AMC 10 · 2007 · #2

Grade 5 geometry-3d
volume-rectangular-prismarea-rectanglesidentify-subproblems identify-subproblems ↑ Prerequisites: volume-rectangular-prism
📏 Short solution 💡 2 insights
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Problem
A tank with a rectangular base holds water part way up. A solid brick is set on the bottom. Find how far the water surface rises.

Pick an answer.

(A)
0.5
(B)
1
(C)
1.5
(D)
2
(E)
2.5
How to solve
Strategy Draw a Diagram

A quick cross-section drawing (Tool #1, Draw a Diagram) shows the key fact: the brick sits fully under the water and shoves aside a chunk of water equal to its own volume, and that water can only go up, forming a thin slab across the entire base. Once that picture is clear, the number splits into two clean subproblems (Tool #7): first the brick's volume, then the height of a slab with that volume spread over the tank's base. Tool #8 (Analyze the Units) confirms the last step, since a volume in cm³ divided by a base area in cm² must give a height in cm.

1STEP 1

See what the brick does to the water

The brick displaces water instead of removing it.

2STEP 2

Find the brick's volume

The displaced amount is the brick's volume.

40×20×10=8000 cm³
3STEP 3

Spread that volume across the base

Spreading it over the base gives a rise of 2, choice (D).

4000 × h = 8000 → h = 8000/4000=2 cm → (D)
Answer
2
The answer is small and positive, which fits — a modest brick should nudge a wide tank's water up only a little. Check the size: the brick's footprint is 40×20=800 cm², one fifth of the tank's 4000 cm² base, and the brick is 10 cm tall, so spreading that 10 cm of solid over a base five times wider gives about 10×800/4000=2 cm — matching. Also the tank does not overflow: the level goes from 40 cm to 42 cm, still under the 50 cm rim, so the whole brick really does stay submerged and the reasoning holds. Answer 2 cm is (D).
💡Key takeaway

A sunk object pushes up its own volume of water, so spread that volume over the tank's base to find how high the water climbs.

  • See what the brick does to the water
  • Find the brick's volume
  • Spread that volume across the base