### Brain Teasers

# Chess-o-mania !!!

Fun: (2.17)
Difficulty: (2.69)
Puzzle ID: #19275

Submitted By: lesternoronha1 Corrected By: shenqiang

Submitted By: lesternoronha1 Corrected By: shenqiang

In a chess tournament conducted by the Romankachi World Chess Federation, every person played one game with every other person in the group. The total number of games that men played between themselves exceeded those played by men with women by 18. If there were 4 women on the tournament, totally how many games were played in the tournament?

### Answer

Totally 120 games were played in the tournament.Let N be the number of men in the tournament.

Then the number of games played by men between themselves

= N*(N-1)/2

The number of games played by men with women

= 4*N

(since there were 4 women in the tournament)

It is given that the total number of games that men played between themselves exceeded those played by men with women by 18.

Hence

N*(N-1)/2 - 4*N = 18

N*(N-1) - 8*N = 36

N*N - N - 8*N = 36

N^2 - 9*N - 36 = 0

Factorizing we get,

N^2 - 12*N + 3*N - 36 = 0

N*(N - 12) + 3*(N - 12) = 0

(N - 12)(N + 3) = 0

Hence N = 12 or -3 (not possible).

Hence the number of men in the chess tournament

= N = 12

Therefore the total number of people participating in the chess tournament = 12 + 4 = 16.

The total number of games played in the tournament

= T*(T-1)/2

= 16*15/2

= 120

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