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 player strained 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

Who all were the players trained by Xena?

The correct answer is Player-1, Player-3, Player-4, Player-8.

Step-by-step Explanation:

1. Analyzing the Coaches' Rules and Players
There are 8 players numbered 1 through 8, and 3 coaches: Xena (X), Yuki (Y), and Zara (Z).
- Yuki (Y) trained only even-numbered players.
- Zara (Z) trained only odd-numbered players.
- Xena (X) could train both odd and even-numbered players.
- Each coach trained at least 2 players.

2. Determining the Number of Players per Coach
Let the number of players trained by Xena, Yuki, and Zara be $n_X$, $n_Y$, and $n_Z$ respectively. Since each player was trained by exactly one coach, we have:

$n_X+n_Y+n_Z=8$

From the given information:
- Each coach trained at least 2 players ($n_X,n_Y,n_Z\ge 2$).
- Xena trained more players than Yuki ($n_X>n_Y$).

If $n_Y=2$, then $n_X$ can be 4, which leaves $n_Z=2$. This is a valid scenario that matches the correct option where Xena trains 4 players.

3. Identifying the Group of Players Trained by Each Coach
- Player-5 and Player-7 were trained by the same coach (Clue 3). Since both are odd numbers and Yuki only trains evens, they must be trained by Zara or Xena. In our setup where $n_Z=2$, Zara trained exactly these 2 players: Player-5 and Player-7.
- Player-1 and Player-4 were trained by the same coach (Clue 2). Since Player-4 is even, Zara cannot train them. Since Zara's spots are filled by Player-5 and Player-7, Player-1 and Player-4 must be trained by Xena.
- The coaches who trained Player-2, Player-3, and Player-5 were all different (Clue 2). Since Player-5 is trained by Zara, Player-2 and Player-3 must be trained by Yuki and Xena in some order. Since Player-3 is odd, Yuki cannot train them, so Player-3 is trained by Xena, which means Player-2 is trained by Yuki.
- Since Yuki trains only even players and $n_Y=2$, and Player-2 is one of them, the remaining even players are 4, 6, and 8. Since Player-4 and Player-8 are trained by Xena (as seen from the correct option), Yuki must train the remaining even player: Player-6. Thus, Yuki trained Player-2 and Player-6.
- Therefore, the remaining players, which are Player-1, Player-3, Player-4, and Player-8, were trained by Xena.

4. Verification of Ratings and Mathematical Consistency
Let us verify this assignment with the ratings:
- The average rating of all 8 players is 4, so the sum of all ratings is:

$8\times 4=32$

- Player-5 and Player-7 got the same rating, and all other players got unique ratings from the scale of 1 to 7. This means the ratings represent the set {1, 2, 3, 4, 5, 6, 7} with one value repeated. The sum of the integers from 1 to 7 is 28. To make the total sum 32, the repeated rating (Player-5 and Player-7's rating) must be:

$32-28=4$

Thus, Player-5 and Player-7 each received a rating of 4. Zara's average rating ($A_Z$) is therefore 4.
- According to Clue 6, the average rating of Yuki's players ($A_Y$) is two more than Zara's, so:

$A_Y=4+2=6$

And Yuki's average is twice that of Xena's ($A_X$):

$A_X=6÷2=3$

- Since Yuki trained 2 players, the sum of their ratings must be $2\times 6=12$. The only distinct integers from {1, 2, 3, 5, 6, 7} that sum to 12 are 5 and 7. Since Player-2 got the highest rating (Clue 5), Player-2's rating is 7 and Player-6's rating is 5.
- Xena trained 4 players, so the sum of their ratings is $4\times 3=12$. The remaining available ratings are {1, 2, 3, 6}, which sum to exactly 12.
- Player-4's rating is double of Player-8's and less than Player-5's rating of 4 (Clue 7). From the set {1, 2, 3, 6}, the only ratings that fit this condition are Player-4 having a rating of 2 and Player-8 having a rating of 1. Player-1 and Player-3 take the remaining ratings of 3 and 6.

All conditions are perfectly satisfied, confirming that the players trained by Xena are indeed Player-1, Player-3, Player-4, and Player-8.