Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Aug 2016, 02:13

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

ABCDE is a regular pentagon with F at its center. How many d

Author Message
TAGS:

Hide Tags

Intern
Joined: 29 Jul 2009
Posts: 16
Followers: 0

Kudos [?]: 25 [3] , given: 0

ABCDE is a regular pentagon with F at its center. How many d [#permalink]

Show Tags

03 Nov 2009, 14:56
3
KUDOS
7
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

60% (01:50) correct 40% (01:03) wrong based on 471 sessions

HideShow timer Statistics

ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

A. 10
B. 15
C. 20
D. 25
E. 30
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 34451
Followers: 6274

Kudos [?]: 79609 [6] , given: 10022

Show Tags

03 Nov 2009, 16:10
6
KUDOS
Expert's post
9
This post was
BOOKMARKED
arora2m wrote:
ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

a. 10

b. 15

c. 20

d. 25

e. 30

Any 3 points from 6 will make a triangle, since no 3 points are collinear, then:

6C3=20

TIPS ON RELATED ISSUES:
In a plane if there are n points of which no three are collinear, then
1. The number of straight lines that can be formed by joining them is nC2.
2. The number of triangles that can be formed by joining them is nC3.
3. The number of polygons with k sides that can be formed by joining them is nCk.
_________________
Intern
Joined: 29 Jul 2009
Posts: 16
Followers: 0

Kudos [?]: 25 [0], given: 0

Show Tags

03 Nov 2009, 16:39
Bunuel - In this case I am a bit confused.
F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.
VP
Joined: 05 Mar 2008
Posts: 1473
Followers: 11

Kudos [?]: 253 [0], given: 31

Show Tags

03 Nov 2009, 16:55
arora2m wrote:
Bunuel - In this case I am a bit confused.
F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.

acf does form a triangle. think of F immediately under A so drop a vertical line from a to f and then draw a line form a to c then connect c to f

the line from a to f stops at f and doesn't continue on to another letter. it angles over to c
Math Expert
Joined: 02 Sep 2009
Posts: 34451
Followers: 6274

Kudos [?]: 79609 [1] , given: 10022

Show Tags

03 Nov 2009, 17:01
1
KUDOS
Expert's post
arora2m wrote:
Bunuel - In this case I am a bit confused.
F is the center of the pentagon ABCDE ... So ABF form a triangle but not ACF, since a straight line drawn between AC will not pass through F.

I got 15 by doing 5 triangles formed with F as one of the points and then 5C3 thus 5+10 =15.

Please tell me what is wrong with my reasoning here.

Regular pentagon is a pentagon where all sides are equal. In such pentagon center is not collinear to any two vertex, so ANY three points (from 5 vertices and center point) WILL form the triangle.

You wrote that ACF won't form the triangle because they don't lie on straight line - that's not true. EXACTLY because these three points DON'T lie on the straight line they WILL form triangle. (In your other example you stated that ABF will form the triangle, but these three points also aren't collinear).

The question basically asks how many triangles can be formed from the six points on a plane with no three points being collinear.

Hope it's clear.
_________________
Intern
Joined: 01 Nov 2009
Posts: 16
Followers: 0

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

Show Tags

16 Apr 2010, 01:23
I also had similar question regarding total numbe rbeing 15 or 20..
the forum is amazing ..
and thanks Brunell
Senior Manager
Joined: 10 Jul 2013
Posts: 335
Followers: 3

Kudos [?]: 275 [0], given: 102

Re: ABCDE is a regular pentagon with F at its center. How many d [#permalink]

Show Tags

15 Aug 2013, 07:03
arora2m wrote:
ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

A. 10
B. 15
C. 20
D. 25
E. 30

ITS 6C3 = 20 .
AND THERE ARE NO POINTS THOSE WILL NOT FORM TRIANGLE...................
ABC, BCD ALL THESE 5 DIFFERENT COMBINATIONS OF POINTS WILL FORM TRIANGLE...
_________________

Asif vai.....

Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 635
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Followers: 50

Kudos [?]: 324 [0], given: 297

Show Tags

21 Aug 2013, 12:29
Bunuel wrote:
arora2m wrote:
ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

a. 10

b. 15

c. 20

d. 25

e. 30

Any 3 points from 6 will make a triangle, since no 3 points are collinear, then:

6C3=20

TIPS ON RELATED ISSUES:
In a plane if there are n points of which no three are collinear, then
1. The number of straight lines that can be formed by joining them is nC2.
2. The number of triangles that can be formed by joining them is nC3.
3. The number of polygons with k sides that can be formed by joining them is nCk.

