CAT LR Puzzles Questions
Master CAT LR Puzzles Questions with practice questions and detailed solutions.
Instructions
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.
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.
Question 1.
What was the rating of Player-6?
What was the rating of Player-6?
Text Explanation:
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.
Question 2.
What best can be concluded about the number of players coached by Zara?
What best can be concluded about the number of players coached by Zara?
Text Explanation:
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.
Question 3.
For how many players the ratings can be determined with certainty?
For how many players the ratings can be determined with certainty?
Text Explanation:
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.
Question 4.
What was the rating of Player-7?
What was the rating of Player-7?
Text Explanation:
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.
Question 5.
Who all were the players trained by Xena?
Who all were the players trained by Xena?
Text Explanation:
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.
Instructions
At InnovateX, six employees, Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni, were split into two groups of three each: Elite led by Manager Kuku, and Novice led by Manager Lalu.
At the end of each quarter, Kuku and Lalu handed out ratings to all members in their respective groups. In each group, each employee received a distinct integer rating from 1 to 3.
The score for an employee at the end of a quarter is defined as their cumulative rating from the beginning of the year. At the end of each quarter the employee in Novice with the highest score was promoted to Elite, and the employee in Elite with the minimum score was demoted to Novice. If there was a tie in scores, the employee with a higher rating in the latest quarter was ranked higher.
1. Asha, Bunty, and Chintu were in Elite at the beginning of Quarter 1. All of them were in Novice at the beginning of Quarter 4.
2. Dolly and Falguni were the only employees who got the same rating across all the quarters.
3. The following is known about ratings given by Lalu:
- Bunty received a rating of 1 in Quarter 2.
- Asha and Dolly received ratings of 1 and 2, respectively, in Quarter 3.
At InnovateX, six employees, Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni, were split into two groups of three each: Elite led by Manager Kuku, and Novice led by Manager Lalu.
At the end of each quarter, Kuku and Lalu handed out ratings to all members in their respective groups. In each group, each employee received a distinct integer rating from 1 to 3.
The score for an employee at the end of a quarter is defined as their cumulative rating from the beginning of the year. At the end of each quarter the employee in Novice with the highest score was promoted to Elite, and the employee in Elite with the minimum score was demoted to Novice. If there was a tie in scores, the employee with a higher rating in the latest quarter was ranked higher.
1. Asha, Bunty, and Chintu were in Elite at the beginning of Quarter 1. All of them were in Novice at the beginning of Quarter 4.
2. Dolly and Falguni were the only employees who got the same rating across all the quarters.
3. The following is known about ratings given by Lalu:
- Bunty received a rating of 1 in Quarter 2.
- Asha and Dolly received ratings of 1 and 2, respectively, in Quarter 3.
Question 6.
What was Eklavya's score at the end of Quarter 2?
What was Eklavya's score at the end of Quarter 2?
Text Explanation:
Let's denote Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F respectively for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.
Putting all the given information in the table, we get,
B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.
We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.
Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,
Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.
We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.
If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.
If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.
After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.
Putting all the calculated values, we get the final table as,
Ekalavya's score after quarter 2 is 4.
Hence, the correct answer is 4.
Question 7.
How many employees changed groups more than once up to the beginning of Quarter 4?
How many employees changed groups more than once up to the beginning of Quarter 4?
Text Explanation:
Let's Denote Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F respectively for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.
Putting all the given information in the table, we get,
B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.
We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.
Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,
Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.
We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.
If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.
If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.
After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.
Putting all the calculated values, we get the final table as,
As we can see in the table, no employee has changed their group more than once.
Hence, the correct answer is 0.
Question 8.
What was Bunty's score at the end of Quarter 3?
What was Bunty's score at the end of Quarter 3?
Text Explanation:
Let's Denote Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F respectively for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.
Putting all the given information in the table, we get,
B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.
We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.
Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,
Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.
We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.
If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.
If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.
After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.
Putting all the calculated values, we get the final table as,
Bunty's score after quarter 3 is 5.
Hence, the correct answer is 5.
Question 9.
For how many employees can the scores at the end of Quarter 3 be determined with certainty?
For how many employees can the scores at the end of Quarter 3 be determined with certainty?
Text Explanation:
Let's Denote Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F respectively for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.
Putting all the given information in the table, we get,
B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.
We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.
Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,
Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.
We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.
If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.
If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.
After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.
Putting all the calculated values, we get the final table as,
The cumulative score of 4 employees can be determined with certainty at the end of quarter 3.
Hence, the correct answer is 4.
Question 10.
Which of the following statements is/are NECESSARILY true?
