AMC 8 · 2008 · #16
Grade 6 geometry-3dProblem
A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?
Pick an answer.
AMC 8 2008 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: Seven unit cubes are glued together: one central cube has six neighbors, one stuck to each of its six faces (above, below, left, right, front, back). Find the ratio of the total volume (cubic units) to the total surface area (square units).
Givens: The shape uses exactly $7$ unit cubes; Each unit cube has volume $1$ and $6$ unit-square faces of area $1$; From the figure, one cube is in the center and the other $6$ cubes are attached one to each face of the center cube; Answer choices: (A) $1:6$, (B) $7:36$, (C) $1:5$, (D) $7:30$, (E) $6:25$
Unknowns: The ratio (volume) $:$ (surface area)
Understand
Restated: Seven unit cubes are glued together: one central cube has six neighbors, one stuck to each of its six faces (above, below, left, right, front, back). Find the ratio of the total volume (cubic units) to the total surface area (square units).
Givens: The shape uses exactly $7$ unit cubes; Each unit cube has volume $1$ and $6$ unit-square faces of area $1$; From the figure, one cube is in the center and the other $6$ cubes are attached one to each face of the center cube; Answer choices: (A) $1:6$, (B) $7:36$, (C) $1:5$, (D) $7:30$, (E) $6:25$
Plan
Primary tool: #13 Systematic Count
Secondary: #7 Break into Subproblems, #15 Visualize / Use Symmetry
Tool #7 (Break into Subproblems) splits the ratio into two clean counts: volume first, then surface area. Volume is just the cube count. For surface area, Tool #13 (Systematic Count) handles each of the $7$ cubes in turn, asking how many of its $6$ faces are exposed. Tool #15 (Visualize / Use Symmetry) makes the count almost trivial: the center cube is hidden by all $6$ neighbors, and by symmetry every outer cube is in the same situation — exactly $1$ glued face and $5$ exposed faces.
Execute — Answer: D
5.MD.C.3 Step 1 - Volume first.
- Each unit cube has volume $1$, and there are $7$ of them, so the total volume is $7$ cubic units.
💡 Grade 5 defines volume by counting unit cubes — exactly the situation here.
6.G.A.4 Step 2 - Set up the surface-area count.
- If the cubes were separate, the total face count would be $7 \times 6 = 42$.
- The shape has $6$ glued seams (each outer cube shares one face with the center cube).
- Each seam hides $2$ unit squares from the outside (the face on the center cube and the matching face on the neighbor).
💡 Grade 6 surface-area work: start with all faces, then subtract the ones that touch another cube.
6.G.A.4 Step 3 - Cross-check with a per-cube count using symmetry.
- The center cube has all $6$ faces glued to a neighbor, so it contributes $0$ to the surface.
- Each of the $6$ outer cubes is glued on exactly $1$ face and shows the other $5$, so each contributes $5$.
💡 Symmetry: the six outer cubes are interchangeable, so we only need to count one of them.
6.RP.A.1 Step 4 - Form the ratio.
- Volume to surface area is $7$ to $30$.
- The fraction $\dfrac{7}{30}$ is already in lowest terms because $\gcd(7, 30) = 1$, so the ratio cannot be simplified further.
💡 Grade 6 ratio language: pair the two totals and write them as $a : b$.
5.MD.C.3 Volume first. Each unit cube has volume $1$, and there are $7$ of them, so the t 6.G.A.4 Set up the surface-area count. If the cubes were separate, the total face count 6.G.A.4 Cross-check with a per-cube count using symmetry. The center cube has all $6$ fa 6.RP.A.1 Form the ratio. Volume to surface area is $7$ to $30$. The fraction $\dfrac{7}{3 Review
Reasonableness: Sanity check with a single cube: volume $1$, surface area $6$, ratio $1:6$ — that is choice (A). Adding a second cube doubles the volume to $2$ but only adds $4$ new exposed faces ($6 + 6 - 2 = 10$), so the ratio grows. Each new cube glued on adds $1$ to the volume and $4$ to the surface (gains $5$ new faces, hides $1$ old one). Starting from $1:6$ and adding $6$ cubes the same way gives volume $1 + 6 = 7$ and surface area $6 + 6 \times 4 = 30$. The arithmetic matches, and (D) is consistent. Choices (A) $1:6$ and (B) $7:36$ correspond to forgetting that glued faces hide squares; (C) $1:5$ matches a single cube minus one face, not this shape.
Alternative: Tool #1 (Draw a Picture / Use Coordinates): place the center cube at the origin with neighbors at $(\pm 1, 0, 0)$, $(0, \pm 1, 0)$, $(0, 0, \pm 1)$. For each unit square face, mark it 'inside' if the cube on the other side is also part of the shape; otherwise mark it 'outside'. There are $6$ inside pairs (one per axis direction from the center), so $42 - 12 = 30$ outside faces. Same answer (D) $7 : 30$.
CCSS standards used (min grade 6)
5.MD.C.3Recognize volume as an attribute of solid figures and understand concepts of volume measurement (Counting the $7$ unit cubes to get a total volume of $7$ cubic units.)6.G.A.4Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area (Counting exposed unit-square faces of the $7$-cube shape to get a surface area of $30$ square units.)6.RP.A.1Understand the concept of a ratio and use ratio language to describe a ratio relationship (Writing the volume-to-surface-area comparison as the ratio $7 : 30$.)
⭐ Volume is just the cube count, and surface area is just face count — every glued seam hides exactly $2$ unit squares. Spot that, and a 3D AMC 8 problem becomes a clean Grade 6 ratio.
⭐ Volume is just the cube count, and surface area is just face count — every glued seam hides exactly $2$ unit squares. Spot that, and a 3D AMC 8 problem becomes a clean Grade 6 ratio.
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