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## 3 sets-overlapping! Experts?

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bryan88 Really wants to Beat The GMAT!
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3 sets-overlapping! Experts? Thu Mar 15, 2012 5:59 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?

(A) 1/8

(B) 1/4

(C) 3/8

(D) 1/2

(E) 5/8

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pemdas GMAT Titan
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Thu Mar 15, 2012 8:11 pm
assigning 24 as total we calculate the portions for each set and find that the sum of (12-9)+(18-9)+(15-9) will give us 18 and we would need 15 to complete 15+9=24. Hence one student from each of the three groups given must be in exactly 2 clubs or 3/24=1/8

a
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LalaB GMAT Destroyer!
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Thu Mar 15, 2012 10:13 pm
All three=3/8
A=1/2
B=5/8
C=3/4
Neither=0

exactly two=?

let Total=8
then all three=(3/8)*8=3
A=(1/2)*8=4
B=(5/8)*8=5
C=(3/4)*8=6

Total=A+B+C+N-exactly2-2*Allthree

8=4+5+6+0-exactly2-6

exactly2=1

exactly2/total=
=1/8

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rahul jaiswal Just gettin' started!
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Fri Mar 16, 2012 2:19 am
Hii,
Sorry but i didn't get it ??
Why u subtracted 2*Allthree ??
Could u please explain how to approach step by step when such prblms appear in exam.

LalaB GMAT Destroyer!
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Fri Mar 16, 2012 4:01 am
see the pic of above, shown by pemdas. see that when u count A B C , the 'all -three' part (i.e. 9 on the pic) is counted three times. that is why u need to subtract it 2 times doing this, u get all three only once (as needed).

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GMATGuruNY GMAT Instructor
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Fri Mar 16, 2012 4:11 am
bryan88 wrote:
3/8 of all students at Social High are in all three of the following clubs: Albanian, Bardic, and Checkmate. 1/2 of all students are in Albanian, 5/8 are in Bardic, and 3/4 are in Checkmate. If every student is in at least one club, what fraction of the student body is in exactly 2 clubs?

(A) 1/8

(B) 1/4

(C) 3/8

(D) 1/2

(E) 5/8
Here is the formula for 3 overlapping groups:

T = G1 + G2 + G3 - (those in 2 of the groups) - 2*(those in all 3 groups)

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in club A, everyone in club B, and everyone in club C:
Those in exactly 2 of the clubs are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 clubs are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

In the problem above:
Let T = 8.
G1 = Albanian = (1/2)8 = 4.
G2 = Bardic = (5/8)8 = 5.
G3 = Checkmate = (3/4)8 = 6.
Those in exactly 2 clubs = x.
Those in all 3 clubs = (3/8)8 = 3.

Plugging these values into the formula, we get:
8 = 4 + 5 + 6 - x - 2(3)
x = 1.

Thus:
(Those in exactly 2 of the clubs)/total = 1/8.

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