13.290. If both pipes are opened simultaneously, the pool will be filled in 2 hours and 24 minutes. In reality, however, only the first pipe was opened initially for $1 / 4$ of the time it takes the second pipe to fill the pool on its own. Then, the second pipe was opened for $1 / 4$ of the time it takes the first pipe to fill the pool on its own, after which it turned out that $11/24$ of the pool's full capacity remained to be filled. How much time is required for each pipe to fill the pool individually?
Solution.

Let's take the entire volume of the pool as 1. Let $x, y$ (liters) per hour leak through the first and second pipe, respectively. According to the conditions, we have the system

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1191.jpg?height=246&width=824&top_left_y=1030&top_left_x=149)

Then for the first pipe, it takes $\frac{1}{x}=4$ (hours) to fill the pool, and $\mathrm{a} \frac{1}{-}=6$ (hours) for the second.

Answer: 4 and $6$ hours.