AMC 8 · 2000 · #22
Grade 7 geometry-3dProblem
A cube has edge length 2. Suppose that we glue a cube of edge length 1 on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to
Pick an answer.
AMC 8 2000 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 cube with edge length $2$ has a smaller cube with edge length $1$ glued on its top face so that one full face of the small cube rests on the big cube's top. Count the new total surface area (top, bottom, all sides) and compare it to the original cube's surface area. Which choice is closest to the percent increase?
Givens: Big cube edge length $= 2$, so each face has area $2 \times 2 = 4$; Small cube edge length $= 1$, so each face has area $1 \times 1 = 1$; The small cube sits on the big cube's top face with one of its $1 \times 1$ faces flush against the big cube's top; Surface area means top $+$ bottom $+$ all sides of the combined solid; Answer choices: (A) $10$, (B) $15$, (C) $17$, (D) $21$, (E) $25$
Unknowns: The percent increase in surface area from the original cube to the new solid, rounded to the closest listed integer
Understand
Restated: A cube with edge length $2$ has a smaller cube with edge length $1$ glued on its top face so that one full face of the small cube rests on the big cube's top. Count the new total surface area (top, bottom, all sides) and compare it to the original cube's surface area. Which choice is closest to the percent increase?
Givens: Big cube edge length $= 2$, so each face has area $2 \times 2 = 4$; Small cube edge length $= 1$, so each face has area $1 \times 1 = 1$; The small cube sits on the big cube's top face with one of its $1 \times 1$ faces flush against the big cube's top; Surface area means top $+$ bottom $+$ all sides of the combined solid; Answer choices: (A) $10$, (B) $15$, (C) $17$, (D) $21$, (E) $25$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #1 Draw a Diagram
Recomputing the surface area of the bumpy new solid face by face is slow and error-prone. Tool #16 (Change Focus) flips the question: don't recount everything, just count the net change at the seam. Gluing the small cube on top removes the $1 \times 1$ contact patch from the outside (two faces actually — the small cube's bottom and the matching patch on the big cube's top), and adds the small cube's other five $1 \times 1$ faces. Tool #1 (Draw a Diagram) makes the contact patch and the five newly exposed faces easy to see, so the net change is a simple count: $+5$ minus $-1$ on the top face $= +4$ square units. The percent step is then one division.
Execute — Answer: C
6.G.A.4 Step 1 - Surface area of the original big cube.
- A cube has $6$ congruent faces; each face of the big cube has area $2^2 = 4$, so the original surface area is $6 \times 4 = 24$.
💡 A cube's net is $6$ congruent squares — the Grade 6 "surface area from nets" move. With side $2$, each square has area $4$.
6.G.A.4 Step 2 - Use the complement to find the net change.
- Picture the small cube being lowered onto the big cube's top.
- Two things happen at the glue seam: a $1 \times 1$ patch on the big cube's top is no longer exposed, and the small cube's bottom face is also no longer exposed (it is the same patch from the other side, so we only "remove" the patch once from the outside).
- At the same time, the small cube contributes five newly exposed faces — its top and its four sides — each with area $1$.
💡 Only the change at the seam matters. Every other face of the big cube is untouched, so we don't need to recount it.
6.G.A.4 Step 3 Combine the gain and the loss to get the net change in surface area.
💡 Five small faces appear, one small patch disappears — net $+4$. The new total would be $24 + 4 = 28$, but we don't actually need the new total to answer the percent question.
7.RP.A.3 Step 4 - Compute the percent increase.
- Percent change is the change divided by the original, written as a percent.
💡 Grade 7 percent change: divide the increase by what you started with. $\tfrac{4}{24}$ simplifies to $\tfrac{1}{6}$, which is about $0.1\overline{6}$.
7.RP.A.3 Step 5 - Round to the closest answer choice.
- $16.67\%$ is between $15\%$ and $17\%$, but much closer to $17$ (distance $0.33$) than to $15$ (distance $1.67$).
💡 When a problem says "closest to," compare distances to the two nearest choices; the smaller distance wins.
6.G.A.4 Surface area of the original big cube. A cube has $6$ congruent faces; each face 6.G.A.4 Use the complement to find the net change. Picture the small cube being lowered 6.G.A.4 Combine the gain and the loss to get the net change in surface area. 7.RP.A.3 Compute the percent increase. Percent change is the change divided by the origin 7.RP.A.3 Round to the closest answer choice. $16.67\%$ is between $15\%$ and $17\%$, but Review
Reasonableness: Cross-check with the long way. New total surface area $= 24 + 4 = 28$, so the increase is $28 - 24 = 4$, and $4 \div 24 = 0.1\overline{6} = 16.\overline{6}\%$. That matches the complement count, and rounding to the listed options pins (C) $17$. A sanity check against the choices: $10\%$ would mean an increase of $2.4$ (impossible — the change at the seam is a whole number of unit squares), and $25\%$ would mean an increase of $6$ (more than the entire small cube can offer to the outside, since the small cube only exposes $5$ new faces). So (C) is the only choice that even fits the geometry.
Alternative: Tool #7 (Identify Subproblems): compute the new surface area from scratch instead of via the change. The big cube contributes $6 \times 4 = 24$, minus the $1 \times 1$ patch its top loses, giving $23$. The small cube contributes its top and four sides, $5 \times 1 = 5$. Total $= 23 + 5 = 28$, so the increase is $4$ and the percent increase is $\tfrac{4}{24} \approx 16.67\% \to (C)$. Same answer, more arithmetic — which is exactly why the complement view is the cleaner tool here.
CCSS standards used (min grade 7)
6.G.A.4Represent three-dimensional figures using nets and find surface area (Computing the original cube's surface area as $6 \times 2^2 = 24$ and tracking the $1 \times 1$ unit faces that are added or hidden when the small cube is glued on.)7.RP.A.3Use proportional relationships to solve multi-step percent problems (Converting the surface-area change $+4$ into a percent of the original $24$, getting $\tfrac{4}{24} \approx 16.67\%$, and selecting the closest listed integer ($17\%$).)
⭐ Don't recount the whole new solid. Gluing a $1 \times 1 \times 1$ cube on top hides one $1 \times 1$ patch and exposes five $1 \times 1$ faces, so the surface area grows by $4$. Then $\tfrac{4}{24} \approx 16.7\%$, which rounds to $17\%$ — answer (C).
⭐ Don't recount the whole new solid. Gluing a $1 \times 1 \times 1$ cube on top hides one $1 \times 1$ patch and exposes five $1 \times 1$ faces, so the surface area grows by $4$. Then $\tfrac{4}{24} \approx 16.7\%$, which rounds to $17\%$ — answer (C).
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