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AMC 8 · 2011 · #13

Easy mode Grade 6
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Problem

Picture two squares of the same size, each with sides of length 1515. They are pushed together so they overlap, and together they exactly fill a 1515 by 2525 rectangle named AQRDAQRD.

The two squares are ABCDABCD on the left and PQRSPQRS on the right. The shaded strip in the middle is the part where the two squares overlap.

What percent of the area of rectangle AQRDAQRD is shaded?

Pick an answer.

(A)
15
(B)
18
(C)
20
(D)
24
(E)
25
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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

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)

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.

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

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