sanju09 wrote:The smallest interior angle of an n-sided covex polygon is 120º, and each interior angle after this, taken in counterclockwise manner, is 5º more than the preceding interior angle. What is n?
(A) 49
(B) 16
(C) 13
(D) 12
(E) 9

Sum of interior angles of a convex polygon = (n-2)*180

Sum of the angles which are in AP = n/2(2*a+(n-1)*d) = n/2(2*120+ 5n-5) = n/2(5n+235)

since these two are equal

180n -360 = 5/2 (n^2+ 79n)
72n - 144 = n^2+79n
n^2+7n-144 =0
(n-9)(n+16) =0

[spoiler]n = 9[/spoiler]

ajith wrote:

sanju09 wrote:The smallest interior angle of an n-sided covex polygon is 120º, and each interior angle after this, taken in counterclockwise manner, is 5º more than the preceding interior angle. What is n?
(A) 49
(B) 16
(C) 13
(D) 12
(E) 9

Sum of interior angles of a convex polygon = (n-2)*180

Sum of the angles which are in AP = n/2(2*a+(n-1)*d) = n/2(2*120+ 5n-5) = n/2(5n+235)

since these two are equal

180n -360 = 5/2 (n^2+ 79n)
72n - 144 = n^2+79n
n^2+7n-144 =0
(n-9)(n+16) =0

[spoiler]n = 9[/spoiler]

are you sure with your work and the thoughts behind work?

Each exterior angle would be 60 , 55 , 50 , etc...its AP with - 5 degree

Sum of all angles = n(2*60+(n-1)*(-5))/2 = 360( sum of all exterior angles = 360)

n/2(120-(n-1)*5)) = 360
==>n/2[ 125 - 5n] = 360
==> 5/2*[25 - n] = 360
n(25-n) = 144 .

... quadratic eqn with roots 9 and 16..but n cant be 16 as in that case some interior angles exceed 180 ( which cant be as polygon is convex )

So 9 is the answer

sanju09 wrote:are you sure with your work and the thoughts behind work?

I would say I am positive!

ajith wrote:

sanju09 wrote:are you sure with your work and the thoughts behind work?

I would say I am positive!

Please tell us the meaning of a convex polygon

sanju09 wrote:
Please tell us the meaning of a convex polygon

Help yourselves : https://en.wikipedia.org/wiki/Convex_and ... e_polygons

Now what, in my defense, 9 angles are

120,125,130,135,140,145,150,155 [all of them less than 180 degrees]

So there is nothing to prove that it is not convex

ajith wrote:

sanju09 wrote:The smallest interior angle of an n-sided covex polygon is 120º, and each interior angle after this, taken in counterclockwise manner, is 5º more than the preceding interior angle. What is n?
(A) 49
(B) 16
(C) 13
(D) 12
(E) 9

Sum of interior angles of a convex polygon = (n-2)*180

Sum of the angles which are in AP = n/2(2*a+(n-1)*d) = n/2(2*120+ 5n-5) = n/2(5n+235)

since these two are equal

180n -360 = 5/2 (n^2+ 79n)
72n - 144 = n^2+79n
n^2+7n-144 =0
(n-9)(n+16) =0

[spoiler]n = 9[/spoiler]

I've problem in the following algebra first of all

180n -360 = 5/2 (n^2+ 79n)
72n - 144 = n^2+79n
n^2+7n-144 =0
(n-9)(n+16) =0

Whereas I'm getting it as n^2 - 25 n + 144 = 0, hence two possibilities for n are 9 and 16, and both are there in the choices, fine. But if it mentions that 120º is the smallest angle, then are we sure that the 9th angle in the so called and described counterclockwise manner will be the greatest possibility for the formation of a convex polygon as demanded in the question. Or, did I need to mention "the greatest" before n in stem? Do I really need to edit my post, once?

What's the folly in the following piece of thought?

175º is the greatest possible interior angle in this fashion, such that 175 = 120 + 5 (n - 1) or n = [spoiler]12.[/spoiler]

[spoiler]D just IMO[/spoiler]

sanju09 wrote: Whereas I'm getting it as n^2 - 25 n + 144 = 0, hence two possibilities for n are 9 and 16, and both are there in the choices, fine. But if it mentions that 120º is the smallest angle, then are we sure that the 9th angle in the so called and described counterclockwise manner will be the greatest possibility for the formation of a convex polygon as demanded in the question. Or, did I need to mention "the greatest" before n in stem? Do I really need to edit my post, once?

What's the folly in the following piece of thought?

