Download CAT 2025 DILR Slot 2 Question Paper PDF with Solutions

CAT 2025 DILR Slot 2 Paper With Answers & Explanation

Directions for questions 1 to 4: There are six spherical balls, B1, B2, B3, B4, B5, and B6, and four circular hoops H1, H2, H3, and H4.

Each ball was tested on each hoop once, by attempting to pass the ball through the hoop. If the diameter of a ball is not larger than the diameter of the hoop, the ball passes through the hoop and makes a “ping”. Any ball having a diameter larger than that of the hoop gets stuck on that hoop and does not make a ping.
The following additional information is known:
1. B1 and B6 each made a ping on H4, but B5 did not.
2. B4 made a ping on H3, but B1 did not.
3. All balls, except B3, made pings on H1.
4. None of the balls, except B2, made a ping on H2.

Q. 1 What was the total number of pings made by B1, B2, and B3?

Correct Answer

6

Explanation

From condition (3), All balls except B3 pinged on H1 ⇒ H1 is larger than B1, B2, B4, B5, B6 ⇒ B3 > H1 From condition (4), Only B2 pinged on H2 ⇒ B2 ⇒ H2, and B1, B3, B4, B5, B6 > H2 ⇒ H2 is the smallest hoop.
From condition (1), B1 and B6 pinged on H4, but B5 did not ⇒ B1, B6 ⇒ H4 < B5
From condition (2), B4 pinged on H3, but B1 did not ⇒ B4 ⇒ H3 < B1
From above we can determine the order of the hoops.
B2 < H2 < B4 < H3 < B1, B6 ⇒ H4 < B5 ⇒ H1 < B3
Also the only consistent hoop order is:
H2 < H3 < H4 < H1.
This information can be represented in a table as follows:

Sum of the number of pings by B1, B2 and B3 = 2 + 4 + 0 = 6.

Q. 2 Which of the following statements about the relative sizes of the balls is NOT NECESSARILY true?

1) B2 < B1 < B5

2) B1 < B5 < B3

3) B4 < B5 < B3

4) B1 < B6 < B3

Correct Answer

4

Explanation

From condition (3), All balls except B3 pinged on H1 ⇒ H1 is larger than B1, B2, B4, B5, B6 ⇒ B3 > H1 From condition (4), Only B2 pinged on H2 ⇒ B2 ⇒ H2, and B1, B3, B4, B5, B6 > H2 ⇒ H2 is the smallest hoop.
From condition (1), B1 and B6 pinged on H4, but B5 did not ⇒ B1, B6 ⇒ H4 < B5
From condition (2), B4 pinged on H3, but B1 did not ⇒ B4 ⇒ H3 < B1
From above we can determine the order of the hoops.
B2 < H2 < B4 < H3 < B1, B6 ⇒ H4 < B5 ⇒ H1 < B3
Also the only consistent hoop order is:
H2 < H3 < H4 < H1.
This information can be represented in a table as follows:

There is no information, on the basis of which B6 and B1 can be compared.
Hence, B1 < B6 < B3 is not necessarily true.

Q. 3 Which of the following statements about the relative sizes of the hoops is true?

1) H2 < H4 < H3 < H1

2) H1 < H4 < H3 < H2

3) H2 < H3 < H4 < H1

4) H1 < H3 < H4 < H2

Correct Answer

3

Explanation

From condition (3), All balls except B3 pinged on H1 ⇒ H1 is larger than B1, B2, B4, B5, B6 ⇒ B3 > H1 From condition (4), Only B2 pinged on H2 ⇒ B2 ⇒ H2, and B1, B3, B4, B5, B6 > H2 ⇒ H2 is the smallest hoop.
From condition (1), B1 and B6 pinged on H4, but B5 did not ⇒ B1, B6 ⇒ H4 < B5
From condition (2), B4 pinged on H3, but B1 did not ⇒ B4 ⇒ H3 < B1
From above we can determine the order of the hoops.
B2 < H2 < B4 < H3 < B1, B6 ⇒ H4 < B5 ⇒ H1 < B3
Also the only consistent hoop order is:
H2 < H3 < H4 < H1.
This information can be represented in a table as follows:

We can see that H2 < H3 < H4 < H1 is true.

