At a party, everyone must shake the hand of the other. If the total number of handshakes is 66, there are — people at the party?
Solution
Correct Answer: Option C
Let the number of persons= n
Number of handshakes by first person= n- 1
(As he cannot handshake with himself)
Number of handshakes by second person = n- 2
(As he has already handshaked with first person)
Number of handshakes by third person= n- 3
.
.
Number of handshakes by second last i.e. ( n- 1)th person= n- (n- 1)= 1
Number of handshakes by last i.e. nth person= n- n= 0
So,
Total number of handshakes=(n- 1)+ (n- 2)+ (n- 3)+ 1+ 0
= 0+ 1+ 2+ 3+ ...........(n- 1)
= Sum of first (n- 1) terms
= (n- 1)(n- 1+ 1)/2
= n(n- 1)/ 2
So, Total number of handshakes= n(n- 1)/2
According to question,
n(n- 1)/ 2= 66
=> n^2 - n= 132
=> n^2- n- 132= 0
=> n^2- 12n+ 11n- 132= 0
=> n(n- 12) + 11(n- 12)= 0
=> (n- 12)(n+ 11) = 0
n= 12 or n= - 11(which is not possible as no. Of persons cannot be negat
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