CAT 2024 Slot 2 · DILR · Set 1: Gymnastics Players and Three Coaches
The full set and all 5 questions (Q1–Q5) from CAT 2024 Slot 2, with answers and solutions.
Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches - Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7, with 1 being the lowest rating and 7 the highest.
The following additional information is known.
1. Xena trained more players than Yuki.
2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
4. The average of the ratings of all the players was 4.
5. Player-2 got the highest rating.
6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
7. Player-4's rating was double of Player-8's and less than Player-5's.
Question 1 · Sports and Scores Based Sets
What best can be concluded about the number of players coached by Zara?
- A
Exactly 2
- B
Exactly 3
- C
Either 2 or 3 or 4
- D
Either 2 or 3
Answer & solution
Answer: Option A
Solving the set
Step 1 – Coaches who can train each player. Yuki trains only even players and Zara only odd players. Xena can train anyone.
• Player-1 (odd) and Player-4 (even) have the same coach (fact 2). Yuki cannot take Player-1 and Zara cannot take Player-4, so both are with Xena.
• Players 2, 3 and 5 have three different coaches (fact 2). Player-2 can be with Xena or Yuki; Players 3 and 5 can be with Xena or Zara. Since only one of them can be with Xena, Players 3 and 5 take Xena and Zara between them, and Player-2 goes to Yuki.
Step 2 – Players 5 and 7 have the same coach (fact 3). If Player-5 were with Xena, Player-7 would also be with Xena and Player-3 with Zara. Zara could then take no other odd player (1, 5 and 7 are with Xena), so Zara would train only one player, which is not allowed. Hence Players 5 and 7 are with Zara and Player-3 is with Xena. Zara trains exactly Players 5 and 7.
Step 3 – Ratings. There are 8 players but only 7 possible ratings, and only Players 5 and 7 share a rating. So the ratings are 1, 2, …, 7 once each, plus one repeated value v for Players 5 and 7. The total is 8 × 4 = 32 (fact 4), so 28 + v = 32 and v = 4. Player-5 = Player-7 = 4.
Step 4 – Player-2 has the highest rating, 7 (fact 5). Player-4 = 2 × Player-8 and Player-4 < Player-5 = 4, so Player-4 = 2 and Player-8 = 1.
Step 5 – Averages (fact 6). Zara's average = (4 + 4)/2 = 4, so Yuki's average = 4 + 2 = 6 and Xena's average = 6/2 = 3.
Step 6 – Yuki has Player-2 (7) and needs at least one more even player: 6 or 8. Including Player-8 (rating 1) cannot give an average of 6 (with Player-6 as well, 7 + 1 + P6 = 18 would need P6 = 10). So Yuki has Players 2 and 6, with (7 + P6)/2 = 6, i.e., Player-6 = 5. Player-8 goes to Xena.
Step 7 – Xena has Players 1, 3, 4 and 8 (4 players, more than Yuki's 2 ✓). The ratings left are 3 and 6 for Players 1 and 3. Xena's total = 3 + 6 + 2 + 1 = 12, average 3 ✓, whichever way 3 and 6 are assigned.
|
Player |
Coach |
Rating |
Reason |
|
1 |
Xena |
3 or 6 |
Same coach as Player-4 (fact 2) |
|
2 |
Yuki |
7 |
Highest rating (fact 5) |
|
3 |
Xena |
6 or 3 |
Different coach from 2 and 5 |
|
4 |
Xena |
2 |
2 × Player-8 and less than 4 (fact 7) |
|
5 |
Zara |
4 |
Repeated rating (facts 3, 4) |
|
6 |
Yuki |
5 |
Makes Yuki's average 6 |
|
7 |
Zara |
4 |
Same coach and rating as Player-5 |
|
8 |
Xena |
1 |
Half of Player-4's rating |
|
Coach |
Players |
Ratings |
Total |
Average |
|
Xena |
1, 3, 4, 8 |
3/6, 6/3, 2, 1 |
12 |
3 |
|
Yuki |
2, 6 |
7, 5 |
12 |
6 |
|
Zara |
5, 7 |
4, 4 |
8 |
4 |
This question
From Step 2, Zara trains exactly Players 5 and 7. The other odd players (1 and 3) are with Xena, so no other player can be with Zara.
Hence, the answer is option 1.
Question 2 · Sports and Scores Based Sets
What was the rating of Player-7?
Answer & solution
Answer: 4
Solving the set
Step 1 – Coaches who can train each player. Yuki trains only even players and Zara only odd players. Xena can train anyone.
• Player-1 (odd) and Player-4 (even) have the same coach (fact 2). Yuki cannot take Player-1 and Zara cannot take Player-4, so both are with Xena.
