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elements in common

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elements in common

by mberkowitz » Sat Sep 13, 2008 3:48 pm
Sets A, B, and 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.

OA is A
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Source: — Data Sufficiency |
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Re: elements in common

by parallel_chase » Sat Sep 13, 2008 11:46 pm
mberkowitz wrote:Sets A, B, and 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.

OA is A
The answer is indeed A.

Use this formula

P(AuBuC) : P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

We just have to find the value of P(AnBnC), Statement I gives us exactly that.
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by mberkowitz » Sun Sep 14, 2008 7:23 am
how do we know p(a) + p(b) + p(c) given sI?
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by tendays2go » Sun Sep 14, 2008 9:00 am
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

But, we don't know the other distribution...so we can't find the common elements to all A, B and C

Hence solution is (A)
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by Thouraya » Sun May 29, 2011 6:45 am
Didnt get why B is not sufficient. Thanks!
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