Aquarium Water Level Rise
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.
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Solution
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}} $.
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