CAT 2024 Slot 2 LRDI Question Paper

We are given that the numbers keep increasing from left to right (clue 1), and the number keeps decreasing from top to bottom (clue 2)

The key takeaway from this is that 10 must be placed in Row 1, column 4, as placing it anywhere else would mean that the number above it, or right to it, must be greater than 10, which is not an option. 

Clue 3 says that one is either in the same row or the same column as 10. 
The same logic that we used for 10 applies for 1; it must be either in Row 1, column 1, or Row 4, column 4; we don't know which one yet. 

We are given that 2 and 3 are not in the same column or row as 10, meaning that they must occupy two spots from (Row 2, Col. 2), (Row 2, Col. 3), and (Row 3, Col. 3)

2 and 3 must be present in Row 2, column 2 or Row 3, column 3, as there is no number smaller than it to be in the cell left to it. 

Clue 6 says that 4 and 6 are in the same row; this can be rows 1, 2, or 3. 

Clue 5 is a good starting point. 
Once we found the position of 10, the only positions possible for 9 are R1C3 or R2C4

If 9 is placed in R2C4, 7 and 8 must be placed in row 1. 
Clue 6 says that 4 and 6 are in the same row, but with 7 and 8 in row 1, no rows are left with two spaces. Hence, 9 can not be in R2C4, and the arrangement must be:

7 and 8 must be in column 4, occupying R3C4 and R2C4, respectively. 

Hence, 4, 6 must be in row 1:

The lowest sum is of the 4th row. 

Therefore,  is the correct answer. 

We are given that the numbers keep increasing from left to right (clue 1), and the number keeps decreasing from top to bottom (clue 2)

The key takeaway from this is that 10 must be placed in Row 1, column 4, as placing it anywhere else would mean that the number above it, or right to it, must be greater than 10, which is not an option. 

Clue 3 says that one is either in the same row or the same column as 10. 
The same logic that we used for 10 applies for 1; it must be either in Row 1, column 1, or Row 4, column 4; we don't know which one yet. 

We are given that 2 and 3 are not in the same column or row as 10, meaning that they must occupy two spots from (Row 2, Col. 2), (Row 2, Col. 3), and (Row 3, Col. 3)

2 and 3 must be present in Row 2, column 2 or Row 3, column 3, as there is no number smaller than it to be in the cell left to it. 

Clue 6 says that 4 and 6 are in the same row; this can be rows 1, 2, or 3. 

Clue 5 is a good starting point. 
Once we found the position of 10, the only positions possible for 9 are R1C3 or R2C4

If 9 is placed in R2C4, 7 and 8 must be placed in row 1. 
Clue 6 says that 4 and 6 are in the same row, but with 7 and 8 in row 1, no rows are left with two spaces. Hence, 9 can not be in R2C4, and the arrangement must be:

7 and 8 must be in column 4, occupying R3C4 and R2C4, respectively. 

Hence, 4, 6 must be in row 1:

We can see that the positions 10 and 1 are fixed and agree with the given statements. 
Both the statements are true. 
Therefore, Option A is the correct answer. 

We are given that the numbers keep increasing from left to right (clue 1), and the number keeps decreasing from top to bottom (clue 2)

The key takeaway from this is that 10 must be placed in Row 1, column 4, as placing it anywhere else would mean that the number above it, or right to it, must be greater than 10, which is not an option. 

Clue 3 says that one is either in the same row or the same column as 10. 
The same logic that we used for 10 applies for 1; it must be either in Row 1, column 1, or Row 4, column 4; we don't know which one yet. 

We are given that 2 and 3 are not in the same column or row as 10, meaning that they must occupy two spots from (Row 2, Col. 2), (Row 2, Col. 3), and (Row 3, Col. 3)

2 and 3 must be present in Row 2, column 2 or Row 3, column 3, as there is no number smaller than it to be in the cell left to it. 

Clue 6 says that 4 and 6 are in the same row; this can be rows 1, 2, or 3. 

Clue 5 is a good starting point. 
Once we found the position of 10, the only positions possible for 9 are R1C3 or R2C4

If 9 is placed in R2C4, 7 and 8 must be placed in row 1. 
Clue 6 says that 4 and 6 are in the same row, but with 7 and 8 in row 1, no rows are left with two spaces. Hence, 9 can not be in R2C4, and the arrangement must be:

7 and 8 must be in column 4, occupying R3C4 and R2C4, respectively. 

