13.069. In one pool there are 200 m $^{3}$ of water, and in another - 112 m $^{3}$. Taps are opened, through which the pools are filled. After how many hours will the amount of water in the pools be the same, if 22 m $^{3}$ more water is poured into the second pool per hour than into the first?
Solution.

Let the amount of water in the pools become equal after $x$ hours. Let $y$ m$^{3}$ of water be added to the first pool per hour, and $y+22$ m$^{3}$ - to the second. According to the condition, $200+x y=112+x(y+22)$, from which $x=4$.

Answer: in 4 hours.