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Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation PDF Download

The game of QUIET is played between two teams. Six teams, numbered 1, 2, 3, 4, 5, and 6, play in a QUIET tournament. These teams are divided equally into two groups. In the tournament, each team plays every other team in the same group only once, and each team in the other group exactly twice. The tournament has several rounds, each of which consists of a few games. Every team plays exactly one game in each round.

The following additional facts are known about the schedule of games in the tournament.

  1. Each team played against a team from the other group in Round 8.
  2. In Round 4 and Round 7, the match-ups, that is the pair of teams playing against each other, were identical. In Round 5 and Round 8, the match-ups were identical.
  3. Team 4 played Team 6 in both Round 1 and Round 2.
  4. Team 1 played Team 5 ONLY once and that was in Round 2.
  5. Team 3 played Team 4 in Round 3. Team 1 played Team 6 in Round 6.
  6. In Round 8, Team 3 played Team 6, while Team 2 played Team 5.

Q1: How many rounds were there in the tournament?

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationView Answer  Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

Ans: 8
We are told that there are six teams, that are divided into two groups.
Teams in the same group will play each other only once, and teams in different group will play each other twice.
Calculating the combinations, there is going to be 3C2 games among teams in the same group among them, and since there is two groups, total such games will be 6.
Now, teams in different group play each other twice. Calculating the combinations for this,
From the first group, a team can be chosen in three ways, and from the second group, a team can be chosen in three ways. Total ways two teams from different groups can play each other is 3 x 3 which is 9. And since they play each other twice, that is 9 + 9 games of this combination.
Total number of games is 18 + 6 = 24
It is given that each team plays one game in each round, that means there is going to be 3 matchups in each round. And given, there is 24 games to played in this format, the number of rounds will be 24 / 3=8
The tournament will have 8 rounds.
Now that we know there are going to be 8 rounds in the tournament, let us identify the teams in a particular group, that will help us build the matchups.
We are told that Round 8 teams from different groups play each other, and Teams 1 and 5 play only once. This means, 1 and 5 have to be on the same group. It is also told that 4 and 6 play each other twice, that means 4 and 6 have to be in different groups. Looking at the matches from Round 8 that is given to us, 3 played 6 and 2 played 5. We already know 1 and 5 are in the same group, so 2 must be in the other group. Among 3 and 6, if we were to place 3 in the group with 1 and 5, 4 and 6 would have to be in the same group, which is not possible, hence 6 is with 1 and 5, giving us the final combination of groups.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, using the given information to build the matchups for the 8 rounds.

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationRounds marked with the same colour represent the fact that the matchups are identical. Now we know that each team plays a game in each round, we know 2 out of the 3 matches for Round 2 and 8, and we can identify the third matchups as well. Giving us this resulting table.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

We are told that Round 4 and 7 are identical, that means they are the matchups between teams from two different groups,
We look at the matchups that are remaining among the 6 teams where both the games are left to play.
Right away we identify that 6 is yet to play 2 twice and 5 once. We are looking for teams playing twice, so both Round 4 and 7 has a matchups between 2 and 6. This means 6 will play 5 in Round 3 and using that we can identify the third matchup in Round 3 as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, we can identify that 1 is yet to play 3 both the times, 2 once. And we are looking for teams playing each other twice, so we can fill in the remaining values in the table as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationThis is the final table.
The tournament will have 8 rounds.

Q2: What is the number of the team that played Team 1 in Round 5?

