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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Which of the following could be the number of diagonals of a tagged by: Max@Math Revolution ##### This topic has 5 expert replies and 0 member replies ### GMAT/MBA Expert ## Which of the following could be the number of diagonals of a ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [GMAT math practice question] Which of the following could be the number of diagonals of an n-polygon? A. 3 B. 6 C. 8 D. 12 E. 20 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$149 for 3 month Online Course
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Max@Math Revolution wrote:
[GMAT math practice question]

Which of the following could be the number of diagonals of an n-polygon?

A. 3
B. 6
C. 8
D. 12
E. 20
To form a diagonal, we must choose 2 vertices.
But any pair of vertices that forms a side of the polygon cannot serve to form a diagonal.
Thus:
Number of diagonals = (number of ways to choose 2 vertices) - (number of sides of the polygon)

When the prompt includes the phrase which of the following, the correct answer is likely to be D or E.

E: 20 diagonals
For 20 diagonals to be formed, the number of sides must be relatively large.
For an 8-sided polygon, the number of diagonals = 8C2 - (8 sides of the polygon) = (8*7)/(2*1) - 8 = 28 - 8 = 20.
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Max@Math Revolution wrote:
[GMAT math practice question]

Which of the following could be the number of diagonals of an n-polygon?

A. 3
B. 6
C. 8
D. 12
E. 20
$$?\,\,\,:\,\,\,\# \,d\,\,\underline {{\rm{could}}\,\,{\rm{be}}} \,\,\,\,\left( {N \ge 3\,\,{\mathop{\rm int}} \,\,\left( * \right)\,,\,\,N{\rm{ - polygon}}} \right)$$
$$d\,\, \to \,\,\,\left\{ \matrix{ \,{\rm{each}}\,\,{\rm{vertex}}\,\,{\rm{with}}\,\,\left( {{\rm{not}}\,\,{\rm{itself}}} \right)\,\,{\rm{nor}}\,\,\left( {{\rm{next \,\, to}}\,\,{\rm{it}}\,\,{\rm{vertex}}} \right) \hfill \cr \,\left[ {A - C} \right]\,\,{\rm{diagonal}}\,\,{\rm{is}}\,\,{\rm{the}}\,\,{\rm{same}}\,\,{\rm{of}}\,\,\left[ {C - A} \right]\,\,{\rm{diagonal}} \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,d = {{N\left( {N - 3} \right)} \over 2}$$
$$\left. \matrix{ \left( A \right)\,\,\,{{N\left( {N - 3} \right)} \over 2} = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} \right) = 6\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\rm{impossible}} \hfill \cr \left( B \right)\,\,\,{{N\left( {N - 3} \right)} \over 2} = 6\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} \right) = 12\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\rm{impossible}} \hfill \cr \left( C \right)\,\,\,{{N\left( {N - 3} \right)} \over 2} = 8\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} \right) = 16\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\rm{impossible}} \hfill \cr \left( D \right)\,\,\,{{N\left( {N - 3} \right)} \over 2} = 12\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} \right) = 24\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\rm{impossible}}\,\,\, \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( E \right)\,\,{\rm{by}}\,\,{\rm{exclusion}}\,\,\,\,\left( {**} \right)$$
$$\left( {**} \right)\,\,\,\,\left( E \right)\,\,\,{{N\left( {N - 3} \right)} \over 2} = 20\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} \right) = 40\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,N = 8$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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GMATGuruNY wrote:
Number of diagonals = (number of ways to choose 2 vertices) - (number of sides of the polygon)
LetÂ´s connect MitchÂ´s nice argument with mine (above):
$$d = C\left( {N,2} \right) - N = {{N!} \over {2!\left( {N - 2} \right)!}} - N = {{N\left( {N - 1} \right)} \over 2} - {{2N} \over 2} = {{N\left( {N - 3} \right)} \over 2}$$
Regards,
Fabio.

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The number of diagonals of an n-polygon is nC2 - n = n(n-1)/2 - n=n(n-3)/2.
If n = 4, then the number of diagonals is 4C2 - 4 = 6 - 4 = 2.
If n = 5, then the number of diagonals is 5C2 - 5 = 10 - 5 = 5.
If n = 6, then the number of diagonals is 6C2 - 6 = 15 - 6 = 9.
If n = 7, then the number of diagonals is 7C2 - 7 = 21 - 7 = 14.
If n = 8, then the number of diagonals is 8C2 - 8 = 28 - 8 = 20.

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Only $149 for 3 month Online Course Free Resources-30 day online access & Diagnostic Test Unlimited Access to over 120 free video lessons-try it yourself Email to : info@mathrevolution.com ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2791 messages Followed by: 18 members Upvotes: 43 Max@Math Revolution wrote: [GMAT math practice question] Which of the following could be the number of diagonals of an n-polygon? A. 3 B. 6 C. 8 D. 12 E. 20 Since every two vertices of a polygon either form a side or a diagonal, the total number of sides and diagonals an n-polygon has is nC2. Since the polygon has n sides, then the number of diagonals must be nC2 - n. Now letâ€™s test some values of n. We can begin with n = 4 (since a triangle has no diagonals). Let d = the number of diagonals, we have: If n = 4, d = 4C2 - 4 = 6 - 4 = 2. If n = 5, d = 5C2 - 5 = 10 - 5 = 5. If n = 6, d = 6C2 - 6 = 15 - 6 = 9. If n = 7, d = 7C2 - 7 = 21 - 7 = 14. If n = 8, d = 8C2 - 8 = 28 - 8 = 20. Answer: E _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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