AMC 8 · 2010 · #10

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-circlesratio-proportionfraction-arithmetic identify-subproblems ↑ Prerequisites: area-circlesfraction-arithmetic

📏 Medium solution 💡 3 insights

Problem

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.

(A)

$\frac 12$

(B)

$\frac 23$

(C)

$\frac 34$

(D)

$\frac 56$

(E)

$\frac 78$

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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

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

#1 Draw a Diagram 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.

$$\text{pepperoni diameter} = \dfrac{12}{6} = 2 \text{ in} \;\Rightarrow\; r_{\text{pep}} = 1 \text{ in}$$

💡 Dividing $12$ in equally among $6$ pepperonis is a Grade 4 'how big is each share?' division.

#1 Draw a Diagram 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.

$$A_{\text{pizza}} = \pi \cdot 6^2 = 36\pi \text{ in}^2$$

💡 Plugging the radius into $A = \pi r^2$ is exactly the Grade 7 area-of-a-circle formula.

#1 Draw a Diagram 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.

$$A_{\text{pep,total}} = 24 \cdot \pi \cdot 1^2 = 24\pi \text{ in}^2$$

💡 Same circle-area formula, scaled by the count — the 'no overlap' rule is what lets us simply add.

#11 Look for Symmetry / Structure 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.

$$\dfrac{A_{\text{pep,total}}}{A_{\text{pizza}}} = \dfrac{24\pi}{36\pi} = \dfrac{24}{36}$$

💡 Recognizing that a common factor (here $\pi$) divides out of a ratio is the Grade 6 ratio-reasoning move.

#11 Look for Symmetry / Structure 4.NF.A.1 Step 5

Simplify the fraction.
Both $24$ and $36$ are divisible by $12$, giving $\tfrac{2}{3}$.

$$\dfrac{24}{36} = \dfrac{24 \div 12}{36 \div 12} = \dfrac{2}{3} \;\Rightarrow\; \textbf{(B)}$$

💡 Reducing a fraction by a common factor is a Grade 4 equivalent-fractions skill.

[1] #1 4.OA.A.2 Draw the picture. The pizza is a circle of diameter $12$ in. Six pepperoni circl

[2] #1 7.G.B.4 Find the pizza's area. The pizza has radius $6$ in, so its area is $\pi r^2 = \p

[3] #1 7.G.B.4 Find one pepperoni's area, then multiply by $24$. Each pepperoni has radius $1$

[4] #11 6.RP.A.1 Take the ratio. The fraction covered is $\dfrac{24\pi}{36\pi}$. The $\pi$ on top

[5] #11 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.2 Multiply 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.1 Explain 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.1 Understand 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.4 Know 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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