Which of the following could be the number of diagonals of a

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Mon Mar 11, 2019 11:28 pm

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[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

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Mar 12, 2019 3:12 am

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.
Success!

The correct answer is E.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Tue Mar 12, 2019 2:34 pm

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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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Tue Mar 12, 2019 2:59 pm

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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Thu Mar 14, 2019 12:27 am

=>

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.

Therefore, E is the answer.
Answer: E

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by Scott@TargetTestPrep » Thu Mar 14, 2019 3:55 pm

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 let'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

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