CAT Question Bank - Games & Tournaments
Q27) There are 16 teams and they are divided into 2 pools of 8 each. Each team in a group plays against one another on a round-robin basis. Draws in the competition are not allowed. The top four teams from each group will qualify for the next round i.e round 2. In case of teams having the same number of wins, the team with better run-rate would be ranked ahead.
- Minimum number of wins required to qualify for the next round ?
- Minimum number of wins required to guarantee qualification in the next round ?
Rowdy Rathore last edited by zabeer
1 group is consisting of 8 teams. So each team will play 7 match each. Suppose each of the 8 teams were seeded and we consider the case where a higher seeded team will always win.
So the number of wins for the 8 teams would be 7,6,5,4,3,2,1,0 with highest seeded team winning all and lowest seeded team losing all.
For minimum number of wins we allow 3 teams to win maximum number of matches. Of the remaining 5 teams just find out the mean of their number of wins.
In this case it would be (4+3+2+1+0)/5=2.
So 5 teams can end up with 2 wins each and a team with better run rate will qualify with 2 wins.
In this case consider the mean of first 5 higher seeded teams
So it may be the case that 5 teams can end up having 5 wins each. And hence 1 team will miss the second round birth. So minimum number of wins to guarantee a place would be 6.
Q28) Two teams P and Q contest in a game of tug of war. Team P has ‘n’ players with an average weight of ‘a’ kgs, ‘a’ being an integer. Team Q also has ‘n’ players with an average weight of ‘a’ kgs. The rule of the game is that the team that wins two rounds consecutively wins the game. However, once a team wins a round, the other team adds one more player to their team such that their average weight increases by 1 kg for the next round. The game progressed as follows.
Round 1: Team Q won
Round 2: Team P won
Round 3: Team Q won
Round 4: Team P won
Round 5: Team Q won
Round 6: Team P won
Round 7: Team P won
Which of the following can be the sum of the weights of the new players added in team P?
First player to be added brings in an excess of 1 kg for each player (including himself) over the average, a.
This means his weight is a+n+1.
The second player to be added brings in an excess of 1 for all including himself over the previous average, a+1. This means his weight is a+1+n+2
Third player weighs a+2+n+3.
Total weight is 3a + 3n + 9. This is divisible by 3. So the total can be 249.
Q29) Ram and Shyam were playing a game that involved picking up marbles from a table. Each player in his turn has to pick up at least 2 or at most 7 marbles, except when there is only one marble left on the table in which case, he has to pick up that marble. The game ends when all the marbles on the table are picked up. If the game starts with 62 marbles and it is Ram's turn to play, then how many marbles should he pick up so as to ensure his win?
Assume that both play intelligently so as to win the game.
Assume that the player who picks the last marble LOSES the game.
a. 4 or 3
b. 6 or 7
c. 5 or 4
d. 2 or 1
Q30) In a soccer tournament n teams play against one another exactly once. The win fetches 3 points, draw 1 each and loss 0. After all the matches were played, it was noticed that the top team had unique number of maximum points and unique least number of wins. What can be the minimum possible value of n?
e) none of these
Amim Fatmi last edited by
@rowdy-rathore Shouldnt it be smallest value of n greater than 5 , or any other limiting factor?
Amim Fatmi last edited by
@rowdy-rathore We select two points out of 98 to make a chord, given by 98C2. This gives 4753 , which leaves a remainder on dividing by 3 .thus anthony wins.
Kushal Khandelwal last edited by Kushal Khandelwal
@rowdy-rathore The rth match in Round 1 will be played between Seed ‘r’ and Seed ‘n+1-r’
so 7th match will be played between seed 7 and seed 32+1-7=26
and 10 th match will be played between seed 10 and seed 32+1-10=23
Surbhi Sinha last edited by
@rowdy-rathore please provide solution