I. Asha received a rating of 2 in Quarter 1.
II. Asha received a rating of 1 in Quarter 2.
Which of the following statements is/are NECESSARILY true?
I. Asha received a rating of 2 in Quarter 1.
II. Asha received a rating of 1 in Quarter 2.
Text Explanation:
Let's Denote Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F respectively for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.
Putting all the given information in the table, we get,
B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.
We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.
Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,
Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.
We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.
If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.
If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.
After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.
Putting all the calculated values, we get the final table as,
I. Asha received a rating of 2 in Quarter 1 - This is not necessarily true, as it can also be 3 in Quarter 1.
II. Asha received a rating of 1 in Quarter 2. - This is definitely true, as Asha received a rating of 1 in Quarter 1.
Only II is definitely correct.
Hence, the correct answer is option C.
Instructions
The two most populous cities and the non-urban region (NUR) of each of three states, Whimshire, Fogglia, and Humbleset, are assigned Pollution Measures (PMs). These nine PMs are all distinct multiples of 10, ranging from 10 to 90. The six cities in increasing order of their PMs are: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo.
The Pollution Index (PI) of a state is a weighted average of the PMs of its NUR and cities, with a weight of 50% for the NUR, and 25% each for its two cities.
There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset.
The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.
The two most populous cities and the non-urban region (NUR) of each of three states, Whimshire, Fogglia, and Humbleset, are assigned Pollution Measures (PMs). These nine PMs are all distinct multiples of 10, ranging from 10 to 90. The six cities in increasing order of their PMs are: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo.
The Pollution Index (PI) of a state is a weighted average of the PMs of its NUR and cities, with a weight of 50% for the NUR, and 25% each for its two cities.
There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset.
The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.
Question 11.
What is the PI of Whimshire?
What is the PI of Whimshire?
Text Explanation:
We are also told there exists a single pair of NUR and city such that the PM of NUR is greater than the city and the NUR, city of the pair belongs to Humbleset. So, we can conclude that PMs of NURs of Whimshire and Fogglia are 10 and 20 in some order.
We also know that out of all the cities, Blusterburg has the lowest PM. So, the PM of Blusterberg has to be 30, and the PM of NUR of Humbleset has to be 40 for the above condition to be satisfied, and we can also conclude that Blusterberg is a city of Humbleset.
We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.
We are given that in the PI calculation, cities have a contribution of 25% and NUR has a contribution of 50%. So, we can calculate the contribution of each city in the calculation of PI by dividing their PI value by 4, and the contribution of NUR can be calculated by dividing the PM value by 2.
We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.For the PI value to be an integer, the cities with decimal contributions must be paired with another city with decimal contributions. Since all the decimals are 0.5, when two cities with 0.5 are added, the sum becomes 1, making the PI an integer.
The cities Splutterville and Mumpypoe must be from the same state; both have integral contributions towards the PI, and if they are added to a decimal, then the result will be a decimal, which should not be the case. These two cities have to belong to either the state of Whimshire or Fogglia, but their combined contribution towards PI is 15 + 20 = 35.
We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.
There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.
CASE 1: Noodleton
If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.
So, we can eliminate the case of Noodleton being the second city of Humbleset.
CASE 2: Quackford
If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.
So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.
We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.
So, we can eliminate the case of Quackford being the second city of Humbleset.
CASE 3: Zingaloo
If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
If the PI of the other state is 40,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.
We can see that PI of the other two states is 40, which violates one of the conditions given.
If the PI of the other state is 45,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 10 = 45
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.
We can see that all the conditions are satisfied in this case, and we can conclude that Zingaloo is the second city of Humbleset, and also we know that PI is the least for Fogglia. So, the PI of Fogglia is 35, and the PI of Whimshire is 45.
We can now assign the cities and the NURs the states and the PM values uniquely based on our calculations. After assigning the table looks like,
We calculated the PI of Whimshire to be 45.Hence, the correct answer is 45.
Question 12.
What is the PI of Fogglia?
What is the PI of Fogglia?
Text Explanation:
We are also told there exists a single pair of NUR and city such that the PM of NUR is greater than the city and the NUR, city of the pair belongs to Humbleset. So, we can conclude that PMs of NURs of Whimshire and Fogglia are 10 and 20 in some order.
We also know that out of all the cities, Blusterburg has the lowest PM. So, the PM of Blusterberg has to be 30, and the PM of NUR of Humbleset has to be 40 for the above condition to be satisfied, and we can also conclude that Blusterberg is a city of Humbleset.
We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.
