TCS
Company
Numerical Ability
Permutation and Combination
An organisation has 3 committees,
only 2 persons are members of all 3
committee but every pair of committee
has 3 members in common. what is the
least possible number of members on
any one committee?
a) 4
b) 5
c) 6
d) 1
Read Solution (Total 11)
-
- applying set theory.
let x be the number of members in each committee
then:
x-(3-3)-(3-3)-3= x-3
now x-3>0 so the least possible value would be 4 - 10 years agoHelpfull: Yes(13) No(1)
- thinking logically,
2 members r common fr 3 commities,so fr a pair of commitee to have 3 members in common it has to b another mutual member..so for three commitiess there are 3 pairs and two additional members mutual fr each comitte.
so totally 4 members - 10 years agoHelpfull: Yes(10) No(1)
- answer is :-5
committe1:X,Y,Z,T
committe2: X,Y,Z,P
COMMITTE3: X,Y,T,P - 10 years agoHelpfull: Yes(4) No(6)
- 4 is the ans
- 10 years agoHelpfull: Yes(3) No(3)
- correct option:a)
ans is 4
lets say each commite as a b and c
a has two member in common with b and c
also it has 1 member in commen with b who is not in c and 1 member common with c not in b
so they all add up to 4 - 10 years agoHelpfull: Yes(3) No(0)
- 5 will be the ans
- 10 years agoHelpfull: Yes(2) No(3)
- logical..
suppose member A and member B is common in all 3 committees..
but 3 members are common among the 3 committees...so we have to add 2 more different member to make 3 members common among pairs..gropus can be formed as..
ABCD, ABCE, ABDE...these are the 3 committees..
- 10 years agoHelpfull: Yes(2) No(0)
- Giv me d crct sltn plzzzz
- 10 years agoHelpfull: Yes(0) No(0)
- anyone plzzzzzzzzzzzzzzzzzzzzz giv me d sltn
- 10 years agoHelpfull: Yes(0) No(0)
- 4 members at least
- 10 years agoHelpfull: Yes(0) No(0)
- there r already 2 members common in all groups so two pairs have 2 members in common to make third one common b/w pairs we have to add members in every pair so as to make three members common b/w 2 groups
1 1 1
1 1 1
1*1*** 1*1** 1**1***
so answer is 4 - 10 years agoHelpfull: Yes(0) No(0)
TCS Other Question