CAT 2024 Slot 1 · DILR · Set 3: QUIET Tournament Schedule
The full set and all 5 questions (Q10–Q14) from CAT 2024 Slot 1, with answers and solutions.
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.
Question 10 · Games and Tournaments
How many rounds were there in the tournament?
Answer & solution
Answer: 8
Solving the set
Step 1 – Count the rounds. With groups of 3, each group has 3 same-group pairs (played once): 6 games. There are 3 × 3 = 9 cross-group pairs (played twice): 18 games. Total = 24 games. Each round has 3 games (six teams, each playing once), so there are 24/3 = 8 rounds.
Step 2 – Find the groups.
• Teams 1 and 5 played only once (fact 4), so they are in the same group.
• Round 8 has 3-6 and 2-5, so its third game is 1-4. Every Round 8 game is cross-group (fact 1), so 1 and 4 are in different groups.
• Teams 4 and 6 played twice (fact 3), so they are in different groups. Hence 6 is in the same group as 1.
|
Group |
Teams |
|
Group I |
1, 5, 6 |
|
Group II |
2, 3, 4 |
Check: 3-6 and 2-5 in Round 8 are both cross-group, as required.
Step 3 – Place the six same-group games (1-5, 1-6, 5-6, 2-3, 2-4, 3-4).
• Round 2 has 4-6 and 1-5, so its third game is 2-3.
• Round 3 has 3-4. Teams 1, 2, 5 and 6 remain. 1-5 (only in Round 2) and 1-6 (only in Round 6) are not possible, so Round 3 is 1-2 and 5-6.
• Round 6 has 1-6. Teams 2, 3, 4 and 5 remain. 2-3 and 3-4 are already used, so Round 6 is 2-4 and 3-5.
All six same-group games are now in Rounds 2, 3 and 6, so Rounds 1, 4, 5, 7 and 8 contain only cross-group games.
Step 4 – Round 1 has 4-6. Teams 1, 2, 3 and 5 remain. 1-5 is not possible. The other choices are (1-2, 3-5) or (1-3, 2-5). But 2-5 is already played twice (Rounds 5 and 8, since Round 5 = Round 8). So Round 1 is 1-2 and 3-5.
Step 5 – Round 5 = Round 8 = 1-4, 2-5, 3-6. Counting the cross-group games placed so far, the pairs still needing two games each are 1-3, 4-5 and 2-6. These six games fill Rounds 4 and 7, which must be identical: 1-3, 4-5, 2-6.
|
Round |
Game 1 |
Game 2 |
Game 3 |
Type of games |
|
1 |
1 v 2 |
3 v 5 |
4 v 6 |
All cross-group |
|
2 |
1 v 5 |
2 v 3 |
4 v 6 |
Two same-group + one cross |
|
3 |
1 v 2 |
3 v 4 |
5 v 6 |
Two same-group + one cross |
|
4 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
5 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
|
6 |
1 v 6 |
2 v 4 |
3 v 5 |
Two same-group + one cross |
|
7 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
8 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
Step 6 – Check every pair: each same-group pair appears once and each cross-group pair appears twice.
|
Cross-group pair |
Rounds played |
Same-group pair |
Round played |
|
1-2 |
1, 3 |
1-5 |
2 |
|
1-3 |
4, 7 |
1-6 |
6 |
|
1-4 |
5, 8 |
5-6 |
3 |
|
5-2 |
5, 8 |
2-3 |
2 |
|
5-3 |
1, 6 |
2-4 |
6 |
|
5-4 |
4, 7 |
3-4 |
3 |
|
6-2 |
4, 7 |
||
|
6-3 |
5, 8 |
||
|
6-4 |
1, 2 |
This question
From Step 1: 6 same-group games + 18 cross-group games = 24 games, at 3 games per round, so 8 rounds. This also matches the facts, which mention Round 8 as the last round used.
Hence, the answer is 8.
Question 11 · Games and Tournaments
What is the number of the team that played Team 1 in Round 5?
Answer & solution
Answer: 4
Solving the set
Step 1 – Count the rounds. With groups of 3, each group has 3 same-group pairs (played once): 6 games. There are 3 × 3 = 9 cross-group pairs (played twice): 18 games. Total = 24 games. Each round has 3 games (six teams, each playing once), so there are 24/3 = 8 rounds.
Step 2 – Find the groups.
• Teams 1 and 5 played only once (fact 4), so they are in the same group.
• Round 8 has 3-6 and 2-5, so its third game is 1-4. Every Round 8 game is cross-group (fact 1), so 1 and 4 are in different groups.
