nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/24 - Simple Arithmetic/Solution

== by : $24$ ==
;"Simple" Arithmetic
== Solution ==
:$20$ and $64$.
== Proof ==
Let their ages be $a$ and $b$.
Then:
You can either factorise $1280$ into its two factors which differ by $44$, which brings us no further than where we started from, or we use the Quadratic Formula to smash this exquisite gem with a mallet:
:$b = \dfrac {-44 \pm \sqrt {44^2 + 4 \times 1280} } 2 = -22 \pm 42$
and we can pick out what we want from the debris.
== Sources ==
* : Solutions: $24$. -- "Simple" Arithmetic
* : Answers: $35$. "Simple" Arithmetic
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/25 - A Dreamland Clock

== by : $25$ ==
;A Dreamland Clock
:In a dream, I was travelling in a country where they had strange ways of doing things.
:One little incident was fresh in my memory when I awakened.
:I saw a clock and announced the time as it appeared to be indicated.
:: but my guide corrected me.
:He said, "You are apparently not aware that the minute hand always moves in the opposite direction to the hour hand.
:Except for this improvement, our clocks are precisely the same as those you have been accustomed to."
:Now, as the hands were exactly together between the hours of $4$ and $5$ o'clock,
::and they started together at noon,
::what was the real time?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $25$. -- A Dreamland Clock
* : Arithmetical and Algebraical Problems: Clock Puzzles: $46$. A Dreamland Clock
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Clock Puzzles
Henry Ernest Dudeney/Modern Puzzles/25 - A Dreamland Clock/Solution

== by : $25$ ==
;A Dreamland Clock
== Solution ==
:$36 \tfrac {12} {13}$ minutes after $4$.
== Proof ==
We have that:
:the minute hand takes $1$ hour to go $360 \degrees$ around the dial anticlockwise
:the hour hand takes $1$ hour to go $30 \degrees$ around the dial clockwise.
Hence in one minute:
:the minute hand travels $6 \degrees$ anticlockwise
:the hour hand travels $\dfrac 1 2 \degrees$ clockwise.
At $4$ o'clock, the minute hand was on the $12$ and the hour hand was on the $4$.
Let the minute hand and hour hand be coincident at some point in time $t$ minutes after $4$, between $4$ and $5$.
We need to find $t$.
Let $\theta$ degrees be the angle the hour hand has moved (clockwise), in time $t$.
Then we have:
:$\theta = \dfrac t 2 \degrees$
During $t$, the minute hand, moving backwards as it does, has moved anticlockwise from $12$ o'clock to $5$ o'clock plus $30 - \theta \degrees$.
That is, the minute hand has moved $7 \times 30 + \paren {30 - \theta} \degrees$, that is, $\paren {240 - \theta} \degrees$.
Hence:
:$t = \dfrac {240 - \theta} 6$
as the minute hand moves $1 \degrees$ every $\dfrac 1 6$ of a minute.
Hence we have:
That is, the time is $36 \tfrac {12} {13}$ minutes after $4$.
== Sources ==
* : Solutions: $25$. -- A Dreamland Clock
* : Answers: $46$. A Dreamland Clock
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/26 - What is the Time

== by : $26$ ==
;What is the Time?
:At what time are the two hands of a clock so situated that,
::reckoning as minute points past $\textit {XII}$,
::one is exactly the square of the distance of the other?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $26$. -- What is the Time?
* : Arithmetical and Algebraical Problems: Clock Puzzles: $47$. What is the Time?
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Clock Puzzles
Henry Ernest Dudeney/Modern Puzzles/26 - What is the Time/Solution

