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Ten persons numbered 1, 2, . . ., l0 play a chess tournament, each player playing against every other player exactly one game. It is known that no game ends in a draw. If w₁, w₂, .., w₁₀ are the number of games won by players 1, 2, 3, ……, 10, respectively, and l₁, l₂, ..., l₁₀ are the number of games lost by the players 1, 2, ..., 10, respectively, then

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$w_{i} + l_{i} = 9$

${Σw}_{i}^{2} = 81 + {Σl}_{1}^{2}$

${Σw}_{i} = {Σl}_{i} = 45$

${Σw}_{i}^{2} = {Σl}_{i}^{2}$

answer is A, B, D.

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Detailed Solution

Clearly, each player will play 9 games. And total number of games is $^{10} C_{2}$ = 45. Clearly,

$w_{i} + l_{i} = 9 and {Σw}_{i} = {Σl}_{i} = 45$

$\begin{array}{l} \Rightarrow w_{i} = 9 - l_{i} \Rightarrow w_{i}^{2} = 81 + l_{i}^{2} - 181_{i} \\ \Rightarrow {Σw}_{i}^{2} = 81 \times 10 + {Σl}_{i}^{2} - 18 \times 45 = {Σl}_{i}^{2} \end{array}$

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