AkamaiBrah wrote:
A certain league has four divisions. The respective divisions had 9, 10, 11, and 12 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?
(A) 79
(B) 83
(C) 85
(D) 87
(E) 88
Let's name the teams in group 1 as 1,2,3,4,5,6,7,8,9.
Case 1; team1 played with every other team and won all of its matches.
so total number of matchs =8
case 2: team2 , played with team 3,4,5,6,7,8,9 and won all of its matches.
total number of matches =7
after case 1 and case 2 we have only two teams remaining in the group1 which are team 1 and team 2. Now since question asks us for the maximum no. of matches. Therefore we must include the extra case in which team 2 defeated team 1. Now both team 2 and team 1 have 1 loss each. Now in the final match, we will found out about the eventual winner in group 1.
maximum no. of matches in group 1 are 8+7+1(in which team2 defeated team1) + 1 ( final) =17
Similarly in group 2 we have 9 + 8 +1 +1 =19
group 3 = 10+9+1+1 =21
group 4 = 11+10+1+1=23
After this we will have 4 winner from each group. lets name them as w1,w2,w3,w4
Let's assume w1 won all of its matches from the remaining three teams and eventually emerged as a winner. Therefore total matches among four winners=3
Therefore maximum total no. of matches played are 17+19+21+23+3=83