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?
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,
Best Answer
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.