== by : $26$ ==
;What is the Time?
== Solution ==
The time can be:
:$5$ seconds past $\text {XII}$
or:
:$6 \tfrac 3 4$ minutes past $\text {IX}$.
== Proof ==
The minute hand by definition moves $1$ minute point per minute.
The hour hand moves $5$ minute points per $60$ minutes, or $\dfrac 1 {12}$ minute points per minute.
The question does not state whether the hand whose minute point is the square of the distance of the other is the hour hand or the minute hand, so the analysis will be done on both.
Let $t = h : m$ be the time of day in $h$ hours and $m$ minutes of the particular instant of interest.
Thus:
:$m$ is the number of minute points of the minute hand past $\text {XII}$
:$5 h + \dfrac m {12}$ is the number of minute points of the hour hand past $\text {XII}$.
while we are constrained by the inequalities:
:$0 \le m < 60$
:$0 \le h < 12$
and such that $h$ is an integer.
=== Hour Hand Position is Square of Minute Hand Position ===
Let $t = h : m$ be the time of day at which the position of the hour hand is the square of the distance of the minute hand.
We have that:
Hence from the Quadratic Formula:
We can solve this quadratic in $m$ its discriminant $D$ is non-negative.
That is, :
which is in fact any value of $h$ we like.
We can try various $h$ and see what happens.
We start by calculating the discriminant $D$ for each $h$ from $0$ to $11$ and see what happens:
The exposition specifically states:
:one is exactly the square of the distance of the other
which in the naïve language of means that the times must be expressible as rational numbers.
Of the above discriminants, only $1 = 1^2$ and $25921 = 161^2$ are square numbers.
So, when $h = 0$ we have:
This leads to:
:the degenerate solution of $\text {XII}$
:the time of $\dfrac 1 {12}$ of a minute, that is, $5$ seconds, past $\text {XII}$.
 implicitly ignores the degenerate solution, and accepts the fractional minute under protest.
When $h = 9$, we have:
Hence the time is $6 \tfrac 3 4$ minutes past $\text {IX}$.
=== Minute Hand Position is Square of Hour Hand Position ===
Let $t = h : m$ be the time of day at which the position of the minute hand is the square of the distance of the hour hand.
We have that:
We can solve this quadratic in $m$ its discriminant is non-negative.
That is, :
The above can happen only when $h = 0$, which gives us:
But such an $m$ does not satisfy the inequality:
:$0 \le m < 60$
Hence there is no time of day at which the position of the minute hand is the square of the distance of the hour hand. 
== Sources ==
* : Solutions: $26$. -- What is the Time?
* : Answers: $47$. What is the Time?
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/27 - The First-Born's Legacy

== by : $27$ ==
;The First-Born's Legacy
:Mrs. Goodheart gave birth to twins.
:The clock showed clearly that Tommy was born about an hour later than Freddy.
:Mr. Goodheart, who died a few months earlier, had made a will leaving $\pounds 8400$,
::and had taken the precaution to provide for the possibility of there being twins.
:In such a case the money was to be divided in the following proportions:
::two-thirds to the widow,
::one-fifth to the first-born,
::one-tenth to the other twin,
::and one-twelfth to his brother.
:Now, what is the exact amount that should be settled on Freddy?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Clock Puzzles: $27$. -- The First-Born's Legacy
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Clock Puzzles
Category:Testament Problems
Henry Ernest Dudeney/Modern Puzzles/27 - The First-Born's Legacy/Solution

== by : $27$ ==
;The First-Born's Legacy
== Solution ==
Freddy gets $\pounds 800$.
 performs the arithmetic that determines the exact sums to be bestowed:
:Mrs. Goodheart gets $\dfrac {43} {63} \times \pounds 8400 = \pounds 5333 \tfrac 1 3 = \pounds 5333 \ 6 \shillings 8 \oldpence$
:the first-born gets $\dfrac {12} {63} \times \pounds 8400 = \pounds 1600$
:the second-born gets $\dfrac 6 {63} \times \pounds 8400 = \pounds 800$
:his brother gets $\dfrac 5 {63} \times \pounds 8400 = \pounds 666 \tfrac 2 3 = \pounds 666 \ 13 \shillings 4 \oldpence$
He then goes on to say:
:But the sting is in the tail, and the question is indeterminate.
What is "indeterminate" is what is meant by "his brother".
 goes on to explain that the twins were born either side of the time at which the clock went back at the end of British Summer Time.
Hence Tommy was actually the first-born and Freddy the younger of the two.
A pencil-written marginal note in the second-hand copy possessed by prime.mover says:
:Silly
with which prime.mover concurs.
== Proof ==
We first note that $\dfrac 2 3 + \dfrac 1 5 + \dfrac 1 {10} + \dfrac 1 {12} = \dfrac {40 + 12 + 6 + 5} {60} = \dfrac {63} {60}$.
As this is greater than $1$, we must interpret the question as that the legacies are in the appropriate proportions.
Hence:
:Mrs. Goodheart gets $\dfrac {43} {63} \times \pounds 8400 = \pounds 5333 \tfrac 1 3 = \pounds 5333 \ 6 \shillings 8 \oldpence$
:the first-born gets $\dfrac {12} {63} \times \pounds 8400 = \pounds 1600$
:the second-born gets $\dfrac 6 {63} \times \pounds 8400 = \pounds 800$
:his brother gets $\dfrac 5 {63} \times \pounds 8400 = \pounds 666 \tfrac 2 3 = \pounds 666 \ 13 \shillings 4 \oldpence$
According to the question, it appears that in fact Freddy is the first-born.
However, pulls a new fact out of his hat:
:Tommy was actually born when the clock showed just before $2$ a.m. on $21$st September $1924$
:At $2$ a.m. the clocks went back $1$ hour, as British Summer Time ended at that point
:Freddy was born soon after $1$ a.m. by the clock, but before $2$ a.m.
Hence Tommy is actually the first-born, and not Freddy.
So Freddy gets $\pounds 800$.
== Sources ==
* : Solutions: $27$. -- The First-Born's Legacy
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/28 - Hill Climbing