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationView Answer  Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

Ans: 4
We are told that there are six teams, that are divided into two groups.
Teams in the same group will play each other only once, and teams in different group will play each other twice.
Calculating the combinations, there is going to be 3C2 games among teams in the same group among them, and since there is two groups, total such games will be 6.
Now, teams in different group play each other twice. Calculating the combinations for this,
From the first group, a team can be chosen in three ways, and from the second group, a team can be chosen in three ways. Total ways two teams from different groups can play each other is 3 x 3 which is 9. And since they play each other twice, that is 9 + 9 games of this combination.
Total number of games is 18 + 6 = 24
It is given that each team plays one game in each round, that means there is going to be 3 matchups in each round. And given, there is 24 games to played in this format, the number of rounds will be 24 / 3=8
The tournament will have 8 rounds.
Now that we know there are going to be 8 rounds in the tournament, let us identify the teams in a particular group, that will help us build the matchups.
We are told that Round 8 teams from different groups play each other, and Teams 1 and 5 play only once. This means, 1 and 5 have to be on the same group. It is also told that 4 and 6 play each other twice, that means 4 and 6 have to be in different groups. Looking at the matches from Round 8 that is given to us, 3 played 6 and 2 played 5. We already know 1 and 5 are in the same group, so 2 must be in the other group. Among 3 and 6, if we were to place 3 in the group with 1 and 5, 4 and 6 would have to be in the same group, which is not possible, hence 6 is with 1 and 5, giving us the final combination of groups.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, using the given information to build the matchups for the 8 rounds.

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationRounds marked with the same colour represent the fact that the matchups are identical. Now we know that each team plays a game in each round, we know 2 out of the 3 matches for Round 2 and 8, and we can identify the third matchups as well. Giving us this resulting table.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

We are told that Round 4 and 7 are identical, that means they are the matchups between teams from two different groups,
We look at the matchups that are remaining among the 6 teams where both the games are left to play.
Right away we identify that 6 is yet to play 2 twice and 5 once. We are looking for teams playing twice, so both Round 4 and 7 has a matchups between 2 and 6. This means 6 will play 5 in Round 3 and using that we can identify the third matchup in Round 3 as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, we can identify that 1 is yet to play 3 both the times, 2 once. And we are looking for teams playing each other twice, so we can fill in the remaining values in the table as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationThis is the final table.
The team that played Team 1 in Round 5 is 4.

Q3: Which team among the teams numbered 2, 3, 4, and 5 was not part of the same group?
(a) 5
(b) 3
(c) 4
(d) 2

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationView Answer  Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

Ans: (a)
We are told that there are six teams, that are divided into two groups.
Teams in the same group will play each other only once, and teams in different group will play each other twice.
Calculating the combinations, there is going to be 3C2 games among teams in the same group among them, and since there is two groups, total such games will be 6.
Now, teams in different group play each other twice. Calculating the combinations for this,
From the first group, a team can be chosen in three ways, and from the second group, a team can be chosen in three ways. Total ways two teams from different groups can play each other is 3 x 3 which is 9. And since they play each other twice, that is 9 + 9 games of this combination.
Total number of games is 18 + 6 = 24
It is given that each team plays one game in each round, that means there is going to be 3 matchups in each round. And given, there is 24 games to played in this format, the number of rounds will be 24 / 3=8
The tournament will have 8 rounds.
Now that we know there are going to be 8 rounds in the tournament, let us identify the teams in a particular group, that will help us build the matchups.
We are told that Round 8 teams from different groups play each other, and Teams 1 and 5 play only once. This means, 1 and 5 have to be on the same group. It is also told that 4 and 6 play each other twice, that means 4 and 6 have to be in different groups. Looking at the matches from Round 8 that is given to us, 3 played 6 and 2 played 5. We already know 1 and 5 are in the same group, so 2 must be in the other group. Among 3 and 6, if we were to place 3 in the group with 1 and 5, 4 and 6 would have to be in the same group, which is not possible, hence 6 is with 1 and 5, giving us the final combination of groups.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, using the given information to build the matchups for the 8 rounds.

