2008 AMC 10A Problems/Problem 2

Problem

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?

$\mathrm{(A)}\ 12.5\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 75\qquad\mathrm{(E)}\ 87.5$

Solution 1

Since they are asking for the "ratio" of two things, we can say that the side of the square is anything that we want. So if we say that it is $1$, then width of the rectangle is $2$, and the length is $4$, thus making the total area of the rectangle $8$. The area of the square is just $1$. So the answer is just $\frac{1}{8} \times 100 = 12.5 \longrightarrow \fbox{A}$

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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