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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
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 × 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.
Now, using the given information to build the matchups for the 8 rounds
Rounds 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.
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.
Now, 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.
This is the final table.
The team that played Team 1 in Round 7 is 3.