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

It is currently 18 Dec 2014, 06:31

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Set A, B, C have some elements in common. If 16 elements are in both A

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 31 Mar 2006
Posts: 164
Followers: 1

Kudos [?]: 9 [0], given: 0

Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 27 Jun 2006, 02:53
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

31% (02:29) correct 69% (00:55) wrong based on 29 sessions
Set A, B, C have some elements in common. If 16 elements are in both A and B, 17 elements are in both A and C, and 18 elements are in both B and C, how many elements do all three of the sets A, B, and C have in common?

(1) Of the 16 elements that are in both A and B, 9 elements are also in C
(2) A has 25 elements, B has 30 elements, and C has 35 elements.

[color=#ff0000][b]OPEN DISCUSSION OF THIS QUESTION IS HERE: set-a-b-c-have-some-elements-in-common-if-16-elements-are-112041.html[/b][/color]
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Dec 2014, 06:08, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
Kaplan Promo CodeKnewton GMAT Discount CodesVeritas Prep GMAT Discount Codes
Senior Manager
Senior Manager
User avatar
Joined: 07 Jul 2005
Posts: 406
Location: Sunnyvale, CA
Followers: 2

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

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 27 Jun 2006, 08:17
amansingla4 wrote:
Sets A,B and C have some elements in common.If 16 elements are in both A&B,17 in A&C ,18 elements are in B&C, how many elements do all the three A,B,C have in common?
1) Of the 16 elements in A&B,9 are in C.
2) A has 25 elements,B has 30 and c has 35 elements

Please explain.


(A) it is.
I implies 9 elements are common between A, B, C
For II, we still need the total # of elements..
This is what I could deduce from the venn diagram
Total = A + B + C - AB - BC - CA - 2ABC
We have everything but Total and ABC.
Cannot calculate ABC without knowing hte total.
Director
Director
User avatar
Joined: 06 May 2006
Posts: 782
Followers: 3

Kudos [?]: 15 [0], given: 0

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 27 Jun 2006, 12:05
shampoo wrote:
Lets slightly change the statement; what if we were given the total # of elements?
Then the ans would be D right?


Yes.

There is a formula for this. Let us say Common elements of sets A and B is denoted by Int(A,B) [It is called 'Intersection' in Set Theory], combined elements of sets A and B are denoted by Un(A,B) ['Union' in Set Theory], and number of elements in set A is denoted by n[A], then

n[Un(A,B,C)]=n[A]+n[.B]+n[C]-n[Int(A,B)]-n[Int(B,C)]-n[Int(C,A)]+n[Int(A,B,C)]

So here you have n[Un(A,B,C)]=25+30+35-16-17-18+n[Int(A,B,C)]

If we have the total number of elements in A, B, C combined then the answer will be D.

In the absence of that vital piece of data, however, we have A. I cannot see how a Venn diagram can help us with that. Can someone explain?
Director
Director
User avatar
Joined: 28 Dec 2005
Posts: 758
Followers: 1

Kudos [?]: 8 [0], given: 0

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 27 Jun 2006, 12:50
paddyboy wrote:
shampoo wrote:
Lets slightly change the statement; what if we were given the total # of elements?
Then the ans would be D right?


Yes.

There is a formula for this. Let us say Common elements of sets A and B is denoted by Int(A,B) [It is called 'Intersection' in Set Theory], combined elements of sets A and B are denoted by Un(A,B) ['Union' in Set Theory], and number of elements in set A is denoted by n[A], then

n[Un(A,B,C)]=n[A]+n[.B]+n[C]-n[Int(A,B)]-n[Int(B,C)]-n[Int(C,A)]+n[Int(A,B,C)]

So here you have n[Un(A,B,C)]=25+30+35-16-17-18+n[Int(A,B,C)]

If we have the total number of elements in A, B, C combined then the answer will be D.

In the absence of that vital piece of data, however, we have A. I cannot see how a Venn diagram can help us with that. Can someone explain?


A venn diagram does help with visualizing what is needed, but does not help solving. Statement A provides this information upfront.
Senior Manager
Senior Manager
User avatar
Joined: 20 Feb 2006
Posts: 331
Followers: 1

Kudos [?]: 12 [0], given: 0

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 27 Jun 2006, 13:34
paddyboy wrote:
shampoo wrote:
Lets slightly change the statement; what if we were given the total # of elements?
Then the ans would be D right?


Yes.