Q. 4 What BEST can be said about the total number of pings from all the tests undertaken?

1) 13 or 14

2) 12 or 13

3) At least 9

4) 12 or 13 or 14

Correct Answer

2

Explanation

From condition (3), All balls except B3 pinged on H1 ⇒ H1 is larger than B1, B2, B4, B5, B6 ⇒ B3 > H1 From condition (4), Only B2 pinged on H2 ⇒ B2 ⇒ H2, and B1, B3, B4, B5, B6 > H2 ⇒ H2 is the smallest hoop.
From condition (1), B1 and B6 pinged on H4, but B5 did not ⇒ B1, B6 ⇒ H4 < B5
From condition (2), B4 pinged on H3, but B1 did not ⇒ B4 ⇒ H3 < B1
From above we can determine the order of the hoops.
B2 < H2 < B4 < H3 < B1, B6 ⇒ H4 < B5 ⇒ H1 < B3
Also the only consistent hoop order is:
H2 < H3 < H4 < H1.
This information can be represented in a table as follows:

We can see from the table that the total number of pings can be 12 or 13.

Directions for questions 5 to 8: The following charts depict details of research papers written by four authors, Arman, Brajen, Chintan, and Devon. The papers were of four types, single-author, two-author, three-author, and four-author, that is, written by one, two, three, or all four of these authors, respectively. No other authors were involved in writing these papers.

The following additional facts are known.
1. Each of the authors wrote at least one of each of the four types of papers.
2. The four authors wrote different numbers of single-author papers.
3. Both Chintan and Devon wrote more three-author papers than Brajen.
4. The number of single-author and two-author papers written by Brajen were the same.

Q. 5 What was the total number of two-author and threeauthor papers written by Brajen?

Correct Answer

4

Explanation

Step 1:
We know the number of titles authored by each person from the first bar graph. Also the number of single author, two author, three author and four author titles have been mentioned. Note that 2 four author titles would mean that all 4 authors are part of these two titles.
Hence, total count = 2 × 4 = 8. Similarly, for 3 author and 2 author titles.
From condition (1), each of the authors wrote at least one of each of the four types of papers. So none of the cells in the table can be zero and Aman has a total of 5 out of which there is 2 against Aman under 4 author, this means that for Aman each of the other cells will have 1.
From condition (2), each author wrote a different number of single author titles and Aman already has the number 1. So the remaining numbers have to be 2, 3 and 4.
Step 2:
From condition (4), we can conclude that the number of single author and two author titles for Brajen cannot be 4 (as the total has to be 8). It cannot be 3 (as that will lead to zero three author titles for Brajen, which is a contradiction). This means that Brajen will have 2 single author and 2 two author titles. Hence, he will have 2 three author titles also.
From condition (3), under three author titles for both Chintan and Devon we will require a number greater than 2 and the total has to be 9. So only the number 3 satisfies this condition.
We can create a table from the given information as follows:

The total number of two-author and three-author papers written by Brajen = 2 + 2 = 4

Q. 6 Which of the following statements is/are NECESSARILY true?
i. Chintan wrote exactly three two-author papers.
ii. Chintan wrote more single-author papers than Devon.

1) Neither i nor ii

2) Only i

3) Both i and ii

4) Only ii

Correct Answer

1

Explanation

Step 1:
We know the number of titles authored by each person from the first bar graph. Also the number of single author, two author, three author and four author titles have been mentioned. Note that 2 four author titles would mean that all 4 authors are part of these two titles.
Hence, total count = 2 × 4 = 8. Similarly, for 3 author and 2 author titles.
From condition (1), each of the authors wrote at least one of each of the four types of papers. So none of the cells in the table can be zero and Aman has a total of 5 out of which there is 2 against Aman under 4 author, this means that for Aman each of the other cells will have 1.
From condition (2), each author wrote a different number of single author titles and Aman already has the number 1. So the remaining numbers have to be 2, 3 and 4.
Step 2:
From condition (4), we can conclude that the number of single author and two author titles for Brajen cannot be 4 (as the total has to be 8). It cannot be 3 (as that will lead to zero three author titles for Brajen, which is a contradiction). This means that Brajen will have 2 single author and 2 two author titles. Hence, he will have 2 three author titles also.
From condition (3), under three author titles for both Chintan and Devon we will require a number greater than 2 and the total has to be 9. So only the number 3 satisfies this condition.
We can create a table from the given information as follows:

