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Polygon

by raunakrajan » Tue Jul 13, 2010 6:48 am
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.
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by kmittal82 » Tue Jul 13, 2010 6:59 am
Sum of angles of a polygon with sides n = (n-2) x 180

(1) (n-2)*180 = 16k

n = 6 is the only value which satisfies this for n<9. SUFFICIENT

(2) (n-2)*180=15Q => (n-2)*12 = Q

Lots of value of n satisfy this criteria, so (2) is INSUFFICIENT

Hence, answer should be (A)
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by Rahul@gurome » Tue Jul 13, 2010 6:59 am
raunakrajan 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.
Sum of interior angles of polygon = 180º(n - 2), where n is the number of sides.

(1) For an integer to be divisible by 16, it should have four 2's, but 180 (= 2*2*3*3*5) has only two 2's, so we should have two more 2's in (n - 2). Only n = 6 can give two more 2's because [180 * (6 - 2)]/16= 45, an integer.

So, (1) is SUFFICIENT.

(2) 180/15 = 12 implies n can take any value between 2 and 8. So no unique answer.

Hence, (2) is NOT SUFFICIENT.

[spoiler]The correct answer is (A).[/spoiler]
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by kvcpk » Tue Jul 13, 2010 7:00 am
raunakrajan 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.
Sum of interior angles of a polygon is 180(n-2)
The sum of the interior angles of Polygon X is divisible by 16.
180(n-2) is div by 16
4*45*(n-2) is div by 4*4
Hence (n-2) should be a multiple of 4.
n-2 = 4 -> n=6
n-2 =8 -> n=10
But polygon has less than 9 sides.
Hence n=6

Suff

The sum of the interior angles of Polygon X is divisible by 15
180(n-2) is div by 15
15*12*(n-2) is div by 15.
n-2 can be anything.
INSUFF

pick A
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