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Set A has 50 members and set B has 53 members. At least 2 of the membe

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Set A has 50 members and set B has 53 members. At least 2 of the membe  [#permalink]

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New post 23 May 2019, 03:51
1
2
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

56% (02:06) correct 44% (01:58) wrong based on 41 sessions

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Re: Set A has 50 members and set B has 53 members. At least 2 of the membe  [#permalink]

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New post 23 May 2019, 05:12
1
2 ≤ x ≤ 50
Hence, (50-50) ≤ y ≤ (50-2)
or 0 ≤ y ≤ 48

(53-48) ≤ z ≤ (53-0)
5 ≤ z ≤ 53

Bunuel wrote:
Set A has 50 members and set B has 53 members. At least 2 of the members in set A are not in set B. Which of the following could be the number of members in set B that are not in set A ?

I. 5
II. 50
III. 53


A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

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Re: Set A has 50 members and set B has 53 members. At least 2 of the membe  [#permalink]

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New post 23 May 2019, 10:54
1
Bunuel wrote:
Set A has 50 members and set B has 53 members. At least 2 of the members in set A are not in set B. Which of the following could be the number of members in set B that are not in set A ?

I. 5
II. 50
III. 53


A. I only
B. II only
C. III only
D. I and III only
E. I, II and III


draw a 2x2 matrix and we see all given values are possible

---sa----nsa----total
sb-48---5-------53
nsb->2-- 0------0
tot--50---5-----
we can have nsb and sa >=2
IMO E
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Re: Set A has 50 members and set B has 53 members. At least 2 of the membe  [#permalink]

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New post 26 May 2019, 08:24
Bunuel VeritasKarishma
How come the number of members in set B that are not in set A can be '53'... as already it is mentioned that At least 2 of the members in set A are not in set B . According to me, it should be 5 and 50 and not 53. Please clarify.
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Re: Set A has 50 members and set B has 53 members. At least 2 of the membe  [#permalink]

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New post 27 May 2019, 19:21
Bunuel wrote:
Set A has 50 members and set B has 53 members. At least 2 of the members in set A are not in set B. Which of the following could be the number of members in set B that are not in set A ?

I. 5
II. 50
III. 53


A. I only
B. II only
C. III only
D. I and III only
E. I, II and III


If 2 members from set A are not in set B and the other 48 members of set A are also members of set B, then the number of members in set B that are not in set A is 53 - 48 = 5. So Roman numeral I could be true.

If 47 members from set A are not in set B and the other 3 members of set A are also members of set B, then the number of members in set B that are not in set A is 53 - 3 = 50. So Roman numeral II could be true also.

If all 50 members from set A are not in set B, then the number of members in set B that are not in set A is 53. So Roman numeral III could also be true.


Answer; E
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Re: Set A has 50 members and set B has 53 members. At least 2 of the membe   [#permalink] 27 May 2019, 19:21
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