• Players 2, 3 and 5 have three different coaches (fact 2). Player-2 can be with Xena or Yuki; Players 3 and 5 can be with Xena or Zara. Since only one of them can be with Xena, Players 3 and 5 take Xena and Zara between them, and Player-2 goes to Yuki.
Step 2 – Players 5 and 7 have the same coach (fact 3). If Player-5 were with Xena, Player-7 would also be with Xena and Player-3 with Zara. Zara could then take no other odd player (1, 5 and 7 are with Xena), so Zara would train only one player, which is not allowed. Hence Players 5 and 7 are with Zara and Player-3 is with Xena. Zara trains exactly Players 5 and 7.
Step 3 – Ratings. There are 8 players but only 7 possible ratings, and only Players 5 and 7 share a rating. So the ratings are 1, 2, …, 7 once each, plus one repeated value v for Players 5 and 7. The total is 8 × 4 = 32 (fact 4), so 28 + v = 32 and v = 4. Player-5 = Player-7 = 4.
Step 4 – Player-2 has the highest rating, 7 (fact 5). Player-4 = 2 × Player-8 and Player-4 < Player-5 = 4, so Player-4 = 2 and Player-8 = 1.
Step 5 – Averages (fact 6). Zara's average = (4 + 4)/2 = 4, so Yuki's average = 4 + 2 = 6 and Xena's average = 6/2 = 3.
Step 6 – Yuki has Player-2 (7) and needs at least one more even player: 6 or 8. Including Player-8 (rating 1) cannot give an average of 6 (with Player-6 as well, 7 + 1 + P6 = 18 would need P6 = 10). So Yuki has Players 2 and 6, with (7 + P6)/2 = 6, i.e., Player-6 = 5. Player-8 goes to Xena.
Step 7 – Xena has Players 1, 3, 4 and 8 (4 players, more than Yuki's 2 ✓). The ratings left are 3 and 6 for Players 1 and 3. Xena's total = 3 + 6 + 2 + 1 = 12, average 3 ✓, whichever way 3 and 6 are assigned.
|
Player |
Coach |
Rating |
Reason |
|
1 |
Xena |
3 or 6 |
Same coach as Player-4 (fact 2) |
|
2 |
Yuki |
7 |
Highest rating (fact 5) |
|
3 |
Xena |
6 or 3 |
Different coach from 2 and 5 |
|
4 |
Xena |
2 |
2 × Player-8 and less than 4 (fact 7) |
|
5 |
Zara |
4 |
Repeated rating (facts 3, 4) |
|
6 |
Yuki |
5 |
Makes Yuki's average 6 |
|
7 |
Zara |
4 |
Same coach and rating as Player-5 |
|
8 |
Xena |
1 |
Half of Player-4's rating |
|
Coach |
Players |
Ratings |
Total |
Average |
|
Xena |
1, 3, 4, 8 |
3/6, 6/3, 2, 1 |
12 |
3 |
|
Yuki |
2, 6 |
7, 5 |
12 |
6 |
|
Zara |
5, 7 |
4, 4 |
8 |
4 |
This question
From Step 3, the sum of all ratings 32 = 28 + v gives the repeated rating v = 4. Player-7 shares this rating with Player-5.
Hence, the answer is 4.
Question 3 · Sports and Scores Based Sets
What was the rating of Player-6?
Answer & solution
Answer: 5
Solving the set
Step 1 – Coaches who can train each player. Yuki trains only even players and Zara only odd players. Xena can train anyone.
• Player-1 (odd) and Player-4 (even) have the same coach (fact 2). Yuki cannot take Player-1 and Zara cannot take Player-4, so both are with Xena.
• Players 2, 3 and 5 have three different coaches (fact 2). Player-2 can be with Xena or Yuki; Players 3 and 5 can be with Xena or Zara. Since only one of them can be with Xena, Players 3 and 5 take Xena and Zara between them, and Player-2 goes to Yuki.
Step 2 – Players 5 and 7 have the same coach (fact 3). If Player-5 were with Xena, Player-7 would also be with Xena and Player-3 with Zara. Zara could then take no other odd player (1, 5 and 7 are with Xena), so Zara would train only one player, which is not allowed. Hence Players 5 and 7 are with Zara and Player-3 is with Xena. Zara trains exactly Players 5 and 7.
Step 3 – Ratings. There are 8 players but only 7 possible ratings, and only Players 5 and 7 share a rating. So the ratings are 1, 2, …, 7 once each, plus one repeated value v for Players 5 and 7. The total is 8 × 4 = 32 (fact 4), so 28 + v = 32 and v = 4. Player-5 = Player-7 = 4.
Step 4 – Player-2 has the highest rating, 7 (fact 5). Player-4 = 2 × Player-8 and Player-4 < Player-5 = 4, so Player-4 = 2 and Player-8 = 1.
