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old-fashioned star

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old-fashioned star [#permalink] New post 23 Jul 2009, 05:33
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The 5 outer angles of an old-fashioned star are represented as a,b,c,d,e.What is the value of a+b+c+d+e?
options:30,60,90,180,360

OA:180(ball parked figure)

Could someone tell me how to find the exact answer?
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Re: old-fashioned star [#permalink] New post 23 Jul 2009, 06:01
This is a very tough question. I don't know whether I would do at the time of GMAT because it took me a while to do it.

First of all you need to close the extremities of your star in triangles, so that now you have a pentagon inside the star.

The formula (n-2)*180, where n is the number of sides, gives you the sum of the intern angles.

Thus, inside the pentagon you have (5-2)*180 = 3*180. Keep this number.

Now if you pick one of the angles inside the pentagon you will notice that together with a angle inside the triangle, the sum is 180.

Let's call the angles of the pentagon 1º, 2º, 3º, 4º, 5º
And the angles of the triangles, v, w, x, y, z. (Notice that there are two angles v, two angles w, ...)

Summing each one of the angles inside the pentagon with one angle inside the triangle we have.

v + 1º + w + 2º + x + 3º + y + 4º + z + 5º = 5*180
As we know that 1º + 2º + 3º + 4º + 5º = 3*180 (the sum of the angles of the pentagon.),
we have v + w + x + y + z = 2*180

Now if you sum all the angles of all triangles you will have
2v + 2w + 2x + 2y + 2z + a + b + c + d + e = 5*180
2*(v + w + x + y + z) + a + b + c + d + e = 5*180

As we know that v + w + x + y + z = 2*180

2*(2*180) + a + b + c + d + e = 5*180
4*180 + a + b + c + d + e = 5*180
a + b + c + d + e = 180

Ufa!! Ok now I really don't want a question like this on real GMAT... :shock:

Did it work fine? Consider a kudo to assist me in accessing the GMATClub tests ;)

Last edited by coelholds on 23 Jul 2009, 06:36, edited 2 times in total.
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Re: old-fashioned star [#permalink] New post 23 Jul 2009, 06:25
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coelholds, your approach looks similar to mine, but i might need some time to understand, so im giving my approach, which might be different from yours (since ive not read exactly what you have written). I did it in about 1 min, and 1 extra minute for rechecking, because i too did it for first time.

When you draw the star, you get 5 triangles lying on the sides of a regular pentagon.

Sum of exterior angles of a regular polygon is always 360. These exterior angles are also the interior angles for the 5 triangles.

This pentagon has 2 sets of exterior angles, 1 set clockwise, and 2nd one counterclockwise.

ie, consider the 5 ext angles clockwise, and again consider the set of 5 ext angles counterclockwise, you will get 2 sets of exterior angles which are also base angels for all triangles. You are required to find the sum of the remaining apex angles of all triangles.

Sum of all 3 angles of 5 triangles should be 180*5.

Sum of 2 sets of base triangles is 360*2 (360 is the sum for each set)

So sum of 5 apex angles = 180*5 - 360*2 = 180
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Re: old-fashioned star [#permalink] New post 23 Jul 2009, 06:43
rashminet84, I must admit: your solution is much faster. thanks. +1 kudo.
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Re: old-fashioned star [#permalink] New post 23 Jul 2009, 07:01
coelholds wrote:
rashminet84, I must admit: your solution is much faster. thanks. +1 kudo.


thanks :)
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Re: old-fashioned star [#permalink] New post 23 Jul 2009, 10:17
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When I have a geometrical figure, I prefer to play with figure. :evil:

Here's the image I drew to solve this one - (Yes, I still use MSpaint to edit image. :|)

Image

Addition of Interior angles of Pentagon = (2n-4)x90 = 540.
Hence, angle P = (540/5) = 108.

P is exterior angle of triangle. Hence, x+y = 108.
In triangle, x+2y = 180.

Hence, y=72.
x=36.

Answer = 36*5 = 180.
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Re: old-fashioned star [#permalink] New post 23 Jul 2009, 15:24
Great explanation +1
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Re: old-fashioned star [#permalink] New post 24 Jul 2009, 08:15
I kudo to alpeshvc !!!!
Good thinking :-D
Re: old-fashioned star   [#permalink] 24 Jul 2009, 08:15
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