In the figure above, SAND and SURF are squares, and O is the center of

IMO C.

Here is how I arrived the solution.

Let \(x\) and \(y\) be the sides of squares SAND and SURF respectively. Then their areas will be \(x^2\) and \(y^2\). This means \(Q = x^2 + y^2\)

UA is the diameter of the circle and triangle USA is a right angled triangle with the right angle at S (because of Thales Theorem).

The diameter of the circle = \(\sqrt{x^2 + y^2}\) and the area C = \(\pi\)*\(\frac{(x^2 + y^2)}{4}\)

\(\frac{C}{Q}\) = \(\frac{\pi*\frac{x^2 + y^2}{4}}{x^2 + y^2}\) = \(\frac{\pi}{4}\)

Since \(3 < \pi < \frac{7}{2} \) --> \(\frac{3}{4} < \frac{\pi}{4} < \frac{7}{8}\)

xabush's solution above is perfect, and is better than what I'll post, because if you can solve the problem that way, that will give you more flexibility on other problems. But there are two ways we can save some time on this particular question. A GMAT question can only have one right answer. Since the question tells us nothing about the lengths or angles in the diagram, the answer to this question must always be the same, regardless of what those lengths or angles are. So we can assume the radius is 1, and we can also assume, say, that the angle at A is 0 degrees, knowing that whatever answer we get in that one case will be the answer in every case. Then we have a circle of radius 1, with area π*r^2 = π ~ 3.14, and just one square (the square SURF vanishes if angle A is 0 degrees) with an edge of length 2, the diameter of the circle, so with area 2^2 = 4. So the ratio C/Q is roughly 3.14/4, which is bigger than 3/4, but less than 3.5/4 = 7/8, and the answer is C.

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