old-fashioned star

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uptowngirl92: Master | Next Rank: 500 Posts; Posts: 447; Joined: Sun Apr 19, 2009 9:08 pm; Location: Kolkata,India; Thanked: 7 times; GMAT Score:670

by uptowngirl92 » Thu Jul 23, 2009 5:33 am

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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Source: Beat The GMAT — Problem Solving | Thu Jul 23, 2009 5:33 am

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

Re: old-fashioned star

by Stuart@KaplanGMAT » Thu Jul 23, 2009 9:36 am

uptowngirl92 wrote: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?

You will never see a GMAT question with a term such as "old-fashioned star".

Was the question accompanied by a diagram?

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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aanderson04: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Fri Jul 17, 2009 4:41 pm; Thanked: 3 times

by aanderson04 » Thu Jul 23, 2009 4:11 pm

There is a similar question in the OG #10 of the Diagnostic Quiz.

Here is the answer straight from the OG:
(The angles at the points are v,x,y,z,w and the angles inside the polygon are a,b,c,d,e.)

The sum of the interior angles of any polygon is 180(n-2), where n is the number of sides. Thus the angles inside the polygon equal 180(5-2)=180(3)=540.

Each of the interior angles of the pentagon defines a triangle with two of the angles at the points of the star. This gives the following five equations:

a+x+z=180
b+v+y=180
c+x+w=180
d+v+w=180
e+y+w=180

Summing these five equations gives:

a+b+c+d+e+2v+2x+2y+2z+2w=900

Substituting 540 for a+b+c+d+e gives:

540+2v+2x+2y+2z+2w=900

From this:
2v+2x+2y+2z+2w=360

2(v+x+y+z+w)=360

v+x+y+z+w=180

The correct answer is 180 or [size=0][size=18[/size]]C[/size].

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jjk: Senior | Next Rank: 100 Posts; Posts: 43; Joined: Thu Mar 12, 2009 8:56 pmFollowed by: 1 member; Thanked: 10 times; GMAT Score:730

by jjk » Fri Jul 24, 2009 11:59 am

Another way to look at it w/ less addition is to calculate the value of each interior angle of the pentagon that forms the center of the star.

You know that each interior angle of the pentagon measures 108 degrees. The triangles that form the points of the star share lines with the outer edges of the pentagon. You therefore know that 72 degrees will be the measure of one of a triangle's angle. 72 degrees will also be another angle measure in the triangle. 180 - 2(72) = 180 - 144 = 36 degrees as the measure of the triangle's outermost angle.

Multiply 36 by 5 and you get 180 degrees. C.

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gmat740: MBA Student; Posts: 1194; Joined: Sat Aug 16, 2008 9:42 pmFollowed by: 17 members; Location: Paris, France; Thanked: 71 times; GMAT Score:710

Re: old-fashioned star

by gmat740 » Sun Jul 26, 2009 4:57 am

Stuart Kovinsky wrote:
uptowngirl92 wrote: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?
You will never see a GMAT question with a term such as "old-fashioned star".

Was the question accompanied by a diagram?

I second that

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