
;The Two Ships
:A correspondent asks the following question.
:Two ships sail from one port to another -- $200$ nautical miles -- and return.
:The Mary Jane travels outwards at $12$ miles an hour and returns at $8$ miles an hour,
::thus taking $41 \tfrac 2 3$ hours on the double journey.
:The Elizabeth Ann travels both ways at $10$ miles an hour, taking $40$ hours on the double journey.
:Now, seeing that both ships travel at the average speed of $10$ miles per hour, why does the Mary Jane take longer than the Elizabeth Ann?
:Perhaps the reader could explain this little paradox.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $43$. -- The Two Ships
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $71$. The Two Ships
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/43 - The Two Ships/Solution

== by : $43$ ==
;The Two Ships
== Solution ==
The error lies in assuming that the average speeds are equal.
They are not.
The first ship does a mile in $\tfrac 1 {12}$ of an hour outwards and $\tfrac 1 8$ of an hour homewards.
Half of the sum of these fractions is $\tfrac 5 {48}$.
Therefore the ship's average speed for the $400$ miles is a mile in $\tfrac 5 {48}$ of an hour.
The average speed of the second ship is a mile in $\tfrac 1 {10}$ of an hour.
== Sources ==
* : Solutions: $43$. -- The Two Ships
* : Answers: $71$. The Two Ships
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/44 - Find the Distance

== by : $44$ ==
;Find the Distance
:A man named Jones set out to walk from $A$ to $B$,
::and on the road he met his friend Kenward, $10$ miles from $A$, who had left $B$ at exactly the same time.
:Jones executed his commission at $B$ and, without delay, set out on his return journey,
::while Kenward as promptly returned from $A$ to $B$.
:They met $12$ miles from $B$.
:Of course, each walked at a uniform rate throughout.
:Now, how far is $A$ from $B$?
:I will show the reader a simple rule by which the distance may be found by anyone in a few seconds without the use of a pencil.
:In fact, it is quite absurdly easy -- when you know how to do it.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $44$. -- Find the Distance
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $72$. Find the Distance
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/44 - Find the Distance/Solution

== by : $44$ ==
;Find the Distance
== Solution ==
:$18$ miles.
The "absurdly simple" rule being referred to by :
:$3$ times the distance from $A$ that they first cross minus the distance from $B$ when they cross for the second time.
== Proof ==
Let $d$ miles be the distance between $A$ and $B$.
Let $V_J$ and $V_K$ miles per hour be the speeds of Jones and Kenward respectively.
Then we have:
For the general case, let $d_1$ and $d_2$ be the distances from $A$ and $B$ that they crossed respectively.
We have:
So the general rule is: 
:$d = 3 d_1 - d_2$
== Sources ==
* : Solutions: $44$. -- Find the Distance
* : Answers: Speed & Distance Puzzles: $72$. Find the Distance
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/45 - The Man and the Dog

== by : $45$ ==
;The Man and the Dog
:"Yes, when I take my dog for a walk," said a mathematical friend, "he frequently supplies me with some interesting problem to solve.
:One day, for example, he waited, as I left the door, to see which way I should go,
::and when I started he raced to the end of the road, immediately returning to me;
::again racing to the end of the road and again returning.
:He did this four times in all, at a uniform speed,
:then ran at my side the remaining distance, which according to my paces measured $27$ yards.
:I afterwards measured the distance from my door to the end of the road and found it to be $625$ feet.
:Now, if I walk $4$ miles per hour, what is the speed of my dog when racing to and fro?"
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $45$. -- The Man and the Dog
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $73$. The Man and the Dog
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/45 - The Man and the Dog/Solution

== by : $45$ ==
;The Man and the Dog
== Solution ==
The speed of the dog is $16$ miles per hour.
== Proof ==
Recall that $1$ mile is $1760$ yards, while $1$ yard is $3$ feet.
Therefore the dog owner is walking at a pace of $21 \, 120$ feet per hour, with $81$ feet remaining after the four trips.
Suppose the dog is running at a pace of $x$ feet per hour.
Suppose after some trips, the owner still $d$ feet of the road to go.
For the next trip, the owner and the dog would have travelled a total of $2 d$ feet.
With their combined speed of $\paren {x + 21 \, 120}$ feet per hour, this trip would have taken $\dfrac {2 d} {x + 21 \, 120}$ hours.
The owner would have walked an additional $\dfrac {21 \, 120 \times 2 d} {x + 21 \, 120}$ feet, so there would be
:$d - \dfrac {21 \, 120 \times 2d } {x + 21 \, 120} = \dfrac {d \paren {x - 21 \, 120} } {x + 21120}$
feet of road to cover, $\dfrac {x - 21 \, 120} {x + 21 \, 120}$ of the previous distance.
Thus after four trips, the remaining distance is $\paren {\dfrac {x - 21120} {x + 21120} }^4$ of the original.
We have:
Hence the speed of the dog is $84 \, 480$ feet per hour, or $16$ miles per hour.
== Sources ==
* : Solutions: $45$. -- The Man and the Dog
* : Answers: $73$. The Man and the Dog
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions