Question: Eight players participated in the recent BOOM Chess Tournament. Each player played all of the others exactly once. The winner of a game received 1 point and a loser 0; draws are allowed, giving each player 1/2 point. Now, it turned out that everyone received a different number of points. Furthermore, Sujay, who came in second, earned as many points as the four bottom finishers put together. What was the result of the game between Parag, who came in third, and Manav, who came in fifth?
Answer:
Let X(i) be the score of the person who finished in i th place.
It is clear that X(2) cannot be 7. If X(2) = 6.5, then this would imply that X(1) = 7. But then first person would have beaten everyone. Hence, X(2) cannot be 6.5. Then X(2) = 6.
Since it is given that second person earned as many points as the four bottom finishers put together, X(5) + X(6) + X(7) + X(8) = X(2) = 6
Now players 5, 6, 7 and 8 played exactly 6 games among themselves. It means that neither player 5, 6, 7 or 8 could have beaten or tied any of players 1, 2, 3 or 4. Otherwise, this sum would be greater than 6.
Thus, it is obvious that Parag beat Manav.