Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 31 Mar 2006
Posts: 162

Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
27 Jun 2006, 03:53
Question Stats:
48% (02:10) correct
52% (01:17) wrong based on 105 sessions
HideShow timer Statistics
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: setabchavesomeelementsincommonif16elementsare112041.html[/b][/color]
Official Answer and Stats are available only to registered users. Register/ Login.
Last edited by Bunuel on 10 Dec 2014, 07:08, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.



Senior Manager
Joined: 07 Jul 2005
Posts: 404
Location: Sunnyvale, CA

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
27 Jun 2006, 09:17
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
Joined: 06 May 2006
Posts: 791

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
27 Jun 2006, 13:05
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+35161718+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?
_________________
Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?



Director
Joined: 28 Dec 2005
Posts: 752

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
27 Jun 2006, 13:50
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+35161718+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
Joined: 20 Feb 2006
Posts: 331

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
27 Jun 2006, 14:34
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+35161718+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
Joined: 26 Jun 2006
Posts: 152

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
27 Jun 2006, 18:03
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
Joined: 07 Jul 2005
Posts: 404
Location: Sunnyvale, CA

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
28 Jun 2006, 15:36
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
Joined: 20 Feb 2006
Posts: 331

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
28 Jun 2006, 17:07
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 17.. and then verify..
this will give you the above formula..



CEO
Joined: 20 Nov 2005
Posts: 2894
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD  Class of 2008

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
28 Jun 2006, 17:09
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+CABBCCA.
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+CABBCCA2ABC.
But in subtracting AB, AC and BC we subtracted ABC three times. So need to add 3ABC
Finally we have A+B+CABBCCA2ABC+3ABC i.e
A+B+CABBCCA+ABC
Try it with marking the areas in a Venn Diagram.
Hope this helps.
_________________
SAID BUSINESS SCHOOL, OXFORD  MBA CLASS OF 2008



SVP
Joined: 03 Jan 2005
Posts: 2233

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
28 Jun 2006, 18:53
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
Joined: 31 Mar 2006
Posts: 162

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
29 Jun 2006, 01:40
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
Joined: 03 Jan 2005
Posts: 2233

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
30 Jun 2006, 08:56
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.



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16032

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
10 Dec 2014, 06:49
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



Math Expert
Joined: 02 Sep 2009
Posts: 39759

Re: Set A, B, C have some elements in common. If 16 elements are in both A [#permalink]
Show Tags
10 Dec 2014, 07:08




Re: Set A, B, C have some elements in common. If 16 elements are in both A
[#permalink]
10 Dec 2014, 07:08








Similar topics 
Author 
Replies 
Last post 
Similar Topics:


1


Both a set S and a set P have the same number of elements. Is a standa

MathRevolution 
5 
02 Apr 2017, 05:09 

1


Two sets, A and B, have the same number of elements

Aristocrat 
3 
03 May 2013, 17:06 

5


Is the probability of an element in Set B also being an elem

tabsang 
4 
17 Jul 2014, 08:58 

1


If set B is a subset of Set A, how many elements are in set

anon1 
11 
26 Jan 2013, 00:41 

11


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

gmatpapa 
19 
23 Mar 2017, 16:20 



