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? Options:

The correct option is Exactly 2.

Let us analyze the information given in the problem step-by-step to understand why Zara trained exactly 2 players.

We have 8 players numbered 1 to 8. They are trained by three coaches: Xena (X), Yuki (Y), and Zara (Z).
Let us denote the set of players trained by Xena, Yuki, and Zara as $P_X$, $P_Y$, and $P_Z$ respectively. Let the number of players trained by them be $n_X$, $n_Y$, and $n_Z$ respectively. Since each player is trained by at least one coach, and we assume each player has exactly one coach (implied by ratings and assignments):
$n_X+n_Y+n_Z=8$

From the constraints, we have:
- Each coach trained at least two players: $n_X,n_Y,n_Z\ge 2$.
- Yuki (Y) trained only even-numbered players: $P_Y\subseteq \{2,4,6,8\}$.
- Zara (Z) trained only odd-numbered players: $P_Z\subseteq \{1,3,5,7\}$.
- Therefore, Xena (X) can train both even and odd players.
- Information 1: Xena trained more players than Yuki: $n_X>n_Y$.
- Information 2: Player-1 and Player-4 were trained by the same coach. Since Player-4 is even, Zara cannot train them. Since Player-1 is odd, Yuki cannot train them. Thus, Player-1 and Player-4 must be trained by Xena. (So, $\{1,4\}\subseteq P_X$).
- The coaches who trained Player-2, Player-3, and Player-5 were all different. Since Player-2 is even, Yuki or Xena must train them. Since Player-3 and Player-5 are odd, Zara or Xena must train them.

Let's check the number of players for each coach. Since $n_X+n_Y+n_Z=8$ and each is at least 2, and $n_X>n_Y$, the possible distributions $(n_X,n_Y,n_Z)$ are:
Case A: $(4,2,2)$
Case B: $(3,2,3)$
Case C: $(4,3,1)$ (not possible since $n_Z\ge 2$)

Let us analyze the average ratings in Information 6:
Let $A_X$, $A_Y$, and $A_Z$ be the average ratings of players trained by Xena, Yuki, and Zara, respectively. We are given:
$A_Y=2A_X$
$A_Y=A_Z+2$

Since the overall average rating of all 8 players is 4 (Information 4), the sum of all ratings is:
$8\times 4=32$
Also, the sum of ratings is:
$n_X{A}_X+n_Y{A}_Y+n_Z{A}_Z=32$

Let us substitute $A_X=\frac{A_Y}{2}$ and $A_Z=A_Y-2$ into the sum equation:
$n_X\left(\frac{A_Y}{2}\right)+n_Y{A}_Y+n_Z(A_Y-2)=32$
$A_Y(\frac{n_X}{2}+n_Y+n_Z)-2n_Z=32$

Let's evaluate this for both possible cases of player counts:

If Case B is true: $(n_X,n_Y,n_Z)=(3,2,3)$
$A_Y(1.5+2+3)-2\left(3\right)=32$
$6.5A_Y-6=32\Rightarrow 6.5A_Y=38\Rightarrow A_Y=\frac{76}{13}\approx 5.85$
Since Yuki only trained 2 players ($n_Y=2$), and all ratings are integers (1 to 7), the sum of Yuki's players' ratings must be an integer. Thus, $2\times A_Y$ must be an integer. But , which is not an integer. Therefore, Case B is mathematically impossible.

If Case A is true: $(n_X,n_Y,n_Z)=(4,2,2)$
$A_Y(2+2+2)-2\left(2\right)=32$
$6A_Y-4=32\Rightarrow 6A_Y=36\Rightarrow A_Y=6$
This yields a perfectly consistent integer average rating:
- Average for Yuki ($A_Y$) = 6
- Average for Xena ($A_X$) = 3
- Average for Zara ($A_Z$) = 4

Thus, the number of players coached by Zara ($n_Z$) must be exactly 2.