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Set A, B, C have some elements in common. If 16 elements are [#permalink]
07 Apr 2011, 11:50
2
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
75% (hard)
Question Stats:
55% (02:09) correct
45% (01:15) wrong based on 287 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.
S2 Very insufficient information. We still need the total number of elements.
Hence A.
gmatpapa wrote:
Set A, B, C have some elements in common. If 16 elements are in both A and B, 17 elements are in both Aand 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.
Re: Set A, B, C have some elements in common. If 16 elements are [#permalink]
17 Jul 2012, 20:08
1
This post received KUDOS
smartmanav wrote:
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.
Re: Set A, B, C have some elements in common. If 16 elements are [#permalink]
11 Jul 2013, 07:07
1. Statement 1 is sufficient because we know that A and B have 16 elements in common. Among these 16 elements, 9 are also in C. 2. Not sufficient since we still don't know if there's any element that's not belong to any of the 3 groups: A, B and C. The answer is A.
Re: Set A, B, C have some elements in common. If 16 elements are [#permalink]
11 Jul 2013, 08:06
Expert's post
fozzzy wrote:
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.
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:
Re: Set A, B, C have some elements in common. If 16 elements are [#permalink]
14 Jul 2013, 02:17
Expert's post
fozzzy wrote:
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? _________________
Re: Set A, B, C have some elements in common. If 16 elements are [#permalink]
02 Oct 2013, 09:33
Expert's post
gmatpapa wrote:
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.
fozzzy wrote:
Hi, Could you please explain this particular question? Thanks in Advance!
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?
Re: Set A, B, C have some elements in common. If 16 elements are [#permalink]
03 Jan 2015, 09:31
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Re: Set A, B, C have some elements in common. If 16 elements are [#permalink]
21 Jan 2016, 04:07
gmatpapa wrote:
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.
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 _________________
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Share some Kudos, if my posts help you. Thank you !
Re: Set A, B, C have some elements in common. If 16 elements are [#permalink]
21 Jan 2016, 09:42
Expert's post
BrainLab wrote:
gmatpapa wrote:
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.
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.
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