1. In an aquarium shaped like a rectangular prism, with a length of $75 \mathrm{~cm}$, a width of $28 \mathrm{~cm}$, and a height of $40 \mathrm{~cm}$, water is present up to a height of $25 \mathrm{~cm}$. A stone is dropped into the aquarium, and the water level rises by a quarter of its height. Fish are then added to the aquarium, occupying 1 \% of the aquarium's volume. The allowed water level in the aquarium must be 10 \% lower than the height of the aquarium to prevent the fish from jumping out. How many liters of water can still be added to the aquarium after the stone and fish have been added?
## First Solution.

If we drop a stone into the aquarium, the water level rises by a quarter of its height, i.e., $25: 4=6.25 \mathrm{~cm}$, so the new water height is $31.25 \mathrm{~cm}$.

If the fish occupy $1 \%$ of the aquarium's volume, then by height they occupy the 100th part of $40 \mathrm{~cm}$ or $0.4 \mathrm{~cm}$.

After the stone and fish are placed in the aquarium, the water height reaches $31.25+0.4=31.65 \mathrm{~cm}$.

$10 \%$ of the aquarium's height is $0.1 \cdot 40=4 \mathrm{~cm}$, so the maximum allowed water height in the aquarium is $40-4=36$ cm, and the difference to this height is $36-31.65=4.35 \mathrm{~cm}$.

The volume of water we can add is $V=a \cdot b \cdot c=75 \cdot 28 \cdot 4.35=9135 \mathrm{~cm}^{3}=9.135$ liters.