## Task B-1.2.

Two pipes fill a pool in 4 hours. One of these pipes fills the pool by itself in 5 hours and 15 minutes. How long does it take for the other pipe to fill the pool by itself?
## Solution.

If two pipes can fill the pool in 4 hours, then they will fill $\frac{1}{4}$ of the pool in 1 hour.

If one pipe can fill the pool in $5 \frac{1}{4}$ hours, then that pipe will fill $\frac{1}{5 \frac{1}{4}}=\frac{4}{21}$ of the pool in 1 hour.

Let the second pipe fill the pool in
Then $\frac{4}{21}+\frac{1}{t}=\frac{1}{4}$

1 point

$\frac{1}{t}=\frac{1}{4}-\frac{4}{21}=\frac{5}{84}$

1 point

$t=\frac{84}{5}=16 \frac{4}{5}$ hours $=16$ hours and 48 minutes.

1 point

Note:

The student can, after determining the fraction of the pool filled by both pipes or just one in 1 hour (2 points), conclude that the second pipe fills $\frac{1}{4}-\frac{4}{21}=\frac{5}{84}$ of the pool in 1 hour. This conclusion is worth 2 points.

After that, the conclusion that the time required for the second pipe to fill the pool is the reciprocal of the obtained fraction is worth 1 point. The final solution, i.e., the calculated time required for the second pipe to fill the pool $\frac{84}{5}=16 \frac{4}{5}$ hours $=16$ hours and 48 minutes, is worth 1 point.

Time recorded in hours or in hours and minutes is accepted.