Locomotion and Speed Puzzles

Every puzzle in this short chapter turns on one line of school arithmetic: distance is speed multiplied by time. The difficulty is never the formula. It lies in keeping clear which quantity is shared by two journeys and which is not, and most of the traps Dudeney sets depend on a reader averaging speeds that ought never to be averaged.

The units are the old ones. A mile is 1,760 yards, and speeds are in miles an hour unless a puzzle says otherwise. The money in the last puzzle is twenty shillings to the pound, so £3 is sixty shillings.

No. 67. Average Speed

No. 68. The Two Trains

No. 69. The Three Villages

No. 70. Drawing Her Pension

No. 71. Sir Edwyn de Tudor

No. 72. The Hydroplane Question

No. 73. Donkey Riding

No. 74. The Basket of Potatoes

No. 75. The Passenger’s Fare

Average Speed

Average speed means total distance over total time, and the two halves of the trip share their distance, not their time. Let the route be $d$ miles. The outward run takes $d/10$ hours and the return $d/15$, so the whole trip of $2d$ miles takes $$\frac{d}{10}+\frac{d}{15}=\frac{3d+2d}{30}=\frac{d}{6}\ \text{hours},$$ and the average speed is $2d\div(d/6)=12$ miles an hour, whatever $d$ is. With Dudeney’s sixty miles: six hours out, four back, 120 miles in ten hours.

The hasty answer of $12\tfrac12$ averages the two speeds as though each were kept up for the same time. The car spends longer at the slower speed, so the slow speed carries more weight. The correct figure, $2\cdot 10\cdot 15/(10+15)$, is the harmonic mean of the two speeds, and it is always below their ordinary mean unless they are equal. Dudeney is right.

Answer 12 miles an hour

The Two Trains

The crux is that after the meeting each train runs over exactly the ground the other has already covered. Say the trains meet $t$ hours after starting, the train that finishes an hour after the meeting runs at $u$ and the other at $v$. The first train still has to cover the stretch the second ran in $t$ hours, and it does so in one hour; the second train covers the first’s stretch in four hours. So $$vt=u\cdot 1,\qquad ut=v\cdot 4 .$$ The first gives $t=u/v$ and the second $t=4v/u$. Setting these equal, $(u/v)^2=4$, so $u=2v$, and the trains met after $t=2$ hours. The train that arrives an hour after the meeting runs exactly twice as fast as the other, which agrees with Dudeney.

The same argument with times $a$ and $b$ after the meeting gives the ratio $\sqrt{b/a}$, which is why the stationmaster could answer so quickly: one and four are both squares.

Answer One train runs twice as fast as the other.

The Three Villages

Drop the perpendicular from Cheesebury $C$ to the direct road $AB$; it is 12 miles long and cuts the triangle into two right-angled triangles sharing that side. Write $b=AC$ and $a=CB$, with $a+b=35$ and $b<a$, and let the foot of the perpendicular be $p$ miles from $A$ and $q$ miles from $B$, so $b^2=144+p^2$ and $a^2=144+q^2$. The figure shows the answer the argument will reach.

The crux is that $p$ and $q$ must themselves be whole numbers. Since the foot lies on the road, $AB=p+q=n$, a whole number. Then $p-q=(p^2-q^2)/n=(b^2-a^2)/n$ is rational, so $p$ is rational; and a rational number whose square is the whole number $b^2-144$ must be a whole number. The same goes for $q$. So $b^2-144$ and $a^2-144$ are both perfect squares.

Now $12\le b<17\tfrac12$.

Running through $b=12,\dots,17$, the values of $b^2-144$ are 0, 25, 52, 81, 112 and 145, so $b$ is 12, 13 or 15, with $a=23$, 22 or 20. The corresponding $a^2-144$ are 385, 340 and 256, and only the last is a square. Hence $b=15$, $a=20$, $p=9$, $q=16$, and the direct road is $9+16=25$ miles. Acrefield to Cheesebury is 15 miles, Cheesebury to Butterford 20, and Acrefield to Butterford 25, a 3, 4, 5 triangle scaled by five. This is Dudeney’s answer, and it is the only one.

A reader who takes “twelve miles to the left of the road” to mean twelve miles from the line of the road, extended if need be, finds a second triangle. With the foot of the perpendicular 9 miles beyond Acrefield, on the far side from Butterford, the direct road is $16-9=7$ miles, and the sides 15, 20, 7 also satisfy every number in the puzzle. But then the nearest point of the road to Cheesebury is Acrefield itself, 15 miles away, so Cheesebury is not twelve miles from the road at all. Measured to the road that exists, the answer is unique.

Answer Acrefield to Cheesebury 15 miles, Cheesebury to Butterford 20, Acrefield to Butterford 25

Drawing Her Pension

The crux is to price one mile of hill, up and back. Climbing a mile at $1\tfrac12$ miles an hour takes $\tfrac23$ of an hour, and coming down it at $4\tfrac12$ takes $\tfrac29$. One mile there and back therefore costs her $$\tfrac23+\tfrac29=\tfrac89\ \text{of an hour},$$ and in six hours she manages $6\div\tfrac89=\tfrac{27}{4}=6\tfrac34$ miles of hill. She spends $4\tfrac12$ hours climbing and $1\tfrac12$ coming down. Dudeney gives the same $6\tfrac34$ miles. Her average speed over the round trip is $13\tfrac12\div 6=2\tfrac14$ miles an hour, the harmonic mean of her two speeds, as in No. 67.