• Teams 4 and 6 played twice (fact 3), so they are in different groups. Hence 6 is in the same group as 1.
|
Group |
Teams |
|
Group I |
1, 5, 6 |
|
Group II |
2, 3, 4 |
Check: 3-6 and 2-5 in Round 8 are both cross-group, as required.
Step 3 – Place the six same-group games (1-5, 1-6, 5-6, 2-3, 2-4, 3-4).
• Round 2 has 4-6 and 1-5, so its third game is 2-3.
• Round 3 has 3-4. Teams 1, 2, 5 and 6 remain. 1-5 (only in Round 2) and 1-6 (only in Round 6) are not possible, so Round 3 is 1-2 and 5-6.
• Round 6 has 1-6. Teams 2, 3, 4 and 5 remain. 2-3 and 3-4 are already used, so Round 6 is 2-4 and 3-5.
All six same-group games are now in Rounds 2, 3 and 6, so Rounds 1, 4, 5, 7 and 8 contain only cross-group games.
Step 4 – Round 1 has 4-6. Teams 1, 2, 3 and 5 remain. 1-5 is not possible. The other choices are (1-2, 3-5) or (1-3, 2-5). But 2-5 is already played twice (Rounds 5 and 8, since Round 5 = Round 8). So Round 1 is 1-2 and 3-5.
Step 5 – Round 5 = Round 8 = 1-4, 2-5, 3-6. Counting the cross-group games placed so far, the pairs still needing two games each are 1-3, 4-5 and 2-6. These six games fill Rounds 4 and 7, which must be identical: 1-3, 4-5, 2-6.
|
Round |
Game 1 |
Game 2 |
Game 3 |
Type of games |
|
1 |
1 v 2 |
3 v 5 |
4 v 6 |
All cross-group |
|
2 |
1 v 5 |
2 v 3 |
4 v 6 |
Two same-group + one cross |
|
3 |
1 v 2 |
3 v 4 |
5 v 6 |
Two same-group + one cross |
|
4 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
5 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
|
6 |
1 v 6 |
2 v 4 |
3 v 5 |
Two same-group + one cross |
|
7 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
8 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
Step 6 – Check every pair: each same-group pair appears once and each cross-group pair appears twice.
|
Cross-group pair |
Rounds played |
Same-group pair |
Round played |
|
1-2 |
1, 3 |
1-5 |
2 |
|
1-3 |
4, 7 |
1-6 |
6 |
|
1-4 |
5, 8 |
5-6 |
3 |
|
5-2 |
5, 8 |
2-3 |
2 |
|
5-3 |
1, 6 |
2-4 |
6 |
|
5-4 |
4, 7 |
3-4 |
3 |
|
6-2 |
4, 7 |
||
|
6-3 |
5, 8 |
||
|
6-4 |
1, 2 |
This question
From Step 5, Round 5 is identical to Round 8: 1-4, 2-5 and 3-6. So Team 1 played Team 4 in Round 5.
Hence, the answer is 4.
Question 12 · Games and Tournaments
Which team among the teams numbered 2, 3, 4, and 5 was not part of the same group?
- A
3
- B
2
- C
4
- D
5
Answer & solution
Answer: Option D
Solving the set
Step 1 – Count the rounds. With groups of 3, each group has 3 same-group pairs (played once): 6 games. There are 3 × 3 = 9 cross-group pairs (played twice): 18 games. Total = 24 games. Each round has 3 games (six teams, each playing once), so there are 24/3 = 8 rounds.
Step 2 – Find the groups.
• Teams 1 and 5 played only once (fact 4), so they are in the same group.
• Round 8 has 3-6 and 2-5, so its third game is 1-4. Every Round 8 game is cross-group (fact 1), so 1 and 4 are in different groups.
• Teams 4 and 6 played twice (fact 3), so they are in different groups. Hence 6 is in the same group as 1.
|
Group |
Teams |
|
Group I |
1, 5, 6 |
|
Group II |
2, 3, 4 |
Check: 3-6 and 2-5 in Round 8 are both cross-group, as required.
Step 3 – Place the six same-group games (1-5, 1-6, 5-6, 2-3, 2-4, 3-4).
• Round 2 has 4-6 and 1-5, so its third game is 2-3.
• Round 3 has 3-4. Teams 1, 2, 5 and 6 remain. 1-5 (only in Round 2) and 1-6 (only in Round 6) are not possible, so Round 3 is 1-2 and 5-6.
• Round 6 has 1-6. Teams 2, 3, 4 and 5 remain. 2-3 and 3-4 are already used, so Round 6 is 2-4 and 3-5.
All six same-group games are now in Rounds 2, 3 and 6, so Rounds 1, 4, 5, 7 and 8 contain only cross-group games.