Hence, 4, 6 must be in row 1:

Positions 2 and 3 can be switched. 
Either of them can be in column 2 or 3 

Neither of the statements is necessarily true. 
Therefore, Option C is the correct answer. 

We are given that the numbers keep increasing from left to right (clue 1), and the number keeps decreasing from top to bottom (clue 2)

The key takeaway from this is that 10 must be placed in Row 1, column 4, as placing it anywhere else would mean that the number above it, or right to it, must be greater than 10, which is not an option. 

Clue 3 says that one is either in the same row or the same column as 10. 
The same logic that we used for 10 applies for 1; it must be either in Row 1, column 1, or Row 4, column 4; we don't know which one yet. 

We are given that 2 and 3 are not in the same column or row as 10, meaning that they must occupy two spots from (Row 2, Col. 2), (Row 2, Col. 3), and (Row 3, Col. 3)

2 and 3 must be present in Row 2, column 2 or Row 3, column 3, as there is no number smaller than it to be in the cell left to it. 

Clue 6 says that 4 and 6 are in the same row; this can be rows 1, 2, or 3. 

Clue 5 is a good starting point. 
Once we found the position of 10, the only positions possible for 9 are R1C3 or R2C4

If 9 is placed in R2C4, 7 and 8 must be placed in row 1. 
Clue 6 says that 4 and 6 are in the same row, but with 7 and 8 in row 1, no rows are left with two spaces. Hence, 9 can not be in R2C4, and the arrangement must be:

7 and 8 must be in column 4, occupying R3C4 and R2C4, respectively. 

Hence, 4, 6 must be in row 1:

Placements of all the numbers except 2 and 3 can be determined. 

Therefore, 2 is the correct answer. 

We are given that the numbers keep increasing from left to right (clue 1), and the number keeps decreasing from top to bottom (clue 2)

The key takeaway from this is that 10 must be placed in Row 1, column 4, as placing it anywhere else would mean that the number above it, or right to it, must be greater than 10, which is not an option. 

Clue 3 says that one is either in the same row or the same column as 10. 
The same logic that we used for 10 applies for 1; it must be either in Row 1, column 1, or Row 4, column 4; we don't know which one yet. 

We are given that 2 and 3 are not in the same column or row as 10, meaning that they must occupy two spots from (Row 2, Col. 2), (Row 2, Col. 3), and (Row 3, Col. 3)

2 and 3 must be present in Row 2, column 2 or Row 3, column 3, as there is no number smaller than it to be in the cell left to it. 

Clue 6 says that 4 and 6 are in the same row; this can be rows 1, 2, or 3. 

Clue 5 is a good starting point. 
Once we found the position of 10, the only positions possible for 9 are R1C3 or R2C4

If 9 is placed in R2C4, 7 and 8 must be placed in row 1. 
Clue 6 says that 4 and 6 are in the same row, but with 7 and 8 in row 1, no rows are left with two spaces. Hence, 9 can not be in R2C4, and the arrangement must be:

7 and 8 must be in column 4, occupying R3C4 and R2C4, respectively. 

Hence, 4, 6 must be in row 1:

The sum of numbers in the 4th column is 10+8+7+1 = 26

Therefore, 26 is the correct answer. 

The first thing to realise here is the lengths of the paths.

KN should be equal to LM, giving the length of KO using Pythagoras theorem as 500m 

Similarly, the length of HJ=OP=150m and length of GJ=EL=200m, giving the length of HG as 250 m

The shortest path from L to B and then to P (or the other way around would involve) using these hypotenuses as much as possible instead of the two adjacent sides. The shortest can be visualised as shown below or multitude of others variations, as there are multiple ways that would make one travel the shortest distance)

The below figure is the simplest one for visualisation. 

Other possible paths are L-E-F-C-.. and following the same path. 

The shortest distance in each of these instance would be LE+ED+DC+CB+BG+GI+IP+PO+OK+KL

Which would be 200+400+300+300+400+250+400+150+500+300 = 3200

Therefore, Option B is the correct answer. 

Since we can only walk along the side of lakes, that drastically reduces the paths we can take. 
The diagram below shows the path with the maximum distance travelled. 