== by : $28$ ==
;Hill Climbing
:Weary Willie went up a certain hill at the rate of $1 \tfrac 1 2$ miles per hour
::and came down at the rate of $4 \tfrac 1 2$ miles per hour,
::so that it took him just $6$ hours to make the double journey.
:Now, how far was it to the top of the hill?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $28$. -- Hill Climbing
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $56$. Hill Climbing
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/28 - Hill Climbing/Solution

== by : $28$ ==
;Hill Climbing
== Solution ==
:$6 \tfrac 3 4$ miles.
He goes up in $4 \tfrac 1 2$ hours and back down again in $1 \tfrac 1 2$ hours.
== Proof ==
Let $d$ miles be the distance to the top of the hill.
Let $t_1$ be the time taken to reach the top.
Let $t_2$ be the time taken to reach the bottom again.
The assumption has to be that no time is taken to rest at the top.
We have:
== Sources ==
* : Solutions: $28$. -- Hill Climbing
* : Answers: $56$. Hill Climbing
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/29 - Timing the Motor-car

== by : $29$ ==
;Timing the Motor-car
:"I was walking along the road at $3 \tfrac 1 2$ miles an hour," said Mr. Pipkins,
::"when the motor-car dashed past me and only missed me by a few inches."
:"Do you know what speed it was going?" asked his friend.
:"Well, from the moment it passed me to its disappearance round a corner I took $27$ steps, and walking on reached that corner with $135$ steps more."
:"Then, assuming you walked, and the car ran, each at a uniform rate, we can easily work out the speed."
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $29$. -- Timing the Motor-car
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $57$. Timing the Motor-car
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/29 - Timing the Motor-car/Solution

== by : $29$ ==
;Timing the Motor-car
== Solution ==
:$21$ miles an hour.
== Proof ==
Let $v$ be the speed of the motor-car.
Let $d$ be the distance to the corner from where the near-accident happened.
Let $t$ be the time it takes Pipkins to take $1$ step.
We have:
:$v = \dfrac d {27 t}$
while Pipkins' speed gives:
:$3 \tfrac 1 2 = \dfrac d {\paren {27 + 135} t}$
which leads us to:
:$v = 6 \times 3 \tfrac 1 2$
and the answer.
== Sources ==
* : Arithmetical and Solutions: $29$. -- Timing the Motor-car
* : Answers: $57$. Timing the Motor-car
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/2 - Pocket-money

== by : $2$ ==
;Pocket-money
:I went down the street with a certain amount of money in my pocket,
:and when I returned home I discovered that I had spent just half of it,
:and that I now had just as many shillings as I previously had pounds,
:and half as many pounds as I then had shillings.
:How much money had I spent?
=== Click here for solution ===
== Also see ==
* Dollars and Cents, which follows exactly the same pattern.
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $2$. -- Pocket-money
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/2 - Pocket-money/Solution

== by : $2$ ==
;Pocket-money
== Solution ==
The amount spent was $\pounds 9 \ 19 \shillings$
== Proof ==
Let $S$ be the amount I started with: $S_L$ pounds and $S_s$ shillings.
Let $F$ be the amount I finished with: $F_L$ pounds and $F_s$ shillings.
We have that:
:$2 F = S$
and so the amount spent is equal to $2 F - F = F$, that is, the amount I finished with.
We recall the conversion factors:
:$20$ shillings make one pound.
Hence any shilling quantities in either $S$ or $F$ cannot be greater than $19$.
That is:
:$S_s < 20$
:$F_s < 20$
It is assumed that $S_s$ and $F_s$ are both integers, that is: both $S$ and $F$ are a whole number of shillings.
We are given that:
We have that:
The smallest values of $S_L$ and $S_s$ that satisfy the above equation are:
As $S_s \le 19$ it follows that there can be no other solution.
Hence:
which is equal to the amount spent.
== Sources ==
* : Solutions: $2$. -- Pocket-money
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/30 - The Staircase Race

