BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

polygon interior angles

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 154
Joined: Tue Aug 26, 2008 12:59 pm
Location: Canada
Thanked: 4 times

polygon interior angles

by canuckclint » Fri Oct 31, 2008 5:49 pm
If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

(1) The sum of the interior angles of Polygon X is divisible by 16.

(2) The sum of the interior angles of Polygon X is divisible by 15.
Top
Source: — Problem Solving |
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 am
Thanked: 146 times
Followed by:2 members

Re: polygon interior angles

by parallel_chase » Fri Oct 31, 2008 10:28 pm
canuckclint wrote:If Polygon X has fewer than 9 sides, how many sides does Polygon X have?

(1) The sum of the interior angles of Polygon X is divisible by 16.

(2) The sum of the interior angles of Polygon X is divisible by 15.
minimum sides of a polygon = 3, maximum sides = 8

Statement I
The sum of the interior angles of Polygon X is divisible by 16
sum of the interior angles = (n-2)*180
sides=4, 2*180 = 360, not divisible by 16
sides=5, 3*180 = 540, not divisible by 16
sides=6, 4*180 = 720, divisible by 16
sides=7, 5*180 = 900, not divisible by 16
sides=8, 6*180 = 1080, not divisible by 16

Sufficient. Sides = 6

Statement II
The sum of the interior angles of Polygon X is divisible by 15
sum of the interior angles = (n-2)*180
sides=3, 1*180 = 180, divisible by 15
sides=4, 2*180 = 360, divisible by 15
sides=5, 3*180 = 540, divisible by 15
sides=6, 4*180 = 720, divisible by 15
sides=7, 5*180 = 900, divisible by 15
sides=8, 6*180 = 1080, divisible by 15

Insufficient.


Hence A.

OA?
No rest for the Wicked....
Top
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

by cramya » Fri Oct 31, 2008 10:40 pm
It should be A)

The other way of looking at A) would for 16 we need 4 2's. 18 provides only 2 so we need 2 more 2's in n-2

The only value of n that could provide this is 6

(6-2) * 180/ 16 = integer (i.e. divisible by 16)

SUFF

B) 180 is divisible by 15 so for any value of 2<n< 8 it will work INSUFF
Top
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

by cramya » Fri Oct 31, 2008 10:42 pm
It should be A)

Stmt I)

The other way of looking at A) would be for 16 in dr to divide the nr exactly we need four 2's. 180 provides only two 2's (factorize 180) so we need two more 2's in n-2

The only value of n that could provide this is 6

(6-2) * 180/ 16 = integer (i.e. divisible by 16)

SUFF

Stmt II
180 is divisible by 15 so for any value of 2<n< 8 it will work INSUFF
Top
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Fri Apr 06, 2012 5:28 am
Thanked: 3 times
Followed by:2 members
GMAT Score:500

by BTG14 » Fri Nov 16, 2012 6:41 pm
Is this regular polygon, Because I was informed that , we can use (2n-4)*90-Sum of interior angles for regular polygon.

Please somebody correct me.
Top
Post new topic Post Reply
• Page 1 of 1