2008 AMC 10A Problem 2

1. The problem gives only ratios, not actual lengths, so you are free to pick an easy size. Let the square have side $1$.
2. The rectangle's width is twice the square's side, so the width is $2$.
3. The rectangle's length is twice its width, so the length is $2 \times 2 = 4$.
4. Now compare areas. The rectangle has area $2 \times 4 = 8$. The square has area $1 \times 1 = 1$.
5. So the square covers $1$ out of $8$ parts of the rectangle:
$$\frac{1}{8} = 0.125 = 12.5\%.$$

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

A percent comparison stays the same if you blow the whole picture up or shrink it down. That is why you may choose the square's side to be $1$. Follow the chain of ratios one link at a time (square side, then width, then length) and you get real numbers to work with.

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 $\boxed{\textbf{(A)}\ 12.5}$.