== by : $30$ ==
;The Staircase Race
:This is a rough sketch of a race up a staircase in which $3$ men took part.
:Ackworth, who is leading, went up $3$ risers at a time, as arranged;
:Barnden, the second man, went $4$ risers at a time,
:and Croft, who is last, went $5$ at a time.
::
:Undoubtedly Ackworth wins.
:But the point is,
:How many risers are there in the stairs, counting the top landing as a riser?
:I have shown only the top of the stairs.
:There may be scores, or hundreds, of risers below the line.
:It was not necessary to draw them, as I only wanted to show the finish.
:But it is possible to tell from the evidence the fewest possible risers in that staircase.
:Can you do it?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $30$. -- The Staircase Race
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $58$. The Staircase Race
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/30 - The Staircase Race/Solution

== by : $30$ ==
;The Staircase Race
== Solution ==
There are $19$ risers.
== Proof ==
We refer to Ackworth, Barnden and Croft as $A$, $B$ and $C$.
Let $N$ be the number of risers.
The diagram shows that:
:$A$ has $1$ odd step at the top
:$B$ will have $3$ such odd steps
:$C$ will have $4$ such steps.
Thus we have:
:$N \equiv 1 \pmod 3$
:$N \equiv 3 \pmod 4$
:$N \equiv 4 \pmod 5$
Notice that:
:$N + 1 \equiv 0 \pmod 4$
:$N + 1 \equiv 0 \pmod 5$
so $N + 1$ must be a multiple of $20$.
The smallest such $N$ is $19$, and we see that it satisfies the first condition as well.
== Sources ==
* : Solutions: $30$. -- The Staircase Race
* : Answers: $58$. The Staircase Race
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/31 - A Walking Puzzle

== by : $31$ ==
;A Walking Puzzle
:A man set out at noon to walk from Appleminster to Boneyham,
::and a friend of his started at $2$ p.m. on the same day to walk from Boneyham to Appleminster.
:They met on the road at $5$ minutes past $4$ o'clock
::and each man reached his destination at exactly the same time.
:Can you say what time they both arrived?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $31$. -- A Walking Puzzle
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $59$. A Walking Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/31 - A Walking Puzzle/Solution

== by : $31$ ==
;A Walking Puzzle
== Solution ==
:$19:00$, that is, $7$ p.m.
== Proof ==
Let Appleminster and Boneyham be referred to as $A$ and $B$.
Let $t$ be the number of hours after $12:00$ at which both men arrived.
Let $d_1$ be the distance from $A$ to $B$.
Let $d_2$ be the distance from $A$ at which the two met.
Let $v_1$ be the speed of the first man.
Let $v_2$ be the speed of the second man.
We have:
The solution $t = \dfrac 7 6$ leads to a time of $13:10$ which is invalid.
Hence we have $t = 7$, that is, $19:00$ or $7$ p.m.
== Sources ==
* : Solutions: $31$. -- A Walking Puzzle
* : Answers: $59$. A Walking Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/32 - Riding in the Wind

== by : $32$ ==
;Riding in the Wind
:A man on a bicycle rode a mile in $3$ minutes with the wind at his back,
::but it took him $4$ minutes to return against the wind.
:How long would it take him to ride a mile if there was no wind?
:Some will say that the average of $3$ and $4$ is $3 \tfrac 1 2$, and it would take him $3 \tfrac 1 2$ minutes.
:That answer is entirely wrong.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $32$. -- Riding in the Wind
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $60$. Riding in the Wind
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Travelling With and Against the Flow Puzzles
Henry Ernest Dudeney/Modern Puzzles/32 - Riding in the Wind/Solution

== by : $32$ ==
;Riding in the Wind
== Solution ==
:$3 \tfrac 3 7$ minutes.
== Proof ==
Let $v_w$ miles per minute be the speed of the wind.
Let $v_m$ miles per minute be the speed of the man without the wind.
Let $t$ minutes be the time taken to ride $1$ mile with no wind.
We have:
== Sources ==
* : Solutions: $32$. -- Riding in the Wind
* : Answers: $60$. Riding in the Wind
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/33 - A Rowing Puzzle

== by : $33$ ==
;A Rowing Puzzle
:A crew can row a certain course upstream in $8 \tfrac 4 7$ minutes,
::and, if there were no stream, they could row it in $7$ minutes less than it takes them to drift down the stream.
:How long would it take to row down with the stream?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $33$. -- A Rowing Puzzle
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $61$. A Rowing Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Travelling With and Against the Flow Puzzles
Henry Ernest Dudeney/Modern Puzzles/33 - A Rowing Puzzle/Solution