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationRounds marked with the same colour represent the fact that the matchups are identical. Now we know that each team plays a game in each round, we know 2 out of the 3 matches for Round 2 and 8, and we can identify the third matchups as well. Giving us this resulting table.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

We are told that Round 4 and 7 are identical, that means they are the matchups between teams from two different groups,
We look at the matchups that are remaining among the 6 teams where both the games are left to play.
Right away we identify that 6 is yet to play 2 twice and 5 once. We are looking for teams playing twice, so both Round 4 and 7 has a matchups between 2 and 6. This means 6 will play 5 in Round 3 and using that we can identify the third matchup in Round 3 as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, we can identify that 1 is yet to play 3 both the times, 2 once. And we are looking for teams playing each other twice, so we can fill in the remaining values in the table as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationThis is the final table.
2, 3 and 4 were part of the same group. 5 is the answer. 

Q4: What is the number of the team that played Team 1 in Round 7?

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationView Answer  Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

Ans: 3
We are told that there are six teams, that are divided into two groups.
Teams in the same group will play each other only once, and teams in different group will play each other twice.
Calculating the combinations, there is going to be 3C2 games among teams in the same group among them, and since there is two groups, total such games will be 6.
Now, teams in different group play each other twice. Calculating the combinations for this,
From the first group, a team can be chosen in three ways, and from the second group, a team can be chosen in three ways. Total ways two teams from different groups can play each other is 3 x 3 which is 9. And since they play each other twice, that is 9 + 9 games of this combination.
Total number of games is 18 + 6 = 24
It is given that each team plays one game in each round, that means there is going to be 3 matchups in each round. And given, there is 24 games to played in this format, the number of rounds will be 24 / 3 = 8
The tournament will have 8 rounds.
Now that we know there are going to be 8 rounds in the tournament, let us identify the teams in a particular group, that will help us build the matchups.
We are told that Round 8 teams from different groups play each other, and Teams 1 and 5 play only once. This means, 1 and 5 have to be on the same group. It is also told that 4 and 6 play each other twice, that means 4 and 6 have to be in different groups. Looking at the matches from Round 8 that is given to us, 3 played 6 and 2 played 5. We already know 1 and 5 are in the same group, so 2 must be in the other group. Among 3 and 6, if we were to place 3 in the group with 1 and 5, 4 and 6 would have to be in the same group, which is not possible, hence 6 is with 1 and 5, giving us the final combination of groups.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, using the given information to build the matchups for the 8 rounds.

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationRounds marked with the same colour represent the fact that the matchups are identical. Now we know that each team plays a game in each round, we know 2 out of the 3 matches for Round 2 and 8, and we can identify the third matchups as well. Giving us this resulting table.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

We are told that Round 4 and 7 are identical, that means they are the matchups between teams from two different groups,
We look at the matchups that are remaining among the 6 teams where both the games are left to play.
Right away we identify that 6 is yet to play 2 twice and 5 once. We are looking for teams playing twice, so both Round 4 and 7 has a matchups between 2 and 6. This means 6 will play 5 in Round 3 and using that we can identify the third matchup in Round 3 as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, we can identify that 1 is yet to play 3 both the times, 2 once. And we are looking for teams playing each other twice, so we can fill in the remaining values in the table as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationThis is the final table.
The team that played Team 1 in Round 7 is 3.

Q5: What is the number of the team that played Team 6 in Round 3?

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationView Answer  Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

Ans: 5
We are told that there are six teams, that are divided into two groups.
Teams in the same group will play each other only once, and teams in different group will play each other twice.
Calculating the combinations, there is going to be 3C2 games among teams in the same group among them, and since there is two groups, total such games will be 6.
Now, teams in different group play each other twice. Calculating the combinations for this,
From the first group, a team can be chosen in three ways, and from the second group, a team can be chosen in three ways. Total ways two teams from different groups can play each other is 3 x 3 which is 9. And since they play each other twice, that is 9 + 9 games of this combination.
Total number of games is 18 + 6 = 24
It is given that each team plays one game in each round, that means there is going to be 3 matchups in each round. And given, there is 24 games to played in this format, the number of rounds will be 24 / 3=8
The tournament will have 8 rounds.
Now that we know there are going to be 8 rounds in the tournament, let us identify the teams in a particular group, that will help us build the matchups.
We are told that Round 8 teams from different groups play each other, and Teams 1 and 5 play only once. This means, 1 and 5 have to be on the same group. It is also told that 4 and 6 play each other twice, that means 4 and 6 have to be in different groups. Looking at the matches from Round 8 that is given to us, 3 played 6 and 2 played 5. We already know 1 and 5 are in the same group, so 2 must be in the other group. Among 3 and 6, if we were to place 3 in the group with 1 and 5, 4 and 6 would have to be in the same group, which is not possible, hence 6 is with 1 and 5, giving us the final combination of groups.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, using the given information to build the matchups for the 8 rounds.

Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationRounds marked with the same colour represent the fact that the matchups are identical. Now we know that each team plays a game in each round, we know 2 out of the 3 matches for Round 2 and 8, and we can identify the third matchups as well. Giving us this resulting table.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation

We are told that Round 4 and 7 are identical, that means they are the matchups between teams from two different groups,
We look at the matchups that are remaining among the 6 teams where both the games are left to play.
Right away we identify that 6 is yet to play 2 twice and 5 once. We are looking for teams playing twice, so both Round 4 and 7 has a matchups between 2 and 6. This means 6 will play 5 in Round 3 and using that we can identify the third matchup in Round 3 as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationNow, we can identify that 1 is yet to play 3 both the times, 2 once. And we are looking for teams playing each other twice, so we can fill in the remaining values in the table as well.
Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT PreparationThis is the final table.
Team that played Team 6 in Round 3 was Team 5.

The document Practice Question - 56 (Games and Tournaments) | 100 DILR Questions for CAT Preparation is a part of the CAT Course 100 DILR Questions for CAT Preparation.
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FAQs on Practice Question - 56 (Games and Tournaments) - 100 DILR Questions for CAT Preparation

1. What are the different types of games and tournaments commonly encountered in competitive scenarios?
Ans. In competitive scenarios, various types of games and tournaments are prevalent, including round-robin tournaments, single-elimination tournaments, double-elimination tournaments, and Swiss-system tournaments. Round-robin tournaments involve each participant playing against every other participant, while single-elimination tournaments eliminate participants after a loss. Double-elimination tournaments allow participants to stay in the competition until they lose twice. Swiss-system tournaments pair participants with similar scores, ensuring that no one is eliminated while still determining a winner.
2. How is the scoring system typically structured in tournament games?
Ans. The scoring system in tournament games can vary widely based on the type of game being played. Commonly, points are awarded for wins, with additional points for draws or ties in some games. In knockout tournaments, a win in a match may allow a player or team to advance to the next round, while in round-robin formats, total points accumulated across all matches determine rankings. Additionally, tie-breakers may be used to resolve situations where participants have equal points.
3. What are some strategies for success in tournaments?
Ans. Successful strategies in tournaments often include thorough preparation and practice, understanding opponents’ strengths and weaknesses, and effective time management. Players should also focus on maintaining mental and physical stamina throughout the competition. Analyzing previous matches can provide insights into effective strategies, while adaptability during games can help overcome unexpected challenges.
4. What is the significance of seeding in tournaments?
Ans. Seeding in tournaments is significant as it determines the initial matchups between participants based on their skill levels or past performances. Higher-seeded players are typically matched against lower-seeded players in the early rounds, which helps prevent the strongest competitors from facing each other too soon. This structure not only enhances the competitiveness of the tournament but also aims to provide a more exciting experience for both players and spectators.
5. How do organizers ensure fairness in tournament play?
Ans. Organizers ensure fairness in tournament play through several measures, including establishing clear rules and regulations, ensuring impartial refereeing, and implementing standardized scoring systems. Additionally, they may use technology to monitor matches and prevent cheating. Randomized seeding and scheduling can also help maintain fairness by preventing biases based on previous performances or rankings.
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