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# Eleven members of Club A are also members of Club B, and five members

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Math Expert
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Eleven members of Club A are also members of Club B, and five members  [#permalink]

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13 Feb 2015, 08:47
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55% (hard)

Question Stats:

61% (01:44) correct 39% (02:00) wrong based on 352 sessions

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Eleven members of Club A are also members of Club B, and five members of Club B are also members of Club C. How many members of Club A are also members of Club C?

(1) Club B has 16 members.
(2) Exactly two people from Club C are also members of Club A.

Kudos for a correct solution.

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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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16 Feb 2015, 02:14
Ans - E

Stmnt 1 - Clearly insuff
Stmnt 2 - Not suff

Combining, also notsuff
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Posts: 6957
Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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16 Feb 2015, 04:01
buddyisraelgmat wrote:
Ans - E

Stmnt 1 - Clearly insuff
Stmnt 2 - Not suff

Combining, also notsuff

hi,
why should statement 2 not be sufficient?
It tells exactly what we want and that is how many members are common between the two groups..
ans B
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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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16 Feb 2015, 06:34
Bunuel wrote:
Eleven members of Club A are also members of Club B, and five members of Club B are also members of Club C. How many members of Club A are also members of Club C?

(1) Club B has 16 members.
(2) Exactly two people from Club C are also members of Club A.

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION

Explanation:

(1) It is possible that the 11 members of Club A who belong to Club B include the 5 members who also belong to Club C. It is also possible that no one who belongs to Club A also belongs to Club C. Accordingly, statement 1 is insufficient.

(2) This statement answers the question directly, indicating that two people are members of both Club A and Club C. Accordingly, this statement is sufficient. Since statement (2) alone is sufficient to answer the question, the correct answer is B.
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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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27 Aug 2015, 07:39
1
Does Statement 2 comprises of both common between A & C and common among A, B, & C. By the term exactly, I understood as if they are just common between A & C and since we do not know those who are common among all three clubs the information is insufficient; and hence Answer is E.
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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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22 Mar 2017, 00:46
Bunuel wrote:
Eleven members of Club A are also members of Club B, and five members of Club B are also members of Club C. How many members of Club A are also members of Club C?

(1) Club B has 16 members.
(2) Exactly two people from Club C are also members of Club A.

Kudos for a correct solution.

This question asks us if we have enough information to determine how many members of club A are also member of club C- it is important to note that conditional logic doesn't necessarily apply in set theory- in other words, we if know how many members of club C are also members of club A then we know how many members of club A are members of club c. If this were critical reasoning, for example, then the argument if all A are B then all B are A would be fallacious ( All humans are mammals but not all mammals are humans). Though if you look at this scenario on a three circle venn diagram you can see that whatever numbers falls into the intersection of Group A and C ( the circle that intersects those groups) this number applies both ways.

Statement (1) tells us the size of club B which in itself does not allow us to find the number of members in club C. Insufficient

Statement (2) answers the question. Sufficient.
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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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27 Aug 2017, 05:04
scorpionkapoor77 wrote:
Does Statement 2 comprises of both common between A & C and common among A, B, & C. By the term exactly, I understood as if they are just common between A & C and since we do not know those who are common among all three clubs the information is insufficient; and hence Answer is E.

Hi Bunuel

Even i had a similar problem. Can you please explain the flaw in our understanding ?
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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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27 Aug 2017, 05:17
1
Kevinjoshi wrote:
scorpionkapoor77 wrote:
Eleven members of Club A are also members of Club B, and five members of Club B are also members of Club C. How many members of Club A are also members of Club C?

(1) Club B has 16 members.
(2) Exactly two people from Club C are also members of Club A.

Does Statement 2 comprises of both common between A & C and common among A, B, & C. By the term exactly, I understood as if they are just common between A & C and since we do not know those who are common among all three clubs the information is insufficient; and hence Answer is E.

Hi Bunuel

Even i had a similar problem. Can you please explain the flaw in our understanding ?

(2) Exactly two people from Club C are also members of Club A means that the overlap between C and A is 2. Below is one of the possible cases:
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Untitled.png [ 5.55 KiB | Viewed 1299 times ]

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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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27 Aug 2017, 05:54
Bunuel wrote:
Kevinjoshi wrote:
scorpionkapoor77 wrote:
Eleven members of Club A are also members of Club B, and five members of Club B are also members of Club C. How many members of Club A are also members of Club C?

(1) Club B has 16 members.
(2) Exactly two people from Club C are also members of Club A.

Does Statement 2 comprises of both common between A & C and common among A, B, & C. By the term exactly, I understood as if they are just common between A & C and since we do not know those who are common among all three clubs the information is insufficient; and hence Answer is E.

Hi Bunuel

Even i had a similar problem. Can you please explain the flaw in our understanding ?

(2) Exactly two people from Club C are also members of Club A means that the overlap between C and A is 2. Below is one of the possible cases:

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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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26 Oct 2017, 07:24
From St1, we don't know the number of members of club A in C. It may or may not be an overlapping set.
From St2, Its the other way of projecting the answer. If exactly 2 from Club C are present in Club A, the same is true in vice-versa.

So ans is B.
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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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12 Nov 2017, 13:32
scorpionkapoor77 wrote:
Does Statement 2 comprises of both common between A & C and common among A, B, & C. By the term exactly, I understood as if they are just common between A & C and since we do not know those who are common among all three clubs the information is insufficient; and hence Answer is E.

Exactly, language is imprecise and hence there gotta be an inference to solve the question...
Veritas' inferece is that AuC excludes B, we (cause I made the same inference) is that not necessary does, because the problem doesn't says so...

I've find Veritas practice kinda annoying because of that
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Re: Eleven members of Club A are also members of Club B, and five members  [#permalink]

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06 Sep 2018, 07:04
Bunuel wrote:
Eleven members of Club A are also members of Club B, and five members of Club B are also members of Club C. How many members of Club A are also members of Club C?

(1) Club B has 16 members.
(2) Exactly two people from Club C are also members of Club A.

We consider the image attached.

$$? = x + y$$

$$\left. \begin{gathered} y + z = 11 \hfill \\ y + w = 5 \hfill \\ \end{gathered} \right\}\,\,\,\,\left( * \right)$$

$$\left( 1 \right)\,\,\underline {B = 16} \,\,\,\left\{ \begin{gathered} \,\,{\text{Take}}\,\,\left( {x,\underline {y,z,w,k} } \right) = \,\,\left( {0,\underline {4,6,1,5} } \right)\,\,\,{\text{viable}}\,\,\,\,\, \Rightarrow \,\,\,\,? = 0 + 4 = 4 \hfill \\ \,\,{\text{Take}}\,\,\left( {x,\underline {y,z,w,k} } \right) = \,\,\left( {1,\underline {4,6,1,5} } \right)\,\,\,{\text{viable}}\,\,\,\,\, \Rightarrow \,\,\,\,? = 1 + 4 = 5 \hfill \\ \end{gathered} \right.$$

$$\left( 2 \right)\,\,x + y = 2\,\,\,\, \Rightarrow \,\,\,\,? = 2\,\,\,\,\left( {{\text{immediately}}} \right)\,\,\,\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Re: Eleven members of Club A are also members of Club B, and five members &nbs [#permalink] 06 Sep 2018, 07:04
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