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2007 AMC 10A Problem 3

2007 AMC 10A Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Solid Geometry · difficulty 1 of 5 · about 1 min · position P1-10

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?

A)
$0.5$
B)
$1$
C)
$1.5$
D)
$2$
E)
$2.5$

Answer

Key insight

The submerged brick displaces its own volume, so the rise equals brick volume divided by the aquarium's base area.

Solution

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

Why this works

Water is incompressible: adding a submerged solid raises the surface by (volume added) $\div$ (surface area). The aquarium's height only matters as a sanity check that the water does not overflow. Always confirm the object is fully submerged before using its whole volume.

The trap

Dividing the brick's volume by the base area of the brick, or using the aquarium's height 50 as if it mattered.

Common mistakes

Dividing the brick's volume by the base area of the brick, or using the aquarium's height $50$ as if it mattered.
Forgetting to check that the brick is fully submerged; if it were taller than the water, only part of its volume would count.

Techniques

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