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2008 AMC 10A Problem 2

2008 AMC 10A Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is $2:1$ . The ratio of the rectangle's length to its width is $2:1$ . What percent of the rectangle's area is inside the square?

A)
$12.5$
B)
$25$
C)
$50$
D)
$75$
E)
$87.5$

Answer

Key insight

Set the square's side to 1; the rectangle is then 2 by 4, so the square covers 1/8 of it.

Solution

Only ratios are given, so pick a convenient size: let the square have side $1$ .

The rectangle's width is twice the square's side, so the width is $2$ . The rectangle's length is twice its width, so the length is $4$ .

Rectangle area: $2 \cdot 4 = 8$ . Square area: $1$ . The square occupies
$\frac{1}{8} = 12.5\%$
of the rectangle.

The answer is $\boxed{\textbf{(A)}\ 12.5}$ .

Why this works

When a problem gives only ratios, the answer is scale-free, so you may fix one length to $1$ and compute actual numbers. The two ratios chain together: side $\to$ width $\to$ length, each doubling, so the area ratio is $1 : (2 \cdot 4)$ .

The trap

Using the rectangle's length as twice the square's side (getting 25%) instead of twice the width.

Common mistakes

Using the rectangle's length as twice the square's side (getting 25%) instead of twice the width.
Comparing side lengths ( $1$ to $2$ ) rather than areas and answering $50$ .

Techniques

Test small/specific values or special cases to find or verify the answer