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Interesting DS - Three sets

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by rahulvsd » Tue Oct 18, 2011 7:56 am
Image

This table lists enrollment in an afterschool program by activity. There are 30 total students enrolled in the entire program. Students may participate in one, two, or three activities. How many students participate in all three activities?

(1) 21 students only participate in one activity.

(2) 6 students participate in both basketball and math.

OA: [/img]A[/img]
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Source: — Data Sufficiency |
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by sl750 » Tue Oct 18, 2011 12:19 pm
a b c represent each of the clubs respectively
ab - represent students in clubs a and b
bc - represent students in clubs b and c
ca - represent students in clubs ca
abc- represent students in all three clubs
30 = 19+12+11 - (ab+bc+ca) - 2abc ---------1

Statement 1
30 = Exactly 1 club + Exactly 2 clubs +Exactly 3 clubs --------2
Exactly 1 club = 21
Exactly 2 clubs = ab+bc+ca
Exactly 3 clubs = abc

12 = ab+bc+ca + 2abc ----- 3
9 = ab+bc+ca + abc ----- 4

Subtract equation 4 from equation 3
3 = abc . Sufficient

Statement 2
ab = 6 . Insufficient
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by bpdulog » Fri Oct 21, 2011 12:06 pm
12 = ab+bc+ca + 2abc ----- 3

How did you get this?
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by sl750 » Fri Oct 21, 2011 12:20 pm
Simplify equation 1
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by HSPA » Fri Nov 11, 2011 10:18 pm
I guess you are well aware of formuala:

(AuBuC) = A+ B +C - AB - BC -CA + ABC ---> mark it equ 1
30 = 19+12+11 -(AB+BC+CA) +ABC

Now option 1 says that we have total of 21 who are exactly in one set only
A only ---- C only ---- B only
(A - AB - CA)+ (C - CA - BC) + (B - BC - BA) + 3ABC = 21 --> mark it 2

solving these 2 you will have the answer.
(hope you understood how 3ABC is added - we removed from A twice the intersections AB,CA, so ABC was added once in each case to make it balance.. thus 3ABC)
bpdulog wrote:12 = ab+bc+ca + 2abc ----- 3

How did you get this?
First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.
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