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Logical Reasoning
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19 people (a1,a2,......,a19) are invited to a party.The host and the hostess shake hands with each invitee and also with each other.The invitees shake hands in a circular fashion i.e. the pairs (a1,a2),(a2,a3),....,(a18,a19),(a19,a1) shake hands.So,in all there are 58 handshakes.let S be a set of people such that every pair in S shakes hands.The size of S is at most
a)2
b)3
c)1
d)4
Read Solution (Total 7)
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- ans is 3
proof: since the 19 invitees are shaking hands in circular fashion the total number of handshakes b/w them is 19..then the host and hostess shake hands with each other (1 handshake) finally host and hostess both shakes hand with each invitee individually i.e.19+19
hence total no of handshakes are 19+1+19+19=58
all the above given conditions are satisfied
the set S ={(guests),(host),(hostess)} - 13 years agoHelpfull: Yes(18) No(2)
- ans=4 because the set will be(r1,r2)(a1,a2)(R1,a1)(r2,a2)
- 13 years agoHelpfull: Yes(9) No(3)
- 3 will be answer
hence a option - 13 years agoHelpfull: Yes(1) No(5)
- can anyone explain me more clearly
- 12 years agoHelpfull: Yes(0) No(0)
- trick - remb that when circular given then it will 3 and st line then 2
- 11 years agoHelpfull: Yes(0) No(0)
- each other+all guest
1+(58-1)/19
1+3=4 - 9 years agoHelpfull: Yes(0) No(0)
- answer is 3 i.e guest,host&hostess
- 9 years agoHelpfull: Yes(0) No(0)
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