There is a formula for this. Let us say Common elements of sets A and B is denoted by Int(A,B) [It is called 'Intersection' in Set Theory], combined elements of sets A and B are denoted by Un(A,B) ['Union' in Set Theory], and number of elements in set A is denoted by n[A], then

n[Un(A,B,C)]=n[A]+n[.B]+n[C]-n[Int(A,B)]-n[Int(B,C)]-n[Int(C,A)]+n[Int(A,B,C)]

So here you have n[Un(A,B,C)]=25+30+35-16-17-18+n[Int(A,B,C)]

If we have the total number of elements in A, B, C combined then the answer will be D.

In the absence of that vital piece of data, however, we have A. I cannot see how a Venn diagram can help us with that. Can someone explain?


Shouldn't it be :
n[Un(A,B,C)]=n[A]+n[.B]+n[C]-n[Int(A,B)]-n[Int(B,C)]-n[Int(C,A)]+2*n[Int(A,B,C)]
Manager
Manager
avatar
Joined: 26 Jun 2006
Posts: 152
Followers: 1

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

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 27 Jun 2006, 17:03
Have to agree with majority -- I think it is (A)

(I) is obvious, but got confused a little bit by (II). But once you write the formula

Total = A + B + C - AB - AC -BC + ABC

everything gets into places. Assuining statement (II), in the above formula we know everything but Total and ABC, so don't have enough info to determine ABC.
Senior Manager
Senior Manager
User avatar
Joined: 07 Jul 2005
Posts: 406
Location: Sunnyvale, CA
Followers: 2

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

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 28 Jun 2006, 14:36
v1rok wrote:
Have to agree with majority -- I think it is (A)

(I) is obvious, but got confused a little bit by (II). But once you write the formula

Total = A + B + C - AB - AC -BC + ABC

everything gets into places. Assuining statement (II), in the above formula we know everything but Total and ABC, so don't have enough info to determine ABC.


I am still confused by the above eqn.
It appears that,
total shd be = A + B + C - AB - AC -BC -2ABC

When we subtract AB, BC, CA from total area, we remove the overlapping part, but still the area common to all three of them is counted thrice .. We need to subtract it twice to get the real total.

Anyone who can explain which one is correct and why ?
Senior Manager
Senior Manager
User avatar
Joined: 20 Feb 2006
Posts: 331
Followers: 1

Kudos [?]: 12 [0], given: 0

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 28 Jun 2006, 16:07
sgrover wrote:
v1rok wrote:
Have to agree with majority -- I think it is (A)

(I) is obvious, but got confused a little bit by (II). But once you write the formula

Total = A + B + C - AB - AC -BC + ABC

everything gets into places. Assuining statement (II), in the above formula we know everything but Total and ABC, so don't have enough info to determine ABC.


I am still confused by the above eqn.
It appears that,
total shd be = A + B + C - AB - AC -BC -2ABC

When we subtract AB, BC, CA from total area, we remove the overlapping part, but still the area common to all three of them is counted thrice .. We need to subtract it twice to get the real total.

Anyone who can explain which one is correct and why ?



Total = A + B + C - AB - AC -BC + ABC - is correct..
Draw a Venn Diagram, name each part 1-7.. and then verify..
this will give you the above formula..
1 KUDOS received
CEO
CEO
User avatar
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 17

Kudos [?]: 93 [1] , given: 0

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 28 Jun 2006, 16:09
1
This post received
KUDOS
sgrover wrote:
v1rok wrote:
Have to agree with majority -- I think it is (A)

(I) is obvious, but got confused a little bit by (II). But once you write the formula

Total = A + B + C - AB - AC -BC + ABC

everything gets into places. Assuining statement (II), in the above formula we know everything but Total and ABC, so don't have enough info to determine ABC.


I am still confused by the above eqn.
It appears that,
total shd be = A + B + C - AB - AC -BC -2ABC

When we subtract AB, BC, CA from total area, we remove the overlapping part, but still the area common to all three of them is counted thrice .. We need to subtract it twice to get the real total.

Anyone who can explain which one is correct and why ?


No this is correct.

In A+B+C we included AB, AC and BC twice so we need to subtract each of these once.
Now we have A+B+C-AB-BC-CA.

But in A+B+C we we also included ABC three times so we need to subtract ABC two times.
Now we have A+B+C-AB-BC-CA-2ABC.

But in subtracting AB, AC and BC we subtracted ABC three times. So need to add 3ABC

Finally we have A+B+C-AB-BC-CA-2ABC+3ABC i.e
A+B+C-AB-BC-CA+ABC

Try it with marking the areas in a Venn Diagram.

