Latest TCS Aptitude Question SOLUTION: An organization has 4 committees. Only 3 persons are members of all four committees, but every pair of committees has 4 members in common. What is the LEAST possible number of the

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An organization has 4 committees. Only 3 persons are members of all four committees, but every pair of committees has 4 members in common. What is the LEAST possible number of the members on any one committee?

Asked In TCS sujitha (10 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

kamna sabharwal  (10 years ago)

6 least possible no. of members in any one committee.
C1 C2 C3 C4
x x x x
y y y y
z z z z
w w v t
r v t u
s u s r

now every two committee has 4 members in common.
ans : 6

Is this solution Helpfull? Yes (26) | No (7)  

JANARDHAN  (10 years ago)

4 with help of set diagram

Is this solution Helpfull? Yes (10) | No (1)  

avishek sen  (9 years ago)

let w,x,y,z are four committees....A,B,C are common in all committee...

w x y z
__________

A A A A

B B B B

C C C C

D D _ D

E E E _

F _ F F
__________

pairs of the commiittee -->AB,AC,AD,BC,BD,CD

IN AB-->A,B,C,D are common
IN AC-->A,B,C,E are common
IN AD-->A,B,C,F are common
IN BC-->A,B,C,E are common
IN BD-->A,B,C,D are common
IN CD-->A,B,C,F are common

SO,the LEAST possible number of the members on any one committee is= 5 (ans.)

Is this solution Helpfull? Yes (9) | No (1)  

E.Vyshnavi  (10 years ago)

4 with the help of ven diagram

Is this solution Helpfull? Yes (5) | No (2)  

saiharsha  (10 years ago)

the least possible no of member of members on any one committee = 4 .. option a)

In all 4 committees, say W,X,Y,Z, 3 persons say A,B and c are common.
D is common between W and X.
E is common between X and Y.
F is common between Y and Z.
G is common between Z and W.

so W committee has ABCDG. X committee has ABCDE. Y committee has ABCEF.
Z committee has ABCFG.

Is this solution Helpfull? Yes (3) | No (4)  

Sangeethab  (9 years ago)

1 2 3 4
A A A A
B B B B
C C C C
e e f d
d f d
Here,1 2 3 4 are the committees,and A B C are the members common to all the committees,A B C e is common to 1 2,A B C f is comm to 2 3,A B C d is comm to 3 4, A B C d is comm to 4 3 so ans is the 4th committee which has least mem 4

Is this solution Helpfull? Yes (2) | No (1)  

Rajat Jain  (9 years ago)

(ABC)DGI
(ABC)DFH
(ABC)EGH
(ABC)EFI

Is this solution Helpfull? Yes (2) | No   

Jasmeet  (10 years ago)

8 with help of set diagram

Is this solution Helpfull? Yes (1) | No (8)  

OUJASWAEE DAS  (9 years ago)

6 people using venn diagram

Is this solution Helpfull? Yes (1) | No (1)  

avishek sen  (9 years ago)

let w,x,y,z are four committees....A,B,C are common in all committee...

w x y z
__________

A A A A

B B B B

C C C C

D D _ D

E E E _

F _ F F
__________

pairs of the commiittee -->AB,AC,AD,BC,BD,CD

IN WX-->A,B,C,D are common
IN WY-->A,B,C,E are common
IN WZ-->A,B,C,F are common
IN XY-->A,B,C,E are common
IN XZ-->A,B,C,D are common
IN YZ-->A,B,C,F are common

SO,the LEAST possible number of the members on any one committee is= 5 (ans.)

Is this solution Helpfull? Yes (1) | No   

pavani payala  (10 years ago)

0 is the answer cause here 3 members are there for all 4 committees

Is this solution Helpfull? Yes | No (6)  

prasannalatha  (9 years ago)

Let the committies be 1,2,3 and 4 then,
A,B,C are common to all the four commities(since given in the que that only 3 persons are the members of all the four committies)
now, D is common to 1&2
E is common to 2&3
F is common to 3&1
so, the least possible number are 4
ANSWER: 4

Is this solution Helpfull? Yes | No (1)  

ILAIYANILA  (9 years ago)

let the committees be 1234
1 2 3 4
a a a a
b b b b
c c c c (since 3 mem common in all four committee)
d d
e e
f f
g g
i i
h h (each pair of committee has 4 members in common)
therefore least possible number of mem in any group is 6

Is this solution Helpfull? Yes | No   

Akshayyha Krishnamurthy  (6 years ago)

6 is the answer.
Lets us assume A,B,C,D are 4 committees.
now using Venn diagram ,
A ∩ B ∩ C ∩ D =3
Now every pair must have 4 members in common
possible pairs are(each containing +1 member since there're already 3 common members) :
A ∩ B - 1
A ∩ C - 1
A ∩ D - 1
B ∩ C - 1
B ∩ D - 1
C ∩ D - 1
Now if we take A alone there are
3+ 1 (in A ∩ B)+1 (in A ∩ C ) + 1 (in A ∩ D)= 6
Similarly if u check for all , there will be 6 members in each.(minimum possibility)
For even more clear understanding , apply this in a Venn diagram.

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