The path is CD-DE-EF-FK-KL-LM-MN-NK-KJ-JG-GF-FC

(the reverse of this path is also valid)

We can either manually add the lengths or use shorter methods to note that in the path we travel walkways of length 400 m 4 times (DE, LM, KN, EF), walkways of length 300m 6 times (CD, EF, KL, MN, KJ, GF) and walkways of length 200 m 2 times (FK, JG)
Giving the total length to be 1600+1800+400 = 3800

Therefore, Option C is the correct answer. 

Counter to the first question, we should minimize the use of those hypotenuse walkways as they reduce the distance we travel.

The longest possible route would include then travelling through the edges of the rectangles and sqaures, and not touch any point more than once. 
Best case scenario would be us touching each point exactly once. 

This can be visualized as:

Where we do not use any of the hypotenuse and cross through each gate exactly once. 

The total distance would be:

AH+HI+ IP = 400+200+400 = 1000

DE+EL+LM = 400+200+400 = 1000

PO+JK+NM = 150+300+300 = 750

DC+FG+BA = 150+300+300 = 750

JO+KN = 400+400 = 800

CF+GB = 400+400 = 800

Adding up to: 5100m

Therefore, 5100 is the correct answer. 

Similar to the previous question, we should try to avoid the hypotenuse

The total path distance would be 300(CD) + 400(DE) + 300(EF) + 200(FK) + 400(KN) + 300(NO) + 150(OP) + 400(PI) + 150(IJ) + 200(JG) + 400(GB) + 300(BC)

Adding up to 3,500 meters

Therefore, 3500 is the correct answer. 

From the given chart, we can find the average rating on day 2:

$$\frac{\left(5\times\ 1\right)+\left(10\times\ 2\right)+\left(5\times\ 3\right)+\left(20\times\ 4\right)+\left(10\times\ 5\right)}{5+10+5+20+10}=\frac{170}{50}=3.4$$

We are given the cumulative average of day 1 and day 2 as 3.1, and the average at the end of day 1 is 3. 

Let's take the number of ratings received on day 1 as $$x$$; using this overall average and the average on day 2, we get the equation:

$$\frac{3x+\left(50\times\ 3.4\right)}{x+50}=3.1$$
$$3x+170=3.1x+155$$
$$15\ =\ 0.1x$$
$$x=150$$

Therefore, the number of ratings received on day 1 is 150. 

We are given that on day 3, a total of 100 ratings came in

The modes were 4 and 5, meaning that an equal number of 4 and 5 ratings came in; let it be 10b (since we are given that all ratings were non-zero multiples of 10)

Let's take the number of 1 and 2 ratings as 10a each, giving the number of 3 ratings as 20a

Adding all of these up: 10a+10a+20a+10b+10b = 100
40a+20b = 100
2a+b = 5

The only integer combination for a and b, without them being zero, is a being 1 and b being 3 or a=2 and b=1

But taking b as 1 would make 3 as the mode rating, so we can not consider the latter case. 

We are giving 10 - 1 ratings, 10 - 2 ratings, 20 - 3 ratings, 30 - 4 ratings, and 30 - 5 ratings. 

The average would be $$\frac{\left(10\times\ 1\right)+\left(10\times\ 2\right)+\left(20\times\ 3\right)+\left(30\times\ 4\right)+\left(30\times\ 5\right)}{100}=\frac{360}{100}=3.6$$

Therefore, Option A is the correct answer. 

As deduced in the previous question, there were 10 1 ratings, 10 2 ratings, 20 3 ratings, 30 4 ratings, and 30 5 ratings. 

The median would be the 50th and 51st ratings' average when arranged in ascending/descending order.
Both of which would be 4

Therefore, 4 is the correct answer 

The cumulative average rating at the end of day 3 would be $$\frac{\left(3.1\times\ 200\right)+\left(3.6\times\ 100\right)}{200+100}$$

[the cumulative average rating on day 2= 3.1
Number of ratings received by day 2= 200]

$$\frac{360+620}{300}=\frac{980}{300}=3.266$$

The increase in the cumulative average from day 2 to day 3 can be calculated as $$\frac{3.266-3.1}{3.1}\times\ 100\approx\ 5.34$$

Which aligns with the statement given in option C. 

Therefore, Option C is the correct answer. 

Since we are given that the annual growth rate should be taken as constant, we can simply compare the ratio changes in the firm's PAT in 2019 and 2023 and compare them. 

