Same idea as the post above, but using more elementary principles: draw just one circle first. Then draw another - it can intersect the first at 2 points, at most. Then draw a third. It can intersect each of the first two circles at two points, so we can make 4 new intersection points. Similarly the next circle can create 6 new intersection points, and so on. So the maximum total number of intersection points will be
2 + 4 + 6 + ... + 18 + 20
which is an equally spaced sum with 10 terms. The average term in that sum is the average of the smallest and largest terms, so is 11, and since sum = avg*number of terms, the sum is thus 11*10 = 110.
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