We are given that in the PI calculation, cities have a contribution of 25% and NUR has a contribution of 50%. So, we can calculate the contribution of each city in the calculation of PI by dividing their PI value by 4, and the contribution of NUR can be calculated by dividing the PM value by 2.
We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.For the PI value to be an integer, the cities with decimal contributions must be paired with another city with decimal contributions. Since all the decimals are 0.5, when two cities with 0.5 are added, the sum becomes 1, making the PI an integer.
The cities Splutterville and Mumpypoe must be from the same state; both have integral contributions towards the PI, and if they are added to a decimal, then the result will be a decimal, which should not be the case. These two cities have to belong to either the state of Whimshire or Fogglia, but their combined contribution towards PI is 15 + 20 = 35.
We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.
There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.
CASE 1: Noodleton
If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.
So, we can eliminate the case of Noodleton being the second city of Humbleset.
CASE 2: Quackford
If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.
So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.
We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.
So, we can eliminate the case of Quackford being the second city of Humbleset.
CASE 3: Zingaloo
If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
If the PI of the other state is 40,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.
We can see that PI of the other two states is 40, which violates one of the conditions given.
If the PI of the other state is 45,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 10 = 45
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.
We can see that all the conditions are satisfied in this case, and we can conclude that Zingaloo is the second city of Humbleset, and also we know that PI is the least for Fogglia. So, the PI of Fogglia is 35, and the PI of Whimshire is 45.
We can now assign the cities and the NURs the states and the PM values uniquely based on our calculations. After assigning the table looks like,
We calculated the PI of Fogglia to be 35.Hence, the correct answer is 35.
Question 13.
What is the PI of Humbleset?
What is the PI of Humbleset?
Text Explanation:
We are also told there exists a single pair of NUR and city such that the PM of NUR is greater than the city and the NUR, city of the pair belongs to Humbleset. So, we can conclude that PMs of NURs of Whimshire and Fogglia are 10 and 20 in some order.
We also know that out of all the cities, Blusterburg has the lowest PM. So, the PM of Blusterberg has to be 30, and the PM of NUR of Humbleset has to be 40 for the above condition to be satisfied, and we can also conclude that Blusterberg is a city of Humbleset.
We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.
We are given that in the PI calculation, cities have a contribution of 25% and NUR has a contribution of 50%. So, we can calculate the contribution of each city in the calculation of PI by dividing their PI value by 4, and the contribution of NUR can be calculated by dividing the PM value by 2.
We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.For the PI value to be an integer, the cities with decimal contributions must be paired with another city with decimal contributions. Since all the decimals are 0.5, when two cities with 0.5 are added, the sum becomes 1, making the PI an integer.
The cities Splutterville and Mumpypoe must be from the same state; both have integral contributions towards the PI, and if they are added to a decimal, then the result will be a decimal, which should not be the case. These two cities have to belong to either the state of Whimshire or Fogglia, but their combined contribution towards PI is 15 + 20 = 35.
We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.
There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.
CASE 1: Noodleton
If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.
So, we can eliminate the case of Noodleton being the second city of Humbleset.
CASE 2: Quackford
If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.
So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.
We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.
So, we can eliminate the case of Quackford being the second city of Humbleset.
CASE 3: Zingaloo
If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
If the PI of the other state is 40,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.
We can see that PI of the other two states is 40, which violates one of the conditions given.
If the PI of the other state is 45,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 10 = 45
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.
We can see that all the conditions are satisfied in this case, and we can conclude that Zingaloo is the second city of Humbleset, and also we know that PI is the least for Fogglia. So, the PI of Fogglia is 35, and the PI of Whimshire is 45.
We can now assign the cities and the NURs, the states, and the PM values uniquely based on our calculations. After assigning the table looks like,
We calculated the PI of Humbleset to be 50.Hence, the correct answer is 50.
Question 14.
Which pair of cities definitely belong to the same state?
Which pair of cities definitely belong to the same state?
Text Explanation:
We are also told there exists a single pair of NUR and city such that the PM of NUR is greater than the city and the NUR, city of the pair belongs to Humbleset. So, we can conclude that PMs of NURs of Whimshire and Fogglia are 10 and 20 in some order.
We also know that out of all the cities, Blusterburg has the lowest PM. So, the PM of Blusterberg has to be 30, and the PM of NUR of Humbleset has to be 40 for the above condition to be satisfied, and we can also conclude that Blusterberg is a city of Humbleset.
We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.
We are given that in the PI calculation, cities have a contribution of 25% and NUR has a contribution of 50%. So, we can calculate the contribution of each city in the calculation of PI by dividing their PI value by 4, and the contribution of NUR can be calculated by dividing the PM value by 2.
We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.For the PI value to be an integer, the cities with decimal contributions must be paired with another city with decimal contributions. Since all the decimals are 0.5, when two cities with 0.5 are added, the sum becomes 1, making the PI an integer.
The cities Splutterville and Mumpypoe must be from the same state; both have integral contributions towards the PI, and if they are added to a decimal, then the result will be a decimal, which should not be the case. These two cities have to belong to either the state of Whimshire or Fogglia, but their combined contribution towards PI is 15 + 20 = 35.
We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.
There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.
CASE 1: Noodleton
If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.
So, we can eliminate the case of Noodleton being the second city of Humbleset.
CASE 2: Quackford
If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.
So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.
We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.
So, we can eliminate the case of Quackford being the second city of Humbleset.
CASE 3: Zingaloo
If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
If the PI of the other state is 40,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.
We can see that PI of the other two states is 40, which violates one of the conditions given.
If the PI of the other state is 45,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 10 = 45
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.
We can see that all the conditions are satisfied in this case, and we can conclude that Zingaloo is the second city of Humbleset, and also we know that PI is the least for Fogglia. So, the PI of Fogglia is 35, and the PI of Whimshire is 45.
We can now assign the cities and the NURs, the states, and the PM values uniquely based on our calculations. After assigning the table looks like,
We can see that Noodleton and Quackford belong to the same state among the options.Hence, the correct answer is option C.
Question 15.
For how many of the cities and NURs is it possible to identify their PM and the state they belong to?
For how many of the cities and NURs is it possible to identify their PM and the state they belong to?
Text Explanation:
We are also told there exists a single pair of NUR and city such that the PM of NUR is greater than the city and the NUR, city of the pair belongs to Humbleset. So, we can conclude that PMs of NURs of Whimshire and Fogglia are 10 and 20 in some order.
We also know that out of all the cities, Blusterburg has the lowest PM. So, the PM of Blusterberg has to be 30, and the PM of NUR of Humbleset has to be 40 for the above condition to be satisfied, and we can also conclude that Blusterberg is a city of Humbleset.
We can also conclude that the PMs of Noodleton, Splutterville, Quackford, Mumpypore, and Zingaloo are 50, 60, 70, 80, and 90, respectively.
We are given that in the PI calculation, cities have a contribution of 25% and NUR has a contribution of 50%. So, we can calculate the contribution of each city in the calculation of PI by dividing their PI value by 4, and the contribution of NUR can be calculated by dividing the PM value by 2.
We are given that all the PI values are unique integers, and Humbleset has the highest PI value, while Fogglia has the lowest.For the PI value to be an integer, the cities with decimal contributions must be paired with another city with decimal contributions. Since all the decimals are 0.5, when two cities with 0.5 are added, the sum becomes 1, making the PI an integer.
The cities Splutterville and Mumpypoe must be from the same state; both have integral contributions towards the PI, and if they are added to a decimal, then the result will be a decimal, which should not be the case. These two cities have to belong to either the state of Whimshire or Fogglia, but their combined contribution towards PI is 15 + 20 = 35.
We already know that for the state of Humbleset, the PI is highest, and the two contributions towards the PI are 7.5 and 20, totalling 27.5.
There are three possibilities for the other city of Humbleset, which are Noodleton, Quackford and Zingaloo.
CASE 1: Noodleton
If the other city of Humbleset is Noodleton, then its PI would be 27.5 + 12.5 = 40.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value; in either case, this condition is not satisfied.
So, we can eliminate the case of Noodleton being the second city of Humbleset.
CASE 2: Quackford
If the other city of Humbleset is Quackford, then its PI would be 27.5 + 17.5 = 45.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
However, we are told that the PI values are unique, and Humblest has the highest PI value. So, we can eliminate the case of PI of NUR being 10 for the other state.
So, for Humbleset, the PI is 7.5 + 17.5 + 20 = 45.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 22.5 + 10 = 45.
We can see that PI of Humbleton and one of the other two states is 45, which violates one of the conditions given.
So, we can eliminate the case of Quackford being the second city of Humbleset.
CASE 3: Zingaloo
If the other city of Humbleset is Zingaloo, then its PI would be 27.5 + 22.5 = 50.
We already obtained the sum of the PIs of one of the other 2 states to be 35 without including the PI of NUR. We know that the NUR contribution for that city would be either 5 or 10. If it is 5, then the PI of that state would be 40, and if it is 10, then the PI of that state would be 45.
If the PI of the other state is 40,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 5 = 40
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 10 = 40.
We can see that PI of the other two states is 40, which violates one of the conditions given.