175º is the greatest possible interior angle in this fashion, such that 175 = 120 + 5 (n - 1) or n = [spoiler]12.[/spoiler]

[spoiler]D just IMO[/spoiler]

You are indeed right about the mistake I made in the calculation

Sum of interior angles of a convex polygon = (n-2)*180

Sum of the angles which are in AP = n/2(2*a+(n-1)*d) = n/2(2*120+ 5n-5) = n/2(5n+235)

since these two are equal

180n -360 = 5/2 (n^2+ 47n)
72n - 144 = n^2+47n
n^2-25n-144 =0
(n-9)(n-16) =0

Still n cant be 16 because in that case ; the maximum angle will be more than 180 and it is mentioned that it is a convex polygon.

n=9

If indeed the number of sides were 12 as you suggested; the sum of all interior angles should be (n-2)*180 = 10*180 =1800
Now, the sum of AP = 6*(120+175) = 1770 and such a convex polygon cannot exist....
Hence D is not the option

sanju09 wrote: Or, did I need to mention "the greatest" before n in stem? Do I really need to edit my post, once?

If you want to confuse the test taker, why not? [I think all these convex, counterclockwise, AP would have done the trick already for you]

Another simple way to solve this knowing that the greatest angle is a convex polygon has to be less that 180
120+ (n-1) d < 180
120 + 5n -5 < 180
n has to be just < 13, cos if n was 13 the angle will be 180 or st. line
n is 12

kstv wrote:Another simple way to solve this knowing that the greatest angle is a convex polygon has to be less that 180
120+ (n-1) d < 180
120 + 5n -5 < 180
n has to be just < 13, cos if n was 13 the angle will be 180 or st. line
n is 12

I have explained in my previous post how n cannot be 12.

Sanju; Mr Devil's Advocate,

Please direct the mislead souls back to the pack.

ajith wrote:

kstv wrote:Another simple way to solve this knowing that the greatest angle is a convex polygon has to be less that 180
120+ (n-1) d < 180
120 + 5n -5 < 180
n has to be just < 13, cos if n was 13 the angle will be 180 or st. line
n is 12

I have explained in my previous post how n cannot be 12.

Sanju; Mr Devil's Advocate,

Please direct the mislead souls back to the pack.

This indeed is a question that needs a perfect blending of AP and the sum formula limitation of the interior angles of any convex polygon; we cannot ignore one resort for the sake of the other. This is the real folly in my piece of devil's thought. Almost all the choices other than the OA are made with a devil's design in mind just to attract all the mislead souls which are barred by the GMAT.

Taking care at calculations can keep few Angel's Advocate smiling forever...

sanju09 wrote:
This indeed is a question that needs a perfect blending of AP and the sum formula limitation of the interior angles of any convex polygon; we cannot ignore one resort for the sake of the other. This is the real folly in my piece of devil's thought. Almost all the choices other than the OA are made with a devil's design in mind just to attract all the mislead souls which are barred by the GMAT.

Taking care at calculations can keep few Angel's Advocate smiling forever...

To translate,

1. Convex polygon has the sum of interior angles = (n-2)*180
2. Sum of AP should be equal to sum of interior angles
3. Convex polygon has all the internal angles less than 180 degrees

Combining all these, the OA is E

The last line is better kept untranslated.

ajith wrote:

sanju09 wrote: Whereas I'm getting it as n^2 - 25 n + 144 = 0, hence two possibilities for n are 9 and 16, and both are there in the choices, fine. But if it mentions that 120º is the smallest angle, then are we sure that the 9th angle in the so called and described counterclockwise manner will be the greatest possibility for the formation of a convex polygon as demanded in the question. Or, did I need to mention "the greatest" before n in stem? Do I really need to edit my post, once?

What's the folly in the following piece of thought?

175º is the greatest possible interior angle in this fashion, such that 175 = 120 + 5 (n - 1) or n = [spoiler]12.[/spoiler]

[spoiler]D just IMO[/spoiler]

You are indeed right about the mistake I made in the calculation

Sum of interior angles of a convex polygon = (n-2)*180

Sum of the angles which are in AP = n/2(2*a+(n-1)*d) = n/2(2*120+ 5n-5) = n/2(5n+235)

since these two are equal

180n -360 = 5/2 (n^2+ 47n)
72n - 144 = n^2+47n
n^2-25n-144 =0
(n-9)(n-16) =0

Still n cant be 16 because in that case ; the maximum angle will be more than 180 and it is mentioned that it is a convex polygon.

n=9

If indeed the number of sides were 12 as you suggested; the sum of all interior angles should be (n-2)*180 = 10*180 =1800
Now, the sum of AP = 6*(120+175) = 1770 and such a convex polygon cannot exist....
Hence D is not the option

I'm a bit confused of how you got the "AP = n/2(2*a+(n-1)*d)". Can you explain a bit further? And how did you get "180n -360 = 5/2 (n^2+ 47n)" from that? Is it me or is 235/2 not equal to 47???? Or is it just late and I should be understanding this??!??! ahh