## Task B-1.6.

A pool can be filled with two pipes. If both pipes are open for 2 hours, 4200 liters of water will be missing to fill the pool to the top, and if both are kept open for 5 hours, the pool will be full to the top and 3000 liters of water will overflow from the pool. The first pipe releases as much water into the pool in two hours as the second pipe releases in three hours. How many liters of water does the pool hold, and how long does it take for each pipe to fill the pool on its own?

(The rates at which the pipes fill the pool are constant.)
## First solution.

Let $B$ be the amount of water that fits in the pool, $x$ the number of hours needed for the first pipe to fill the pool alone, and $y$ the number of hours needed for the second pipe to fill the pool alone.

The ratio of the filling time (when both pipes are open) is equal to the ratio of the amount of water in the pool, that is,

$$
\frac{B-4200}{B+3000}=\frac{2}{5}
$$

It follows that the amount of water that fits in the pool is $B=9000$ liters.

Then, in one hour, the pipes together discharge

$$
\frac{9000-4200}{2}=\frac{4800}{2}=2400 \text { liters. }
$$

According to the condition of the problem, the ratio of the time needed for each pipe to fill the pool separately is $\frac{x}{y}=\frac{2}{3}$, so $y=\frac{3}{2} x$.

The first pipe fills $\frac{1}{x}$ of the full pool in one hour, or $\frac{B}{x}$ liters.

The second pipe fills $\frac{1}{y}$ of the full pool in one hour, or $\frac{B}{y}$ liters.

Since both pipes together discharge 2400 liters of water into the pool in one hour, we have $B\left(\frac{1}{x}+\frac{1}{y}\right)=2400$

so, due to $(*) B\left(\frac{1}{x}+\frac{2}{3 x}\right)=2400$.

We get $\frac{5}{3 x} \cdot 9000=2400$.

Thus, the time needed for the first pipe to fill the pool alone is $x=\frac{25}{4}=6.25$ hours, and the second pipe will fill the pool in $y=\frac{3}{2} x=\frac{75}{8}=9.375$ hours.