Firm A: 3000 to 3900; an increase of 900, giving the growth rate to be $$\frac{900}{3000}=\frac{3}{10}=0.3$$

Firm C: 2400 to 3000; an increase of 600, giving the growth rate to be $$\frac{600}{2400}=\frac{1}{4}=0.25$$

Firm E: 2400 to 3500; an increase of 1100, giving the growth rate to be $$\frac{1100}{2400}=\frac{11}{24}\approx\ 0.45$$

Firm B: 2800 to 3800; an increase of 1000, giving the growth to be $$\frac{1000}{2800}=\frac{5}{14}\approx\ 0.35$$

We can see that of the given ratios, the growth rate is highest for firm E (closest to 0.5). Therefore, the highest ARG, among the given options, will be for firm E. 

Therefore, Option C is the correct answer. 

This can be a little difficult to solve since the bubbles are not closed within a fixed boundary, but in such instances, we should approximate the most visible cases.

Like the bubble of C in 2019 is almost 3 units vertically long in diameter, while that in 2023 is 2 units long. 

The ratio of the amount spend on R&D on 2019 to 2024 would be given by: $$\frac{2400\pi\left(1.5\right)^2\ }{3000\pi\ \left(1\right)^2}=\frac{8}{10}\times\ \frac{9}{4}=9:5$$

Which is option D. 

The Pat per employee for the firm in 2023 will be:

Firm E:$$\frac{3500}{1400}=\frac{35}{14}=\frac{5}{2}$$

Firm A: $$\frac{3900}{1300}=3$$

Firm F: $$\frac{3200}{1000}=3.2$$

Firm C: $$\frac{3000}{800}=\frac{15}{4}=3.75$$

The value if maximum for firm C, therefore, Option D is the correct answer. 

We calculated the PAT per employee for F, C and E in the last question, using that information and calculating the PAT per employee for firm D to be $$\frac{2400}{800}=3$$

For firms F, C and D, we can see that the area is the same hence the only factor to consider is PAT per employee
F is 3.2, C is 3.75, and D is 3

This leaves D as the minimum. 

Comparing E and D (taking the proportionality constant as k)

E has PAT per employee as 2.5 with a radius of 1.5 units, the PRD value per employee would be $$\frac{5}{2}k\pi\ \left(\frac{3}{2}\right)^2=k\pi\ \times\ \frac{5\times\ 9}{8}$$

D has PAT per employee as 3, with a radius of 1 unit.the PRD value per employee would be $$3k\pi\ \left(1\right)^2=3k\pi\ $$

We can see that the value would be greater for E, therefore firm D will have the lowest R&D spending per employee in 2023

Therefore, Option B is the correct answer. 

We are told that each coach had at least two players. Clue 1 says that Xena trained more people than Yuki. 
It must be the case that Yuki trained only two people. If Yuki had trained three, then Xena would have trained at least four players, leaving only one for Zara. 

Hence, Yuki trained two people. 

Coming to the scores given to the players themselves:

We are given that only 5 and 7 received a sam rating; everyone else received a distinct rating. 
We are also given their average to be 4 (clue 4), giving the sum of all the scores they got to be 32
The sum of all numbers from 1 to 7 would be $$\frac{8}{2}\times\ 7=28$$, hence the repeated score must be 32-28 = 4

Thus, the score of 5 and 7 was 4

Clue 5 says that player 2 got the highest score, which is 7

Clue 7 says that player 4 got a score double that of player 8 but less than player 5; the only possibility is player 4 getting 2 and player 8 getting 1. 

Now, considering clues 2, 3 and 6:

We are given the same coach trained in 5 and 7. And 2, 3, and 5 were trained by different coaches, with 1 and 4 being trained by the same coach. 
Cheer 6 informs us about the average number of players the coaches train. 

We know that Yuki trained only two players. 
Let's take the average score of Yuki's players to be x

The average of Xena's players would be x/2, and that of Zara's players would be x-2

The total score of Yuki's players would be 2x. 
The total score of Xena's players would be 3x/2 or 2x. 
The total score of Zara's players would be 3x-6 or 2x-4

We have two cases to consider:

Let's say Xena and Zara had three players each. 
The total score of Yuki's players would be 2x.
The total score of Xena's players would be 3x/2.
The total score of Zara's players would be 3x-6. 

The sum of all these scores would be $$\frac{13x}{2}-6$$ which should be equal to 32
This would give us the value of x as $$\frac{13x}{2}=38$$
Which would give a non-integral value of 2s, that is, the sum of Yuki's player's score. 

Hence, this must not be the case. 

