AMC 8 · 2006 · #21
Grade 5 geometry-3dProblem
An aquarium has a rectangular base that measures cm by cm and has a height of cm. The aquarium is filled with water to a depth of cm. A rock with volume is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A rectangular aquarium has a base of $100 \text{ cm} \times 40 \text{ cm}$ and a height of $50$ cm. It is filled with water to a depth of $37$ cm. A rock of volume $1000 \text{ cm}^3$ is dropped in and fully submerged. By how many centimeters does the water level rise?
Givens: Base of the aquarium: $100 \text{ cm} \times 40 \text{ cm}$; Aquarium height: $50$ cm; initial water depth: $37$ cm; Rock volume: $1000 \text{ cm}^3$, and the rock is fully submerged; Answer choices: (A) $0.25$, (B) $0.5$, (C) $1$, (D) $1.25$, (E) $2.5$
Unknowns: The rise in water level, in centimeters
Understand
Restated: A rectangular aquarium has a base of $100 \text{ cm} \times 40 \text{ cm}$ and a height of $50$ cm. It is filled with water to a depth of $37$ cm. A rock of volume $1000 \text{ cm}^3$ is dropped in and fully submerged. By how many centimeters does the water level rise?
Givens: Base of the aquarium: $100 \text{ cm} \times 40 \text{ cm}$; Aquarium height: $50$ cm; initial water depth: $37$ cm; Rock volume: $1000 \text{ cm}^3$, and the rock is fully submerged; Answer choices: (A) $0.25$, (B) $0.5$, (C) $1$, (D) $1.25$, (E) $2.5$
Plan
Primary tool: #10 Create a Physical Representation
Secondary: #7 Identify Subproblems
The story is physical: a rock goes in, the water has nowhere to go but up. Tool #10 (Create a Physical Representation) makes the key insight concrete — picture the rock sliding in and shoving an equal volume of water upward. That displaced water spreads across the same rectangular base, forming a thin slab whose volume equals the rock's volume. Tool #7 (Identify Subproblems) splits the work into two clean pieces: (i) compute the base area, (ii) divide the rock's volume by that area to get the slab's thickness, which is the rise. The given numbers $37$ cm and $50$ cm are deliberately included to confirm the rock fits and the water does not overflow — they do not enter the calculation.
Execute — Answer: A
5.NBT.B.5 Step 1 - Compute the base area.
- The aquarium's base is a $100 \text{ cm} \times 40 \text{ cm}$ rectangle, and the water sits on top of this same rectangle no matter how deep it gets.
💡 Multiplying length by width for a rectangle is the Grade 5 multi-digit multiplication move — and the base area stays the same as the water level changes.
5.MD.C.5 Step 2 - Apply the displacement idea.
- When the rock is fully submerged, it pushes up exactly its own volume of water.
- Picture that pushed-up water sitting on top of the existing water as a thin slab.
- The slab covers the same $4000 \text{ cm}^2$ base, and its volume equals the rock's volume.
💡 Think of the rock as silently swapping places with water: where rock now sits, water used to sit, and that water has to go somewhere — up.
5.MD.C.5 Step 3 - Use the rectangular-prism volume formula on the slab.
- Its volume is base area $\times$ thickness, where the thickness $h$ is exactly the rise we want.
💡 Any thin slab with a rectangular base has volume "area times thickness" — the same Grade 5 volume formula as for a full rectangular prism.
5.NBT.B.7 Step 4 Solve for $h$ by dividing both sides by $4000$.
💡 Dividing $1000$ by $4000$ and writing the result as the decimal $0.25$ is Grade 5 decimal arithmetic — a quarter of a centimeter rise.
5.NBT.B.5 Compute the base area. The aquarium's base is a $100 \text{ cm} \times 40 \text{ 5.MD.C.5 Apply the displacement idea. When the rock is fully submerged, it pushes up exac 5.MD.C.5 Use the rectangular-prism volume formula on the slab. Its volume is base area $\ 5.NBT.B.7 Solve for $h$ by dividing both sides by $4000$. Review
Reasonableness: Sanity check the size. The base area $4000 \text{ cm}^2$ is huge compared with the rock's volume $1000 \text{ cm}^3$, so spreading $1000$ over $4000$ should give a thin film — and indeed $0.25$ cm (a quarter of a centimeter) is tiny. Also, the new water level would be $37 + 0.25 = 37.25$ cm, well below the $50$ cm aquarium height, so nothing overflows — the given height and initial depth are red herrings the problem includes to confirm the setup, not to enter the formula. The other answer choices come from likely mistakes: $0.5$ from using $50 \times 40 = 2000$ instead of $100 \times 40$, $1$ and $1.25$ from miscomputing the area, $2.5$ from $1000 / 400$.
Alternative: Tool #13 (Convert to Algebra) gives the same path more compactly. Let $h$ be the rise in cm. The displaced water occupies a rectangular prism of dimensions $100 \times 40 \times h$, with volume equal to the rock's $1000 \text{ cm}^3$. Then $100 \cdot 40 \cdot h = 1000$, so $4000h = 1000$ and $h = 0.25$ cm — answer (A). The two approaches agree because Tool #10 just walks through what the algebra is modeling step by step.
CCSS standards used (min grade 5)
5.NBT.B.5Fluently multiply multi-digit whole numbers using the standard algorithm (Computing the rectangular base area as $100 \times 40 = 4000 \text{ cm}^2$.)5.MD.C.5Relate volume to the operations of multiplication and addition and solve real-world and mathematical problems involving volume (Modeling the displaced water as a rectangular prism with volume = (base area) $\times$ (height rise), and setting that volume equal to the rock's $1000 \text{ cm}^3$.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Solving $4000h = 1000$ to get $h = \tfrac{1}{4} = 0.25$ cm in decimal form.)
⭐ A submerged rock pushes up its own volume of water, which spreads across the aquarium's base as a thin slab — divide the rock's volume by the base area and you have the rise.
⭐ A submerged rock pushes up its own volume of water, which spreads across the aquarium's base as a thin slab — divide the rock's volume by the base area and you have the rise.