Step 4 – Round 1 has 4-6. Teams 1, 2, 3 and 5 remain. 1-5 is not possible. The other choices are (1-2, 3-5) or (1-3, 2-5). But 2-5 is already played twice (Rounds 5 and 8, since Round 5 = Round 8). So Round 1 is 1-2 and 3-5.
Step 5 – Round 5 = Round 8 = 1-4, 2-5, 3-6. Counting the cross-group games placed so far, the pairs still needing two games each are 1-3, 4-5 and 2-6. These six games fill Rounds 4 and 7, which must be identical: 1-3, 4-5, 2-6.
|
Round |
Game 1 |
Game 2 |
Game 3 |
Type of games |
|
1 |
1 v 2 |
3 v 5 |
4 v 6 |
All cross-group |
|
2 |
1 v 5 |
2 v 3 |
4 v 6 |
Two same-group + one cross |
|
3 |
1 v 2 |
3 v 4 |
5 v 6 |
Two same-group + one cross |
|
4 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
5 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
|
6 |
1 v 6 |
2 v 4 |
3 v 5 |
Two same-group + one cross |
|
7 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
8 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
Step 6 – Check every pair: each same-group pair appears once and each cross-group pair appears twice.
|
Cross-group pair |
Rounds played |
Same-group pair |
Round played |
|
1-2 |
1, 3 |
1-5 |
2 |
|
1-3 |
4, 7 |
1-6 |
6 |
|
1-4 |
5, 8 |
5-6 |
3 |
|
5-2 |
5, 8 |
2-3 |
2 |
|
5-3 |
1, 6 |
2-4 |
6 |
|
5-4 |
4, 7 |
3-4 |
3 |
|
6-2 |
4, 7 |
||
|
6-3 |
5, 8 |
||
|
6-4 |
1, 2 |
This question
From Step 2, the groups are {1, 5, 6} and {2, 3, 4}. Among 2, 3, 4 and 5, Teams 2, 3 and 4 form one group and Team 5 belongs to the other.
Hence, the answer is option 4.
Question 13 · Games and Tournaments
What is the number of the team that played Team 1 in Round 7?
Answer & solution
Answer: 3
Solving the set
Step 1 – Count the rounds. With groups of 3, each group has 3 same-group pairs (played once): 6 games. There are 3 × 3 = 9 cross-group pairs (played twice): 18 games. Total = 24 games. Each round has 3 games (six teams, each playing once), so there are 24/3 = 8 rounds.
Step 2 – Find the groups.
• Teams 1 and 5 played only once (fact 4), so they are in the same group.
• Round 8 has 3-6 and 2-5, so its third game is 1-4. Every Round 8 game is cross-group (fact 1), so 1 and 4 are in different groups.
• Teams 4 and 6 played twice (fact 3), so they are in different groups. Hence 6 is in the same group as 1.
|
Group |
Teams |
|
Group I |
1, 5, 6 |
|
Group II |
2, 3, 4 |
Check: 3-6 and 2-5 in Round 8 are both cross-group, as required.
Step 3 – Place the six same-group games (1-5, 1-6, 5-6, 2-3, 2-4, 3-4).
• Round 2 has 4-6 and 1-5, so its third game is 2-3.
• Round 3 has 3-4. Teams 1, 2, 5 and 6 remain. 1-5 (only in Round 2) and 1-6 (only in Round 6) are not possible, so Round 3 is 1-2 and 5-6.
• Round 6 has 1-6. Teams 2, 3, 4 and 5 remain. 2-3 and 3-4 are already used, so Round 6 is 2-4 and 3-5.
All six same-group games are now in Rounds 2, 3 and 6, so Rounds 1, 4, 5, 7 and 8 contain only cross-group games.
Step 4 – Round 1 has 4-6. Teams 1, 2, 3 and 5 remain. 1-5 is not possible. The other choices are (1-2, 3-5) or (1-3, 2-5). But 2-5 is already played twice (Rounds 5 and 8, since Round 5 = Round 8). So Round 1 is 1-2 and 3-5.
Step 5 – Round 5 = Round 8 = 1-4, 2-5, 3-6. Counting the cross-group games placed so far, the pairs still needing two games each are 1-3, 4-5 and 2-6. These six games fill Rounds 4 and 7, which must be identical: 1-3, 4-5, 2-6.
|
Round |
Game 1 |
Game 2 |
Game 3 |
Type of games |
|
1 |
1 v 2 |
3 v 5 |
4 v 6 |
All cross-group |
|
2 |
1 v 5 |
2 v 3 |
4 v 6 |
Two same-group + one cross |
|
3 |
1 v 2 |
3 v 4 |
5 v 6 |
Two same-group + one cross |
|
4 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
5 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
|
6 |
1 v 6 |
2 v 4 |
3 v 5 |
Two same-group + one cross |
|
7 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
8 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
Step 6 – Check every pair: each same-group pair appears once and each cross-group pair appears twice.
