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parveen110
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Where did you get this question, parveen110?
Unless I'm missing something, it's wayyyy out of scope for the GMAT.
Cheers,
Brent
Geometry
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Brent@GMATPrepNow
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parveen110
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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.
I just picked up this question from another forum.
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Brent@GMATPrepNow
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Brent Hanneson - Creator of GMATPrepNow.com
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Hi parveen110,
Brent is correct. This isn't just "a level higher", it's absolutely outside the boundaries of what the GMAT will test.
While it's tempting to use randomly discovered (and free) resources when studying for the GMAT, part of your success depends on the realism of the resources in question. This question is not worth your time.
GMAT assassins aren't born, they're made,
Rich
Brent is correct. This isn't just "a level higher", it's absolutely outside the boundaries of what the GMAT will test.
While it's tempting to use randomly discovered (and free) resources when studying for the GMAT, part of your success depends on the realism of the resources in question. This question is not worth your time.
GMAT assassins aren't born, they're made,
Rich
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GMATGuruNY
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This problem is WAY beyond the scope of the GMAT.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
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]
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
