CAT 2024 Slot 2 Question Paper

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 $$\frac{5}{2}k\pi\ \left(\frac{3}{2}\right)^2=k\pi\ \times\ \frac{25\times\ 9}{8}$$

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. 

Let the cost price of the item be C
We are given that Bina sells this at 19% loss or at (1 - 0.19)C = 0.81C at 4860
This gives us the value of C at Rs. 6000

If Bina had sold this at 17% profit, the selling price would have been $$1.17\times\ 6000\ =\ 7020$$

So Shyam bought the product at 4860 and sold it to Hari at 7020
Giving the profit made by Shyam to be $$7020-4860\ =\ 2160$$

Therefore, 2160 is the correct answer.

Upon drawing a rough sketch of the coordinates given, we realise that this is a right-angled triangle.

The three side lengths are 6, 8 and 10 units

The inradius of a circle can be calculated using the formula: $$rs=Area$$, where s is the semi-perimeter of the triangle 

The Area would be $$\frac{1}{2}\times\ 6\times\ 8=24$$
and the semi-perimeter would be $$\frac{10+6+8}{2}=12$$

Giving the inradius to be $$\frac{24}{12}=2$$ units

Therefore, 2 is the correct answer. 

(Note: This was a repeated question that appeared in both CAT 2023 and CAT 2024)

Let us assume the initial stock of all the fruits is S.

Let us take we have 'b' and 'a' mangoes initially.

Stock of Mangoes = 40% of S = 2S/5

The total number of fruits sold are Mangoes Sold + Apples Sold + Bananas Sold

= 2S/10 + 96 + 4a/10 = S/2 (Given)

=> S/5 + 96 + 2a/5 = S/2

=> S = $$\dfrac{\left(4a+960\right)}{3}$$

=> $$\dfrac{4a}{3}+320$$

'a' has to be a multiple of 3 for the above term to be an integer.

But 'a' has to be a multiple of 5 for 4a/10 to be an integer.

=> The smallest value of 'a' satisfying both conditions is 15.

=> $$\dfrac{4a}{3}+320=\dfrac{4\left(15\right)}{3}+320$$ = 340

Therefore, 340 is the correct answer.

The first term of the expression can be rewritten as $$\frac{\frac{1}{3}\log_2\left(a+b\right)}{\log_2c}$$
Using the property $$\frac{m}{n}\log_ab=\log_ab^{\frac{m}{n}}$$ this can be rewritten as

$$\frac{\log_2\left(a+b\right)^{\frac{1}{3}}}{\log_2c}$$

And finally using the property $$\frac{\log_ba}{\log_bc}=\log_ca$$, we can rewrite the expression as 

$$\log_c\left(a+b\right)^{\frac{1}{3}}$$

Doing identical operations in the second term, we get the entire left-hand side to be:

$$\log_c\left(a+b\right)^{\frac{1}{3}}+\log_c\left(a-b\right)^{\frac{1}{3}}$$
Using property $$\log_ca+\log_cb=\log_c\left(ab\right)$$ we get

$$\log_c\left[\left(a+b\right)^{\frac{1}{3}}\left(a-b\right)^{\frac{1}{3}}\right]$$
$$\log_c\left[\left(a+b\right)\left(a-b\right)\right]^{\frac{1}{3}}$$
$$\log_c\left[\left(a^2-b^2\right)\right]^{\frac{1}{3}}$$

This expression is given to be equal to 2/3
Using the definition of log: $$\log_cN=a\ $$ which is $$c^a=N$$
we get:$$c^{\frac{2}{3}}=\left(a^2-b^2\right)^{\frac{1}{3}}$$

Cubing both sides:
$$c^2=a^2-b^2$$
Finally giving $$a^2=b^2+c^2$$

We have upper limits on b and c as 10, and we want to maximize the value of a squared. 
This can be thought of as a right-angled triangle, and the value of a will be maximum when both b and c are equal to 10, giving $$a^2=200$$, but this would not give an integer value of a
We need to adjust $$a^2$$ to the biggest square less than 200, which is 196
Giving the value of $$a$$ as 14. 

Therefore, 14 is the correct answer.