|
Cross-group pair |
Rounds played |
Same-group pair |
Round played |
|
1-2 |
1, 3 |
1-5 |
2 |
|
1-3 |
4, 7 |
1-6 |
6 |
|
1-4 |
5, 8 |
5-6 |
3 |
|
5-2 |
5, 8 |
2-3 |
2 |
|
5-3 |
1, 6 |
2-4 |
6 |
|
5-4 |
4, 7 |
3-4 |
3 |
|
6-2 |
4, 7 |
||
|
6-3 |
5, 8 |
||
|
6-4 |
1, 2 |
This question
From Step 5, Rounds 4 and 7 are identical: 1-3, 4-5 and 2-6. So Team 1 played Team 3 in Round 7.
Hence, the answer is 3.
Question 14 · Games and Tournaments
What is the number of the team that played Team 6 in Round 3?
Answer & solution
Answer: 5
Solving the set
Step 1 – Count the rounds. With groups of 3, each group has 3 same-group pairs (played once): 6 games. There are 3 × 3 = 9 cross-group pairs (played twice): 18 games. Total = 24 games. Each round has 3 games (six teams, each playing once), so there are 24/3 = 8 rounds.
Step 2 – Find the groups.
• Teams 1 and 5 played only once (fact 4), so they are in the same group.
• Round 8 has 3-6 and 2-5, so its third game is 1-4. Every Round 8 game is cross-group (fact 1), so 1 and 4 are in different groups.
• Teams 4 and 6 played twice (fact 3), so they are in different groups. Hence 6 is in the same group as 1.
|
Group |
Teams |
|
Group I |
1, 5, 6 |
|
Group II |
2, 3, 4 |
Check: 3-6 and 2-5 in Round 8 are both cross-group, as required.
Step 3 – Place the six same-group games (1-5, 1-6, 5-6, 2-3, 2-4, 3-4).
• Round 2 has 4-6 and 1-5, so its third game is 2-3.
• Round 3 has 3-4. Teams 1, 2, 5 and 6 remain. 1-5 (only in Round 2) and 1-6 (only in Round 6) are not possible, so Round 3 is 1-2 and 5-6.
• Round 6 has 1-6. Teams 2, 3, 4 and 5 remain. 2-3 and 3-4 are already used, so Round 6 is 2-4 and 3-5.
All six same-group games are now in Rounds 2, 3 and 6, so Rounds 1, 4, 5, 7 and 8 contain only cross-group games.
Step 4 – Round 1 has 4-6. Teams 1, 2, 3 and 5 remain. 1-5 is not possible. The other choices are (1-2, 3-5) or (1-3, 2-5). But 2-5 is already played twice (Rounds 5 and 8, since Round 5 = Round 8). So Round 1 is 1-2 and 3-5.
Step 5 – Round 5 = Round 8 = 1-4, 2-5, 3-6. Counting the cross-group games placed so far, the pairs still needing two games each are 1-3, 4-5 and 2-6. These six games fill Rounds 4 and 7, which must be identical: 1-3, 4-5, 2-6.
|
Round |
Game 1 |
Game 2 |
Game 3 |
Type of games |
|
1 |
1 v 2 |
3 v 5 |
4 v 6 |
All cross-group |
|
2 |
1 v 5 |
2 v 3 |
4 v 6 |
Two same-group + one cross |
|
3 |
1 v 2 |
3 v 4 |
5 v 6 |
Two same-group + one cross |
|
4 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
5 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
|
6 |
1 v 6 |
2 v 4 |
3 v 5 |
Two same-group + one cross |
|
7 |
1 v 3 |
4 v 5 |
2 v 6 |
All cross-group |
|
8 |
1 v 4 |
2 v 5 |
3 v 6 |
All cross-group |
Step 6 – Check every pair: each same-group pair appears once and each cross-group pair appears twice.
|
Cross-group pair |
Rounds played |
Same-group pair |
Round played |
|
1-2 |
1, 3 |
1-5 |
2 |
|
1-3 |
4, 7 |
1-6 |
6 |
|
1-4 |
5, 8 |
5-6 |
3 |
|
5-2 |
5, 8 |
2-3 |
2 |
|
5-3 |
1, 6 |
2-4 |
6 |
|
5-4 |
4, 7 |
3-4 |
3 |
|
6-2 |
4, 7 |
||
|
6-3 |
5, 8 |
||
|
6-4 |
1, 2 |
This question
From Step 3, Round 3 is 3-4, 1-2 and 5-6. So Team 6 played Team 5 in Round 3.
Hence, the answer is 5.
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