The Quest for 700: Weekly GMAT Challenge (Answer)

The first trick is to draw this complicated picture correctly!

 

 

Next, work your way from the inside out. Pick the radius of circle F as 1. Notice that this is the distance from the center of triangle G to a vertex of G. For now, don’t worry about finding the area of triangle G—we’ll see a shortcut later.

 

Now, what is the radius of the next circle out, circle D? 1 is the distance from the center of square E to the center of any side of that square. So we can draw a 45-45-90 triangle and find that the “half-diagonal” of square E is √2. This is also the radius of circle D.

 

Applying the same reasoning, we can see that √2 is the distance from the center of square C to the center of any side of that square. We can draw another 45-45-90 triangle and find that the “half-diagonal” of square C is 2. This is also the radius of circle B.

 

Finally, we can draw a 30-60-90 triangle within equilateral triangle A and see that the distance from the center to any vertex of triangle A is 4.

 

Now, rather than calculate each area and divide, we can use a huge shortcut. The distance from center to vertex for triangle G was 1; the same distance for triangle A is 4. Since these two triangles are similar, this means that every “distance” ratio for the two triangles will be 4 : 1. For instance, their side lengths will be in a 4 : 1 ratio. And since areas are squares of distances, the ratio of areas will be 4² to 1², or 16 : 1.