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Sajjad1994
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­
Hope this helps, s_pranita

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SALAKSHYA
Another solution of the problem would be B=0, C=1 1/2
­How? See the pairs are 6 AB AC AD (won by all A) BC BD (Wone by B) and CD won by C
B total would be = \(2-\frac{1}{2}\)
and C total = \(1-\frac{1}{2}-\frac{1}{2}\)­
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Each player will play 3 matches.

B:
1: B x A. B looses, B has -0,5 points
2: B x C. B wins, B got +1 point, beeing 0,5 points
3: B x D. B wins, B got +1 point, beeing 1,5 points in total

C:
1: C x A. C looses, C has -0,5 points
2: C x B. C looses, C got -0,5 point, beeing -1 points
3: C x D. C wins, B got +1 point, beeing 0 points in total
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Sajjad1994
Persons A, B, C, and D and no other players compete in a tournament. In each match of the tournament, one person plays against another, and over the course of the tournament, each person plays against every other person exactly once. A player gains 1 point for a win, loses 1/2 point for a loss, and neither gains nor loses any points for a draw. In every match in which Person A played, Person A won. In every match in which Person D played, Person D lost. When Person B played Person C, Person B won.

In the table, select Person B's tournament score and Person C's tournament score such that the two scores are consistent with the information above. Make only two selections, one in each column.

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