If 10 circles, all with different radii, are positioned in the same

If you're faced with a GMAT question that is different from anything you've seen before, one approach you can use to answer it is to try some different scenarios and look for a pattern.

In the case of this question, we can look for a pattern to how circles of different sizes can intersect.

Start with two large circles of similar but different sizes. They can intersect at two points, and there's no way for them to intersect at more than two points.



Then, if we add a third, slightly smaller circle, it can intersect the other two at 4 points. So, with three circles, we now have 6 intersections.



Each successive circle can intersect each of the others at two points. So the fourth circle can intersect the first three at 6 points.



So, we see the pattern.

A circle can intersect any other circle at 2 points maximum. So, at maximum, each successive circle intersects the others at two more points than the previous circle intersected the others at.

Thus, we have the following:

The first circle intersects no others at no points.

The second intersects 1 other at 2 points.

The third intersects 2 others at 4 points.

The fourth intersects 3 others at 6 points.

The fifth intersects 4 others at 8 points.

The sixth circle intersects 5 others at 10 points.

The seventh intersects 6 other at 12 points.

The eighth intersects 7 others at 14 points.

The ninth intersects 8 others at 16 points.

The tenth intersects 9 others at 18 points.

2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 = 90

The correct answer is (A).