It is currently 21 Oct 2017, 02:22

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# By sharing adjacent sides , exactly n regular pentagons can

Author Message
Director
Joined: 01 Jan 2008
Posts: 504

Kudos [?]: 54 [0], given: 0

### Show Tags

22 Jun 2008, 05:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

By sharing adjacent sides , exactly n regular pentagons can form a circular ring . What is the value of n= ?

A 8
B 9
C 10
D 12
E 16

Kudos [?]: 54 [0], given: 0

Director
Joined: 01 Jan 2008
Posts: 504

Kudos [?]: 54 [1], given: 0

### Show Tags

22 Jun 2008, 10:19
1
KUDOS
Walker Atleast u tell me how to solve this.

Kudos [?]: 54 [1], given: 0

Senior Manager
Joined: 26 Mar 2008
Posts: 317

Kudos [?]: 91 [0], given: 4

Location: Washington DC

### Show Tags

22 Jun 2008, 11:00
Ans is 10.
Each interior angle of pentagon =108, factors=6.6.3
In order to create the complete circular all angle must be equal to 360.
Now see the factors of 360=6.6.10
Now lets do the highest common factor=6.6, that means we need to split the each angle of pentagon into 3 parts and use those sides to form some other pentagon then only circular ring can be formed.
that means 10 pentagon are required to create circular ring.

Let me know the source of the question, its really difficult and quite difficult to solve in the exam. +1 to you mate.

Kudos [?]: 91 [0], given: 4

Director
Joined: 01 Jan 2008
Posts: 504

Kudos [?]: 54 [0], given: 0

### Show Tags

22 Jun 2008, 20:41
Thanks . The Answer is 10

I got this in my friends GMAT notes. But couldnt understand the last step

Let me mention the technque

Interior angle of a pentagon = 108 deg by using formula ( n-2)x180 / n ; where n is no of sides

Now I didnt follow hence forth

360 - ( 108 x2 ) = 144 ( Why )

Hence interior angle of pentagon in middle is 144.

Now using same formula n-2 (180) /n = 144

we get n = 10.

Hence there are 10 pentagons needed

Kudos [?]: 54 [0], given: 0

GMAT Instructor
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 335 [0], given: 0

### Show Tags

23 Jun 2008, 02:46
The circular ring encloses a n-sided regular polygon. Since the sum of the angles of a k-sided polygon is (n-2)180º, each angle of a regular pentagon has a degree measure of 540/5 = 108º. Thus the degree measure of each angle of the n-sided regular polygon is 360 - 2(108) = 144º. But we know that the degree measure of each angle of the n-sided regular polygon is also 180(n-2)/n, so n=10.

It is also true that a 144º angle causes a change in direction of 36º. 10 such changes are needed to yield the entire 360º needed
Attachments

gmatclub.jpg [ 8.2 KiB | Viewed 2697 times ]

Kudos [?]: 335 [0], given: 0

Director
Joined: 01 Jan 2008
Posts: 504

Kudos [?]: 54 [0], given: 0

### Show Tags

23 Jun 2008, 11:10
[quote="kevincan"]Thus the degree measure of each angle of the n-sided regular polygon is 360 - 2(108) = 144º.

Please explain this step.. Is it a formula.

Kudos [?]: 54 [0], given: 0

GMAT Instructor
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 335 [0], given: 0

### Show Tags

23 Jun 2008, 12:54
Perhaps the diagram will help!

Kudos [?]: 335 [0], given: 0

Senior Manager
Joined: 26 Mar 2008
Posts: 317

Kudos [?]: 91 [0], given: 4

Location: Washington DC

### Show Tags

23 Jun 2008, 17:06
Well I am not sure if I could follow your solution. I used simpler method..
108=6.6.3, while 360=6.6.10, so if we create circular ring using pentagons then interior angle of pentagon must be able to divide 360. Thats possible only when we divide the pentagon into 3 parts, ie creating angle =36. In that case we need 10 such pentagons.

Kudos [?]: 91 [0], given: 4

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4587 [2], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

### Show Tags

29 Jun 2008, 11:43
2
KUDOS
Expert's post
the line between centers of neighbor pentagons change its direction by 1/2*360/5=36. Therefore, we need 360/36=10 pentagons to complete a circle.
Attachments

7-t65948.png [ 2.49 KiB | Viewed 2566 times ]

_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Kudos [?]: 4587 [2], given: 360

SVP
Joined: 30 Apr 2008
Posts: 1867

Kudos [?]: 615 [1], given: 32

Location: Oklahoma City
Schools: Hard Knocks

### Show Tags

29 Jun 2008, 12:49
1
KUDOS
I think everyone is using the same formula but doing certain steps in their head and now writing the steps down on paper and that is what is confusing. Start with the basic formula.

If we want to find the value of an interior angle of any polygon the formula is:

$$\frac{180(n-2)}{n}$$

When n = 5 for a pentagon, you get

$$\frac{180(5-2)}{5} = 108$$

So in the picture below, A & B each = 108. If you think of A + B + C as being inside a complete circle (imaginary circle) you know the total must = 360. So 360 - 108 - 108 = 144.

So Angle C in the picture is 144 degrees. This is an interior angle of a polygon formed by all the pentagons being joined. We now have to answer the question: A polygon with how many sides has interior angles of 144? Now we know the answer, but we don't have n. Before we had n =5 (pentagon) but we didn't have the answer. This is basic alegbra.

$$\frac{180(N-2)}{n}=144$$
180(N-2) = 144N
180N - 360 = 144N
36N - 360 = 0
36N = 360
N = 10

So we have created a 10-sided polygon by joining all of those pentagons together.

Attachment:

InteriorAngle.gif [ 4.99 KiB | Viewed 2543 times ]

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Kudos [?]: 615 [1], given: 32 SVP Joined: 30 Apr 2008 Posts: 1867 Kudos [?]: 615 [2], given: 32 Location: Oklahoma City Schools: Hard Knocks Re: Regular Pentagon [#permalink] ### Show Tags 29 Jun 2008, 13:15 2 This post received KUDOS Because I was bored this afternoon and I like Photoshop way too much Attachment: 10Pentagons.gif [ 25.91 KiB | Viewed 2539 times ] _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 615 [2], given: 32

Director
Joined: 01 Jan 2008
Posts: 504

Kudos [?]: 54 [0], given: 0

### Show Tags

02 Jul 2008, 02:08
+3 to all of you . Thanks

Kudos [?]: 54 [0], given: 0

Director
Joined: 01 Jan 2008
Posts: 504

Kudos [?]: 54 [0], given: 0

### Show Tags

02 Jul 2008, 02:18
+3 to all of you . Thanks

Kudos [?]: 54 [0], given: 0

Senior Manager
Joined: 07 Jan 2008
Posts: 398

Kudos [?]: 293 [0], given: 0

### Show Tags

02 Jul 2008, 03:36
I've got it. Thanks! +Kudos for both of you.

Kudos [?]: 293 [0], given: 0

Re: Regular Pentagon   [#permalink] 02 Jul 2008, 03:36
Display posts from previous: Sort by