33. In a constructed school boarding house, several rooms were double-occupancy, and the rest were triple-occupancy. In total, the boarding house is designed for more than 30 but fewer than 70 places. When $\frac{1}{5}$ of the available places were occupied, the number of unoccupied places turned out to be the same as the total number of places would be if the number of double-occupancy rooms was the same as the actual number of triple-occupancy rooms, and the number of triple-occupancy rooms was the same as the actual number of double-occupancy rooms. For how many places is the boarding house built?
33. Let there be $x$ two-person rooms and $y$ three-person rooms in the boarding school. Then the total number of places in the boarding school is $2 x + 3 y$. If $\frac{1}{5}$ of the places were occupied, then $\frac{4}{5}$ of the places remained unoccupied, i.e., $\frac{4}{5}(2 x + 3 y)$. According to the condition, $\frac{4}{5}(2 x + 3 y) = 3 x + 2 y$, from which we get $2 y = 7 x$. Considering that $x$ and $y$ are natural numbers, we will have $x_{1} = 2 ; y_{1} = 7$; $x_{2} = 4 ; y_{2} = 14$ and so on. If $x = 2$ and $y = 7$, then the total number of places is 25; if $x = 4, y = 14$, then the number of places is 50; if $x = 6 ; y = 21$, then the number of places is 75. Since only $50(30 < 50 < 70)$ satisfies the condition, the total number of places in the boarding school was 50.