Henry Ernest Dudeney/Modern Puzzles/33 - A Rowing Puzzle/Solution

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Modern Puzzles by Henry Ernest Dudeney: $33$

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?


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:

\(\text {(1)}: \quad\) \(\ds v_u\) \(=\) \(\ds \dfrac d {t_u}\)
\(\text {(2)}: \quad\) \(\ds v\) \(=\) \(\ds \dfrac d t\)
\(\text {(3)}: \quad\) \(\ds v_d\) \(=\) \(\ds \dfrac d {t_d}\)
\(\text {(4)}: \quad\) \(\ds v_s\) \(=\) \(\ds \dfrac d {t_s}\)
\(\text {(5)}: \quad\) \(\ds v_u\) \(=\) \(\ds v - v_s\)
\(\text {(6)}: \quad\) \(\ds v_d\) \(=\) \(\ds v + v_s\)
\(\text {(7)}: \quad\) \(\ds t_u\) \(=\) \(\ds 8 \tfrac 4 7\)
\(\text {(8)}: \quad\) \(\ds t_s - t\) \(=\) \(\ds 7\)


Thus:

\(\ds t_u\) \(=\) \(\ds \frac d {v_u}\) from $(1)$
\(\ds \) \(=\) \(\ds \frac d {v - v_s}\) from $(5)$
\(\ds \) \(=\) \(\ds \frac {60} 7\) from $(7)$
\(\ds t - t_s\) \(=\) \(\ds \frac d v - \frac d {v_s}\) from $(2)$ and $(4)$
\(\ds \) \(=\) \(\ds d \paren {\frac {v_s - v} {v v_s} }\)
\(\ds \) \(=\) \(\ds 7\) from $(8)$


Multiplying the two results above:

\(\ds 60\) \(=\) \(\ds \frac {60} 7 \times 7\)
\(\ds \) \(=\) \(\ds \frac d {v - v_s} \times d \paren {\frac {v_s - v} {v v_s} }\)
\(\ds \) \(=\) \(\ds \frac {d^2} {v v_s}\)
\(\ds \) \(=\) \(\ds t t_s\) from $(2)$ and $(4)$


Substituting $t = t_s + 7$:

\(\ds t t_s\) \(=\) \(\ds t_s \paren {t_s + 7}\)
\(\ds \) \(=\) \(\ds 60\)
\(\ds \leadsto \ \ \) \(\ds t_s^2 + 7 t_s - 60\) \(=\) \(\ds 0\)
\(\ds \paren {t_s + 12} \paren {t_s - 5}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds t_s\) \(=\) \(\ds 5 \text{ or } -12\)

We reject $t_s = -12$ as $t_s > 0$.

Thus $t_s = 5$ and $t = 12$.


Therefore:

\(\ds t_d\) \(=\) \(\ds \frac d {v_d}\) from $(3)$
\(\ds \) \(=\) \(\ds \frac d {v + v_s}\) from $(6)$
\(\ds \) \(=\) \(\ds \frac d {\frac d t + \frac d {t_s} }\) from $(2)$ and $(4)$
\(\ds \) \(=\) \(\ds \paren {\frac 1 t + \frac 1 {t_s} }^{-1}\)
\(\ds \) \(=\) \(\ds \paren {\frac 1 {12} + \frac 1 5 }^{-1}\)
\(\ds \) \(=\) \(\ds \frac {60} {17}\)
\(\ds \) \(=\) \(\ds 3 \tfrac 9 {17}\)

Hence the result.

$\blacksquare$


Sources

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