The other possibility:

The total score of Yuki's players would be 2x.
The total score of Xena's players would be 2x.
The total score of Zara's players would be 2x-4.

The sum of all these scores would be 6x-4, which should be equal to 32
This would give the value of x as 6

Hence, the sum of all of Yuki's players would be 12
The sum of all of Xena's players would be 12
The sum of all of Zara's players would be 8

Yuki has only two players whose scores add up to 12; the only combination possible is scores 7+5, where seven were scored by player 2. Hence, player two must be under Yuki. 

Zara got a total of 8 scores, with 7 and 5 gone. The combinations that could get this score are 2+6 and 4+4
The score of 2 is obtained by player 4, which must come with player 1
It is possible that 1 could have gotten a score 6
But then we run into a contradiction: players 3 and 5 would end up under the same coach. 

Hence, Zara must have gotten 8 through 4+4 with players 5 and 7. 

All the remaining scores must be with Xena, adding up to 12, which is the case. 

4 and 1 must be present together, and 3 must be present in Xena as well, giving us the arrangement. 

We can see that Zara had two players. 

Therefore, Option B is the correct answer. 

We are told that each coach had at least two players. Clue 1 says that Xena trained more people than Yuki. 
It must be the case that Yuki trained only two people. If Yuki had trained three, then Xena would have trained at least four players, leaving only one for Zara. 

Hence, Yuki trained two people. 

Coming to the scores given to the players themselves:

We are given that only 5 and 7 received a sam rating; everyone else received a distinct rating. 
We are also given their average to be 4 (clue 4), giving the sum of all the scores they got to be 32
The sum of all numbers from 1 to 7 would be $$\frac{8}{2}\times\ 7=28$$, hence the repeated score must be 32-28 = 4

Thus, the score of 5 and 7 was 4

Clue 5 says that player 2 got the highest score, which is 7

Clue 7 says that player 4 got a score double that of player 8 but less than player 5; the only possibility is player 4 getting 2 and player 8 getting 1. 

Now, considering clues 2, 3 and 6:

We are given the same coach trained in 5 and 7. And 2, 3, and 5 were trained by different coaches, with 1 and 4 being trained by the same coach. 
Cheer 6 informs us about the average number of players the coaches train. 

We know that Yuki trained only two players. 
Let's take the average score of Yuki's players to be x

The average of Xena's players would be x/2, and that of Zara's players would be x-2

The total score of Yuki's players would be 2x. 
The total score of Xena's players would be 3x/2 or 2x. 
The total score of Zara's players would be 3x-6 or 2x-4

We have two cases to consider:

Let's say Xena and Zara had three players each. 
The total score of Yuki's players would be 2x.
The total score of Xena's players would be 3x/2.
The total score of Zara's players would be 3x-6. 

The sum of all these scores would be $$\frac{13x}{2}-6$$ which should be equal to 32
This would give us the value of x as $$\frac{13x}{2}=38$$
Which would give a non-integral value of 2s, that is, the sum of Yuki's player's score. 

Hence, this must not be the case. 

The other possibility:

The total score of Yuki's players would be 2x.
The total score of Xena's players would be 2x.
The total score of Zara's players would be 2x-4.

The sum of all these scores would be 6x-4, which should be equal to 32
This would give the value of x as 6

Hence, the sum of all of Yuki's players would be 12
The sum of all of Xena's players would be 12
The sum of all of Zara's players would be 8

Yuki has only two players whose scores add up to 12; the only combination possible is scores 7+5, where seven were scored by player 2. Hence, player two must be under Yuki. 

Zara got a total of 8 scores, with 7 and 5 gone. The combinations that could get this score are 2+6 and 4+4
The score of 2 is obtained by player 4, which must come with player 1
It is possible that 1 could have gotten a score 6
But then we run into a contradiction: players 3 and 5 would end up under the same coach. 

Hence, Zara must have gotten 8 through 4+4 with players 5 and 7. 

All the remaining scores must be with Xena, adding up to 12, which is the case. 

4 and 1 must be present together, and 3 must be present in Xena as well, giving us the arrangement. 

Score of player 7 was 4

Therefore, 4 is the correct answer. 

We are told that each coach had at least two players. Clue 1 says that Xena trained more people than Yuki. 
It must be the case that Yuki trained only two people. If Yuki had trained three, then Xena would have trained at least four players, leaving only one for Zara. 

Hence, Yuki trained two people. 

