## Task 4 - 320724

In an indoor swimming pool, there is also a paddling pool for children. It can be filled up to a marked height by a hot water pipe and a cold water pipe. If only the hot water pipe were used, it would take $12 \frac{1}{2}$ minutes for the water level to reach this height. With only the cold water pipe, it would take 10 minutes.

To achieve a specified water temperature, both pipes were initially used for $2 \frac{1}{2}$ minutes; then the hot water pipe was closed.

Calculate the time that was still needed afterward to fill the rest of the pool to the marked height using only the cold water pipe!
The hot water pipe fills the basin in $12 \frac{1}{2}$ minutes, i.e., in $\frac{25}{2}$ minutes, so it fills $\frac{2}{25}$ of the basin in one minute.

Similarly, the cold water pipe fills $\frac{1}{10}$ of the basin in one minute. Therefore, both pipes together fill $\frac{2}{25} + \frac{1}{10} = \frac{9}{50}$ of the basin in one minute; in $2 \frac{1}{2}$ minutes, $\frac{5}{2} \cdot \frac{9}{50} = \frac{9}{20}$ of the basin.

Thus, the remaining part to be filled is $1 - \frac{9}{20} = \frac{11}{20}$ of the basin. Since $\frac{11}{20} : \frac{1}{10} = \frac{11}{2}$, this remaining part is $\frac{11}{2}$ times as large as $\frac{1}{10}$ of the basin, i.e., the part of the basin that can be filled by the cold water pipe alone in one minute.

Therefore, it took an additional $\frac{11}{2} = 5 \frac{1}{2}$ minutes to fill the remaining part.

Solutions of the 2nd Round 1992 adopted from [5]

### 4.34.3 3rd Round 1992, Class 7