30 teams in a hockey tournment.A team is out of the tournment if it lost 2 games.what is the maximum number of games to be played to decide one winner
65
59
61
30
34

• maximum number of games to be played to decide one winner = one less than twice the no. of participating teams = 59
• 13 years agoHelpfull: Yes(11) No(2)
• firstly there are 30 teams so we have 15 matches in round 1...now 15 teams won and 15 loses....now in order to maximise the no. of games the teams that have lost in round 1 will have to win and the tams winning round 1 will lose the match....now we have 30 teams each winning a game and losing one...now in round 3 again when these teams play against each other 15 teams win and 15 loses now the 15 teams that loses are out of the tournament.....now we are left with 15 teams having lost one match....in round 4 there will be 7 matches and one team doesn't gets the chance to play...after this round we are left with 4 teams...in round 5 two teams gets knocked out..and only two teams remain....and now we have the final match which gives us our winner...ON ADDING THE TOTAL NO. OF MATCHES IN ALL ROUNDS WE HAVE 15+15+15+7+4+2+1=59.
• 13 years agoHelpfull: Yes(6) No(1)
• twice the no of teams ie 30*2=60
  but we had counted 1 match twice sol will be 60-1=59
• 13 years agoHelpfull: Yes(0) No(0)