2007 AMC 10A Problem 3

P1-10~1 minPrint

Solid Geometry Set up the equation/formula and compute; no special trick needed

An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. It is filled with water to a height of $40$ cm. A brick with a rectangular base that measures $40$ cm by $20$ cm and a height of $10$ cm is placed in the aquarium. By how many centimeters does the water rise?

Show solution without answering

Key insight

The brick sits fully under water and pushes up exactly its own volume of water. Spread that volume over the tank's floor to get the rise.

Solution

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 $\boxed{\textbf{(D)}\ 2}$ .

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(D)}\ 2}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Geometry, Chapter 15: Curved Surfaces