polygon with n sides is 180(n – 2) degrees

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arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Wed Dec 29, 2010 3:20 am

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

[spoiler]the answer could be both B & E, but why only B is given the answer ?[/spoiler]

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Source: Beat The GMAT — Problem Solving | Wed Dec 29, 2010 3:20 am

shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Wed Dec 29, 2010 3:26 am

Just put the formula and get the answer...

180(p-2) = 3*360 => p = 8, where p is number of sides of the polygon

Can u plz tell how did u get 14?

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arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Wed Dec 29, 2010 3:31 am

shovan85 wrote:Just put the formula and get the answer...

180(p-2) = 3*360 => p = 8, where p is number of sides of the polygon

Can u plz tell how did u get 14?

180(14-2) = 180*12

now when we divide by 3 we get

180 * 4 = 180(6-2) thus n=6, so I believe E also fits the list.

shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Wed Dec 29, 2010 4:06 am

arora007 wrote:
180 * 4 = 180(6-2) thus n=6, so I believe E also fits the list.

But when n=6 then it is a Hexagon not a Quadrilateral.

If the problem is Easy Respect it, if the problem is tough Attack it

diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

by diebeatsthegmat » Wed Dec 29, 2010 6:54 pm

arora007 wrote:
shovan85 wrote:Just put the formula and get the answer...

180(p-2) = 3*360 => p = 8, where p is number of sides of the polygon

Can u plz tell how did u get 14?
180(14-2) = 180*12

now when we divide by 3 we get

180 * 4 = 180(6-2) thus n=6, so I believe E also fits the list.

nope, babe, it should be B
180(n-2)=3*360
n-2=6 so n = 8

arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Thu Dec 30, 2010 1:41 am

shovan85 wrote:
arora007 wrote:
180 * 4 = 180(6-2) thus n=6, so I believe E also fits the list.
But when n=6 then it is a Hexagon not a Quadrilateral.

Thanks dude... quadrilateral(4 sided polygon). How much more silly can it get ?

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