## Task 2 - 320832

To be able to fill a container with water, a number of pipes are to be laid. The water should flow evenly through each pipe (i.e., the same amount of water in the same time). In one hour, the same amount of water should flow through each pipe as through any other pipe.

There are three suggestions for the number of pipes. According to the second suggestion, using two pipes fewer than in the first suggestion would take two hours longer to fill the container compared to the first. According to the third suggestion, using four pipes more than in the first suggestion would take two hours less to fill the container compared to the first.

Determine from these data the number of pipes and the time to fill the container in the first suggestion!
For the first proposal, let $n$ pipes and a filling time of $t$ hours be planned. Each pipe then fills $\frac{1}{n \cdot t}$ of the container volume per hour.

The $n-2$ pipes of the second proposal fill $\frac{n-2}{n \cdot t}$ of the container volume per hour; thus, the container is filled in $\frac{n \cdot t}{n-2}$ hours according to this proposal. Since this is 2 hours more than in the first proposal, we have

$$
\frac{n \cdot t}{n-2}=t+2 \quad \text { and therefore } \quad t=n-2
$$

Accordingly, it follows that: According to the third proposal, the container is filled in $\frac{n \cdot t}{n+4}$ hours, and we have

$$
\frac{n \cdot t}{n+4}=t-2 \quad ; \quad t=\frac{n}{2}+2
$$

From (1) and (2), we get $n-2=\frac{n}{2}+2$ and thus, from (1), $n=8$ and $t=6$.

Therefore, the first proposal planned to achieve a filling time of 6 hours with 8 pipes.