AMC 8 · 2006 · #18
Grade 6 geometry-3dProblem
A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A $3\times 3\times 3$ cube is built from $27$ unit cubes. The $8$ corner unit cubes are black; the other $19$ are white. What fraction of the big cube's outer surface is white?
Givens: The big cube has edge length $3$, made of $27$ unit cubes; The $8$ corner unit cubes are black; The remaining $19$ unit cubes are white; Answer choices: (A) $\tfrac{1}{9}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{4}{9}$, (D) $\tfrac{5}{9}$, (E) $\tfrac{19}{27}$
Unknowns: The fraction of the big cube's surface area that is white
Understand
Restated: A $3\times 3\times 3$ cube is built from $27$ unit cubes. The $8$ corner unit cubes are black; the other $19$ are white. What fraction of the big cube's outer surface is white?
Givens: The big cube has edge length $3$, made of $27$ unit cubes; The $8$ corner unit cubes are black; The remaining $19$ unit cubes are white; Answer choices: (A) $\tfrac{1}{9}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{4}{9}$, (D) $\tfrac{5}{9}$, (E) $\tfrac{19}{27}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #17 Visualize Spatial Relationships, #7 Identify Subproblems
Counting white unit-squares directly is messy — the edge and center cubes contribute different numbers of faces. Tool #16 (Count the Complement) flips the problem: count the black unit-squares instead. Black cubes sit only at the $8$ corners, and Tool #17 (Visualize Spatially) tells us each corner cube shows exactly $3$ faces on the surface. That makes the black count a single multiplication. Tool #7 (Identify Subproblems) keeps the work organized: (a) find the total surface, (b) find the black surface, (c) subtract, (d) form the fraction.
Execute — Answer: D
6.G.A.4 Step 1 - Count the total number of unit-square faces on the surface.
- The big cube has $6$ faces, and each face is a $3\times 3$ grid of $9$ unit squares.
💡 Grade 6 surface area: a cube's surface is $6$ congruent square faces, and each face's area in unit squares is $\text{side}^2 = 9$.
3.OA.A.1 Step 2 - Switch focus to the black squares.
- The $8$ black cubes sit at the corners of the big cube.
- Each corner of a cube touches $3$ of the cube's faces, so each corner unit cube shows exactly $3$ of its own faces on the surface.
💡 Visualize one corner: pick up a cube, look at the corner — three faces meet there. Grade 3 multiplication does the rest: $8$ corners, $3$ faces each.
3.OA.A.1 Step 3 - Use the complement to get the white count.
- Every surface square is either black or white, so subtract the black squares from the total.
💡 Counting the complement avoids tracking edge cubes and center cubes separately — black is small and easy, so subtract.
4.NF.A.1 Step 4 - Form the white fraction and simplify.
- Divide both numerator and denominator by their common factor $6$.
💡 Grade 4 equivalent fractions: $30 = 6 \times 5$ and $54 = 6 \times 9$, so $\tfrac{30}{54} = \tfrac{5}{9}$.
6.G.A.4 Count the total number of unit-square faces on the surface. The big cube has $6$ 3.OA.A.1 Switch focus to the black squares. The $8$ black cubes sit at the corners of the 3.OA.A.1 Use the complement to get the white count. Every surface square is either black 4.NF.A.1 Form the white fraction and simplify. Divide both numerator and denominator by t Review
Reasonableness: Cross-check by counting white directly using the symmetry of one face. One face of the big cube is a $3\times 3$ grid. The $4$ corners of that grid are corners of the big cube and are black; the $4$ edge squares and the $1$ center square are white. So each face shows $5$ white and $4$ black unit squares, giving $\tfrac{5}{9}$ per face — and therefore $\tfrac{5}{9}$ overall. This matches answer (D). Magnitude is sensible: $19$ of $27$ cubes are white but they sit mostly inside, so the white surface fraction $\tfrac{5}{9}$ is much smaller than the white cube fraction $\tfrac{19}{27}$, which rules out (E).
Alternative: Tool #1 (Draw a Diagram): sketch one face as a $3\times 3$ grid, shade the $4$ corner squares black. Five squares per face are white, six faces total, so $\tfrac{6 \times 5}{6 \times 9} = \tfrac{5}{9}$. By symmetry every face looks the same, so the whole-cube fraction equals the per-face fraction.
CCSS standards used (min grade 6)
6.G.A.4Represent three-dimensional figures using nets and use the nets to find the surface area (Treating the cube's surface as $6$ congruent $3\times 3$ square faces, totaling $54$ unit squares.)3.OA.A.1Interpret products of whole numbers (Computing $6 \times 9 = 54$ total unit squares and $8 \times 3 = 24$ black unit squares.)4.NF.A.1Explain why a fraction $a/b$ is equivalent to a fraction $(n\times a)/(n\times b)$ (Simplifying $\tfrac{30}{54}$ to $\tfrac{5}{9}$ by dividing numerator and denominator by their common factor $6$.)
⭐ When most of the surface is one color, count the other color instead — each of the $8$ black corner cubes shows just $3$ faces, so $24$ black squares out of $54$ leaves $\tfrac{5}{9}$ white.
⭐ When most of the surface is one color, count the other color instead — each of the $8$ black corner cubes shows just $3$ faces, so $24$ black squares out of $54$ leaves $\tfrac{5}{9}$ white.
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