Competition · AMC preparation · step 4 of 4
AMC 8 · 2018 · #9
Grade 3 geometry-2dPick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The floor is a compound region: a thin border of small tiles wrapped around a big inner rectangle of large tiles. Tool #7 (Identify Subproblems) is perfect — solve the border count and the inner count as two independent area problems, then add. Tool #1 (Draw a Diagram) makes the split visible: sketch the 12 × 16 rectangle, shade the one-foot border, and label the inner rectangle as 10 × 14. That picture also makes it obvious that the inner sides are even, so 2 × 2 tiles tile it perfectly.
Mark off the inner region
Sketch the room and frame off the one-foot border; it shrinks each side by 1 ft, so the inner rectangle is 14 by 10 feet.
A picture of the room with a one-foot frame around it shows the inner box is 14 by 10 — a Grade 3 perimeter/border reasoning move.
3.MD.D.8Draw A DiagramCount the border tiles
Subproblem A — border area is whole minus inner; each 1-ft tile covers 1 sq ft, so the tile count equals that area: 52.
Treating the border as (big rectangle area) - (inner rectangle area) is Grade 3 area-by-multiplication and subtraction.
The one-foot border tiles number exactly 52.
▸ Why?
The border covers 52 square feet, and each small tile covers exactly one square foot with no gaps or overlaps, so the number of tiles equals the number of square feet.
▸ Why?
The border is the whole floor with the inner rectangle taken out, so its area is the whole area minus the inner area: 192 - 140 = 52 square feet.
▸ Why?
The floor breaks into two non-overlapping pieces — the border frame and the inner rectangle — with no gaps, so the two pieces' areas add back to the whole floor, which lets us find one piece by subtracting the other from the whole.
▸ Why?
Each rectangle's area is its length times its width (192 = 16 × 12 for the whole floor, 140 = 14 × 10 for the inside), because area counts equal rows of unit squares.
▸ Why?
Every 1 ft × 1 ft tile sits on exactly one square foot of the border, so matching tiles to square feet one for one makes the two counts equal.
Count the inner tiles
Subproblem B — the inner area 140 divided by each 2-ft tile's 4 sq ft gives 35 tiles (check: 7 × 5).
Dividing 140 by 4 and multiplying 7 × 5 are basic Grade 3 multiplication/division facts within 100.
3.OA.C.7Identify SubproblemsAdd both tile counts
Combine the subproblems — add the 52 border tiles and 35 interior tiles for a total of 87.
Combining two sub-answers with a single addition is the Grade 3 two-step word-problem skill.
3.OA.D.8Identify SubproblemsThis AMC 8 problem only needs Grade 3 area and multiplication you already know!
- Mark off the inner region
- Count the border tiles
- Count the inner tiles
- Add both tile counts
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