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# What is the number of sides of a regular polygon in which 1/3rd of the

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Intern
Joined: 30 Dec 2016
Posts: 18
Location: India
GMAT 1: 480 Q22 V22
What is the number of sides of a regular polygon in which 1/3rd of the  [#permalink]

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23 Aug 2018, 00:42
3
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Difficulty:

55% (hard)

Question Stats:

53% (02:06) correct 47% (01:12) wrong based on 40 sessions

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What is the number of sides of a regular polygon in which 1/3rd of the sum of exterior angle is equal to the each interior angle ?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 8

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Dread it, Run from it, Destiny still arrives.
Math Expert
Joined: 02 Aug 2009
Posts: 7996
Re: What is the number of sides of a regular polygon in which 1/3rd of the  [#permalink]

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23 Aug 2018, 00:52
2
1
Thanos7 wrote:
What is the number of sides of a regular polygon in which 1/3rd of the sum of exterior angle is equal to the each interior angle ?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 8

sum of interior angles = $$(n-2)*180$$...... so each interior angle = $$\frac{(n-2)*180}{n}$$.
Sum of exterior angles = 360

thus $$\frac{(n-2)*180}{n}=\frac{1}{3}*360.............(n-2)*180=120n..........180n-120n=360......n=6$$.

D
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Posts: 189
Concentration: Operations, General Management
Re: What is the number of sides of a regular polygon in which 1/3rd of the  [#permalink]

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27 Aug 2018, 07:18
chetan2u wrote:
Thanos7 wrote:
What is the number of sides of a regular polygon in which 1/3rd of the sum of exterior angle is equal to the each interior angle ?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 8

sum of interior angles = $$(n-2)*180$$...... so each interior angle = $$\frac{(n-2)*180}{n}$$.
Sum of exterior angles = 360

thus $$\frac{(n-2)*180}{n}=\frac{1}{3}*360.............(n-2)*180=120n..........180n-120n=360......n=6$$.

D

chetan2u I may sound silly while asking this-
How to deduce the sum of exteriors=360?
Math Expert
Joined: 02 Aug 2009
Posts: 7996
Re: What is the number of sides of a regular polygon in which 1/3rd of the  [#permalink]

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27 Aug 2018, 07:37
1
siddreal wrote:
chetan2u wrote:
Thanos7 wrote:
What is the number of sides of a regular polygon in which 1/3rd of the sum of exterior angle is equal to the each interior angle ?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 8

sum of interior angles = $$(n-2)*180$$...... so each interior angle = $$\frac{(n-2)*180}{n}$$.
Sum of exterior angles = 360

thus $$\frac{(n-2)*180}{n}=\frac{1}{3}*360.............(n-2)*180=120n..........180n-120n=360......n=6$$.

D

chetan2u I may sound silly while asking this-
How to deduce the sum of exteriors=360?

Hi siddreal

Any polygon having n sides will have n angles and when you see each angle , the interior and exterior angles make a straight line and thus are equal to 180.

Therefore sum of all n interior angles and exterior angles is 180*n
But what is the sum of all interior angles of n sides = 180*(n-2)
So Sum of all exterior angles+180(n-2)=180n...
Sum of all exterior angles = 180n-180(n-2)=180n-180n+360=360
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Re: What is the number of sides of a regular polygon in which 1/3rd of the   [#permalink] 27 Aug 2018, 07:37
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