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Retired Moderator V
Joined: 22 Jun 2014
Posts: 1097
Location: India
Concentration: General Management, Technology
GMAT 1: 540 Q45 V20 GPA: 2.49
WE: Information Technology (Computer Software)
The sum of the interior angles of any polygon with n sides  [#permalink]

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Difficulty:   15% (low)

Question Stats: 77% (01:17) correct 23% (01:45) wrong based on 87 sessions

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The sum of the interior angles of any polygon with n sides is 180(n - 2) degrees. If the sum of the interior angles of polygon P is three times the sum of the interior angles of quadrilateral Q, how many sides does P have?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 14

Project PS Butler : Question #96

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Intern  B
Joined: 24 Jun 2018
Posts: 35
Re: The sum of the interior angles of any polygon with n sides  [#permalink]

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Answer is B ie. 8.

Sum of the internal angles of a quadrilateral is 360.
{Even if during the exam, the above fact isn't known by the examinee, one can simply substitute 4 in the formula stated in the question to get 360. 180(n-2) => 180(4-2) => 360}

Question states that sum of internal angles of a certain polygon P is three times the sum of internal angles of quadrilateral Q.

Hence, the Sum of internal angles(S) = 3 x 360 = 1080.

S=1080=180(n-2)

Hence, n=8
GMAT Club Legend  D
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Concentration: Sustainability, Marketing
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Re: The sum of the interior angles of any polygon with n sides  [#permalink]

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180(n-2)=3*360

n= 8

IMO B

HKD1710 wrote:
The sum of the interior angles of any polygon with n sides is 180(n - 2) degrees. If the sum of the interior angles of polygon P is three times the sum of the interior angles of quadrilateral Q, how many sides does P have?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 14

Project PS Butler : Question #96

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Senior PS Moderator V
Joined: 26 Feb 2016
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GPA: 3.12
Re: The sum of the interior angles of any polygon with n sides  [#permalink]

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By definition, a quadrilateral has 4 sides and the sum of interior angles is 360.

From the question stem, we have been told that the sum of the interior angles of
the polygon is thrice the sum of the interior angles of the polygon($$3*360 = 1080$$)

Solving for n, we get $$180(n - 2) = 1080$$ -> $$n - 2 = \frac{1080}{180} = 6$$ -> $$n = 8$$

Therefore, Polygon P has 8(Option B) sides whose interior angles sum thrice that of a quadrilateral.
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You've got what it takes, but it will take everything you've got Re: The sum of the interior angles of any polygon with n sides   [#permalink] 24 Dec 2018, 01:12
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