## Solution.

Determine the base of the system in which the task is possible.

$$
\begin{aligned}
\frac{1}{4} 112_{(b)} & =\frac{1}{3} 53_{(b)} \\
3\left(b^{2}+b+2\right) & =4(5 b+3) \\
3 b^{2}-17 b & -6=0 \\
b & =6
\end{aligned}
$$

The task is possible in a system with base 6.

Let the number of girls be $x$. Then,

$$
\begin{array}{r}
\frac{1}{3} 53_{(6)}=x \\
5 \cdot 6+3=3 x \\
x=11
\end{array}
$$

In the decimal system, the number of girls is 11. In the system with base 6, the number of girls is 15.

(1 point)

Accept either of these two answers.
Task B-4.2. In one group, every fourth person out of a total of 112 students is a girl. In the second group, there are a total of 53 students, and the ratio of boys to girls is 2:1. In both groups, the number of girls is the same. How is this possible and how many girls are there in each group?