Answer $6\tfrac34$ miles

Sir Edwyn de Tudor

The two rides differ by two hours, from an hour early to an hour late, and the crux is to see how much of that difference each mile contributes. A mile takes $\tfrac1{15}$ of an hour at the faster speed and $\tfrac1{10}$ at the slower, a difference of $$\tfrac1{10}-\tfrac1{15}=\tfrac1{30}\ \text{of an hour}.$$ Two hours of difference therefore needs $2\div\tfrac1{30}=60$ miles. At fifteen miles an hour the ride takes four hours, an hour less than the time allowed, so he has five hours, and must ride at $60\div 5=12$ miles an hour. To arrive at five o’clock he sets out at noon. Dudeney’s answer is the same. Note once more that the right speed, 12, is the harmonic mean of 10 and 15 and not their average: the two wrong rides are equally wrong in time, not in speed.

Answer 60 miles (leaving at noon and riding at 12 miles an hour)

The Hydroplane Question

The crux is that the engine’s own speed is exactly halfway between the ground speeds with and against the wind, since the wind adds to one and subtracts the same amount from the other. Work in miles a minute. With the wind he covers five miles in ten minutes, $\tfrac12$ a mile a minute; against it, five miles in sixty minutes, $\tfrac1{12}$. So $$\text{engine}=\tfrac12\bigl(\tfrac12+\tfrac1{12}\bigr)=\tfrac7{24},\qquad \text{wind}=\tfrac12\bigl(\tfrac12-\tfrac1{12}\bigr)=\tfrac5{24}$$ miles a minute, that is $17\tfrac12$ and $12\tfrac12$ miles an hour. In a calm the ten miles take $$10\div\tfrac7{24}=\tfrac{240}{7}=34\tfrac27\ \text{minutes}.$$ This matches Dudeney. It is worth noticing that the windy round trip took seventy minutes, half as long again. A wind along the course always slows a round trip, because the machine spends longer being held back than being helped. The whole calculation assumes, as Dudeney says, that the wind blows along the line of flight.

Answer $34\tfrac27$ minutes

Donkey Riding

Call the four quarter-mile times $q_1,q_2,q_3,q_4$ minutes. The crux is that the second and third facts together make the first half-mile worth two of the last quarters: $q_1+q_2=q_3+q_4=2q_3$. The first fact then reads $$q_1+q_2+q_3=2q_3+q_3=3q_3=6\tfrac34,$$ so $q_3=q_4=2\tfrac14$, the first half-mile took $4\tfrac12$ minutes, and the whole mile took $4\tfrac12+2\tfrac14+2\tfrac14=9$ minutes.

There are four unknowns and only three facts, so something must be left open, and it is the split of the first half-mile: any $q_1$ between 0 and $4\tfrac12$ works, with $q_2=4\tfrac12-q_1$. The judges’ facts happen to fix the total regardless. Dudeney says exactly this, and his nine minutes is right.

Answer 9 minutes

The Basket of Potatoes

The crux is where the potatoes lie. The gaps are the odd numbers $1,3,5,\dots$, and the first $k$ odd numbers add up to $k^2$ (each new odd number turns a $k\times k$ square of dots into a $(k+1)\times(k+1)$ one). So the potato after $k$ gaps lies $k^2$ yards from the basket, and the fifty potatoes lie at $0,1,4,9,\dots,49^2$ yards; the last is 2,401 yards out.

For each potato the boy walks out and back, twice its distance. His total walk is $$2\bigl(0^2+1^2+\dots+49^2\bigr)=2\cdot\frac{49\cdot 50\cdot 99}{6}=80{,}850\ \text{yards},$$ using $1^2+\dots+m^2=m(m+1)(2m+1)/6$. With $n$ potatoes the same sum is $n(n-1)(2n-1)/3$, which is Dudeney’s rule. Since $80{,}850=45\times1{,}760+1{,}650$ and $1{,}650=\tfrac{15}{16}\times1{,}760$, the walk is $45\tfrac{15}{16}$ miles, in agreement with Dudeney.

Answer 80,850 yards, or $45\tfrac{15}{16}$ miles

The Passenger’s Fare

The trip has four equal legs, Addleford to Bakenham, Bakenham to Clinkerville, back to Bakenham, and back to Addleford, so each leg costs fifteen shillings. Tompkins rides on the middle two, thirty shillings’ worth, and Smithers rides on all four. The question has no single forced answer, since “reasonable” is a matter of agreement, and that is presumably why the two men argued. The crux is to name the rule of sharing, and then the arithmetic is easy.

Dudeney’s rule is that those in the car share the cost of each leg equally. Smithers pays the two outer legs alone and half of each middle leg; Tompkins pays half of thirty shillings, which is fifteen shillings. A second argument lands in the same place. If Tompkins had hired the car alone for his part of the journey, it would have cost him thirty shillings; if he joins a hire Smithers has already made, he adds nothing to the bill. Neither man is entitled to be counted as first, so split the difference: Tompkins pays the average of thirty and nothing, fifteen shillings. (This averaging over the order of joining is what economists call the Shapley value.)

A third rule, charging by the mile travelled, divides the £3 between Smithers’s four legs and Tompkins’s two, and asks twenty shillings of Tompkins. It overcharges him, since it takes no account of the fact that the car had to make his journey anyway: Smithers’s bill would fall from sixty shillings to forty, although carrying Tompkins cost nothing extra. Dudeney’s fifteen shillings is the fairer answer, and the one both sound arguments give.

Answer 15 shillings