AMC 8 · 2010 · #10
Grade 7 geometry-2dProblem
Six pepperoni circles will exactly fit across the diameter of a 12-inch pizza when placed. If a total of 24 circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?
Pick an answer.
AMC 8 2010 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 pizza is a circle with diameter $12$ inches. Six pepperoni circles, all the same size, fit edge-to-edge exactly across that diameter. Twenty-four such pepperoni circles are placed on the pizza without any two overlapping. What fraction of the pizza's area is covered by pepperoni?
Givens: Pizza diameter $= 12$ inches, so pizza radius $= 6$ inches; $6$ pepperoni circles fit edge-to-edge across the diameter; Total of $24$ pepperoni circles on the pizza, no overlap; Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{2}{3}$, (C) $\tfrac{3}{4}$, (D) $\tfrac{5}{6}$, (E) $\tfrac{7}{8}$
Unknowns: The fraction $\dfrac{\text{total pepperoni area}}{\text{pizza area}}$
Understand
Restated: A pizza is a circle with diameter $12$ inches. Six pepperoni circles, all the same size, fit edge-to-edge exactly across that diameter. Twenty-four such pepperoni circles are placed on the pizza without any two overlapping. What fraction of the pizza's area is covered by pepperoni?
Givens: Pizza diameter $= 12$ inches, so pizza radius $= 6$ inches; $6$ pepperoni circles fit edge-to-edge across the diameter; Total of $24$ pepperoni circles on the pizza, no overlap; Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{2}{3}$, (C) $\tfrac{3}{4}$, (D) $\tfrac{5}{6}$, (E) $\tfrac{7}{8}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #11 Look for Symmetry / Structure
Tool #1 (Draw a Diagram) turns the words into a labeled picture: a big circle (the pizza) with $6$ small circles lined up across its diameter. That picture immediately exposes the key length — each pepperoni's diameter — by simple division. Tool #11 (Look for Symmetry / Structure) then handles the finish: both the pizza area and the total pepperoni area are multiples of $\pi$, so the $\pi$ cancels and the answer is just a clean number ratio. No algebra needed.
Execute — Answer: B
4.OA.A.2 Step 1 - Draw the picture.
- The pizza is a circle of diameter $12$ in.
- Six pepperoni circles sit edge-to-edge across that diameter.
- So the $6$ pepperoni diameters together span $12$ in, which means each pepperoni has diameter $12 \div 6 = 2$ in and therefore radius $1$ in.
💡 Dividing $12$ in equally among $6$ pepperonis is a Grade 4 'how big is each share?' division.
7.G.B.4 Step 2 - Find the pizza's area.
- The pizza has radius $6$ in, so its area is $\pi r^2 = \pi \cdot 6^2 = 36\pi$ square inches.
💡 Plugging the radius into $A = \pi r^2$ is exactly the Grade 7 area-of-a-circle formula.
7.G.B.4 Step 3 - Find one pepperoni's area, then multiply by $24$.
- Each pepperoni has radius $1$ in, area $\pi \cdot 1^2 = \pi$.
- With $24$ non-overlapping pepperonis the total covered area is $24\pi$ square inches.
💡 Same circle-area formula, scaled by the count — the 'no overlap' rule is what lets us simply add.
6.RP.A.1 Step 4 - Take the ratio.
- The fraction covered is $\dfrac{24\pi}{36\pi}$.
- The $\pi$ on top and bottom cancel — this is the structure Tool #11 was watching for — leaving a plain number ratio.
💡 Recognizing that a common factor (here $\pi$) divides out of a ratio is the Grade 6 ratio-reasoning move.
4.NF.A.1 Step 5 - Simplify the fraction.
- Both $24$ and $36$ are divisible by $12$, giving $\tfrac{2}{3}$.
💡 Reducing a fraction by a common factor is a Grade 4 equivalent-fractions skill.
4.OA.A.2 Draw the picture. The pizza is a circle of diameter $12$ in. Six pepperoni circl 7.G.B.4 Find the pizza's area. The pizza has radius $6$ in, so its area is $\pi r^2 = \p 7.G.B.4 Find one pepperoni's area, then multiply by $24$. Each pepperoni has radius $1$ 6.RP.A.1 Take the ratio. The fraction covered is $\dfrac{24\pi}{36\pi}$. The $\pi$ on top 4.NF.A.1 Simplify the fraction. Both $24$ and $36$ are divisible by $12$, giving $\tfrac{ Review
Reasonableness: The pepperoni are circles packed into a circle — circle packing always leaves gaps, so the covered fraction must be less than $1$. Also, the $24$ small circles each have radius $\tfrac{1}{6}$ of the pizza's radius, so each pepperoni's area is $\left(\tfrac{1}{6}\right)^2 = \tfrac{1}{36}$ of the pizza. Twenty-four of them give $\tfrac{24}{36} = \tfrac{2}{3}$ — same answer. $\tfrac{2}{3}$ sits sensibly between the other choices and rules out the larger ones $\tfrac{3}{4}, \tfrac{5}{6}, \tfrac{7}{8}$.
Alternative: Tool #8 (Analyze the Units) — scale-free version. Measure all lengths in 'pepperoni radii'. The pizza radius is $6$ pepperoni radii, so the pizza area is $6^2 = 36$ pepperoni-area units. There are $24$ pepperonis, each $1$ unit of pepperoni area. Ratio $= \tfrac{24}{36} = \tfrac{2}{3}$. By choosing the natural unit, the $\pi$ never even appears.
CCSS standards used (min grade 7)
4.OA.A.2Multiply or divide to solve word problems involving multiplicative comparison (Dividing the $12$-inch diameter equally among $6$ pepperonis to get each pepperoni's diameter $= 2$ in.)4.NF.A.1Explain why a fraction $a/b$ is equivalent to $(n \times a)/(n \times b)$ and use this to generate equivalent fractions (Reducing $\tfrac{24}{36}$ to $\tfrac{2}{3}$ by dividing numerator and denominator by the common factor $12$.)6.RP.A.1Understand the concept of a ratio and use ratio language to describe a ratio relationship (Setting up the covered-area fraction as a ratio $\tfrac{\text{pepperoni area}}{\text{pizza area}}$ and canceling the common factor $\pi$.)7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Applying $A = \pi r^2$ to compute both the pizza area ($36\pi$) and each pepperoni's area ($\pi$).)
⭐ This AMC 8 problem just needs the Grade 7 circle-area formula $A = \pi r^2$ — and the $\pi$ cancels out, leaving a clean fraction!
⭐ This AMC 8 problem just needs the Grade 7 circle-area formula $A = \pi r^2$ — and the $\pi$ cancels out, leaving a clean fraction!
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