Task B-1.3. A pool is filled by two pumps. If both pumps are open for 20 minutes, 1000 liters would still be missing to fill the pool completely, and if both are open for 70 minutes, 250 liters would overflow from the full pool. The first pump can fill in 2 minutes the same part of the pool that the second pump can fill in 3 minutes. How much time does each of them need to fill the entire pool on their own?
## Solution.

## Let

$x$ - the time (in minutes) required for the first pump to fill the pool alone

$y$ - the time (in minutes) required for the second pump to fill the pool alone

$V$ - the volume of water the pool can hold

Then $\frac{V}{x}$ is the volume of water the first pump fills in one minute, and $\frac{V}{y}$ is the volume of water the second pump fills in one minute. From the conditions in the problem, we get the system of equations:

$$
\left\{\begin{array}{l}
20 \cdot\left(\frac{V}{x}+\frac{V}{y}\right)=V-1000 \\
70 \cdot\left(\frac{V}{x}+\frac{V}{y}\right)=V+250 \\
2 \cdot \frac{V}{x}=3 \cdot \frac{V}{y}
\end{array}\right.
$$

From the last equation, we get

$$
\frac{V}{x}=\frac{3}{2} \cdot \frac{V}{y}, \text { or } x=\frac{2 y}{3}
$$

so the system becomes

$$
\left\{\begin{array}{l}
20 \cdot\left(\frac{5}{2} \cdot \frac{V}{y}\right)=V-1000 \\
70 \cdot\left(\frac{5}{2} \cdot \frac{V}{y}\right)=V+250
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
50 \cdot \frac{V}{y}=V-1000 \\
175 \cdot \frac{V}{y}=V+250
\end{array}\right.
$$

Dividing these two equations, we get

$$
\frac{2}{7}=\frac{V-1000}{V+250}
$$

or

$$
\begin{aligned}
& 7 V-7000=2 V+500 \\
& V=1500
\end{aligned}
$$

Then

$$
50 \cdot \frac{1500}{y}=500
$$

so the time required for the second pump to fill the pool alone is $y=150 \mathrm{~min}$.

The time for the first pump is $x=\frac{2 y}{3}=100 \mathrm{~min}$.