Any Circle intersects another circle at maximum two intersection points.
With two circles : greatest number of intersection points = 2
With third circle introduced to the first two (already intersecting at two points), the third circle can intersect both the circles at two different intersection points, introducing
4 more intersection points ( third circle cuts each of the first two at two different poitns) . in addition to the two intersection points of first two.
i.e., with three circles , max # of intersection points = 2 + 2*2 = 6
with 4th circle introduced, it cuts the first three circles at two intersection points each i.e, 2*3 additional intersection points introduced now.
with 4 circles , max # of intersection points = (2 + 2*2)
(intersection points for 3 circles) + 2*3
(additional intersection points introduced by 4th circle) = 2 + 2*2 + 2*3
Similarily , with 5th circle . total = 2 + 2*2 + 2*3 + 2*4
we see a series emerging here
greatest number of of intersection points for m circles = 2 + 2*2 + 2*3 + .....+ 2*(m-1)
Applying the series extrapolation for 11 circles, greatest number of intersection points = 2 + 2*2 + 2*3 + .....+ 2*10
= 2 * (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 2 * (110)/2 = 110
( calculation for the sum shown below)-- Calculation for the sum.
since the sum in the paranthesis is a sum of consecutive integers , we can calculate the average by (last + first) / 2
Sum in paranthesis = average * # of terms = 10*((1 + 10)/2) = 110/2
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Correct Answer : E
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