Subfield: Algebra


Question: Amrita and Zhang cross a lake in a straight line with the help of a one-seat kayak. Each can paddle the kayak at $7 \mathrm{~km} / \mathrm{h}$ and swim at $2 \mathrm{~km} / \mathrm{h}$. They start from the same point at the same time with Amrita paddling and Zhang swimming. After a while, Amrita stops the kayak and immediately starts swimming. Upon reaching the kayak (which has not moved since Amrita started swimming), Zhang gets in and immediately starts paddling. They arrive on the far side of the lake at the same time, 90 minutes after they began. Determine the amount of time during these 90 minutes that the kayak was not being paddled.

Solution:
Suppose that Amrita paddles the kayak for $p \mathrm{~km}$ and swims for $s \mathrm{~km}$.

Since Amrita leaves the kayak in the lake and it does not move, then Zhang swims $p \mathrm{~km}$ and paddles the kayak for $s \mathrm{~km}$.

Note that each paddles at $7 \mathrm{~km} / \mathrm{h}$ and each swims at $2 \mathrm{~km} / \mathrm{h}$ and each takes exactly 90 minutes (or 1.5 hours) to complete the trip.

If $s<p$, then Amrita would paddle farther and swim less distance than Zhang and so would reach the other side in less time than Zhang.

If $s>p$, then Zhang would paddle farther and swim less distance than Amrita and so would reach the other side in less time than Amrita.

Since they each take 90 minutes, then we must have $s=p$.

Alternatively, since each paddles at $7 \mathrm{~km} / \mathrm{h}$ and each swims at $2 \mathrm{~km} / \mathrm{h}$ and each takes exactly 90 minutes (or 1.5 hours) to complete the trip, then we obtain the two equations

$$
\frac{p}{7}+\frac{s}{2}=1.5 \quad \frac{p}{2}+\frac{s}{7}=1.5
$$

Using the fact that the right sides of these equations are equal, we obtain

$$
\begin{aligned}
\frac{p}{7}+\frac{s}{2} & =\frac{s}{7}+\frac{p}{2} \\
\frac{s}{2}-\frac{s}{7} & =\frac{p}{2}-\frac{p}{7} \\
s\left(\frac{1}{2}-\frac{1}{7}\right) & =p\left(\frac{1}{2}-\frac{1}{7}\right) \\
s & =p
\end{aligned}
$$

Therefore, $\frac{p}{7}+\frac{p}{2}=1.5$ or $\frac{9}{14} p=1.5=\frac{3}{2}$ and so $p=\frac{7}{3}$.

For Amrita to paddle these $\frac{7}{3} \mathrm{~km}$ at $7 \mathrm{~km} / \mathrm{h}$, it takes $\frac{7 / 3}{7}=\frac{1}{3}$ hour, or 20 minutes.

For Zhang to swim these $\frac{7}{3} \mathrm{~km}$ at $2 \mathrm{~km} / \mathrm{h}$, it takes $\frac{7 / 3}{2}=\frac{7}{6}$ hour, or 70 minutes.

The kayak is not being paddled for the period of time from when Amrita stops paddling to the time when Zhang stops swimming, which is a period of $70-20=50$ minutes.
Let $t_{1}$ hours be the length of time during which Amrita paddles and Zhang swims.

Let $t_{2}$ hours be the length of time during which Amrita swims and Zhang swims; the kayak is not moving during this time.

Let $t_{3}$ hours be the length of time during which Amrita swims and Zhang paddles.

Let $d \mathrm{~km}$ be the total distance across the lake.

Since Amrita paddles at $7 \mathrm{~km} / \mathrm{h}$ and swims at $2 \mathrm{~km} / \mathrm{h}$, then $7 t_{1}+2 t_{2}+2 t_{3}=d$.

Since Zhang paddles at $7 \mathrm{~km} / \mathrm{h}$ and swims at $2 \mathrm{~km} / \mathrm{h}$, then $2 t_{1}+2 t_{2}+7 t_{3}=d$.

Since the kayak travels at $7 \mathrm{~km} / \mathrm{h}$ and does not move while both Amrita and Zhang are swimming, then $7 t_{1}+0 t_{2}+7 t_{3}=d$.

Since Amrita and Zhang each take 90 minutes ( $\frac{3}{2}$ hours) to cross the lake, then the total time gives $t_{1}+t_{2}+t_{3}=\frac{3}{2}$.

From $7 t_{1}+2 t_{2}+2 t_{3}=d$ and $2 t_{1}+2 t_{2}+7 t_{3}=d$, we obtain $7 t_{1}+2 t_{2}+2 t_{3}=2 t_{1}+2 t_{2}+7 t_{3}$ or $5 t_{1}=5 t_{3}$ and so $t_{1}=t_{3}$.

Since $7 t_{1}+2 t_{2}+2 t_{3}=d$ and $7 t_{1}+0 t_{2}+7 t_{3}=d$ and $t_{1}=t_{3}$, then $7 t_{1}+2 t_{2}+2 t_{1}=7 t_{1}+7 t_{1}$ or $2 t_{2}=5 t_{1}$ or $t_{2}=\frac{5}{2} t_{1}$.

Since $t_{1}+t_{2}+t_{3}=\frac{3}{2}$, then $t_{1}+\frac{5}{2} t_{1}+t_{1}=\frac{3}{2}$ or $\frac{9}{2} t_{1}=\frac{3}{2}$ and so $t_{1}=\frac{1}{3}$.

Thus, $t_{2}=\frac{5}{2} \cdot \frac{1}{3}=\frac{5}{6}$ hours (or 50 minutes) is the period of time that the kayak is not moving.

Final answer: 50 minutes