Let us check each statement:
i. The statement, Chintan wrote exactly three two-author papers may not necessarily be true.
ii. The statement, Chintan wrote more single-author papers than Devon may not necessarily be true.
Hence, neither i nor ii is definitely true.

Q. 7 Which of the following statements is/are NECESSARILY true?
i. Arman wrote three-author papers only with Chintan and Devon.
ii. Brajen wrote three-author papers only with Chintan and Devon.

1) Neither i or ii

2) Both i and ii

3) Only ii

4) Only i

Correct Answer

2

Explanation

Step 1:
We know the number of titles authored by each person from the first bar graph. Also the number of single author, two author, three author and four author titles have been mentioned. Note that 2 four author titles would mean that all 4 authors are part of these two titles.
Hence, total count = 2 × 4 = 8. Similarly, for 3 author and 2 author titles.
From condition (1), each of the authors wrote at least one of each of the four types of papers. So none of the cells in the table can be zero and Aman has a total of 5 out of which there is 2 against Aman under 4 author, this means that for Aman each of the other cells will have 1.
From condition (2), each author wrote a different number of single author titles and Aman already has the number 1. So the remaining numbers have to be 2, 3 and 4.
Step 2:
From condition (4), we can conclude that the number of single author and two author titles for Brajen cannot be 4 (as the total has to be 8). It cannot be 3 (as that will lead to zero three author titles for Brajen, which is a contradiction). This means that Brajen will have 2 single author and 2 two author titles. Hence, he will have 2 three author titles also.
From condition (3), under three author titles for both Chintan and Devon we will require a number greater than 2 and the total has to be 9. So only the number 3 satisfies this condition.
We can create a table from the given information as follows:

There are 3 three author papers and both Chintan and Devon wrote 3 three author papers whereas Aman wrote
1 and Brajen wrote 2. So the only possible combination will be {(Chintan, Devon, Aman), (Chintan, Devon, Brajen), (Chintan, Devon, Brajen)}. Hence, the statement, Arman wrote three-author papers only with Chintan and Devon, is true. Brajen wrote three-author papers only with Chintan and Devon is also true. Hence, both (i) and (ii) are true.

Q. 8 If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?

Correct Answer

3

Explanation

Step 1:
We know the number of titles authored by each person from the first bar graph. Also the number of single author, two author, three author and four author titles have been mentioned. Note that 2 four author titles would mean that all 4 authors are part of these two titles.
Hence, total count = 2 × 4 = 8. Similarly, for 3 author and 2 author titles.
From condition (1), each of the authors wrote at least one of each of the four types of papers. So none of the cells in the table can be zero and Aman has a total of 5 out of which there is 2 against Aman under 4 author, this means that for Aman each of the other cells will have 1.
From condition (2), each author wrote a different number of single author titles and Aman already has the number 1. So the remaining numbers have to be 2, 3 and 4.
Step 2:
From condition (4), we can conclude that the number of single author and two author titles for Brajen cannot be 4 (as the total has to be 8). It cannot be 3 (as that will lead to zero three author titles for Brajen, which is a contradiction). This means that Brajen will have 2 single author and 2 two author titles. Hence, he will have 2 three author titles also.
From condition (3), under three author titles for both Chintan and Devon we will require a number greater than 2 and the total has to be 9. So only the number 3 satisfies this condition.
We can create a table from the given information as follows:

If Devon wrote more than one two-author papers, then the number of two-author papers written by Chintan is 3.

Directions for questions 9 to 13: 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.

Q. 9 In which of the following years were Ananya and Bhaskar Gurubhai?

1) 2020

2) 2018

3) 2021

4) 2014

Correct Answer

1

Explanation

Step 1:
From condition (1), we can say that each of the 4 students trained for 3 years under Ustad Samiran as he never trained more than one of these students in the same year.
From condition (2), we have a gap of 4 years from 2015-2018 and another gap of 4 years from 2021-24 for Acharya Raghunath when he does not train any student. So Acharya Ragunath trains each student for 2 years.
From condition (4), Ananya and Bhaskar started their training under Pandit Meghnath and under Ustad Samiran, respectively in 2013. As none of them train under more than one Guru at the same time, it means that Charu and Devendra started their training under Acharya Raghunath in 2013 and we can also conclude that Ananya and Bhaskar started training under Acharya Raghunath in 2019 and it lasted for 2 years. It is given that each guru trains for 2, 3 or 4 years and we have already concluded that Acharya Raghunath and Pandit Meghnath train for 2 and 3 years respectively. Hence, Ustad Samiran trains each student for 4 years. Also Ananya began in 2013. Since any student cannot train under 2 gurus at the same time, Bhaskar has to train under Pandit Meghnath from 2021 to 2024 only.
From condition (3), we can say that among the six combinations of the 4 students : AB, AC, AD, BC, BD, CD, it is said that AD and BC are not valid. Also each of the other pairs are gurubhais for exactly 2 years.
Hence, under Pandit Meghnath the combination of Ananya and Charu will be there for 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.
Step 2:
Under Ustad Samiran (each one has a distinct 3 year training period without overlap), So Ananya will have to train during 2022 to 2024 as any other 3 year period will lead to an overlap with her training under the other two gurus.
Finally we can conclude that Devendra will train under Ustad Samiran from 2016 to 2018 and Charu will train under Ustad Samiran from 2019 to 2021.
Hence, we have the following table:

Ananya and Bhaskar were Gurubhai in 2020.

Q. 10 In which year did Charu begin her training under Pandit Meghnath?

1) 2017

2) 2015

3) 2016

4) 2021

Correct Answer

2

Explanation

Step 1:
From condition (1), we can say that each of the 4 students trained for 3 years under Ustad Samiran as he never trained more than one of these students in the same year.
From condition (2), we have a gap of 4 years from 2015-2018 and another gap of 4 years from 2021-24 for Acharya Raghunath when he does not train any student. So Acharya Ragunath trains each student for 2 years.
From condition (4), Ananya and Bhaskar started their training under Pandit Meghnath and under Ustad Samiran, respectively in 2013. As none of them train under more than one Guru at the same time, it means that Charu and Devendra started their training under Acharya Raghunath in 2013 and we can also conclude that Ananya and Bhaskar started training under Acharya Raghunath in 2019 and it lasted for 2 years. It is given that each guru trains for 2, 3 or 4 years and we have already concluded that Acharya Raghunath and Pandit Meghnath train for 2 and 3 years respectively. Hence, Ustad Samiran trains each student for 4 years. Also Ananya began in 2013. Since any student cannot train under 2 gurus at the same time, Bhaskar has to train under Pandit Meghnath from 2021 to 2024 only.
From condition (3), we can say that among the six combinations of the 4 students : AB, AC, AD, BC, BD, CD, it is said that AD and BC are not valid. Also each of the other pairs are gurubhais for exactly 2 years.
Hence, under Pandit Meghnath the combination of Ananya and Charu will be there for 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.
Step 2:
Under Ustad Samiran (each one has a distinct 3 year training period without overlap), So Ananya will have to train during 2022 to 2024 as any other 3 year period will lead to an overlap with her training under the other two gurus.
Finally we can conclude that Devendra will train under Ustad Samiran from 2016 to 2018 and Charu will train under Ustad Samiran from 2019 to 2021.
Hence, we have the following table:

Charu began her training under Pandit Meghnath in 2015.