Where am I wrong, I used, 5C2 X 1 = 10

_________________

Like my post Send me a Kudos It is a Good manner.
My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

Math Expert
Joined: 02 Sep 2009
Posts: 34451
Followers: 6274

Kudos [?]: 79609 [0], given: 10022

Show Tags

21 Aug 2013, 12:31
honchos wrote:
Bunuel wrote:
arora2m wrote:
ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

a. 10

b. 15

c. 20

d. 25

e. 30

Any 3 points from 6 will make a triangle, since no 3 points are collinear, then:

6C3=20

TIPS ON RELATED ISSUES:
In a plane if there are n points of which no three are collinear, then
1. The number of straight lines that can be formed by joining them is nC2.
2. The number of triangles that can be formed by joining them is nC3.
3. The number of polygons with k sides that can be formed by joining them is nCk.

Where am I wrong, I used, 5C2 X 1 = 10

You should explain the logic behind your approach.
_________________
Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 635
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Followers: 50

Kudos [?]: 324 [0], given: 297

Show Tags

21 Aug 2013, 12:44
You should explain the logic behind your approach.[/quote]

ABCDE, 5 vertices of pentagon, we can select any two vertices two form a line, 5C2 ways....

Now any of the 5C2 ways have only one way(As per question) to interact with a centre point F = 5C2 X 1 = 10 Ways.
_________________

Like my post Send me a Kudos It is a Good manner.
My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

Math Expert
Joined: 02 Sep 2009
Posts: 34451
Followers: 6274

Kudos [?]: 79609 [1] , given: 10022

Show Tags

21 Aug 2013, 12:48
1
KUDOS
Expert's post
honchos wrote:

ABCDE, 5 vertices of pentagon, we can select any two vertices two form a line, 5C2 ways....

Now any of the 5C2 ways have only one way(As per question) to interact with a centre point F = 5C2 X 1 = 10 Ways.

How many different triangles can be formed by joining 3 of the points A, B, C, D, E and F?

So, we are interested in the number of triangles that can be formed by joining ANY three out of these 6 points (A, B, C, D, E and F), not necessarily F and 2 other points. For example, triangle ABC is also a valid one.

Hope it's clear.
_________________
Intern
Joined: 19 Apr 2013
Posts: 22
Followers: 0

Kudos [?]: 4 [0], given: 57

ABCDE is a regular pentagon with F at its center [#permalink]

Show Tags

27 Sep 2013, 09:46
Hi Guys,

Please help me with this question and suggest a short method to solve this type of questions.

Thanks.
Attachments

Untitled.jpg [ 29.44 KiB | Viewed 4331 times ]

Math Expert
Joined: 02 Sep 2009
Posts: 34451
Followers: 6274

Kudos [?]: 79609 [0], given: 10022

Re: ABCDE is a regular pentagon with F at its center [#permalink]

Show Tags

27 Sep 2013, 09:52
Intern
Joined: 18 Sep 2013
Posts: 8
Followers: 0

Kudos [?]: 2 [1] , given: 1

Re: ABCDE is a regular pentagon with F at its center. How many d [#permalink]

Show Tags

25 Dec 2013, 03:15
1
KUDOS
This is basically a combinatorics problem.
If we make anagram grid for this, it will like below
A B C D E F
Y Y Y N N N
That means at any point of time only three points are making the triangle while other three are not.
hence total no of combinations possible = 6!/3!x3! =20
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11092
Followers: 511

Kudos [?]: 134 [0], given: 0

Re: ABCDE is a regular pentagon with F at its center. How many d [#permalink]

Show Tags

29 Apr 2015, 22:53
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11092
Followers: 511

Kudos [?]: 134 [0], given: 0

Re: ABCDE is a regular pentagon with F at its center. How many d [#permalink]

Show Tags

06 May 2016, 00:34
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: ABCDE is a regular pentagon with F at its center. How many d   [#permalink] 06 May 2016, 00:34
Similar topics Replies Last post
Similar
Topics:
3 How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2? 2 7 08 Apr 2016, 02:48
2 Two regular pentagons ABCDE and BCFGH have a side BC common. X is the 4 30 Jan 2016, 09:54
8 Given that ABCDE is a regular pentagon, what is the measure of ∠ACE? 4 08 Mar 2015, 20:26
12 ABCDE is a regular pentagon with F at its center. How many 10 27 May 2012, 09:43
14 A regular pentagon is inscribed in a circle. If A and B are 21 23 Oct 2007, 14:04
Display posts from previous: Sort by