If the PI of the other state is 45,
Then, for Humbleset, the PI is 7.5 + 22.5 + 20 = 50.
For one of the other 2 states, the PI is 15 + 20 + 10 = 45
The PI of the other state can be calculated by adding all the leftover PIs, which is 12.5 + 17.5 + 5 = 35.
We can see that all the conditions are satisfied in this case, and we can conclude that Zingaloo is the second city of Humbleset, and also we know that PI is the least for Fogglia. So, the PI of Fogglia is 35, and the PI of Whimshire is 45.
We can now assign the cities and the NURs, the states, and the PM values uniquely based on our calculations. After assigning the table looks like,
We can see that for all 6 cities and 3 NURs, the PMs and the city to which they belong can be identified uniquely. So, in total, we can identify the values uniquely for all 9 of them.Hence, the correct answer is 9.
Instructions
Ananya Raga, Bhaskar Tala, Charu Veena, and Devendra Sur are four musicians.
Each of them started and completed their training as students under each of three Gurus Pandit Meghnath, Ustad Samiran, and Acharya Raghunath between 2013 and 2024, including both the years. Each Guru trains any student for consecutive years only, for a span of 2, 3, or 4 years, with each Guru having a different span. During some of these years, a student may not have trained under these Gurus; however, they never trained under multiple Gurus in the same year.
In none of these years, any of these Gurus trained more than two of these students at the same time. When two students train under the same Guru at the same time, they are referred to as Gurubhai, irrespective of their gender.
The following additional facts are known:
1. Ustad Samiran never trained more than one of these students in the same year.
2. Acharya Raghunath did not train any of these students during 2015-2018, as well as during 2021-24.
3. Ananya and Devendra were never Gurubhai; neither were Bhaskar and Charu. All other pairs of musicians were Gurubhai for exactly 2 years.
4. In 2013, Ananya and Bhaskar started their trainings under Pandit Meghnath and under Ustad Samiran, respectively.
Ananya Raga, Bhaskar Tala, Charu Veena, and Devendra Sur are four musicians.
Each of them started and completed their training as students under each of three Gurus Pandit Meghnath, Ustad Samiran, and Acharya Raghunath between 2013 and 2024, including both the years. Each Guru trains any student for consecutive years only, for a span of 2, 3, or 4 years, with each Guru having a different span. During some of these years, a student may not have trained under these Gurus; however, they never trained under multiple Gurus in the same year.
In none of these years, any of these Gurus trained more than two of these students at the same time. When two students train under the same Guru at the same time, they are referred to as Gurubhai, irrespective of their gender.
The following additional facts are known:
1. Ustad Samiran never trained more than one of these students in the same year.
2. Acharya Raghunath did not train any of these students during 2015-2018, as well as during 2021-24.
3. Ananya and Devendra were never Gurubhai; neither were Bhaskar and Charu. All other pairs of musicians were Gurubhai for exactly 2 years.
4. In 2013, Ananya and Bhaskar started their trainings under Pandit Meghnath and under Ustad Samiran, respectively.
Question 16.
In which of the following years were Ananya and Bhaskar Gurubhai?
In which of the following years were Ananya and Bhaskar Gurubhai?
Text Explanation:
We are given that AR did not train during the periods of 2015-2018 and 2021-2024.
Other information provided is that PM, US, and AR had a span of 2, 3, and 4 consecutive years for each student, in some order. We also know that all the gurus trained each of the students for exactly one span.
In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.
Putting all the known information in the table, we get,
Now, if we look at the table, we can see that AR is training for only 4 years, with two years each, separated by a few years. Therefore, the span of AR must be 2 years because we know that he is not teaching continuously for 3 or 4 years.Since we know that each guru taught all the students, and AR only has 4 years to teach all the students, he must definitely teach 2 students at a time each year, as we are told that no guru taught more than 2 students in a single year. We also know that A and B were taught by PM and US in 2013, respectively. Therefore, we can conclude that C and D were taught by AR during both 2013 and 2014, as AR's span is 2 years.
We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.
We are also given in clue 1 that US did not train more than 1 student in any year, and there are a total of 12 years from 2013 to 2024. If the span of US is 4 years, then he would need a total of 16 years to teach every student, as he is not teaching more than 1 student in any year, but he only taught for 12 years, so the only possibility is for US to have a span of 3 years and PM to have a span of 4 years. We can also conclude that US taught B from 2013 to 2015, as he has a span of 3 years.
