2007 AMC 10A Problem 3

When you drop a solid object under water, the water level goes up by exactly enough to make room for the object.

1. Is the brick fully under water? The water is $40$ cm deep and the brick is only $10$ cm tall. Yes, the whole brick is covered.
2. Brick volume. $40 \times 20 \times 10 = 8000$ cubic cm. That is how much "extra" volume now sits below the water surface.
3. Floor area of the tank. $100 \times 40 = 4000$ square cm.
4. Rise. Spreading $8000$ cubic cm over $4000$ square cm raises the level by
$$\frac{8000}{4000} = 2 \text{ cm}.$$
5. Check for spilling. The new depth is $42$ cm, less than the $50$ cm height, so no water spills out.

The answer is $\boxed{\textbf{(D)}\ 2}$.

Height of a layer of water = volume $\div$ floor area. Putting a fully submerged object in the tank is the same as adding its volume of water. The tank's height only matters to check that nothing overflows.

The brick's volume is $40 \cdot 20 \cdot 10 = 8000$ cubic centimeters. The water is $40$ cm deep and the brick is only $10$ cm tall, so the brick is completely under water and pushes exactly $8000$ cubic centimeters of water upward.

That extra volume spreads over the aquarium's full base, $100 \cdot 40 = 4000$ square centimeters. If the level rises by $h$, then
$$4000\,h = 8000 \quad\Longrightarrow\quad h = 2.$$
The new depth $42$ cm is still below the $50$ cm rim, so nothing spills.

The answer is $\boxed{\textbf{(D)}\ 2}$.