Q. 11 In which of the following years were Bhaskar and Devendra Gurubhai?

1) 2015

2) 2022

3) 2018

4) 2020

Correct Answer

2

Explanation

Step 1:
From condition (1), we can say that each of the 4 students trained for 3 years under Ustad Samiran as he never trained more than one of these students in the same year.
From condition (2), we have a gap of 4 years from 2015-2018 and another gap of 4 years from 2021-24 for Acharya Raghunath when he does not train any student. So Acharya Ragunath trains each student for 2 years.
From condition (4), Ananya and Bhaskar started their training under Pandit Meghnath and under Ustad Samiran, respectively in 2013. As none of them train under more than one Guru at the same time, it means that Charu and Devendra started their training under Acharya Raghunath in 2013 and we can also conclude that Ananya and Bhaskar started training under Acharya Raghunath in 2019 and it lasted for 2 years. It is given that each guru trains for 2, 3 or 4 years and we have already concluded that Acharya Raghunath and Pandit Meghnath train for 2 and 3 years respectively. Hence, Ustad Samiran trains each student for 4 years. Also Ananya began in 2013. Since any student cannot train under 2 gurus at the same time, Bhaskar has to train under Pandit Meghnath from 2021 to 2024 only.
From condition (3), we can say that among the six combinations of the 4 students : AB, AC, AD, BC, BD, CD, it is said that AD and BC are not valid. Also each of the other pairs are gurubhais for exactly 2 years.
Hence, under Pandit Meghnath the combination of Ananya and Charu will be there for 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.
Step 2:
Under Ustad Samiran (each one has a distinct 3 year training period without overlap), So Ananya will have to train during 2022 to 2024 as any other 3 year period will lead to an overlap with her training under the other two gurus.
Finally we can conclude that Devendra will train under Ustad Samiran from 2016 to 2018 and Charu will train under Ustad Samiran from 2019 to 2021.
Hence, we have the following table:

Bhaskar and Devendra were Gurubhai in 2022.

Q. 12 Which of the following statements is TRUE?

1) Ananya was training under Ustad Samiran in 2015.

2) Charu was training under Ustad Samiran in 2019.

3) Ananya was training under Ustad Samiran in 2018.

4) Charu was training under Ustad Samiran in 2018.

Correct Answer

2

Explanation

Step 1:
From condition (1), we can say that each of the 4 students trained for 3 years under Ustad Samiran as he never trained more than one of these students in the same year.
From condition (2), we have a gap of 4 years from 2015-2018 and another gap of 4 years from 2021-24 for Acharya Raghunath when he does not train any student. So Acharya Ragunath trains each student for 2 years.
From condition (4), Ananya and Bhaskar started their training under Pandit Meghnath and under Ustad Samiran, respectively in 2013. As none of them train under more than one Guru at the same time, it means that Charu and Devendra started their training under Acharya Raghunath in 2013 and we can also conclude that Ananya and Bhaskar started training under Acharya Raghunath in 2019 and it lasted for 2 years. It is given that each guru trains for 2, 3 or 4 years and we have already concluded that Acharya Raghunath and Pandit Meghnath train for 2 and 3 years respectively. Hence, Ustad Samiran trains each student for 4 years. Also Ananya began in 2013. Since any student cannot train under 2 gurus at the same time, Bhaskar has to train under Pandit Meghnath from 2021 to 2024 only.
From condition (3), we can say that among the six combinations of the 4 students : AB, AC, AD, BC, BD, CD, it is said that AD and BC are not valid. Also each of the other pairs are gurubhais for exactly 2 years.
Hence, under Pandit Meghnath the combination of Ananya and Charu will be there for 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.
Step 2:
Under Ustad Samiran (each one has a distinct 3 year training period without overlap), So Ananya will have to train during 2022 to 2024 as any other 3 year period will lead to an overlap with her training under the other two gurus.
Finally we can conclude that Devendra will train under Ustad Samiran from 2016 to 2018 and Charu will train under Ustad Samiran from 2019 to 2021.
Hence, we have the following table:

Let us check each statement:
1. Ananya was training under Ustad Samiran in 2015, is not true.
2. Charu was training under Ustad Samiran in 2019, is true.
3. Ananya was training under Ustad Samiran in 2018, is not true
4. Charu was training under Ustad Samiran in 2018, is not true.

Q. 13 In how many of the years between 2013-24, were only two of these four musicians training under these three Gurus?

Correct Answer

4

Explanation

Step 1:
From condition (1), we can say that each of the 4 students trained for 3 years under Ustad Samiran as he never trained more than one of these students in the same year.
From condition (2), we have a gap of 4 years from 2015-2018 and another gap of 4 years from 2021-24 for Acharya Raghunath when he does not train any student. So Acharya Ragunath trains each student for 2 years.
From condition (4), Ananya and Bhaskar started their training under Pandit Meghnath and under Ustad Samiran, respectively in 2013. As none of them train under more than one Guru at the same time, it means that Charu and Devendra started their training under Acharya Raghunath in 2013 and we can also conclude that Ananya and Bhaskar started training under Acharya Raghunath in 2019 and it lasted for 2 years. It is given that each guru trains for 2, 3 or 4 years and we have already concluded that Acharya Raghunath and Pandit Meghnath train for 2 and 3 years respectively. Hence, Ustad Samiran trains each student for 4 years. Also Ananya began in 2013. Since any student cannot train under 2 gurus at the same time, Bhaskar has to train under Pandit Meghnath from 2021 to 2024 only.
From condition (3), we can say that among the six combinations of the 4 students : AB, AC, AD, BC, BD, CD, it is said that AD and BC are not valid. Also each of the other pairs are gurubhais for exactly 2 years.
Hence, under Pandit Meghnath the combination of Ananya and Charu will be there for 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.
Step 2:
Under Ustad Samiran (each one has a distinct 3 year training period without overlap), So Ananya will have to train during 2022 to 2024 as any other 3 year period will lead to an overlap with her training under the other two gurus.
Finally we can conclude that Devendra will train under Ustad Samiran from 2016 to 2018 and Charu will train under Ustad Samiran from 2019 to 2021.
Hence, we have the following table:

Between 2013-24, there were 4 years when only two of these four musicians were training under these three Gurus.
These years were 2017, 2018, 2023 and 2024.

Directions for questions 14 to 17: The Sustainability Index (SI) of a country at a point in time is an integer between 1 and 100. This question is related to SI of six countries – A, B, C, D, E, and F – at three different points in time – 2016, 2020, and 2024. The plot represents the exact changes in their SI, with X-coordinate representing % increase in 2020 from 2016, i.e., (SI in 2020 minus SI in 2016) / (SI in 2016), and Y-coordinate representing % increase in 2024 from 2020. At any point in time, the country with highest SI is ranked 1, while the country with the lowest SI is ranked 6. The following additional facts are known.

1. In 2016, B, C, E, and A had ranks 1, 2, 3, and 4 respectively.
2. F had lower SI than any other country in 2016, 2020, and 2024.
3. In 2024, E was the only country with SI of 90.
4. The range of SI of the six countries was 60 in 2016 as well as in 2024.

Q. 14 What was the SI of E in 2016?

Correct Answer

60

Explanation

Step 1:
Note that the questions require the calculations for B, C, E and F only.
In 2016, B had the highest rank and F had the lowest rank. Also the SI of B in 2016 was 60 more than that of F.
Let the SI of F in 2016 be x and that of B be (x + 60).
From the percentages in the graph, we get the SI of F in 2020 and 2024 as 2x and 1.5x respectively.
In 2024, the SI of E was 90. Using the percentages in the graph, we can find the SI for E in 2020 and 2016 as 75 and 60 respectively.
Given that the range in 2024 is 60. We can say that 1.5x is at least 30. And consequently we can say that in 2016 the value of x is at least 20. Hence, (x + 60) would be at least 80.
Let us look at the values of percentage changes in SI for B. The SI of B decreases by ¾ in 2020 wrt 2016 and by ¾ in 2024 wrt 2020. This gives us a multiplying factor of 9/16 for 2024. Now all the SI values are integers. This means that (x + 60) should be a multiple of 16 and we already know that it is at least 80. So possible values of (x + 60) are 80 and 96.
Now, for (x + 60) = 96 we have x = 36, and the SI of B in 2020 will be ¾ × 96 = 72. Also the SI of F in 2020 will become 2 × 36 = 72. But this contradicts the condition that F had the lowest SI in 2020. Hence, we discard the value 96.
For (x + 60) = 80, we get x = 20 and 2x = 40, which does not contradict any condition. We get the SI of B in 2020 and 2024 as 80 × ¾ = 60 and 80 × 9/16 = 45.
Step 2:
Note that in 2016, the rank of C lies between B and E, so the SI of C in 2016 will lie between 60 and 80 and from the graph the SI of C in 2020 decreases to 4/5 of the SI in 2016 and in 2024 it increases to 7/5 of the SI in 2020. This gives us a multiplying factor of 28/25 in 2024, which means that the SI of C has to be a multiple of 25 in 2016. The only multiple of 25 between 60 and 80 is 75. This gives us the values of the SI of C in 2020 and 2024 as 60 and 84 respectively.
The above information can be tabulated as follows:

