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A steamboat leaving pier $1$ takes $20$ hours to go against the current upriver to pier $2$. It can return downriver with the current from pier $2$ to pier $1$ in $15$ hours. If there were no current, how long would it take for the steamboat to travel between the $2$ piers?

I am told that I cannot divide the sun of $20$ and $15$ by two to get $17.5$. I am also told that I can find the distance between the two piers by:

$$\left(\dfrac x{20}\right) + \left(\dfrac x{15}\right) = 35$$ where $x$ is the distance between the two piers.

But I’m so confused by this. $\dfrac x{20}$ and $\dfrac x{15}$ are rates of speed on the way up and back, respectively, so why would their sum need to equal $35$, the total number of hours it took to go up and back?

Question, transcribed above Solution, page 1, transcribed below Solution, page 2, transcribed below

SOLUTION

On being asked this question, many people will add $20$ and $15$ together and then divide the result by $2$, obtaining $17.5$. They will then say that it would take the steamboat $17.5$ hours to complete the trip if there were no tide.

It comes as a surprise to many to learn that this result is incorrect.

The surprising answer is that if there were no tide, the steamboat would complete the trip between piers in $17.142857$ hours.

On the upriver trip, the steamboat would travel three-quarters of the journey in $15$ hours. On the downriver trip, the steamboat traveled the entire journey in 15 hours. Since the steamboat is traveling with the tide and against the tide for the same period of time, the effect of the tide cancels. Therefore, the steamboat would cover 1 3/4 trips in 30 hours. It would therefore do one entire trip in 30/1 3/4), or 17.142857, hours if there were no tide.

This puzzle is interesting because, in addition to obtaining the required solution, we can also learn the distance between the 2 piers and the speed of the flow of the river.

We know from the question that the steamboat took a total of 35 hours to do the upriver trip and the downriver journey.

Let $x$ equal the distance from pier to pier. This allows us to write the following equation:

$$\frac{x}{20}+\frac{x}{15}=35$$

This may be rewritten as $$15x+20x=10500$$ $$x=300$$

Therefore, the distance between both piers is 300 miles. The total journey upriver and downriver is 600 miles, and this is accomplished in 35 hours. Thus, the average speed is 600 divided by 35, or 17.142857 miles per hour. Therefore, if there were no current, the steamboat would be traveling at 17.142857 miles per hour, which confirms the answer that we have already arived at. It would therefore complete the 600-mile trip in 600/17.142857, or 35, hours.

What is the speed of the river? From the question, we know that the steamboat took 20 hours to go 300 miles upriver and 15 hours to go 300 miles downriver. The steamboat traveling at 17.142857 miles per hour would do the upriver trip in 17.5 hours if there were no tide. But it takes the steamboat 20 hours, or an extra 2.5 hours, to do this journey with the tide flowing against it. Thus, the tide is flowing at the rate of 2.5 miles per hour. On the downriver trip, the steamboat traveling at the rate 17.142857 miles per hour in no tide would do the journey in 17.5 hours, but with the tide in its favor, it does the trip in 2.5 hours less, or 15 hours.

SOURCE

Henry Ernest Dudeney, More Puzzles and Curious Problems, edited by Martin Gardner (London: Fontana Books, 1970), puzzle 42, pp. 20, 118,

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  • $\begingroup$ You cannot equate rates with times. What you can do is equate the distances traveled upstream and downstream. How can you find the distance traveled in terms of the boat's speed in still water and the river's current? $\endgroup$ Jul 12, 2023 at 2:36
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    $\begingroup$ There's not enough information to determine the absolute distance or the absolute boat speed or river current, but only the ratio between them that matters. Maybe the solution in your book is choosing an arbitrary value for one of the above in order to determine the others. It would help if you included more of the book's solution. $\endgroup$
    – Karl
    Jul 12, 2023 at 3:00
  • $\begingroup$ But there is a way to determine the absolute boat speed. See answer below. But I still don’t get the equation in the second photo because it implies 300 is the only distance when there are clearly many distances possible where it takes 20 hours up and 15 back. $\endgroup$
    – Mick
    Jul 12, 2023 at 3:55
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    $\begingroup$ For starters, x has a unit of 'square hours', not miles, which should immediately tell you something has gone wrong. Nowhere in the problem does the word 'mile' appear until they proclaim that the answer 300 is in miles somehow. $\endgroup$
    – mm201
    Jul 13, 2023 at 15:21
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    $\begingroup$ It's incredible how many bad books with wrong formulas are entering the market. I read a book last year in which I found >100 issues just by accident while reading it during my holidays. Back home, I took a closer look at those issues and identified even more - so many in fact that I stopped documenting them. $\endgroup$ Jul 13, 2023 at 17:42

6 Answers 6

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The book is making a mistake. That entire second half is nonsense, and the formula is incorrect.