We are also given that A, D and B, C were never Gurubhai, and all the other pairs were Gurubhai for exactly 2 years. Therefore, there must be two years each of {A, B}, {A, C}, {B, D}, and {C, D} with classes together. We have already completed the 2 years of {A, B} and {C, D}, so the ones left are two years of {A, C} and {B, D}. We know that the US did not take classes of 2 students together, and we have already filled the schedule of AR, so the only one left is PM, and he must have the other two left.Since the span of PM is 4, he taught A from 2013 to 2016 and during this span, there must be 2 years of {A, C}, as we know that A will not be taught again after 2016. To have an exact span of 2 years together, C must start in 2015, as it is the only possibility, and he will teach C until 2018, as he has a span of 4 years.
There must be 2 years of {B, D} in the remaining time period of PM. In 2019, A and B were already enrolled with AR, and C had already completed his time with PM. So, the only person possible to have class in 2019 with PM is D, and D will have classes with PM until 2022. Now for 2 years of {B, D}, D must have his classes start from 2021, as it is the only possibility.
There cannot be any gap in the case of US, and in 2016, the only possibility is D as A, C were having classes with PM, and B already had his classes with US. D had classes from 2016 to 2018. In 2019, A, B and D were already having classes, so the only possibility is for C to have classes with US from 2019 to 2021, and finally A will have his classes from 2022 to 2024 with US, as he is the only person left to have classes with US.The final table looks like,
Ananya and Bhaskar were Gurubhai from 2019 to 2020. Since 2020 is the only option present during this period, it has to be the answer.Hence, the correct answer is option A.
Question 17.
In which year did Charu begin her training under Pandit Meghnath?
In which year did Charu begin her training under Pandit Meghnath?
Text Explanation:
We are given that AR did not train during the periods of 2015-2018 and 2021-2024.
Other information provided is that PM, US, and AR had a span of 2, 3, and 4 consecutive years for each student, in some order. We also know that all the gurus trained each of the students for exactly one span.
In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.
Putting all the known information in the table, we get,
Now, if we look at the table, we can see that AR is training for only 4 years, with two years each, separated by a few years. Therefore, the span of AR must be 2 years because we know that he is not teaching continuously for 3 or 4 years.Since we know that each guru taught all the students, and AR only has 4 years to teach all the students, he must definitely teach 2 students at a time each year, as we are told that no guru taught more than 2 students in a single year. We also know that A and B were taught by PM and US in 2013, respectively. Therefore, we can conclude that C and D were taught by AR during both 2013 and 2014, as AR's span is 2 years.
We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.
We are also given in clue 1 that US did not train more than 1 student in any year, and there are a total of 12 years from 2013 to 2024. If the span of US is 4 years, then he would need a total of 16 years to teach every student, as he is not teaching more than 1 student in any year, but he only taught for 12 years, so the only possibility is for US to have a span of 3 years and PM to have a span of 4 years. We can also conclude that US taught B from 2013 to 2015, as he has a span of 3 years.
We are also given that A, D and B, C were never Gurubhai, and all the other pairs were Gurubhai for exactly 2 years. Therefore, there must be two years each of {A, B}, {A, C}, {B, D}, and {C, D} with classes together. We have already completed the 2 years of {A, B} and {C, D}, so the ones left are two years of {A, C} and {B, D}. We know that the US did not take classes of 2 students together, and we have already filled the schedule of AR, so the only one left is PM, and he must have the other two left.Since the span of PM is 4, he taught A from 2013 to 2016 and during this span, there must be 2 years of {A, C}, as we know that A will not be taught again after 2016. To have an exact span of 2 years together, C must start in 2015, as it is the only possibility, and he will teach C until 2018, as he has a span of 4 years.
There must be 2 years of {B, D} in the remaining time period of PM. In 2019, A and B were already enrolled with AR, and C had already completed his time with PM. So, the only person possible to have class in 2019 with PM is D, and D will have classes with PM until 2022. Now for 2 years of {B, D}, D must have his classes start from 2021, as it is the only possibility.
There cannot be any gap in the case of US, and in 2016, the only possibility is D as A, C were having classes with PM, and B already had his classes with US. D had classes from 2016 to 2018. In 2019, A, B and D were already having classes, so the only possibility is for C to have classes with US from 2019 to 2021, and finally A will have his classes from 2022 to 2024 with US, as he is the only person left to have classes with US.The final table looks like,
Charu began her training under Pandit Meghnath in 2015.Hence, the correct answer is option B.
Question 18.
In which of the following years were Bhaskar and Devendra Gurubhai?
In which of the following years were Bhaskar and Devendra Gurubhai?
Text Explanation:
We are given that AR did not train during the periods of 2015-2018 and 2021-2024.