== by : $33$ ==
;A Rowing Puzzle
== Solution ==
:$3 \tfrac 9 {17}$ minutes.
== Proof ==
Let the course be $d$ miles long.
Let $v$ miles per minute be the speed the crew would be able to row in still water.
Let $v_s$ miles per minute be the speed of the stream.
Let $v_u$ miles per minute be the speed upstream.
Let $v_d$ miles per minute be the speed downstream.
Let $t$ minutes be the time taken to row $d$ miles in still water.
Let $t_u$ minutes be the time taken to row $d$ miles upstream.
Let $t_d$ minutes be the time taken to row $d$ miles downstream.
Let $t_s$ minutes be the time taken to drift $d$ miles downstream without rowing.
We have:
Thus:
Multiplying the two results above:
Substituting $t = t_s + 7$:
We reject $t_s = -12$ as $t_s > 0$.
Thus $t_s = 5$ and $t = 12$.
Therefore:
Hence the result.
== Sources ==
* : Solutions: $33$. -- A Rowing Puzzle
* : Answers: $61$. A Rowing Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/34 - The Moving Stairway

== by : $34$ ==
;The Moving Stairway
:On one of the moving stairways on the London Tube Railway I find that if I walk down $26$ steps I require $30$ seconds to get to the bottom,
::but if I take $34$ steps I require only $18$ seconds to reach the bottom.
:What is the height of the stairway in steps?
:The time is measured from the moment the top step begins to descend to the time I step off the last step at the bottom onto the level platform.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $34$. -- The Moving Stairway
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $62$. The Escalator
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Travelling With and Against the Flow Puzzles
Henry Ernest Dudeney/Modern Puzzles/34 - The Moving Stairway/Solution

== by : $34$ ==
;The Moving Stairway
== Solution ==
:$46$ steps.
== Proof ==
Let $S$ be the number of steps on the stairway.
Let $v$ be the speed of descent of the stairway in steps per second.
We are given that:
== Sources ==
* : Solutions: $34$. -- The Moving Stairway
* : Answers: $62$. The Escalator
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/35 - Sharing a Bicycle

== by : $35$ ==
;Sharing a Bicycle
:Two brothers had to go on a journey and arrive at the same time.
:They had only a single bicycle, which they rode in turns, each rider leaving it in the hedge when he dismounted for the one walking behind to pick up,
::and walking ahead himself, to be again overtaken.
:What was the best way of arranging their distances?
:As their walking and riding speeds were the same, it is extremely easy.
:Simply divide the route into any even number of equal stages and drop the bicycle at every stage, using the cyclometer.
:Each man would then walk half-way and ride half-way.
:But here is a case that will require a little more thought.
:Anderson and Brown have to go $20$ miles and arrive at exactly the same time.
:They have only one bicycle.
:Anderson can only walk $4$ miles an hour,
::while Brown can walk $5$ miles an hour,
::but Anderson can ride $10$ miles an hour to Brown's $8$ miles an hour.
:How are they to arrange the journey?
:Each man always either walks or rides at the speeds mentioned, without any rests.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Locomotion and Speed Puzzles: $35$. -- Sharing a Bicycle
* : Arithmetical and Algebraical Problems: Speed & Distance Puzzles: $63$. Sharing a Bicycle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/35 - Sharing a Bicycle/Solution

== by : $35$ ==
;Sharing a Bicycle
== Solution ==
Anderson cycles $11 \tfrac 1 9$ miles, drops the bike, and walks the rest of the way, which is $8 \tfrac 8 9$ miles.
Thus he cycles for $1 \tfrac 1 9$ hours and walks for $2 \tfrac 2 9$ hours.
Brown thus walks $11 \tfrac 1 9$ miles, picks up the bike, and walks the rest of the way, which is $8 \tfrac 8 9$ miles.
Thus he also walks for $2 \tfrac 2 9$ hours and cycles for $1 \tfrac 1 9$ hours.
== Proof ==
Let Anderson and Brown be denoted by $A$ and $B$ respectively.
To keep it simple, we will assume one changeover, and that $A$ starts by cycling.
Let $d$ miles be the distance from the start to where $A$ dismounts to start walking.
Let $t$ hours be the time taken to do the total journey.
Let $t_a$ hours be the time taken by $A$ to travel $d$.
Let $t_b$ hours be the time taken by $B$ to travel $d$.
We have:
== Sources ==
* : Solutions: $35$. -- Sharing a Bicycle
* : Answers: $63$. Sharing a Bicycle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions