Competition · AMC preparation · step 4 of 4
AMC 8 · 2000 · #22
Grade 7 geometry-3d
Pick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 × 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 × 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.
Find the original surface area
The original big cube has 6 congruent faces of area 2² = 4, so its surface area is 6 × 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.4Draw A DiagramFind the gain and the loss
At the seam a 1 × 1 patch on the big top is hidden while the small cube reveals its top and four sides — five new 1 × 1 faces.
Only the change at the seam matters. Every other face of the big cube is untouched, so we don't need to recount it.
Placing the unit cube on top hides exactly one 1 × 1 patch of the big cube's top face and newly exposes exactly five 1 × 1 faces of the small cube.
▸ Why?
The small cube's bottom sits flush on the top, so the only place outside surface disappears is the single 1 × 1 square where the two faces press together — that patch of the big cube's top is now sealed inside instead of facing out.
▸ Why?
A solid's surface is just its outward-facing pieces added up with no gaps and no overlaps, so where two faces press flat together that shared square stops being on the outside and drops out of the surface count.
▸ Why?
The small cube keeps its shape when it is set down, so its bottom stays a 1 × 1 square and covers exactly a 1 × 1 square of the top — no more, no less.
▸ Why?
The small cube has six 1 × 1 faces and only its bottom is pressed against the big cube, so the other five faces — the top and the four sides — are now out in the open and add to the surface.
▸ Why?
The small cube's whole surface is its six equal square faces added together, so setting aside the one hidden bottom leaves exactly five faces still facing outward.
Combine into the net change
Combine gain and loss: +5 - 1 = +4 square units net.
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.
6.G.A.4Change Focus Count The ComplementCompute the percent increase
Percent change is the increase over the original: = ≈ 16.67%.
Grade 7 percent change: divide the increase by what you started with. 4/24 simplifies to 1/6, which is about 0.16.
7.RP.A.3Change Focus Count The ComplementRound to the closest choice
16.67% sits between 15% and 17%, but it is far closer to 17% — choice (C).
When a problem says "closest to," compare distances to the two nearest choices; the smaller distance wins.
7.RP.A.3Change Focus Count The ComplementDon't recount the whole new solid. Gluing a 1 × 1 × 1 cube on top hides one 1 × 1 patch and exposes five 1 × 1 faces, so the surface area grows by 4. Then ≈ 16.7%, which rounds to 17% — answer (C).
- Find the original surface area
- Find the gain and the loss
- Combine into the net change
- Compute the percent increase
- Round to the closest choice
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