Hope this helps.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

SVP
SVP
User avatar
Joined: 03 Jan 2005
Posts: 2251
Followers: 13

Kudos [?]: 210 [0], given: 0

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 28 Jun 2006, 17:53
See if this helps (from the basic priniciple sticky):

HongHu wrote:
Formula:

Total = N(A) + N(B) + N(C) - N(A n B) - N(A n C) - N(C n B) + N(A n B n C)

If instead of numbers for (A n B) and (A n C) and (C n B), what is given is the total number of people who choose exactly two items, then the formula becomes:

Total = N(A) + N(B) + N(C) - (N(choose exactly two items)) - 2N(choose all three items)

Also,
Total = N(A) + N(B) + N(C) - (N(choose at least two items)) - N(choose all three items)

_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Manager
Manager
avatar
Joined: 31 Mar 2006
Posts: 164
Followers: 1

Kudos [?]: 9 [0], given: 0

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 29 Jun 2006, 00:40
HongHu wrote:
See if this helps (from the basic priniciple sticky):

HongHu wrote:
Formula:

Total = N(A) + N(B) + N(C) - N(A n B) - N(A n C) - N(C n B) + N(A n B n C)

If instead of numbers for (A n B) and (A n C) and (C n B), what is given is the total number of people who choose exactly two items, then the formula becomes:

Total = N(A) + N(B) + N(C) - (N(choose exactly two items)) - 2N(choose all three items)

Also,
Total = N(A) + N(B) + N(C) - (N(choose at least two items)) - N(choose all three items)



Can you please explain these two formulas?

Regards,
Aman
SVP
SVP
User avatar
Joined: 03 Jan 2005
Posts: 2251
Followers: 13

Kudos [?]: 210 [0], given: 0

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 30 Jun 2006, 07:56
amansingla4 wrote:
HongHu wrote:
See if this helps (from the basic priniciple sticky):

HongHu wrote:
Formula:

Total = N(A) + N(B) + N(C) - N(A n B) - N(A n C) - N(C n B) + N(A n B n C)

If instead of numbers for (A n B) and (A n C) and (C n B), what is given is the total number of people who choose exactly two items, then the formula becomes:

Total = N(A) + N(B) + N(C) - (N(choose exactly two items)) - 2N(choose all three items)

Also,
Total = N(A) + N(B) + N(C) - (N(choose at least two items)) - N(choose all three items)



Can you please explain these two formulas?

Regards,
Aman


Hmmm let me see. Say, total is 100 people, 60 bought Apples, 50 bought Bananas, 35 bought Cranberries. If we know that number of people who bought all three is 10, then we know 25 people bought exactly two of the three, and that 35 people bought more than one fruit (or at least two fruits).

However, in this case we do not know how many people bought A&B, A&C and B&C exactly. It might be the case that 15 people bought A&B, 20 people bought B&C and 20 people bought A&C. The point that needs to be noticed is that when we say 20 people bought B&C they may have or have not bought A as well. In our case 10 of the 20 actually bought all three. I know this sometimes can be very confusing. You just need to make sure if you are talking about "exactly two" or "at least two". Making a Venn gram will help most of the time.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

1 KUDOS received
CEO
CEO
User avatar
Joined: 09 Sep 2013
Posts: 3492
Followers: 231

Kudos [?]: 45 [1] , given: 0

Premium Member
Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 10 Dec 2014, 05:49
1
This post received
KUDOS
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 Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 24584
Followers: 3801

Kudos [?]: 32758 [0], given: 3558

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink] New post 10 Dec 2014, 06:08
Expert's post
Re: Set A, B, C have some elements in common. If 16 elements are in both A   [#permalink] 10 Dec 2014, 06:08
    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic Set A, B, C have some elements in common. If 16 elements are gmatpapa 12 07 Apr 2011, 11:50
3 Sets A, B and C have some elements in common. If 16 elements boeinz 8 26 Sep 2009, 04:02
Set A, B, C have some elements in common. If 16 elements are marcodonzelli 4 11 Mar 2008, 11:30
Sets A, B, and C have some elements in common Googlore 4 06 Dec 2006, 03:08
Sets A, B and C have some elements in common. If 16 elements rigger 4 26 Oct 2005, 21:08
Display posts from previous: Sort by

Set A, B, C have some elements in common. If 16 elements are in both A

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.