a) Helena and Gabriel jump simultaneously from a raft into a river and swim in opposite directions. Gabriel swims downstream with the current, and Helena swims upstream against the current, possibly at a different speed. After 5 minutes, they turn around and swim back to the raft, each maintaining a constant speed throughout the time. Who arrives first? (During this time, the raft was anchored, i.e., it remained stationary despite the current.)

b) At his usual speed, a man descends a river that is $15 \mathrm{~km}$ long in 5 hours less than the time he spends swimming the same river in the opposite direction. If he doubles his usual speed, he spends only 1 hour less descending the river than the time spent on the return trip. Assuming the speed of the current remains constant during the trips, what is its value in $\mathrm{km} / \mathrm{h}$?
Solution

a) The distance Gabriel traveled in 5 minutes was $D_{G}=\left(V_{G}+V_{C}\right) \cdot 5$, where $V_{G}$ and $V_{C}$ are the speeds of Gabriel and the current, respectively. Helena traveled $D_{H}=\left(V_{H}-V_{C}\right) \cdot 5$, with $V_{H}$ being her speed. On the return trip, the time $t_{G}$ for Gabriel will be $D_{G}=\left(V_{G}-V_{C}\right) \cdot t_{G}$ and $t_{H}$ for Helena will be $D_{H}=\left(V_{H}+V_{C}\right) \cdot t_{H}$. Notice that

$$
\begin{aligned}
\left(V_{H}-V_{C}\right) \cdot 5 & =\left(V_{H}+V_{C}\right) \cdot t_{H} \\
t_{H} & =5 \cdot \frac{V_{H}-V_{C}}{V_{H}+V_{C}} \\
& 5
$$

Since $t_{H}<t_{G}$, Helena arrives first.

b) Let $c$ and $v$ be the speeds of the river current and the man, respectively, in $\mathrm{km} / \mathrm{h}$. The man takes $\frac{15}{v+c}$ minutes swimming with the current and $\frac{15}{v-c}$ minutes swimming against the current. Thus, the data in the statement can be translated into the following system:

$$
\left\{\begin{array}{l}
\frac{15}{v+c}=\frac{15}{v-c}-5 \\
\frac{15}{2 v+c}=\frac{15}{2 v-c}-1
\end{array}\right.
$$

Multiplying the first equation by $\left(v^{2}-c^{2}\right)$ and the second by $\left(4 v^{2}-c^{2}\right)$, we get

$$
\left\{\begin{array}{l}
15(v-c)=15(v+c)-5\left(v^{2}-c^{2}\right) \\
15(2 v-c)=15(2 v+c)-\left(4 v^{2}-c^{2}\right)
\end{array}\right.
$$

By comparison, it follows that

$$
5\left(v^{2}-c^{2}\right)=\left(4 v^{2}-c^{2}\right)
$$

or,

$$
v^{2}=4 c^{2}
$$

Using one of the equations, we arrive at

$$
\begin{aligned}
15(v-c) & =15(v+c)-5\left(v^{2}-c^{2}\right) \\
-15 c & =15 c-5 \cdot 3 c^{2} \\
15 c^{2}-30 c & = \\
c^{2}-2 c & =0
\end{aligned}
$$

Therefore, the speed of the current is $2 \mathrm{~km} / \mathrm{h}$ (the only positive solution).