Step 5 – Averages (fact 6). Zara's average = (4 + 4)/2 = 4, so Yuki's average = 4 + 2 = 6 and Xena's average = 6/2 = 3.
Step 6 – Yuki has Player-2 (7) and needs at least one more even player: 6 or 8. Including Player-8 (rating 1) cannot give an average of 6 (with Player-6 as well, 7 + 1 + P6 = 18 would need P6 = 10). So Yuki has Players 2 and 6, with (7 + P6)/2 = 6, i.e., Player-6 = 5. Player-8 goes to Xena.
Step 7 – Xena has Players 1, 3, 4 and 8 (4 players, more than Yuki's 2 ✓). The ratings left are 3 and 6 for Players 1 and 3. Xena's total = 3 + 6 + 2 + 1 = 12, average 3 ✓, whichever way 3 and 6 are assigned.
|
Player |
Coach |
Rating |
Reason |
|
1 |
Xena |
3 or 6 |
Same coach as Player-4 (fact 2) |
|
2 |
Yuki |
7 |
Highest rating (fact 5) |
|
3 |
Xena |
6 or 3 |
Different coach from 2 and 5 |
|
4 |
Xena |
2 |
2 × Player-8 and less than 4 (fact 7) |
|
5 |
Zara |
4 |
Repeated rating (facts 3, 4) |
|
6 |
Yuki |
5 |
Makes Yuki's average 6 |
|
7 |
Zara |
4 |
Same coach and rating as Player-5 |
|
8 |
Xena |
1 |
Half of Player-4's rating |
|
Coach |
Players |
Ratings |
Total |
Average |
|
Xena |
1, 3, 4, 8 |
3/6, 6/3, 2, 1 |
12 |
3 |
|
Yuki |
2, 6 |
7, 5 |
12 |
6 |
|
Zara |
5, 7 |
4, 4 |
8 |
4 |
This question
From Step 6, Yuki trains Players 2 and 6 and averages 6: (7 + Player-6)/2 = 6, so Player-6 = 5.
Hence, the answer is 5.
Question 4 · Sports and Scores Based Sets
For how many players the ratings can be determined with certainty?
Answer & solution
Answer: 6
Solving the set
Step 1 – Coaches who can train each player. Yuki trains only even players and Zara only odd players. Xena can train anyone.
• Player-1 (odd) and Player-4 (even) have the same coach (fact 2). Yuki cannot take Player-1 and Zara cannot take Player-4, so both are with Xena.
• Players 2, 3 and 5 have three different coaches (fact 2). Player-2 can be with Xena or Yuki; Players 3 and 5 can be with Xena or Zara. Since only one of them can be with Xena, Players 3 and 5 take Xena and Zara between them, and Player-2 goes to Yuki.
Step 2 – Players 5 and 7 have the same coach (fact 3). If Player-5 were with Xena, Player-7 would also be with Xena and Player-3 with Zara. Zara could then take no other odd player (1, 5 and 7 are with Xena), so Zara would train only one player, which is not allowed. Hence Players 5 and 7 are with Zara and Player-3 is with Xena. Zara trains exactly Players 5 and 7.
Step 3 – Ratings. There are 8 players but only 7 possible ratings, and only Players 5 and 7 share a rating. So the ratings are 1, 2, …, 7 once each, plus one repeated value v for Players 5 and 7. The total is 8 × 4 = 32 (fact 4), so 28 + v = 32 and v = 4. Player-5 = Player-7 = 4.
Step 4 – Player-2 has the highest rating, 7 (fact 5). Player-4 = 2 × Player-8 and Player-4 < Player-5 = 4, so Player-4 = 2 and Player-8 = 1.
Step 5 – Averages (fact 6). Zara's average = (4 + 4)/2 = 4, so Yuki's average = 4 + 2 = 6 and Xena's average = 6/2 = 3.
Step 6 – Yuki has Player-2 (7) and needs at least one more even player: 6 or 8. Including Player-8 (rating 1) cannot give an average of 6 (with Player-6 as well, 7 + 1 + P6 = 18 would need P6 = 10). So Yuki has Players 2 and 6, with (7 + P6)/2 = 6, i.e., Player-6 = 5. Player-8 goes to Xena.
