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Each side of a given polygon is parallel to either the X or

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Each side of a given polygon is parallel to either the X or [#permalink] New post 29 Apr 2010, 05:10
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Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90o or concave if the internal angle is 270o. If the number of convex corners in such a polygon is 25, the number of concave corners must be:

(1)20
(2)0
(3)21
(4)22
(5)23
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Re: polygon [#permalink] New post 29 Apr 2010, 05:58
Is the answer C-21?
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Re: polygon [#permalink] New post 29 Apr 2010, 06:17
I also think the answer is 21...

In the figure attached

1,2,3,4 so on represent convex angles...

1`,2`,3`... so on represent concave angles... which will be 21..
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Re: polygon [#permalink] New post 29 Apr 2010, 06:44
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lanka1 wrote:
Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90o or concave if the internal angle is 270o. If the number of convex corners in such a polygon is 25, the number of concave corners must be:

(1)20
(2)0
(3)21
(4)22
(5)23



Sum of all the angles is 180(n-2).
Let the total no of 90 deg angle be x
Let the total no of 270 deg angle be y

n= x+y, so sum of all the angles is 180(x+y -2)

so, 90x + 270y = 180(x+y -2)
90x + 270y = 180x+ 180y - 360
90y = 90x -360

Given x=25
y= (90*25 - 360)/90
y= 21
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Re: polygon [#permalink] New post 29 Apr 2010, 21:06
nicely explained sir...
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Re: polygon [#permalink] New post 30 Apr 2010, 00:36
nverma wrote:
Sum of all the angles is 180(n-2).
Let the total no of 90 deg angle be x
Let the total no of 270 deg angle be y

n= x+y, so sum of all the angles is 180(x+y -2)

so, 90x + 270y = 180(x+y -2)
90x + 270y = 180x+ 180y - 360
90y = 90x -360

Given x=25
y= (90*25 - 360)/90
y= 21

Very well explained.
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Re: polygon [#permalink] New post 02 May 2010, 08:06
thanks nverma..

This clearly hints that rather than getting bogged down by question person should just apply for basics
Re: polygon   [#permalink] 02 May 2010, 08:06
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