An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume $1000 \text{cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?  Express your answer as a decimal to the nearest 100th.
Using the volume formula $lwh = V$, the volume of water in the aquarium is $100 \times 40 \times 37 = 148{,}000
\text{ cm}^3$. When the rock is put in, the water and the rock will occupy a box-shaped region with volume $148{,}000 + 1000 = 149{,}000
\text{ cm}^3$. The volume of the water and the rock is $100 \times 40
\times h$, where $h$ is the new height of the water. $\text{The new
volume} = 4000h = 149{,}000 \text{ cm}^3$, so the new height is $$h=\frac{149000}{4000}=37.25 \text{ cm}.$$After adding the rock, the water rises $37.25-37=\boxed{0.25\text{ cm}}$.

\[ OR \]Because the shape of the rock is irrelevant, we may assume that the rock is shaped like a rectangular box with base measuring $100
\text{ cm} \times 40 \text{ cm}$ and height $h$ cm. Using the volume formula, $100 \times 40 \times h =1000$, so $h =
\frac{1000}{100\times 40}=0.25 \text{ cm}$. When the rock is put into the aquarium, the water level will rise by $\boxed{0.25\,\text{ cm}}$.