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In a certain club, every member likes red wine or white wine

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In a certain club, every member likes red wine or white wine [#permalink] New post 24 Jun 2012, 02:43
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In a certain club, every member likes red wine or white wine or both. If the number of club members that like red wine and do not like white wine is three times the number of club members that like white wine and do not like red wine, then what is the number of club members that like both red wine and white wine?

(1) The total number of club members is 60.

(2) The number of club members that do not like white wine is three times that number of club members that do like white wine.
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Re: In a certain club, every member likes red wine or white wine [#permalink] New post 24 Jun 2012, 03:10
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In a certain club, every member likes red wine or white wine or both. If the number of club members that like red wine and do not like white wine is three times the number of club members that like white wine and do not like red wine, then what is the number of club members that like both red wine and white wine?

Since in the club every member likes red wine or white wine or both, then there are no members who like neither of the wines, so we have that:
{Total}={Red}+{White}-{Both}.

Also given that {Red}-{Both}=3*({White}-{Both}) --> {Red}=3*{White}-2*{Both}. So, {Total}=(3*{White}-2{Both})+{White}-{Both} --> {Total}=4*{White}-3*{Both}

The question is asks about {Both}=?

(1) The total number of club members is 60 --> 60=4*{White}-3*{Both}. Two unknowns. Not sufficient.

(2) The number of club members that do not like white wine is three times that number of club members that do like white wine --> {Total}-{White}=3*{White} --> {Total}=4*{White}. So, 4*{White}=4*{White}-3*{Both} --> 3*{Both}=0 --> {Both}=0. Sufficient.

Answer: B.

Hope it's clear.
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Re: In a certain club, every member likes red wine or white wine [#permalink] New post 10 Aug 2012, 15:19
Bunuel wrote:
In a certain club, every member likes red wine or white wine or both. If the number of club members that like red wine and do not like white wine is three times the number of club members that like white wine and do not like red wine, then what is the number of club members that like both red wine and white wine?

Since in the club every member likes red wine or white wine or both, then there are no members who like neither of the wines, so we have that:
{Total}={Red}+{White}-{Both}.

Also given that {Red}-{Both}=3*({White}-{Both}) --> {Red}=3*{White}-2*{Both}. So, {Total}=(3*{White}-2{Both})+{White}-{Both} --> {Total}=4*{White}-3*{Both}

The question is asks about {Both}=?

(1) The total number of club members is 60 --> 60=4*{White}-3*{Both}. Two unknowns. Not sufficient.

(2) The number of club members that do not like white wine is three times that number of club members that do like white wine --> {Total}-{White}=3*{White} --> {Total}=4*{White}. So, 4*{White}=4*{White}-3*{Both} --> 3*{Both}=0 --> {Both}=0. Sufficient.

Answer: B.

Hope it's clear.



Doing the set matrix:

------------ red---- not red--- tot
white -------- ? ------ x
not white ---- 3x ------ 0
tot


2)
------------ red---- not red--- tot
white -------- ? ------ x ------ y
not white ---- 3x ----- 0 ----- 3y
tot

So, y=x. and this implies white/red = 0

Is this the right reasoning?
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Re: In a certain club, every member likes red wine or white wine [#permalink] New post 10 Aug 2012, 23:54
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superpus07 wrote:
Bunuel wrote:
In a certain club, every member likes red wine or white wine or both. If the number of club members that like red wine and do not like white wine is three times the number of club members that like white wine and do not like red wine, then what is the number of club members that like both red wine and white wine?

Since in the club every member likes red wine or white wine or both, then there are no members who like neither of the wines, so we have that:
{Total}={Red}+{White}-{Both}.

Also given that {Red}-{Both}=3*({White}-{Both}) --> {Red}=3*{White}-2*{Both}. So, {Total}=(3*{White}-2{Both})+{White}-{Both} --> {Total}=4*{White}-3*{Both}

The question is asks about {Both}=?

(1) The total number of club members is 60 --> 60=4*{White}-3*{Both}. Two unknowns. Not sufficient.

(2) The number of club members that do not like white wine is three times that number of club members that do like white wine --> {Total}-{White}=3*{White} --> {Total}=4*{White}. So, 4*{White}=4*{White}-3*{Both} --> 3*{Both}=0 --> {Both}=0. Sufficient.

Answer: B.

Hope it's clear.



Doing the set matrix:

------------ red---- not red--- tot
white -------- ? ------ x
not white ---- 3x ------ 0
tot


2)
------------ red---- not red--- tot
white -------- ? ------ x ------ y
not white ---- 3x ----- 0 ----- 3y
tot

So, y=x. and this implies white/red = 0

Is this the right reasoning?


Yes, your approach is correct.
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Re: In a certain club, every member likes red wine or white wine [#permalink] New post 12 Aug 2012, 09:42
manulath wrote:

(2) The number of club members that do not like white wine is three times that number of club members that do like white wine.

--x--
Ah, i failed to carefully read (2) where it says at the end "..number of club members that do like white wine", which is the sum of those who like Both + those who like only like White wine (as opposed to those who only like white)
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Re: In a certain club, every member likes red wine or white wine [#permalink] New post 16 Sep 2012, 05:59
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Let a --> no of ppl who like Red wine alone
c -->no of ppl who like white wine alone
b --> no of ppl who like both

from the question a=3c and we have to find b

1) a+b+c = 60.. nothing can be inferred abt b
2) a=3(b+c)

and from the question we know a=3c which implies b=0

hence 2 alone is sufficient
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Re: In a certain club, every member likes red wine or white wine [#permalink] New post 14 Jul 2013, 23:42
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Re: In a certain club, every member likes red wine or white wine [#permalink] New post 22 Jul 2014, 08:44
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Re: In a certain club, every member likes red wine or white wine   [#permalink] 22 Jul 2014, 08:44
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