Coming to the scores given to the players themselves:

We are given that only 5 and 7 received a sam rating; everyone else received a distinct rating. 
We are also given their average to be 4 (clue 4), giving the sum of all the scores they got to be 32
The sum of all numbers from 1 to 7 would be $$\frac{8}{2}\times\ 7=28$$, hence the repeated score must be 32-28 = 4

Thus, the score of 5 and 7 was 4

Clue 5 says that player 2 got the highest score, which is 7

Clue 7 says that player 4 got a score double that of player 8 but less than player 5; the only possibility is player 4 getting 2 and player 8 getting 1. 

Now, considering clues 2, 3 and 6:

We are given the same coach trained in 5 and 7. And 2, 3, and 5 were trained by different coaches, with 1 and 4 being trained by the same coach. 
Cheer 6 informs us about the average number of players the coaches train. 

We know that Yuki trained only two players. 
Let's take the average score of Yuki's players to be x

The average of Xena's players would be x/2, and that of Zara's players would be x-2

The total score of Yuki's players would be 2x. 
The total score of Xena's players would be 3x/2 or 2x. 
The total score of Zara's players would be 3x-6 or 2x-4

We have two cases to consider:

Let's say Xena and Zara had three players each. 
The total score of Yuki's players would be 2x.
The total score of Xena's players would be 3x/2.
The total score of Zara's players would be 3x-6. 

The sum of all these scores would be $$\frac{13x}{2}-6$$ which should be equal to 32
This would give us the value of x as $$\frac{13x}{2}=38$$
Which would give a non-integral value of 2s, that is, the sum of Yuki's player's score. 

Hence, this must not be the case. 

The other possibility:

The total score of Yuki's players would be 2x.
The total score of Xena's players would be 2x.
The total score of Zara's players would be 2x-4.

The sum of all these scores would be 6x-4, which should be equal to 32
This would give the value of x as 6

Hence, the sum of all of Yuki's players would be 12
The sum of all of Xena's players would be 12
The sum of all of Zara's players would be 8

Yuki has only two players whose scores add up to 12; the only combination possible is scores 7+5, where seven were scored by player 2. Hence, player two must be under Yuki. 

Zara got a total of 8 scores, with 7 and 5 gone. The combinations that could get this score are 2+6 and 4+4
The score of 2 is obtained by player 4, which must come with player 1
It is possible that 1 could have gotten a score 6
But then we run into a contradiction: players 3 and 5 would end up under the same coach. 

Hence, Zara must have gotten 8 through 4+4 with players 5 and 7. 

All the remaining scores must be with Xena, adding up to 12, which is the case. 

4 and 1 must be present together, and 3 must be present in Xena as well, giving us the arrangement. 

The rating of player 6 was 5

Therefore, 5 is the correct answer. 

We are told that each coach had at least two players. Clue 1 says that Xena trained more people than Yuki. 
It must be the case that Yuki trained only two people. If Yuki had trained three, then Xena would have trained at least four players, leaving only one for Zara. 

Hence, Yuki trained two people. 

Coming to the scores given to the players themselves:

We are given that only 5 and 7 received a sam rating; everyone else received a distinct rating. 
We are also given their average to be 4 (clue 4), giving the sum of all the scores they got to be 32
The sum of all numbers from 1 to 7 would be $$\frac{8}{2}\times\ 7=28$$, hence the repeated score must be 32-28 = 4

Thus, the score of 5 and 7 was 4

Clue 5 says that player 2 got the highest score, which is 7

Clue 7 says that player 4 got a score double that of player 8 but less than player 5; the only possibility is player 4 getting 2 and player 8 getting 1. 

Now, considering clues 2, 3 and 6:

We are given the same coach trained in 5 and 7. And 2, 3, and 5 were trained by different coaches, with 1 and 4 being trained by the same coach. 
Cheer 6 informs us about the average number of players the coaches train. 

We know that Yuki trained only two players. 
Let's take the average score of Yuki's players to be x

The average of Xena's players would be x/2, and that of Zara's players would be x-2

The total score of Yuki's players would be 2x. 
The total score of Xena's players would be 3x/2 or 2x. 
The total score of Zara's players would be 3x-6 or 2x-4

We have two cases to consider:

Let's say Xena and Zara had three players each. 
The total score of Yuki's players would be 2x.
The total score of Xena's players would be 3x/2.
The total score of Zara's players would be 3x-6. 