In particular, it’s impossible to determine the speed of the river since there is no unit of distance in the original problem. For example, doubling all (non-specified) velocities and distances will give the exact same times.

This does give one nice way to solve the problem. You can choose any nice value of the distance (or do as they did by making up a nonsensical formula to get some random $x$), and then use that to solve for the time.

Here’s an example using that method with different numbers. $120$ is a nice number since $15$ and $20$ divide it and the ratios are even. Let’s define a mile as $1/120$ the distance, so by definition, they are $120$ miles apart. Then, it moves at $120/20=6$ mph against the river and $120/15=8$ mph with the river, so the boat moves $(6+8)/2=7$ mph and the river travels at $(8-6)/2=1$ mph. Thus, without current the trip takes $120/7=17.1428\dots$ hours as desired.

Note that the original problem from Henry Dudeney (compiled by Martin Gardner) specifies the speed, so the exact distances could be computed and were relevant. I suspect that the writer removed them when he saw they weren’t essential to the core of the problem but didn’t properly fix his solution.

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    $\begingroup$ @Mick This answer more directly addresses your query; please treat my answer as supplementary. $\endgroup$
    – ryang
    Jul 12, 2023 at 4:01
  • $\begingroup$ This helps a ton thanks, Eric. And thank you Ryan, yours was helpful too! Maybe there should be a more formal PSA that this book has errors. I’ve found more than this, I think. Unfortunate cause of headaches! $\endgroup$
    – Mick
    Jul 12, 2023 at 4:11
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    $\begingroup$ It seems that all modern textbooks ignore the harmonic mean. The simple answer to this problem is "the harmonic mean of the rates" $2/(1/20+1/15) = 17.1428...$ If this were taught sensibly, most of the "hard" story problems in HS algebra texts would become "easy." $\endgroup$
    – B. Goddard
    Jul 12, 2023 at 12:01
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    $\begingroup$ The original problem was from Henry Earnest Dudeney. Martin Gardner just compiled the book of puzzles the author found it in. $\endgroup$ Jul 12, 2023 at 13:42
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    $\begingroup$ It's pretty entertaining that the book concludes "the distance between both piers is 300 miles" when the unit "miles" hasn't been mentioned yet - it could just as easily conclude the distance is 300 kilometers, 300 inches, or 300 light-years. $\endgroup$ Jul 13, 2023 at 19:49
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Equating the distance between the two piers: $$(b-r)20=(b+r)15\\b=7r.$$ So, on still water, the one-way journey time (distance/speed) is $$\frac{(b+r)15}{b}\text{ hrs}=\frac{120}7\text{ hrs}.$$


Addendum

it doesn’t make sense to me that the distance can only be 300. There are many distances where it can take 20 hours up and 15 hours back, not just one. So how can this equation be true if it yields only one distance?

Yes, that's true. The author's formulation

$\dfrac x{\color\red{20}} + \dfrac x{\color\red{15}} = 35,$ where $x$ is the distance between the two piers

seems to be based on their observation, “on the upriver trip, the steamboat would travel three-quarters of the journey in $15$ hrs [instead of the $20$ hrs taken for the downriver trip]”, that is, $$\text{upriver speed}=(b-r)=\frac{15}{20}(b+r)=\color\red{\frac{15}{20}}\times\text{downriver speed},$$ but this doesn't mean that the respective speeds are $15$ and $20.$

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  • $\begingroup$ If I plug in the two distances in terms of b, it reduces to 2b = 35, and b = 17.5 which the author says is incorrect. $\endgroup$
    – Mick
    Jul 12, 2023 at 3:41
  • $\begingroup$ Also it doesn’t make sense to me that the distance can only be 300. There are many distances where it can take 20 hours up and 15 hours back, not just one. So how can this equation be true if it yields only one distance? $\endgroup$
    – Mick
    Jul 12, 2023 at 3:45
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The book formula is horribly wrong, mistakes were made.