Other information provided is that PM, US, and AR had a span of 2, 3, and 4 consecutive years for each student, in some order. We also know that all the gurus trained each of the students for exactly one span.
In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.
Putting all the known information in the table, we get,
Now, if we look at the table, we can see that AR is training for only 4 years, with two years each, separated by a few years. Therefore, the span of AR must be 2 years because we know that he is not teaching continuously for 3 or 4 years.Since we know that each guru taught all the students, and AR only has 4 years to teach all the students, he must definitely teach 2 students at a time each year, as we are told that no guru taught more than 2 students in a single year. We also know that A and B were taught by PM and US in 2013, respectively. Therefore, we can conclude that C and D were taught by AR during both 2013 and 2014, as AR's span is 2 years.
We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.
We are also given in clue 1 that US did not train more than 1 student in any year, and there are a total of 12 years from 2013 to 2024. If the span of US is 4 years, then he would need a total of 16 years to teach every student, as he is not teaching more than 1 student in any year, but he only taught for 12 years, so the only possibility is for US to have a span of 3 years and PM to have a span of 4 years. We can also conclude that US taught B from 2013 to 2015, as he has a span of 3 years.
We are also given that A, D and B, C were never Gurubhai, and all the other pairs were Gurubhai for exactly 2 years. Therefore, there must be two years each of {A, B}, {A, C}, {B, D}, and {C, D} with classes together. We have already completed the 2 years of {A, B} and {C, D}, so the ones left are two years of {A, C} and {B, D}. We know that the US did not take classes of 2 students together, and we have already filled the schedule of AR, so the only one left is PM, and he must have the other two left.Since the span of PM is 4, he taught A from 2013 to 2016 and during this span, there must be 2 years of {A, C}, as we know that A will not be taught again after 2016. To have an exact span of 2 years together, C must start in 2015, as it is the only possibility, and he will teach C until 2018, as he has a span of 4 years.
There must be 2 years of {B, D} in the remaining time period of PM. In 2019, A and B were already enrolled with AR, and C had already completed his time with PM. So, the only person possible to have class in 2019 with PM is D, and D will have classes with PM until 2022. Now for 2 years of {B, D}, D must have his classes start from 2021, as it is the only possibility.
There cannot be any gap in the case of US, and in 2016, the only possibility is D as A, C were having classes with PM, and B already had his classes with US. D had classes from 2016 to 2018. In 2019, A, B and D were already having classes, so the only possibility is for C to have classes with US from 2019 to 2021, and finally A will have his classes from 2022 to 2024 with US, as he is the only person left to have classes with US.The final table looks like,
Bhaskar and Devendra were Gurubhai from 2021 to 2022. Since 2022 is the only option present during this period, it has to be the answer.Hence, the correct answer is option A.
Question 19.
Which of the following statements is TRUE?
Which of the following statements is TRUE?
Text Explanation:
We are given that AR did not train during the periods of 2015-2018 and 2021-2024.
Other information provided is that PM, US, and AR had a span of 2, 3, and 4 consecutive years for each student, in some order. We also know that all the gurus trained each of the students for exactly one span.
In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.
Putting all the known information in the table, we get,
Now, if we look at the table, we can see that AR is training for only 4 years, with two years each, separated by a few years. Therefore, the span of AR must be 2 years because we know that he is not teaching continuously for 3 or 4 years.Since we know that each guru taught all the students, and AR only has 4 years to teach all the students, he must definitely teach 2 students at a time each year, as we are told that no guru taught more than 2 students in a single year. We also know that A and B were taught by PM and US in 2013, respectively. Therefore, we can conclude that C and D were taught by AR during both 2013 and 2014, as AR's span is 2 years.
We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.
We are also given in clue 1 that US did not train more than 1 student in any year, and there are a total of 12 years from 2013 to 2024. If the span of US is 4 years, then he would need a total of 16 years to teach every student, as he is not teaching more than 1 student in any year, but he only taught for 12 years, so the only possibility is for US to have a span of 3 years and PM to have a span of 4 years. We can also conclude that US taught B from 2013 to 2015, as he has a span of 3 years.
We are also given that A, D and B, C were never Gurubhai, and all the other pairs were Gurubhai for exactly 2 years. Therefore, there must be two years each of {A, B}, {A, C}, {B, D}, and {C, D} with classes together. We have already completed the 2 years of {A, B} and {C, D}, so the ones left are two years of {A, C} and {B, D}. We know that the US did not take classes of 2 students together, and we have already filled the schedule of AR, so the only one left is PM, and he must have the other two left.Since the span of PM is 4, he taught A from 2013 to 2016 and during this span, there must be 2 years of {A, C}, as we know that A will not be taught again after 2016. To have an exact span of 2 years together, C must start in 2015, as it is the only possibility, and he will teach C until 2018, as he has a span of 4 years.
