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

I do not agree at all that A is right ans....

Even if u consider 9 are common among A, B and C

still we dont have any clue that

whether

elements which are common b/w B and C also common with A also.... ?

and

whether

elements which are common b/w C and A also common with B too... ?

without these inf... nothing can be said....

Hiya - the statement reads that "of the 16 elements that are in both A and B, 9 elements are also in C". The first half of this means that there are 16 elements (let's say, 1 to 16) that are in A, and are also in B. The second half of the statement would indicate that of the numbers 1-16, 1-9 are also in C. This allows you to answer the question - there are 9 elements in A, B and C.

Did that clear it up a little?

Just curious about the interpretation of question when it says

If 16 elements are in both A and B

if we draw a venn diagram it means the intersection of all 3 sections and a,b ( hope I made sense)



PS: it isn't the best diagram...

16 elements are in both A and B means sections d and g below:


Hope it helps.

Using statement 1 as statement 2 is insufficient

The answer for the question common elements in all 3 (a,b and c) Statement 1 would be 25

Since C,B =9
A = 7

Correct?

(1) says: of the 16 elements that are in both A and B, 9 elements are also in C --> sets A, B, and C have in 9 elements in common.

Your answer does not make sense: if A and B have 16 elements in common, how can A, B, and C have more elements in common than only A and B?

Dear Fozzzy,
I got your p.m. and I am happy to help.

First, the prompt.
16 elements are in both A and B --- this 16 includes elements that are just in A & B as well as elements in A & B & C.
17 elements are in both A and C --- this 17 includes elements that are just in A & C as well as elements in A & B & C.
18 elements are in both B and C --- this 18 includes elements that are just in B & C as well as elements in A & B & C.

To understand this, think about real world categories (these categories will include more elements than 18). Suppose
A = set of males
B = set of people who hold public office in the United States of America
C = set of people who are African-American.

Some people are just in one of these categories. I'm a member of A, but not B or C. My senators Dianne Feinstein & Barbara Boxer are members of B, but not A or C. Oprah Winfrey & Alice Walker are members of C but not A or B. The US Secretary of State, John Kerry, is a member of sets A & B but not C. By contrast, the US President, Barack Obama, is a member of all three sets. If I say: list people who are members of A & B, then it would be perfectly acceptable to list both Kerry and Obama --- all males who hold public office would be listed, irrespective of their race. The set of people in A & B, male office holders, would include some members who were part of C (such as Obama) and some members who were not part of C (such as Kerry).

Now, the statements.
(1) Of the 16 elements that are in both A and B, 9 elements are also in C
Well, the members of the intersection set A & B includes some elements that are part of C and some elements that are not part of C. The 9 elements of (A & B) who are also included in C are the the nine elements common to all three sets. The remaining 7 would be those elements that, like John Kerry, are members of A & B but not C. Thus, this statement gives us enough information to answer the question, so it is sufficient.

Did you have a question about the second statement as well?

Mike

Hi dear math experts, I'm just trying to refresh my skills for 3-Way-Venn-Diagram, would appreciate some comments on my solution. Thanks.
(1) This gives us straight the solution. A,b,c have 9 elements in common. Sufficient
(2) Clearly not sufficient, as we have no info about the TOTAL and the elements in group NEITHER (see formula: Total=a+b+c-Sum of 2-Group overlaps+All 3+Neither)

Answer A

Dear BrainLab,
I'm happy to respond. My friend, you seem to understand quite well.

If you would like more info on 2-way and 3-way Venn Diagrams, see this post:
https://magoosh.com/gmat/2012/gmat-sets-venn-diagrams/

Best of luck!
Mike

if elements are a member of a and b, there are two possibilities. Either they can be in a, b, AND c. or they can be in ONLY a and b. It is very easy to see with a venn diagram. So let x = belongs to A,B, and C. Let y be ONLY belongs to A and B. Then x+y=16. Y is given as 9 from the first statement. That is what we want. So the number of elements belonging to A, B, AND C is 9.

We can attempt a formula for statement two. 25+35+30-(16+17+18)+x+neither=?
We have two many unknowns.

option(1) =>

n(A u B) = 16 out of which 9 are also there in C, thus these 9 elements are common to all A,B & C, hence the answer and therefore this is sufficient to answer.

option(2) =>

N(A) = 25
N(B) = 30
N(C) = 35

Missing Total and Neither category. Insufficient.

Hence A.

Hi all,

Just wanted to mention that for St-2, we can use the following formula:

Total = A + B + C - (no. in two groups - no. in all) - no. in all + no. neither

Now, we don't have no. in all, no. neither and total. We need atleast two of these to be able to answer. Even if we were told that no. neither is 0, ST-2 would still be insufficient.

Thanks

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.