The SI of E in 2016 = 60.

Q. 15 What was the SI of F in 2020?

Correct Answer

40

Explanation

Step 1:
Note that the questions require the calculations for B, C, E and F only.
In 2016, B had the highest rank and F had the lowest rank. Also the SI of B in 2016 was 60 more than that of F.
Let the SI of F in 2016 be x and that of B be (x + 60).
From the percentages in the graph, we get the SI of F in 2020 and 2024 as 2x and 1.5x respectively.
In 2024, the SI of E was 90. Using the percentages in the graph, we can find the SI for E in 2020 and 2016 as 75 and 60 respectively.
Given that the range in 2024 is 60. We can say that 1.5x is at least 30. And consequently we can say that in 2016 the value of x is at least 20. Hence, (x + 60) would be at least 80.
Let us look at the values of percentage changes in SI for B. The SI of B decreases by ¾ in 2020 wrt 2016 and by ¾ in 2024 wrt 2020. This gives us a multiplying factor of 9/16 for 2024. Now all the SI values are integers. This means that (x + 60) should be a multiple of 16 and we already know that it is at least 80. So possible values of (x + 60) are 80 and 96.
Now, for (x + 60) = 96 we have x = 36, and the SI of B in 2020 will be ¾ × 96 = 72. Also the SI of F in 2020 will become 2 × 36 = 72. But this contradicts the condition that F had the lowest SI in 2020. Hence, we discard the value 96.
For (x + 60) = 80, we get x = 20 and 2x = 40, which does not contradict any condition. We get the SI of B in 2020 and 2024 as 80 × ¾ = 60 and 80 × 9/16 = 45.
Step 2:
Note that in 2016, the rank of C lies between B and E, so the SI of C in 2016 will lie between 60 and 80 and from the graph the SI of C in 2020 decreases to 4/5 of the SI in 2016 and in 2024 it increases to 7/5 of the SI in 2020. This gives us a multiplying factor of 28/25 in 2024, which means that the SI of C has to be a multiple of 25 in 2016. The only multiple of 25 between 60 and 80 is 75. This gives us the values of the SI of C in 2020 and 2024 as 60 and 84 respectively.
The above information can be tabulated as follows:

The SI of F in 2020 = 40.

Q. 16 What was the SI of C in 2024?

Correct Answer

84

Explanation

Step 1:
Note that the questions require the calculations for B, C, E and F only.
In 2016, B had the highest rank and F had the lowest rank. Also the SI of B in 2016 was 60 more than that of F.
Let the SI of F in 2016 be x and that of B be (x + 60).
From the percentages in the graph, we get the SI of F in 2020 and 2024 as 2x and 1.5x respectively.
In 2024, the SI of E was 90. Using the percentages in the graph, we can find the SI for E in 2020 and 2016 as 75 and 60 respectively.
Given that the range in 2024 is 60. We can say that 1.5x is at least 30. And consequently we can say that in 2016 the value of x is at least 20. Hence, (x + 60) would be at least 80.
Let us look at the values of percentage changes in SI for B. The SI of B decreases by ¾ in 2020 wrt 2016 and by ¾ in 2024 wrt 2020. This gives us a multiplying factor of 9/16 for 2024. Now all the SI values are integers. This means that (x + 60) should be a multiple of 16 and we already know that it is at least 80. So possible values of (x + 60) are 80 and 96.
Now, for (x + 60) = 96 we have x = 36, and the SI of B in 2020 will be ¾ × 96 = 72. Also the SI of F in 2020 will become 2 × 36 = 72. But this contradicts the condition that F had the lowest SI in 2020. Hence, we discard the value 96.
For (x + 60) = 80, we get x = 20 and 2x = 40, which does not contradict any condition. We get the SI of B in 2020 and 2024 as 80 × ¾ = 60 and 80 × 9/16 = 45.
Step 2:
Note that in 2016, the rank of C lies between B and E, so the SI of C in 2016 will lie between 60 and 80 and from the graph the SI of C in 2020 decreases to 4/5 of the SI in 2016 and in 2024 it increases to 7/5 of the SI in 2020. This gives us a multiplying factor of 28/25 in 2024, which means that the SI of C has to be a multiple of 25 in 2016. The only multiple of 25 between 60 and 80 is 75. This gives us the values of the SI of C in 2020 and 2024 as 60 and 84 respectively.
The above information can be tabulated as follows:

