AMC 8 · 2006 · #21
Easy mode Grade 5Problem
Picture an aquarium shaped like a box. The bottom is a rectangle that measures 100 cm by 40 cm. The aquarium is 50 cm tall.
Water is poured in. The water reaches a depth of 37 cm.
Now a rock is dropped in, and it sinks all the way underwater. The rock takes up 1000 cm3 of space.
The rock pushes the water up. By how many centimeters does the water level rise?
Pick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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.
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