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by rahul.s » Tue Feb 16, 2010 1:21 am
there are 3 country clubs in town: A, B and C. A has 300 members, B has 400, and C has 450. 30 people belong to both A and B, 40 to both A and C, and 50 to both B and C. 20 people are members of all 3. how many people belong to at least 1 country club?

OA: 1050
Source: mgmat

my solution:
x = (300 + 400 + 450) - (30) - (40) - (50) - 2*(20)
x = 1150 - 120 - 40
x = 1150 - 160
x = 990

where am i making a mistake?
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by ajith » Tue Feb 16, 2010 1:30 am
rahul.s wrote:there are 3 country clubs in town: A, B and C. A has 300 members, B has 400, and C has 450. 30 people belong to both A and B, 40 to both A and C, and 50 to both B and C. 20 people are members of all 3. how many people belong to at least 1 country club?

OA: 1050
Source: mgmat

my solution:
x = (300 + 400 + 450) - (30) - (40) - (50) - 2*(20)
x = 1150 - 120 - 40
x = 1150 - 160
x = 990

where am i making a mistake?
x = (300 + 400 + 450) - (30) - (40) - (50) - 2*(20)

it should have been
x = (300 + 400 + 450) - (30) - (40) - (50) + (20)
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by rahul.s » Tue Feb 16, 2010 1:36 am
is that the formula?

i came across it in Stuart's and papgust's post.

https://www.beatthegmat.com/trouble-with ... tml#223802

https://www.beatthegmat.com/sets-t52148. ... hington-DC
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by ajith » Tue Feb 16, 2010 1:50 am
rahul.s wrote:is that the formula?

i came across it in Stuart's and papgust's post.

https://www.beatthegmat.com/trouble-with ... tml#223802

https://www.beatthegmat.com/sets-t52148. ... hington-DC
n(A U B U C) = n(A) + n(B) + n(C) - [n(A n B) + n(B n C) + n(C n A) ] + n(A n B n C)

is the formula- as for the links I will check and revert
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by ajith » Tue Feb 16, 2010 2:03 am
rahul.s wrote:is that the formula?

i came across it in Stuart's and papgust's post.

https://www.beatthegmat.com/trouble-with ... tml#223802

https://www.beatthegmat.com/sets-t52148. ... hington-DC
In papgust's formula what he refers only A n B ( A n B - (A n B n C) )

So applying his formula in this problem

x = (300 + 400 + 450) - (10) - (20) - (30) - 2*(20) =1050

That is different way of doing it all together

and Stuart explains both the methods

To Quote him
"
True # of objects = (# only A) + (# only B) + (# only C) + (# only AB) + (# only AC) + (# only BC) + (# only ABC)

The second formula is the one we use more often:

True # of objects = (total # A) + (total # B) + (total #C) - (# only AB) - (# only AC) - (# only BC) - 2(# ABC) "

The # only AB in both formulas are different

in first formula it refers to A n B
and in second formula it refers to ( A n B - (A n B n C))

I hope it is clear...
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