Re: This game season, five divisions are going to play. Out of all the tea
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22 Apr 2015, 18:01
# of games in division I:
Total teams qualified \(N\,=\,6\)
number of games to determine top 2 teams \(=\,(total\,teams\,-2)\,*\,2_{losses}\,=\,8\,\); each team has to lose \(2\) games
number of games to pick winner among top two teams \(=\,two\,losses\,+\,one\,win\,=\,3\,\); each team lose one game & a win
so total games \(=\,(N-2)*2\,+\,3\,=\,(6-2)*2\,+\,3\,=\,11\)
applying the above method
Division II: Total games \(=\,(N-2)*2\,+\,3\,=\,(9-2)*2\,+\,3\,=\,17\)
Division III: Total games \(=\,(N-2)*2\,+\,3\,=\,(12-2)*2\,+\,3\,=\,23\)
Division IV: Total games \(=\,(N-2)*2\,+\,3\,=\,(13-2)*2\,+\,3\,=\,25\)
Division V: Total games \(=\,(N-2)*2\,+\,3\,=\,(14-2)*2\,+\,3\,=\,27\)
Total \(=\,11\,+\,17\,+\,23\,+\,25\,+27\,=\,103\)
League Championship:
# of teams qualified \(=\,5\)
# of games \(=\,4\); knock-off system
Maximum Games \(=\,103\,+\,4\,=\,107\)
or
Division I: 6 * 2 losses but one [(6*2) - 1] will give the winner; total = \([(6*2) - 1] = 11\)
Division II: 9 * 2 losses but one [(9*2) - 1] will give the winner; total = \([(9*2) - 1] = 17\)
Division III: 12 * 2 losses but one [(12*2) - 1] will give the winner; total = \([(12*2) - 1] = 23\)
Division IV: 13 * 2 losses but one [(13*2) - 1] will give the winner; total = \([(13*2) - 1] = 25\)
Division V: 14 * 2 losses but one [(14*2) - 1] will give the winner; total = \([(14*2) - 1] = 27\)
League Championship: \(4\) losses for qualified five teams will determine the winner
Maximum Games \(=\,11\,+\,17\,+\,23\,+\,25\,+27\,+\,4\,=\,107\)
Answer D