AMC 8 · 2006 · #21

Easy mode Grade 5

Problem

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 $1000 \text{ cm}^3$ of space.

The rock pushes the water up. By how many centimeters does the water level rise?

Pick an answer.

(A)

0.25

(B)

0.5

(C)

(D)

1.25

(E)

2.5

View mode:

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

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

#7 Identify Subproblems 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.

$$\text{Base area} = 100 \times 40 = 4000 \text{ cm}^2$$

💡 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.

#10 Create a Physical Representation 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.

$$\text{Slab volume} = \text{rock volume} = 1000 \text{ cm}^3$$

💡 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.

#10 Create a Physical Representation 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.

$$\text{Slab volume} = \text{Base area} \times h \;\Rightarrow\; 1000 = 4000 \times h$$

💡 Any thin slab with a rectangular base has volume "area times thickness" — the same Grade 5 volume formula as for a full rectangular prism.

#7 Identify Subproblems 5.NBT.B.7 Step 4

Solve for $h$ by dividing both sides by $4000$.

$$h = \dfrac{1000}{4000} = \dfrac{1}{4} = 0.25 \text{ cm} \;\Rightarrow\; \textbf{(A)}$$

💡 Dividing $1000$ by $4000$ and writing the result as the decimal $0.25$ is Grade 5 decimal arithmetic — a quarter of a centimeter rise.

[1] #7 5.NBT.B.5 Compute the base area. The aquarium's base is a $100 \text{ cm} \times 40 \text{

[2] #10 5.MD.C.5 Apply the displacement idea. When the rock is fully submerged, it pushes up exac

[3] #10 5.MD.C.5 Use the rectangular-prism volume formula on the slab. Its volume is base area $\

[4] #7 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.5 Fluently 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.5 Relate 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.7 Add, 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 5)

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