The sum of all these scores would be $$\frac{13x}{2}-6$$ which should be equal to 32
This would give us the value of x as $$\frac{13x}{2}=38$$
Which would give a non-integral value of 2s, that is, the sum of Yuki's player's score. 

Hence, this must not be the case. 

The other possibility:

The total score of Yuki's players would be 2x.
The total score of Xena's players would be 2x.
The total score of Zara's players would be 2x-4.

The sum of all these scores would be 6x-4, which should be equal to 32
This would give the value of x as 6

Hence, the sum of all of Yuki's players would be 12
The sum of all of Xena's players would be 12
The sum of all of Zara's players would be 8

Yuki has only two players whose scores add up to 12; the only combination possible is scores 7+5, where seven were scored by player 2. Hence, player two must be under Yuki. 

Zara got a total of 8 scores, with 7 and 5 gone. The combinations that could get this score are 2+6 and 4+4
The score of 2 is obtained by player 4, which must come with player 1
It is possible that 1 could have gotten a score 6
But then we run into a contradiction: players 3 and 5 would end up under the same coach. 

Hence, Zara must have gotten 8 through 4+4 with players 5 and 7. 

All the remaining scores must be with Xena, adding up to 12, which is the case. 

4 and 1 must be present together, and 3 must be present in Xena as well, giving us the arrangement. 

We can determine the ratings of all the players except 1 and 3

Therefore, 6 is the correct answer. 

We are told that each coach had at least two players. Clue 1 says that Xena trained more people than Yuki. 
It must be the case that Yuki trained only two people. If Yuki had trained three, then Xena would have trained at least four players, leaving only one for Zara. 

Hence, Yuki trained two people. 

Coming to the scores given to the players themselves:

We are given that only 5 and 7 received a sam rating; everyone else received a distinct rating. 
We are also given their average to be 4 (clue 4), giving the sum of all the scores they got to be 32
The sum of all numbers from 1 to 7 would be $$\frac{8}{2}\times\ 7=28$$, hence the repeated score must be 32-28 = 4

Thus, the score of 5 and 7 was 4

Clue 5 says that player 2 got the highest score, which is 7

Clue 7 says that player 4 got a score double that of player 8 but less than player 5; the only possibility is player 4 getting 2 and player 8 getting 1. 

Now, considering clues 2, 3 and 6:

We are given the same coach trained in 5 and 7. And 2, 3, and 5 were trained by different coaches, with 1 and 4 being trained by the same coach. 
Cheer 6 informs us about the average number of players the coaches train. 

We know that Yuki trained only two players. 
Let's take the average score of Yuki's players to be x

The average of Xena's players would be x/2, and that of Zara's players would be x-2

The total score of Yuki's players would be 2x. 
The total score of Xena's players would be 3x/2 or 2x. 
The total score of Zara's players would be 3x-6 or 2x-4

We have two cases to consider:

Let's say Xena and Zara had three players each. 
The total score of Yuki's players would be 2x.
The total score of Xena's players would be 3x/2.
The total score of Zara's players would be 3x-6. 

The sum of all these scores would be $$\frac{13x}{2}-6$$ which should be equal to 32
This would give us the value of x as $$\frac{13x}{2}=38$$
Which would give a non-integral value of 2s, that is, the sum of Yuki's player's score. 

Hence, this must not be the case. 

The other possibility:

The total score of Yuki's players would be 2x.
The total score of Xena's players would be 2x.
The total score of Zara's players would be 2x-4.

The sum of all these scores would be 6x-4, which should be equal to 32
This would give the value of x as 6

Hence, the sum of all of Yuki's players would be 12
The sum of all of Xena's players would be 12
The sum of all of Zara's players would be 8

Yuki has only two players whose scores add up to 12; the only combination possible is scores 7+5, where seven were scored by player 2. Hence, player two must be under Yuki. 

Zara got a total of 8 scores, with 7 and 5 gone. The combinations that could get this score are 2+6 and 4+4
The score of 2 is obtained by player 4, which must come with player 1
It is possible that 1 could have gotten a score 6
But then we run into a contradiction: players 3 and 5 would end up under the same coach. 

Hence, Zara must have gotten 8 through 4+4 with players 5 and 7. 

All the remaining scores must be with Xena, adding up to 12, which is the case. 

4 and 1 must be present together, and 3 must be present in Xena as well, giving us the arrangement. 

Xena coached players numbered 1, 3, 4 and 8

Therefore, Option B is the correct answer.