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Re: This game season, five divisions are going to play. Out of all the tea
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22 Apr 2015, 19:01
# 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 \(=\,(N2)*2\,+\,3\,=\,(62)*2\,+\,3\,=\,11\)
applying the above method Division II: Total games \(=\,(N2)*2\,+\,3\,=\,(92)*2\,+\,3\,=\,17\) Division III: Total games \(=\,(N2)*2\,+\,3\,=\,(122)*2\,+\,3\,=\,23\) Division IV: Total games \(=\,(N2)*2\,+\,3\,=\,(132)*2\,+\,3\,=\,25\) Division V: Total games \(=\,(N2)*2\,+\,3\,=\,(142)*2\,+\,3\,=\,27\)
Total \(=\,11\,+\,17\,+\,23\,+\,25\,+27\,=\,103\)
League Championship: # of teams qualified \(=\,5\) # of games \(=\,4\); knockoff 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
