Manhattan GMAT Challenge Problem of the Week - 19 May 2010

by , May 19, 2010

Welcome back to this weeks Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

Question

In a rectangular coordinate system, point A has coordinates (d, d), where d > 0. Point A and the origin form the endpoints of a diameter of circle C. What fraction of the area of circle C lies within the first quadrant?

A.

B.

C.

D.

E.

Answer

First, draw the coordinate plane and add in point A, which has coordinates (d,d). Since d is greater than 0, this point lies in the first quadrant, as shown:

Notice that the origin, the point (0, d), the point (d, 0), and point A (d,d) form the corners of a square:

Now, we draw the line segment between (0,0) and (d,d). This is the diameter of a circle, which we also draw:

Notice that the answer does not depend on d. All that matters is that the circle contains a square. The fraction of the circle's area in the first quadrant will be the same, no matter what. Thus, we can drop d as a variable and create a new variable r for the radius of the circle. It will be easier to compute areas in terms of r (which will also cancel out in the end).

The area of the circle is just [pmath]pi r^2[/pmath].

The fraction we are looking for is this:

So we need the shaded area of the circle, which consists of the square ( ) and 2 of the four "leaves of the table" ( ).

The square's area can be found this way:

Now the area of the 4 leaves ( ) is the area of the circle ( ) minus the area of the square ( ). So the area of the 4 leaves is [pmath]pi r^2-2r^2=(pi-2)r^2[/pmath].

This means that the area of 2 leaves is .

The whole shaded area is thus .

Finally, the desired fraction is .

The correct answer is D.

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