66(1182). Two containers with capacities of 144 liters and 100 liters contain some amount of water. If the larger container is filled to the top with water from the smaller one, then the latter will have $\frac{1}{5}$ of its initial amount of water left. If the smaller container is filled to the top from the larger one, then the larger will have $\frac{7}{12}$ of its initial amount of water left. How many liters of water are in each container
Solution. I with assistance. Let there be $x$ liters of water in the larger vessel, and $y$ liters in the smaller one. According to the condition, $\frac{4}{5} y$ is added to the larger vessel, filling it completely, i.e.,

$$
x+\frac{4}{5} y=144
$$

To the smaller vessel, $\frac{5}{12} x$ is added to fill it to the top, hence, $y+\frac{5}{12} x=100$.

We have the system of equations:

$$
\left\{\begin{array}{l}
x+\frac{4}{5} y=144 \\
y+\frac{5}{12} x=100
\end{array}\right.
$$

Solving the system, we find $x=96, y=60$.

II with assistance. Let there be $x$ liters in the larger vessel and $y$ liters in the smaller one. If ( $144-x$ ) liters are added to the first vessel, then ( $y-(144-x)$ ) liters will remain in the second, which, according to the condition, equals $\frac{1}{5} y$.

We have the equation $x+y-144=\frac{1}{5} y$.

If ( $100-y$ ) liters are transferred from the first to the second vessel, then ( $x-(100-y)$ ) liters, i.e., ( $x+y-100$ ) liters, will remain in the first.

According to the condition, $x+y-100=\frac{7}{12} x$.

We get the system of equations:

$$
\left\{\begin{array}{l}
x+y-144=\frac{1}{5} y \\
x+y-100=\frac{7}{12} x
\end{array}\right.
$$

from which $x=96, y=60$.

Answer: 96 liters, 60 liters.