AMC 8 · 2018 · #9

Grade 3 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-rectanglesperimeterspatial-visualization area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglesmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

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Problem

Monica is tiling the floor of her 12-foot by 16-foot living room. She plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will she use?

$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }89\qquad\textbf{(D) }96\qquad \textbf{(E) }120$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

120

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Toolkit + CCSS Solution

Understand

Restated: Monica is tiling a rectangular living room that is $12$ feet by $16$ feet. She places a one-foot-wide border of $1\text{ ft} \times 1\text{ ft}$ square tiles along all four edges, and fills the interior with $2\text{ ft} \times 2\text{ ft}$ square tiles. How many tiles does she use in total?

Givens: Room dimensions: $12$ ft by $16$ ft; Border tiles are $1$ ft $\times$ $1$ ft squares, one tile wide along every edge; Inner region is filled with $2$ ft $\times$ $2$ ft square tiles; Answer choices: (A) $48$, (B) $87$, (C) $89$, (D) $96$, (E) $120$

Unknowns: Total number of tiles used (small border tiles plus large inner tiles)

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram

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 $12 \times 16$ rectangle, shade the one-foot border, and label the inner rectangle as $10 \times 14$. That picture also makes it obvious that the inner sides are even, so $2 \times 2$ tiles tile it perfectly.

Execute — Answer: B

#1 Draw a Diagram 3.MD.D.8 Step 1

Draw the floor and mark off the border.
Shrink each dimension by $1$ ft on each side ($2$ ft total) to find the inner rectangle that gets the big tiles.

$$\text{inner length} = 16 - 2 = 14 \text{ ft},\quad \text{inner width} = 12 - 2 = 10 \text{ ft}$$

💡 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.

#7 Identify Subproblems 3.MD.C.7 Step 2

Subproblem A — count the border tiles.
The border's area equals the whole-room area minus the inner-rectangle area, and each border tile covers $1$ square foot, so the count of border tiles equals the border area in square feet.

$$A_{\text{border}} = 16 \times 12 - 14 \times 10 = 192 - 140 = 52 \text{ sq ft} \;\Rightarrow\; 52 \text{ border tiles}$$

💡 Treating the border as (big rectangle area) $-$ (inner rectangle area) is Grade 3 area-by-multiplication and subtraction.

#7 Identify Subproblems 3.OA.C.7 Step 3

Subproblem B — count the inner tiles.
Each $2 \text{ ft} \times 2 \text{ ft}$ tile covers $4$ square feet, so divide the inner area by $4$.
As a check, the inner sides are $14$ and $10$ — both even — so we can place $14 \div 2 = 7$ tiles along the long side and $10 \div 2 = 5$ tiles along the short side, $7 \times 5 = 35$.

$$\dfrac{140 \text{ sq ft}}{4 \text{ sq ft/tile}} = 35 \text{ tiles} \quad\text{and}\quad 7 \times 5 = 35 \text{ tiles} \;\checkmark$$

💡 Dividing $140$ by $4$ and multiplying $7 \times 5$ are basic Grade 3 multiplication/division facts within $100$.

#7 Identify Subproblems 3.OA.D.8 Step 4

Combine the two subproblems.
Add the border tile count and the inner tile count to get the total number of tiles Monica uses.

$$52 + 35 = 87 \;\Rightarrow\; \textbf{(B)}$$

💡 Combining two sub-answers with a single addition is the Grade 3 two-step word-problem skill.

[1] #1 3.MD.D.8 Draw the floor and mark off the border. Shrink each dimension by $1$ ft on each

[2] #7 3.MD.C.7 Subproblem A — count the border tiles. The border's area equals the whole-room a

[3] #7 3.OA.C.7 Subproblem B — count the inner tiles. Each $2 \text{ ft} \times 2 \text{ ft}$ ti

[4] #7 3.OA.D.8 Combine the two subproblems. Add the border tile count and the inner tile count

Review

Reasonableness: Sanity check the magnitude: the total floor area is $192$ sq ft. If every tile were the small $1 \times 1$ kind, Monica would need $192$ tiles — close to choice (E) $120$ but too many. If every tile were the big $2 \times 2$ kind, she would need $192 \div 4 = 48$ tiles (choice A). Our answer $87$ sits between $48$ and $192$, exactly where a mix-and-match floor should land, and it is the only choice that splits as $52 + 35$ with both pieces matching the border-vs-inner geometry. Choice (B) is right.

Alternative: Tool #1 (Draw a Diagram) used directly: in the picture, count the border tiles around the perimeter without double-counting the corners. The two long sides take $2 \times 16 = 32$ tiles (corners included), and the two short sides need only the $12 - 2 = 10$ middle squares each, giving $2 \times 10 = 20$ more. Border $= 32 + 20 = 52$, inner $= 7 \times 5 = 35$, total $= 87$ — same answer (B).

CCSS standards used (min grade 3)

3.MD.D.8 Solve real-world problems involving perimeters of polygons (Reading the one-foot border off the diagram and shrinking the $12 \times 16$ rectangle to the inner $10 \times 14$ rectangle.)
3.MD.C.7 Relate area to multiplication and addition operations (Computing the total area $16 \times 12 = 192$, the inner area $14 \times 10 = 140$, and the border area as their difference $192 - 140 = 52$.)
3.OA.C.7 Fluently multiply and divide within 100 (Dividing the inner area $140 \div 4 = 35$ and verifying with $7 \times 5 = 35$ to count the $2 \times 2$ tiles.)
3.OA.D.8 Solve two-step word problems using four operations within 100 (Adding the two subproblem answers $52 + 35 = 87$ to get the total tile count.)

⭐ This AMC 8 problem only needs Grade 3 area and multiplication you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 3)

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