Interesting DS - Three sets

[rahulvsd]

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]

[sl750]

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

[bpdulog]

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

How did you get this?

[sl750]

Simplify equation 1

[HSPA]

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)

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

How did you get this?