AMC 8 · 2011 · #13

Easy mode Grade 6

Problem

Picture two squares of the same size, each with sides of length $15$ . They are pushed together so they overlap, and together they exactly fill a $15$ by $25$ rectangle named $AQRD$ .

The two squares are $ABCD$ on the left and $PQRS$ on the right. The shaded strip in the middle is the part where the two squares overlap.

What percent of the area of rectangle $AQRD$ is shaded?

Diagram

Two side-15 squares overlap inside a 15-by-25 rectangle AQRD; the middle strip is the overlap.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Two congruent squares $ABCD$ and $PQRS$, each with side length $15$, overlap so that together they form rectangle $AQRD$ of size $15 \times 25$. The shaded region is their overlap. What percent of the area of $AQRD$ is shaded?

Givens: Square $ABCD$ has side $15$; Square $PQRS$ has side $15$; The two squares overlap to form rectangle $AQRD$ of dimensions $15 \times 25$; The shaded region is the overlap of the two squares; Answer choices: (A) $15$, (B) $18$, (C) $20$, (D) $24$, (E) $25$ (percent)

Unknowns: The percent of rectangle $AQRD$'s area that is shaded (the overlap area)

Understand

Plan

Primary tool: #5 Draw a Picture

Secondary: #11 Look for Symmetry / Structure

The figure is the whole point: two side-$15$ squares slide horizontally until their union is a $15 \times 25$ rectangle. Tool #5 (Draw a Picture) — or just reading the given Asymptote diagram carefully — pins down that the overlap is a rectangle whose height equals the square's side ($15$) and whose width $w$ is what we have to find. Tool #11 (Look for Structure) gives the clean inclusion-exclusion idea along the long edge: the two squares together span length $15 + 15 = 30$, but they actually cover only $25$, so the missing $5$ is exactly the doubly-covered overlap width.

Execute — Answer: C

#5 Draw a Picture 3.MD.C.7 Step 1

Find the area of rectangle $AQRD$ from its given dimensions $15 \times 25$.

$$\text{Area}(AQRD) = 25 \times 15 = 375$$

💡 Area of a rectangle as length times width is a Grade 3 standard.

#5 Draw a Picture 3.MD.C.7 Step 2

Identify the overlap shape.
Both squares have the full height of the rectangle ($15$), so the overlap is a rectangle that is $15$ tall and $w$ wide, where $w$ is the horizontal length shared by the two squares.

$$\text{overlap} = w \times 15$$

💡 Drawing or reading the picture shows the overlap is a vertical strip with the same height as the squares.

#11 Look for Symmetry / Structure 4.OA.A.3 Step 3

Use inclusion-exclusion along the long edge to find $w$.
Laid side by side without overlap, the two squares would span $15 + 15 = 30$.
The actual total length is $25$, so the doubly-covered piece must be the difference.

$$15 + 15 - w = 25 \;\Rightarrow\; w = 30 - 25 = 5$$

💡 When two segments of total length $30$ fit into a span of $25$, the surplus $5$ is exactly the overlap — Grade 4 multi-step reasoning.

#5 Draw a Picture 3.MD.C.7 Step 4

Compute the area of the shaded overlap rectangle.

$$\text{overlap area} = 5 \times 15 = 75$$

💡 Same rectangle-area idea as Step 1, applied to the smaller rectangle.

#11 Look for Symmetry / Structure 6.RP.A.3 Step 5

Convert the area ratio to a percent.

$$\dfrac{75}{375} = \dfrac{1}{5} = 20\% \;\Rightarrow\; \textbf{(C)}$$

💡 Writing a part-to-whole ratio as a percent is Grade 6 ratio reasoning.

[1] #5 3.MD.C.7 Find the area of rectangle $AQRD$ from its given dimensions $15 \times 25$.

[2] #5 3.MD.C.7 Identify the overlap shape. Both squares have the full height of the rectangle (

[3] #11 4.OA.A.3 Use inclusion-exclusion along the long edge to find $w$. Laid side by side witho

[4] #5 3.MD.C.7 Compute the area of the shaded overlap rectangle.

[5] #11 6.RP.A.3 Convert the area ratio to a percent.

Review

Reasonableness: Sanity check on the dimensions: each square is $15 \times 15 = 225$. The two squares together cover area $2 \times 225 - 75 = 375$, which equals the area of $AQRD$. So the inclusion-exclusion bookkeeping is consistent. The overlap $75$ out of $375$ is $\tfrac{1}{5}$, which matches choice (C) $= 20\%$ and rules out (B) $18$ and (D) $24$.

Alternative: Tool #8 (Analyze the Units) — work with ratios instead of areas. The overlap and the full rectangle share the same height $15$, so the area ratio equals the width ratio: $\dfrac{w}{25} = \dfrac{5}{25} = \dfrac{1}{5} = 20\%$. The height cancels, which is why the answer doesn't depend on the actual side length $15$ at all.

CCSS standards used (min grade 6)

3.MD.C.7 Relate area to multiplication; find area of rectangles by multiplying side lengths (Computing the area of rectangle $AQRD$ ($25 \times 15 = 375$) and the area of the shaded overlap rectangle ($5 \times 15 = 75$).)
4.OA.A.3 Solve multi-step word problems using the four operations (Setting up and solving the inclusion-exclusion equation $15 + 15 - w = 25$ to find the overlap width $w = 5$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Expressing the part-to-whole area ratio $\tfrac{75}{375} = \tfrac{1}{5}$ as the percent $20\%$.)

⭐ This AMC 8 problem only needs Grade 6 ratio reasoning: find the overlap width, then turn the part-to-whole ratio into a percent.

Problem

Diagram

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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