Problem 2. A brother and sister wanted to buy a ball. If the brother were to buy the ball alone, he would be short of $\frac{5}{19}$ of the ball's price, and if the sister were to buy it alone, she would be short of $\frac{1}{4}$ of the ball's price. Together, they had 185 denars more than the price of the ball. They bought the ball together, with the brother paying $45 \%$ of the ball's price, and the sister paying the remaining amount. How much money was left with the brother, and how much with the sister?
Solution. If the price of the ball is $x$ denars, then from the condition of the problem it follows that the brother has $\frac{14}{19} x$, and the sister has $\frac{3}{4} x$ denars. According to this,

$$
\begin{array}{ll}
\frac{14}{19} x+\frac{3}{4} x=x+185 & \Leftrightarrow \frac{14}{19} x+\frac{3}{4} x-x=185 \Leftrightarrow \frac{56+57-76}{76} x=185 \Leftrightarrow \\
\frac{37}{76} x=185 & \Leftrightarrow x=380
\end{array}
$$

The brother had $\frac{14}{19} \cdot 380=280$ denars, and the sister had $\frac{3}{4} \cdot 380=285$ denars. For the ball, the brother paid $\frac{45}{100} \cdot 380=171$ denars, and the sister paid $380-171=209$ denars. Therefore, the brother was left with 280-171=109 denars, and the sister was left with $285-209=76$ denars.