CAT 2024 Slot 2 — DILR Question 18
Answer the following questions based on the information given below.
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.
What best can be concluded about the number of players coached by Zara?
Answer & solution
- A
Either 2 or 3
- B
Exactly 3
- C
Either 2 or 3 or 4
Exactly 2
Hard
Fix the coaching assignment first. Since P1 (odd) and P4 (even) share a coach, only Xena can take both. P2, P3, P5 need three different coaches, forcing P2→Yuki. Then Zara is squeezed to exactly two players.
Yuki trains only even players, Zara only odd; each coach ≥ 2 players; Xena > Yuki in count. Solved assignment:
| Coach | Players | Players' ratings |
|---|---|---|
| Xena | P1, P3, P4, P8 | {3,6}, {6,3}, 2, 1 (sum 12) |
| Yuki | P2, P6 | 7, 5 (sum 12) |
| Zara | P5, P7 | 4, 4 (sum 8) |
P1, P4 → Xena; P2 → Yuki. P1 is odd and P4 even, so only Xena can train both (Clue 2). Among P2, P3, P5 needing different coaches, only the even P2 can be Yuki.
Zara gets exactly P5 and P7. If P5→Xena then (Clue 3) P7→Xena too, leaving Zara only P3 — under 2. So P5→Zara, P3→Xena, and P7→Zara. No other odd player is left for Zara (1, 3 taken by Xena).