If Polygon X has fewer than 9 sides, how many sides does : Quant Question Archive [LOCKED]
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# If Polygon X has fewer than 9 sides, how many sides does

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If Polygon X has fewer than 9 sides, how many sides does [#permalink]

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23 Nov 2007, 10:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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23 Nov 2007, 10:25

For any polygon, sum of interior angles = (n-2)*180.

Given: n<9

A) (n-2)*180 is divisible by 16. Only one value of 'n' (n = 6) is possible.
Sufficient.

B) (n-2)*180 is divisible by 15. 180 is divisible by 15. Therefore there are multiple possible values for 'n'.
Insufficient.
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23 Nov 2007, 10:59
jbs wrote:

For any polygon, sum of interior angles = (n-2)*180.

Given: n<9

A) (n-2)*180 is divisible by 16. Only one value of 'n' (n = 6) is possible.
Sufficient.

B) (n-2)*180 is divisible by 15. 180 is divisible by 15. Therefore there are multiple possible values for 'n'.
Insufficient.

Same strategy...only A is sufficient..since (n-2)*180 is always divisible by 15.

1. (N-2)*180/16 = (N-2)*45/4

SO (N-2) should be divisible by 4 and only when N= 6 its possible!!
23 Nov 2007, 10:59
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