## Task B-1.3.

A ship traveling along a river has covered $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream. For this journey, it took half an hour less than for traveling $30 \mathrm{~km}$ upstream and $21 \mathrm{~km}$ downstream, or half an hour more than for traveling $15 \mathrm{~km}$ upstream and $42 \mathrm{~km}$ downstream, assuming that both the ship and the river move uniformly.

Determine the speed of the ship in still water and the speed of the river.
## Solution.

Let $t$ be the time required for the boat to travel $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream, $v_{R}$ the speed of the river, and $v_{B}$ the speed of the boat. When the boat is traveling upstream, its speed is $v_{B}-v_{R}$, and when it is traveling downstream, its speed is $v_{B}+v_{R}$.

Since $t=\frac{s}{v}$, from the given data, we obtain the following system of equations:

$\left\{\begin{array}{l}t=\frac{24}{v_{B}-v_{R}}+\frac{28}{v_{B}+v_{R}} \\ t+0.5=\frac{30}{v_{B}-v_{R}}+\frac{21}{v_{B}+v_{R}} \\ t-0.5=\frac{15}{v_{B}-v_{R}}+\frac{42}{v_{B}+v_{R}}\end{array}\right.$

By introducing new variables $x=\frac{3}{v_{B}-v_{R}}, y=\frac{7}{v_{B}+v_{R}}$, the system transforms into:

$\left\{\begin{array}{l}t=8 x+4 y \\ t+0.5=10 x+3 y \\ t-0.5=5 x+6 y\end{array}\right.$

Substituting $t$ from the first equation into the remaining two, we get:

$\left\{\begin{array}{l}8 x+4 y+0.5=10 x+3 y \\ 8 x+4 y-0.5=5 x+6 y\end{array}\right.$

$\left\{\begin{array}{l}2 x-y=0.5 \\ 3 x-2 y=0.5\end{array}\right.$

The solution to the last system is (0.5, 0.5). Then we have:

$\frac{3}{v_{B}-v_{R}}=0.5$, hence, $v_{B}-v_{R}=6 \mathrm{~and}$

$\frac{7}{v_{B}+v_{R}}=0.5$, hence, $v_{B}+v_{R}=14$.

The speed of the river is $v_{R}=4 \mathrm{~km} / \mathrm{h}$, and the speed of the boat is $v_{B}=10 \mathrm{~km} / \mathrm{h}$.

## Note:

By substituting $x=\frac{1}{v_{B}-v_{R}}, y=\frac{1}{v_{B}+v_{R}} \mathrm{~and}$ following the same procedure, the initial system transforms into the system $\left\{\begin{array}{l}6 x-7 y=0.5 \\ 9 x-14 y=0.5\end{array}\right.$

The solution to this system is $\left(\frac{1}{6}, \frac{1}{14}\right)$.