Yes Brent, I know its a level higher. I tried solving it but couldn't get anywhere closer to the options. So, i thought i'm missing something and may get some help out..because i've seen some solution on these lines on the forum.

I just picked up this question from another forum.

parveen110 wrote:No three diagonals of a convex polygon are concurrent except at its vertices. If the number of points of intersection of the diagonals in the interior of the polygon is 715, then find the number of diagonals of the polygon

(A) 65
(B) 55
(C) 45
(D) 35
(E) 75

This problem is WAY beyond the scope of the GMAT.
Anyone studying for the GMAT should feel free to disregard the explanation below.

Given a polygon with an odd number of sides, any combination of 4 vertices can serve to form an interior point of intersection.
Check my post here for an explanation:
https://www.beatthegmat.com/permutation- ... 73635.html

To determine how many vertices are required to form 715 interior points of intersections, test a few options:
From 9 vertices, the number of combinations of 4 that can be formed = 9C4 = (9*8*7*6)/(4*3*2*1) = 126.
From 11 vertices, the number of combinations of 4 that can be formed = 11C4 = (11*10*9*8)/(4*3*2*1) = 330.
From 13 vertices, the number of combinations of 4 that can be formed = 13C4 = (13*12*11*10)/(4*3*2*1) = 715.
Thus, a polygon whose diagonals can form 715 interior points of intersection has 13 vertices.

A diagonal is formed by connecting two non-adjacent vertices.
From 13 vertices, the number of combinations of 2 that can be formed = 13C2 = (13*12)/(2*1) = 78.
These 78 combinations of 2 include the polygon's 13 sides, which should not be included in our tally of the diagonals.
Subtracting the 13 sides of the polygon, we get:
78-13 = 65.

The correct answer is A.[spoiler][/spoiler]