Step 7 – Xena has Players 1, 3, 4 and 8 (4 players, more than Yuki's 2 ✓). The ratings left are 3 and 6 for Players 1 and 3. Xena's total = 3 + 6 + 2 + 1 = 12, average 3 ✓, whichever way 3 and 6 are assigned.
|
Player |
Coach |
Rating |
Reason |
|
1 |
Xena |
3 or 6 |
Same coach as Player-4 (fact 2) |
|
2 |
Yuki |
7 |
Highest rating (fact 5) |
|
3 |
Xena |
6 or 3 |
Different coach from 2 and 5 |
|
4 |
Xena |
2 |
2 × Player-8 and less than 4 (fact 7) |
|
5 |
Zara |
4 |
Repeated rating (facts 3, 4) |
|
6 |
Yuki |
5 |
Makes Yuki's average 6 |
|
7 |
Zara |
4 |
Same coach and rating as Player-5 |
|
8 |
Xena |
1 |
Half of Player-4's rating |
|
Coach |
Players |
Ratings |
Total |
Average |
|
Xena |
1, 3, 4, 8 |
3/6, 6/3, 2, 1 |
12 |
3 |
|
Yuki |
2, 6 |
7, 5 |
12 |
6 |
|
Zara |
5, 7 |
4, 4 |
8 |
4 |
This question
From the final table, the ratings of Players 2 (7), 4 (2), 5 (4), 6 (5), 7 (4) and 8 (1) are fixed. Players 1 and 3 have 3 and 6 in some order.
So the ratings of 6 players are certain.
Hence, the answer is 6.
Question 5 · Sports and Scores Based Sets
Who all were the players trained by Xena?
- A
Player-1, Player-4, Player-6, Player-8
- B
Player-1, Player-3, Player-4, Player-6
- C
Player-1, Player-3, Player-4
- D
Player-1, Player-3, Player-4, Player-8
Answer & solution
Answer: Option D
Solving the set
Step 1 – Coaches who can train each player. Yuki trains only even players and Zara only odd players. Xena can train anyone.
• Player-1 (odd) and Player-4 (even) have the same coach (fact 2). Yuki cannot take Player-1 and Zara cannot take Player-4, so both are with Xena.
• Players 2, 3 and 5 have three different coaches (fact 2). Player-2 can be with Xena or Yuki; Players 3 and 5 can be with Xena or Zara. Since only one of them can be with Xena, Players 3 and 5 take Xena and Zara between them, and Player-2 goes to Yuki.
Step 2 – Players 5 and 7 have the same coach (fact 3). If Player-5 were with Xena, Player-7 would also be with Xena and Player-3 with Zara. Zara could then take no other odd player (1, 5 and 7 are with Xena), so Zara would train only one player, which is not allowed. Hence Players 5 and 7 are with Zara and Player-3 is with Xena. Zara trains exactly Players 5 and 7.
Step 3 – Ratings. There are 8 players but only 7 possible ratings, and only Players 5 and 7 share a rating. So the ratings are 1, 2, …, 7 once each, plus one repeated value v for Players 5 and 7. The total is 8 × 4 = 32 (fact 4), so 28 + v = 32 and v = 4. Player-5 = Player-7 = 4.
Step 4 – Player-2 has the highest rating, 7 (fact 5). Player-4 = 2 × Player-8 and Player-4 < Player-5 = 4, so Player-4 = 2 and Player-8 = 1.
Step 5 – Averages (fact 6). Zara's average = (4 + 4)/2 = 4, so Yuki's average = 4 + 2 = 6 and Xena's average = 6/2 = 3.
Step 6 – Yuki has Player-2 (7) and needs at least one more even player: 6 or 8. Including Player-8 (rating 1) cannot give an average of 6 (with Player-6 as well, 7 + 1 + P6 = 18 would need P6 = 10). So Yuki has Players 2 and 6, with (7 + P6)/2 = 6, i.e., Player-6 = 5. Player-8 goes to Xena.
Step 7 – Xena has Players 1, 3, 4 and 8 (4 players, more than Yuki's 2 ✓). The ratings left are 3 and 6 for Players 1 and 3. Xena's total = 3 + 6 + 2 + 1 = 12, average 3 ✓, whichever way 3 and 6 are assigned.
|
Player |
Coach |
Rating |
Reason |
|
1 |
Xena |
3 or 6 |
Same coach as Player-4 (fact 2) |
|
2 |
Yuki |
7 |
Highest rating (fact 5) |
|
3 |
Xena |
6 or 3 |
Different coach from 2 and 5 |
|
4 |
Xena |
2 |
2 × Player-8 and less than 4 (fact 7) |
|
5 |
Zara |
4 |
Repeated rating (facts 3, 4) |
|
6 |
Yuki |
5 |
Makes Yuki's average 6 |
|
7 |
Zara |
4 |
Same coach and rating as Player-5 |
|
8 |
Xena |
1 |
Half of Player-4's rating |
|
Coach |
Players |
Ratings |
Total |
Average |
|
Xena |
1, 3, 4, 8 |
3/6, 6/3, 2, 1 |
12 |
3 |
|
Yuki |
2, 6 |
7, 5 |
12 |
6 |
|
Zara |
5, 7 |
4, 4 |
8 |
4 |
This question
From Step 7, Xena trained Players 1, 3, 4 and 8.
Hence, the answer is option 4.
Take this paper as a timed test
Real exam timing, instant score, rank and topic-wise analysis.