67 (1183). One vessel contains 49 liters of water, and the other contains 56 liters. If the first vessel is filled to the top with water from the second vessel, the second vessel will be only half full. If the second vessel is filled to the top with water from the first, the first vessel will be only one-third full. What is the capacity of each vessel?
Solution. Method I. Let the capacity of the first vessel be $(49+x)$ liters, and the capacity of the second vessel be $(56+y)$ liters. If $x$ liters are poured from the second vessel into the first, then $(56-x)$ liters will remain in the second vessel, or $\frac{1}{2}(56+y)$. We have the equation:

$$
56-x=\frac{1}{2}(56+y)
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-073.jpg?height=243&width=463&top_left_y=258&top_left_x=545)

| $y$ |
| :---: |
| 56 |

If $y$ liters are poured from the first vessel into the second, then $(49-y)$ liters will remain in the first vessel. By the condition, $49-y = \frac{1}{3}(49+x)$.

We obtain the system of equations:

$$
\left\{\begin{array}{l}
56-x=\frac{1}{2}(56+y) \\
49-y=\frac{1}{3}(49+x)
\end{array}\right.
$$

After simplification, we have:

$$
\left\{\begin{array}{l}
2 x+y=56 \\
x+3 y=98
\end{array}\right.
$$

from which $x=14, y=28$.

The capacity of the first vessel is 63 liters, and the capacity of the second vessel is 84 liters.

Method II. Let the capacity of the first vessel be $x$ liters, and the capacity of the second vessel be $y$ liters. If the first vessel is filled to the top, i.e., $(x-49)$ liters are added, then $(56-(x-49))$ liters, or $(105-x)$ liters, will remain in the second vessel. By the condition, $105-x=\frac{y}{2}$.
![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-073.jpg?height=240&width=990&top_left_y=1505&top_left_x=542)

If the second vessel is filled to the top, i.e., $(y-56)$ liters are added, then $(49-(y-56))$ liters, or $(105-y)$ liters, will remain in the first vessel. By the condition, $105-y=\frac{x}{3}$.

We have the system of equations:

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

from which $x=63, y=84$.

Answer: 63 liters, 84 liters.

$\square$ 68(1184). A bus and a minibus, which leave $M$ and $K$ at 8:40 AM according to the schedule, usually meet at 8:52 AM. One day, the minibus left 8 minutes late and met the bus at 8:57 AM. Find the speed of the bus and the minibus, given that the distance from $M$ to $K$ is 24 km.

Solution. Let the speed of the bus be $v_{1}$ km/min, and the speed of the minibus be $v_{2}$ km/min. Since the bus and the minibus usually meet after 12 minutes (8:52 AM - 8:40 AM = 12 minutes), we have the equation $12 v_{1} + 12 v_{2} = 24$.

In the case when the minibus left late, it traveled for 9 minutes (8:57 AM - 8:48 AM = 9 minutes) before meeting the bus, and the bus traveled for 17 minutes (8:57 AM - 8:40 AM = 17 minutes) before meeting the minibus. By the condition, $17 v_{1} + 9 v_{2} = 24$.

We obtain the system of equations:

$$
\left\{\begin{array}{l}
12 v_{1} + 12 v_{2} = 24, \\
17 v_{1} + 9 v_{2} = 24,
\end{array}\right.
$$

from which $v_{1} = \frac{3}{4}$ km/min, or 45 km/h; $v_{2} = \frac{5}{4}$ km/min, or 75 km/h.

Remark. If the speed of the bus $v_{1}$ and the speed of the minibus $v_{2}$ are expressed in kilometers per hour, the solution of the problem reduces to the system:

$$
\left\{\begin{array}{l}
\frac{1}{5} v_{1} + \frac{1}{5} v_{2} = 24 \\
\frac{17}{60} v_{1} + \frac{3}{20} v_{2} = 24
\end{array}\right.
$$

Answer: the speed of the bus is 45 km/h, the speed of the minibus is 75 km/h.