Discussion

Explanation:

Total number of matches played = ⁶C₂ = 15
Each team plays 5 matches, one against each of the remaining 5 teams.

Points distributed in a match = 2 (2 + 0 or 1 + 1)
Total points to be distributed in the tournament = 2 × 15 = 30 points.
So far points scored by A, B, C and D = 8 + 6 + 5 + 5 = 24
∴ Points scored by E and F together = 30 – 24 = 6.

E: E lost only 1 match. The minimum number of points E can score is when it draws its remaining matches i.e., 4 points. Since E and F scored less than 5 points hence, E’s score is 4 point when it draws its remaining 4 matches.

A: A did not lose any game, hence it can score 8 points only by winning 3 matches and drawing 2 matches.

B: B lost 2 matches, hence it can score 6 points only by winning the remaining 3 matches.

C: C lost 2 games, hence it can score 5 points only by winning 2 matches and drawing 1 match.

D: D lost only 1 game, hence it can score 5 points by drawing 3 matches and winning 1 match.

So far A, B, C, D and E together have won 9 matches and lost only 6 matches. But the number of wins should be equal to number of loses.
∴ F must have lost 3 matches more than it wins.

F: Since E’s score is 4, F’s score = 6 – 4 = 2 points. F lost 3 matches more than it won. This is possible in 2 cases.

Case (a): F won 1 match and lost 4 matches and 0 draws.
Since E has to play 4 draws, E will play a draw match against F. [E cannot play drawn against B. B has 0 draws.]
Hence, F cannot have 0 draws.
∴ This case is rejected.

Case (b): F won 0 match and lost 3 matches and drew 2 matches.
This is the only possible case.

We can make the following table with the information available.

​​​​​​​

Sum of the draws column = 12 hence there are total 6 drawn matches. [Each draw results in entry of draw for 2 teams.]

Hence, option (d).

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