Solution
In the set we are given that each coach had atleast 2 players.
So, there are only two possibilities:- 3+3+2=8 or 4+2+2=8.
Also, According to clue 1,
Xena has trained more players than Yuki
therefore two cases can be made as follows -
| Xena | Yuki | Zara |
|---|---|---|
| 3 | 2 | 3 |
| 4 | 2 | 2 |
| Player | XENA | YUKI | ZARA | Rating |
|---|---|---|---|---|
| Player 1 | x | |||
| Player 2 | x | |||
| Player 3 | x | |||
| Player 4 | x | |||
| Player 5 | x | |||
| Player 6 | x | |||
| Player 7 | x | |||
| Player 8 | x |
To fill this table we will need to find the coach who had trained a particular player and the rating given by him.
Also, we know from the question, that
Yuki trained only Even numbered players
and Zara trained only odd numbered players.
| Player | XENA | YUKI | ZARA | Rating |
|---|---|---|---|---|
| Player 1 | ✓ | x | x | |
| Player 2 | x | ✓ | x | |
| Player 3 | x | |||
| Player 4 | ✓ | x | x | |
| Player 5 | x | |||
| Player 6 | x | |||
| Player 7 | x | |||
| Player 8 | x |
Using clue 2 -
Player 1 and player 4 was trained by same coach, which can only be Xena.
Also, The coaches who trained Player-2, Player-3 and Player-5 are all different.
But player 3 and player 5 can't be trained by Yuki
hence, player 2 must be trained by yuki.
| Player | XENA | YUKI | ZARA | Rating |
|---|---|---|---|---|
| Player 1 | ✓ | x | x | |
| Player 2 | x | ✓ | x | |
| Player 3 | ✓ | x | x | |
| Player 4 | ✓ | x | x | |
| Player 5 | x | x | ✓ | |
| Player 6 | x | |||
| Player 7 | x | x | ✓ | |
| Player 8 | x |
Now using Clue 3 ,
We know player 5 and player 7 was trained by same coach.
But, we already know that Player 3 and Player 5 cannot be trained by the same Person.
If for example, player 5 and 7 were trained by Xena then player 3 will be trained by Zara,
but it will contradict first clue that all the coaches had trained atleast 2 players.
Therefore, only possiblity is player 5 and 7 are trianed by Zara and player 3 then will be trained by Xena.
Therefore, the number of players trained by the coaches are fixed.
Xena - 4
Yuki - 2
Zara - 2
| Player | XENA | YUKI | ZARA | Rating |
|---|---|---|---|---|
| Player 1 | ✓ | x | x | |
| Player 2 | x | ✓ | x | |
| Player 3 | ✓ | x | x | |
| Player 4 | ✓ | x | x | |
| Player 5 | x | x | ✓ | 4 |
| Player 6 | x | |||
| Player 7 | x | x | ✓ | 4 |
| Player 8 | x |
As given in clue 3 &4 ,
Player 5 and Player 7 got trained by the same coach and both got the same ratings. All other players got a unique rating.
And, The average of the ratings of all the players is 4. Total ratings is 8*4 = 32.
The sum of ratings will be 1 + 2 + 3 + 4 + 5 + 6 + 7 + x = 32.
Therefore x = 4.
Therefore Player 5 and Player 7 got a rating of 4.
| Player | XENA | YUKI | ZARA | Rating |
|---|---|---|---|---|
| Player 1 | ✓ | x | x | |
| Player 2 | x | ✓ | x | 7 |
| Player 3 | ✓ | x | x | |
| Player 4 | ✓ | x | x | 2 |
| Player 5 | x | x | ✓ | 4 |
| Player 6 | x | |||
| Player 7 | x | x | ✓ | 4 |
| Player 8 | x | 1 |
Now, using clue 5,
Player 2 got the highest rating which is 7.
Also clue 7 says, Player 4 rating was double of Player 8 and less than Player 5.
Player 4 should get an even number rating less than 4 which is 2.
Player 4 rating is 2 and Player 8 rating is 1.
| Player | XENA | YUKI | ZARA | Rating |
|---|---|---|---|---|
| Player 1 | ✓ | x | x | |
| Player 2 | x | ✓ | x | 7 |
| Player 3 | ✓ | x | x | |
| Player 4 | ✓ | x | x | 2 |
| Player 5 | x | x | ✓ | 4 |
| Player 6 | x | |||
| Player 7 | x | x | ✓ | 4 |
| Player 8 | x | 1 |
Using clue 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.
We know that players trained by zara are 5 and 7 who has a rating of 4 each.
Therefore, average of ratings given by Zara will be 4.
So, average of rating given by Yuki = 4 + 2 = 6
similarly, Average of ratings given by Xena = 6/2 = 3.
| Player | XENA | YUKI | ZARA | Rating |
|---|---|---|---|---|
| Player 1 | ✓ | x | x | |
| Player 2 | x | ✓ | x | 7 |
| Player 3 | ✓ | x | x | |
| Player 4 | ✓ | x | x | 2 |
| Player 5 | x | x | ✓ | 4 |
| Player 6 | x | ✓ | x | 5 |
| Player 7 | x | x | ✓ | 4 |
| Player 8 | ✓ | x | x | 1 |
We know, Average of ratings given by Yuki to two players is 6 and One of them is Player 2 whose rating is 7.
Therefore, the other player should have got rating 5.
which has to be of one among player 6 or 8.
But player 8 rating is 1,
Hence Player 6’s rating is 5.
Since Player 6 is trained by Yuki, Player 8 has to be trainer by Xena.
| Player | XENA | YUKI | ZARA | Rating |
|---|---|---|---|---|
| Player 1 | ✓ | x | x | 6/3 |
| Player 2 | x | ✓ | x | 7 |
| Player 3 | ✓ | x | x | 3/6 |
| Player 4 | ✓ | x | x | 2 |
| Player 5 | x | x | ✓ | 4 |
| Player 6 | x | ✓ | x | 5 |
| Player 7 | x | x | ✓ | 4 |
| Player 8 | ✓ | x | x | 1 |
Here, we can't determine the exact value of rating given to player 1 and player 3 but can conclude that it will be one among 3 or 6.
In DILR sets many times you will find more than one cases, but that's completely alright, the questions of the set will give you more details, if needed.
The final table will be as follows:-
| Player | XENA | YUKI | ZARA | Rating |
|---|---|---|---|---|
| Player 1 | ✓ | x | x | 6/3 |
| Player 2 | x | ✓ | x | 7 |
| Player 3 | ✓ | x | x | 3/6 |
| Player 4 | ✓ | x | x | 2 |
| Player 5 | x | x | ✓ | 4 |
| Player 6 | x | ✓ | x | 5 |
| Player 7 | x | x | ✓ | 4 |
| Player 8 | ✓ | x | x | 1 |
The rating of player 6 is 5.