## 1. Math Puzzle $6 / 65$

In a lock basin, which is sealed off by gates and has an area of $4000 \mathrm{~m}^{2}$, $80 \mathrm{~m}^{3}$ of sludge (density: $2.5 \frac{\mathrm{t}}{\mathrm{m}^{3}}$) are dredged out and loaded onto a barge.

By how much does the water level drop?
When $V_{1}=80 \mathrm{~m}^{3}$ of sludge is removed from the basin, the water level, with $F$ = basin base area, drops by

$$
h_{1}=\frac{V_{1}}{F}=\frac{80 \mathrm{~m}^{3}}{4000 \mathrm{~m}^{2}}=\frac{1}{50} \mathrm{~m}=2 \mathrm{~cm}
$$

If the $80 \mathrm{~m}^{3}$ of sludge ( $=200 \mathrm{t}$ ) is loaded onto the barge in the basin, the barge will sink by $V_{2}=200 \mathrm{~m}^{3}$, as the water density is $1 \frac{\mathrm{t}}{\mathrm{m}^{3}}$.

If $h$ is the water depth after dredging and $h_{2}$ is the rise in water after loading the barge,

$$
h_{2}=\frac{V_{2}}{F}=\frac{200 \mathrm{~m}^{3}}{4000 \mathrm{~m}^{2}}=\frac{1}{20} \mathrm{~m}=5 \mathrm{~cm}
$$

Both changes overlap. Therefore, at the end of the dredging and after loading the tug barge with the sludge, the water has risen by $3 \mathrm{~cm}$ and not fallen.