There must be 2 years of {B, D} in the remaining time period of PM. In 2019, A and B were already enrolled with AR, and C had already completed his time with PM. So, the only person possible to have class in 2019 with PM is D, and D will have classes with PM until 2022. Now for 2 years of {B, D}, D must have his classes start from 2021, as it is the only possibility.
There cannot be any gap in the case of US, and in 2016, the only possibility is D as A, C were having classes with PM, and B already had his classes with US. D had classes from 2016 to 2018. In 2019, A, B and D were already having classes, so the only possibility is for C to have classes with US from 2019 to 2021, and finally A will have his classes from 2022 to 2024 with US, as he is the only person left to have classes with US.The final table looks like,
Statement 1) Charu was training under Ustad Samiran in 2018. This is false as she was training under Pandit Meghnath in 2018.Statement 2) Ananya was training under Ustad Samiran in 2015. This is false as she was training under Pandit Meghnath in 2015.
Statement 3) Ananya was training under Ustad Samiran in 2018. This is false as she was not training in 2018.
Statement 4) Charu was training under Ustad Samiran in 2019. This is true.
Hence, the correct answer is option D.
Question 20.
In how many of the years between 2013-24, were only two of these four musicians training under these three Gurus?
In how many of the years between 2013-24, were only two of these four musicians training under these three Gurus?
Text Explanation:
We are given that AR did not train during the periods of 2015-2018 and 2021-2024.
Other information provided is that PM, US, and AR had a span of 2, 3, and 4 consecutive years for each student, in some order. We also know that all the gurus trained each of the students for exactly one span.
In clue 4, we are given that A started the training in 2013 under PM, and B started training in 2013 under US.
Putting all the known information in the table, we get,
Now, if we look at the table, we can see that AR is training for only 4 years, with two years each, separated by a few years. Therefore, the span of AR must be 2 years because we know that he is not teaching continuously for 3 or 4 years.Since we know that each guru taught all the students, and AR only has 4 years to teach all the students, he must definitely teach 2 students at a time each year, as we are told that no guru taught more than 2 students in a single year. We also know that A and B were taught by PM and US in 2013, respectively. Therefore, we can conclude that C and D were taught by AR during both 2013 and 2014, as AR's span is 2 years.
We can also conclude that A and B were taught by AR during both 2019 and 2020, as AR's span is 2 years, and they were the only students left for AR's class.
We are also given in clue 1 that US did not train more than 1 student in any year, and there are a total of 12 years from 2013 to 2024. If the span of US is 4 years, then he would need a total of 16 years to teach every student, as he is not teaching more than 1 student in any year, but he only taught for 12 years, so the only possibility is for US to have a span of 3 years and PM to have a span of 4 years. We can also conclude that US taught B from 2013 to 2015, as he has a span of 3 years.
We are also given that A, D and B, C were never Gurubhai, and all the other pairs were Gurubhai for exactly 2 years. Therefore, there must be two years each of {A, B}, {A, C}, {B, D}, and {C, D} with classes together. We have already completed the 2 years of {A, B} and {C, D}, so the ones left are two years of {A, C} and {B, D}. We know that the US did not take classes of 2 students together, and we have already filled the schedule of AR, so the only one left is PM, and he must have the other two left.Since the span of PM is 4, he taught A from 2013 to 2016 and during this span, there must be 2 years of {A, C}, as we know that A will not be taught again after 2016. To have an exact span of 2 years together, C must start in 2015, as it is the only possibility, and he will teach C until 2018, as he has a span of 4 years.
There must be 2 years of {B, D} in the remaining time period of PM. In 2019, A and B were already enrolled with AR, and C had already completed his time with PM. So, the only person possible to have class in 2019 with PM is D, and D will have classes with PM until 2022. Now for 2 years of {B, D}, D must have his classes start from 2021, as it is the only possibility.
There cannot be any gap in the case of US, and in 2016, the only possibility is D as A, C were having classes with PM, and B already had his classes with US. D had classes from 2016 to 2018. In 2019, A, B and D were already having classes, so the only possibility is for C to have classes with US from 2019 to 2021, and finally A will have his classes from 2022 to 2024 with US, as he is the only person left to have classes with US.The final table looks like,
Between 2013 and 2024, the years 2017, 2018, 2023, and 2024 were the only years when only two of these four musicians were training under these three Gurus.Hence, the correct answer is 4.
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