AMC 8 · 2011 · #3
Grade 6 geometry-2dProblem
Extend the square pattern of 8 black and 17 white square tiles by attaching a border of black tiles around the square. What is the ratio of black tiles to white tiles in the extended pattern?
Pick an answer.
AMC 8 2011 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 $5 \times 5$ square of tiles has $8$ black tiles forming a middle-ring border and $17$ white tiles (the $1 \times 1$ center plus the outer $5 \times 5$ ring). Add one more layer of black tiles all the way around. What is the ratio of black tiles to white tiles in the new figure?
Givens: The original pattern is a $5 \times 5$ square of $25$ tiles; It contains $8$ black tiles and $17$ white tiles; A new border of black tiles is wrapped around the outside; Answer choices: (A) $8:17$, (B) $25:49$, (C) $36:25$, (D) $32:17$, (E) $36:17$
Unknowns: The ratio of black tiles to white tiles after the new border is added
Understand
Restated: A $5 \times 5$ square of tiles has $8$ black tiles forming a middle-ring border and $17$ white tiles (the $1 \times 1$ center plus the outer $5 \times 5$ ring). Add one more layer of black tiles all the way around. What is the ratio of black tiles to white tiles in the new figure?
Givens: The original pattern is a $5 \times 5$ square of $25$ tiles; It contains $8$ black tiles and $17$ white tiles; A new border of black tiles is wrapped around the outside; Answer choices: (A) $8:17$, (B) $25:49$, (C) $36:25$, (D) $32:17$, (E) $36:17$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The picture is the whole problem. Tool #1 (Draw a Diagram) lets us see the figure as nested squares — a $1 \times 1$ white center inside a $3 \times 3$ black ring inside a $5 \times 5$ white ring — and then mentally wrap a $7 \times 7$ black ring around it. Tool #7 (Identify Subproblems) splits the counting into two clean pieces: (a) how many black tiles are in the new outer ring (a difference of two square areas), and (b) what are the new totals once we add that ring to the old counts. The white count never changes, so the only real work is counting the new black tiles.
Execute — Answer: D
3.MD.C.7 Step 1 - Draw or imagine the figure as nested squares.
- From the center outward, the layers are: $1 \times 1$ white center, then a black ring that fills it out to $3 \times 3$, then a white ring that fills it out to $5 \times 5$.
- Check the given counts: black tiles $= 3^2 - 1^2 = 9 - 1 = 8$ and white tiles $= 1 + (5^2 - 3^2) = 1 + 16 = 17$.
- The picture matches the problem.
💡 Counting tiles in a square ring by subtracting the inner square's area from the outer square's area is the Grade 3 "area as multiplication" idea.
3.MD.C.7 Step 2 - Find the side length of the new square.
- A border of tiles one thick adds one tile on the left and one on the right (and one on the top and one on the bottom), so the side length grows by $2$.
- The new figure is a $7 \times 7$ square, with $7^2 = 49$ tiles total.
💡 Wrapping a square in a one-tile border always bumps the side length by $2$ — drawing it once makes this stick.
3.OA.A.3 Step 3 Count the new black tiles using the subproblem "how many tiles are in the new ring alone?" That ring is the $7 \times 7$ square with the $5 \times 5$ square cut out, so its tile count is the difference of the two areas.
💡 "Big square minus small square" is the same subproblem pattern we already used in Step 1 — it scales straight from $3$ vs $1$ to $7$ vs $5$.
3.OA.A.3 Step 4 - Update the totals.
- The old $8$ black tiles stay, plus $24$ new black tiles from the border, so the new black total is $8 + 24 = 32$.
- The white count is unchanged at $17$.
- Sanity check: $32 + 17 = 49$, which matches the $7 \times 7$ total.
💡 Separating "old" from "new" turns the count into a small addition — and the total $49$ checks our work.
6.RP.A.1 Step 5 - Write the ratio of black tiles to white tiles.
- Since $\gcd(32, 17) = 1$ (because $17$ is prime and does not divide $32$), the ratio is already in lowest terms.
💡 A ratio just compares two counts; once we have $32$ and $17$, the answer is right there.
3.MD.C.7 Draw or imagine the figure as nested squares. From the center outward, the layer 3.MD.C.7 Find the side length of the new square. A border of tiles one thick adds one til 3.OA.A.3 Count the new black tiles using the subproblem "how many tiles are in the new ri 3.OA.A.3 Update the totals. The old $8$ black tiles stay, plus $24$ new black tiles from 6.RP.A.1 Write the ratio of black tiles to white tiles. Since $\gcd(32, 17) = 1$ (because Review
Reasonableness: The new black ring ($24$ tiles) is much bigger than the original black ring ($8$ tiles), so the black count should jump well past the white count. We get $32 > 17$, which matches. The total $32 + 17 = 49 = 7^2$ confirms no tile was lost or double-counted. Choices that put black below white (A) or use only the original $25$-square (B) or swap the colors (C) all fail this sanity check; only (D) $32 : 17$ is consistent.
Alternative: Tool #3 (Eliminate Possibilities) on the choices is fast here. The new figure has $7^2 = 49$ tiles, and only $17$ of them are white, so black $= 49 - 17 = 32$. The ratio must be $32 : 17$, which is exactly (D). The other ratios either don't sum to $49$ in the right way or use the wrong square's area.
CCSS standards used (min grade 6)
3.MD.C.7Relate area to multiplication and find areas of rectilinear figures by decomposing (Counting tiles in each nested square ring as the difference of two square areas, e.g. $3^2 - 1^2 = 8$ for the original black ring and $7^2 - 5^2 = 24$ for the new black border.)3.OA.A.3Use multiplication and division within $100$ to solve word problems (Combining the old and new black tile counts ($8 + 24 = 32$) and checking the total ($32 + 17 = 49 = 7^2$).)6.RP.A.1Understand the concept of a ratio and use ratio language (Expressing the final answer as the ratio black : white $= 32 : 17$ and recognizing it is already in lowest terms.)
⭐ This AMC 8 problem only needs Grade 6 ratio language built on Grade 3 area thinking — "big square minus small square" — that you already know!
⭐ This AMC 8 problem only needs Grade 6 ratio language built on Grade 3 area thinking — "big square minus small square" — that you already know!
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