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55 people live in an apartment complex with three fitness [#permalink]

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14 Apr 2013, 16:17

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A

B

C

D

E

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75% (hard)

Question Stats:

49% (02:25) correct
51% (01:08) wrong based on 187 sessions

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55 people live in an apartment complex with three fitness clubs (A, B, and C). Of the 55 residents, 40 residents are members of exactly one of the three fitness clubs in the complex. Are any of the 55 residents members of both fitness clubs A and C but not members of fitness club B?

(1) 2 of the 55 residents are members of all three of the fitness clubs in the apartment complex. (2) 8 of the 55 residents are members of fitness club B and exactly one other fitness club in the apartment complex.

Re: 55 people live in an apartment complex with three fitness cl [#permalink]

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14 Apr 2013, 18:55

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Answer is C

I would use the 3 overlapping sets formula to evaluate whether the stat 1 and 2 are sufficient.

Total people = people enrolled in A + people enrolled in B + people enrolled in C - (people enrolled in A & B only + people enrolled in B & C only + people enrolled in A &C only ) - 2 (people enrolled in A, B, and C).

Equation based on question stem :

55 = 40 - (people enrolled in A & B only + people enrolled in B & C only + people enrolled in A &C only ) - 2 (people enrolled in A, B, and C).

Stmt 1 gives only 2 (people enrolled in A, B, and C) - Not Sufficient

Stmt 2 gives only (people enrolled in A & B only + people enrolled in B & C only + people enrolled in A &C only ) - Not Sufficient

Considering Stmt 1 and Stmt 2 people enrolled in A &C only can be calculated.

Re: 55 people live in an apartment complex with three fitness cl [#permalink]

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14 Apr 2013, 22:05

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We know that 55-40 people are numbers of 2 or more clubs. Lets call a the members of only AC, b the members of only AB, c the members of only CB and d the members of all three. And consider n that are the members of none of those clubs! \(n+a+b+c+d=15\) Now the question is: is a >0? there are members in AC only?

1) tells us that d=2. We cannot say anything about a \(m+a+b+2=15\) 2)tells us that b+c=8.Again not sufficient \(n+a+8+d=15\)

1+2)\(n+a+8+2=15\) We cannot say anything about A. E _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: 55 people live in an apartment complex with three fitness cl [#permalink]

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15 Apr 2013, 05:38

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The question does not specify if there are people not member of any clubs. So isn't E a possibility as well?

Assume there are 5 people not a part of any club _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

Re: 55 people live in an apartment complex with three fitness cl [#permalink]

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02 Jun 2014, 06:29

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55 people live in an apartment complex with three fitness [#permalink]

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21 Aug 2015, 10:58

Tagger wrote:

55 people live in an apartment complex with three fitness clubs (A, B, and C). Of the 55 residents, 40 residents are members of exactly one of the three fitness clubs in the complex. Are any of the 55 residents members of both fitness clubs A and C but not members of fitness club B?

(1) 2 of the 55 residents are members of all three of the fitness clubs in the apartment complex. (2) 8 of the 55 residents are members of fitness club B and exactly one other fitness club in the apartment complex.

Ans - D

(since the question does not mention that "every resident belongs to at least one fitness club")

Had the question included the above condition the answer would've been an easy C.

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