No need for arbitrary, just use $1$ or $x$

Speed upstream: $x/20=v-curr$

Speed downstream: $x/15=v+curr$

$2v=x/20+x/15$

$v=35x/600$

$t=x/v=600/35=17+\frac{1}{7}$

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  • $\begingroup$ I'm not entirely sure but I believe the $+curr$ and $-curr$ should be swapped, but yeah that shouldn't matter for the result $\endgroup$
    – Ivo
    Jul 14, 2023 at 8:07
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The problem can be phrased as a nonsingular system consisting of 2 equations with 2 unknowns. Let $d$ denote the distance between the two piers, let $v$ denote the speed of the ferry in the absence of any current and let $w$ denote the speed of the river. Then $$\begin{bmatrix} d \\ d \end{bmatrix} = \begin{bmatrix} 15(v+w) \\ 20(v-w) \end{bmatrix} = \begin{bmatrix} 15 & 15 \\ 20 & -20 \end{bmatrix} \begin{bmatrix} v \\ w \end{bmatrix}.$$ The determinant of the matrix is $-600 \not = 0$. By Cramer's rule we have $$\begin{bmatrix} v \\ w \end{bmatrix} = - \frac{1}{600} \begin{bmatrix} -20 & -15 \\ -20 & 15 \end{bmatrix}\begin{bmatrix} d \\ d \end{bmatrix} = \begin{bmatrix} \frac{35}{600} d \\ \frac{5}{600}d \end{bmatrix}.$$ We observe that if $d$ is given, then $v$ and $w$ are uniquely determined. If $d$ is unknown, then we cannot compute $v$ and $w$, but in any case $v/w = 7$. If there was no current, then the trip from one pier to the other would require $t$ units of time, where $$t = \frac{d}{v} = \frac{600}{35} = \frac{120}{7} = 17 + \frac{1}{7}.$$

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EDIT: all this was said faster and better in @Erics solution

The original question makes no mention of any distance unit. So you can't determine the distance. Just go through the printed solution to the original problem: why does the author introduce miles as the unit? Why not furlongs or leagues or kilometres or parsecs or...?

In many problems like this, you can pick an arbitrary value for the distance. $60$ is convenient for calculation purposes (see below), and call the unit of length a $\text{Len}$.

So, the upstream velocity is $60/20=3\text{ len/hour}$

Downstream velocity is $60/15=4\text{ len/hour}$

So the boat goes at $3.5\text{ len/hour}$, while the current either adds or subtracts $0.5\text{ len/hour}$. Nice neat do-it-in-your-head numbers,

If the current stops, the one way trip would take $60/3.5 = 17.142857 \text{ hours}$

Now, if you were to go back and pick any number for the distance other than $60$, the intermediate calculations would look quite different, but the final no-current time would be the same.

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  • $\begingroup$ Agreed but I still can’t see why the author would equate the sum of the rates to the total travel time to arrive at an arbitrary 300. $\endgroup$
    – Mick
    Jul 12, 2023 at 4:27
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    $\begingroup$ Because the author is wrong. You can't even match the units. The math equation adds velocities to get time. That doesn't work, ever. The sad part is that this wrong equation gives a solution for the distance, and it works, so it must be right, right? The fact that any other number would work isn't apparent. $\endgroup$
    – DJohnM
    Jul 12, 2023 at 4:31
  • $\begingroup$ Precisely what I thought. Shouldn’t these things be peer reviewed before selling at book stores? Another discussion I suppose. $\endgroup$
    – Mick
    Jul 12, 2023 at 4:40
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Let the distance between the piers be d, and the distance the current takes you in an hour c.

So in 20 hours you cover d + 20c going upstream. In 15 hours you cover d - 15c going downstream. In 60 hours you cover either 3d + 60c or 4d - 60c, and in 120 hours you cover the sum 7d. So to cover the distance without any current would take you 120 / 7 hours = 17 1/7 hours.

Note the "60" is the least common multiple of 20 and 15. To get a simple formula you would calculate what happens in 20 x 15 hours. If it takes X hours upstream and Y hours downstream, then in XY hours you go Yd + XY * c upstream or Xd - XY * c downstream, the sum is (X + Y)*d in 2XY hours, so it takes 2XY / (X+Y) hours to cover the distance d without current.

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