The SI of C in 2024 = 84

Q. 17 What was the SI of B in 2024?

1) 54

2) 45

3) 60

4) 80

Correct Answer

2

Explanation

Step 1:
Note that the questions require the calculations for B, C, E and F only.
In 2016, B had the highest rank and F had the lowest rank. Also the SI of B in 2016 was 60 more than that of F.
Let the SI of F in 2016 be x and that of B be (x + 60).
From the percentages in the graph, we get the SI of F in 2020 and 2024 as 2x and 1.5x respectively.
In 2024, the SI of E was 90. Using the percentages in the graph, we can find the SI for E in 2020 and 2016 as 75 and 60 respectively.
Given that the range in 2024 is 60. We can say that 1.5x is at least 30. And consequently we can say that in 2016 the value of x is at least 20. Hence, (x + 60) would be at least 80.
Let us look at the values of percentage changes in SI for B. The SI of B decreases by ¾ in 2020 wrt 2016 and by ¾ in 2024 wrt 2020. This gives us a multiplying factor of 9/16 for 2024. Now all the SI values are integers. This means that (x + 60) should be a multiple of 16 and we already know that it is at least 80. So possible values of (x + 60) are 80 and 96.
Now, for (x + 60) = 96 we have x = 36, and the SI of B in 2020 will be ¾ × 96 = 72. Also the SI of F in 2020 will become 2 × 36 = 72. But this contradicts the condition that F had the lowest SI in 2020. Hence, we discard the value 96.
For (x + 60) = 80, we get x = 20 and 2x = 40, which does not contradict any condition. We get the SI of B in 2020 and 2024 as 80 × ¾ = 60 and 80 × 9/16 = 45.
Step 2:
Note that in 2016, the rank of C lies between B and E, so the SI of C in 2016 will lie between 60 and 80 and from the graph the SI of C in 2020 decreases to 4/5 of the SI in 2016 and in 2024 it increases to 7/5 of the SI in 2020. This gives us a multiplying factor of 28/25 in 2024, which means that the SI of C has to be a multiple of 25 in 2016. The only multiple of 25 between 60 and 80 is 75. This gives us the values of the SI of C in 2020 and 2024 as 60 and 84 respectively.
The above information can be tabulated as follows:

The SI of B in 2024 = 45

Directions for questions 18 to 22: 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 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.
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.

Q. 18 What is the PI of Whimshire?

Correct Answer

45

Explanation

Step 1:
According to the given information the nine PIs are all distinct multiples of 10, ranging from 10 to 90. The values will be {10, 20, 30, …, 80, 90}
Order of PIs of the 6 cities: Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo.
There is only one pair of an NUR and a city (considering all cities and all NURs) where the PI of the NUR is greater than that of the city. This has to be Blusterburg with a PI of 30 and the respective NUR has to have a PI of 40. Any other combination would contradict the condition. Also these two (NUR and Blusterburg) belong to Humbleset.
The remaining cities in increasing order will have PIs from 50 to 90 and the remaining two NURs will have PIs 10 and 20.

Step 2:
Given that, the PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.
For Humbleset to have an integer PI, the remaining city can have a weighted PI of (12.5, 17.5, 22.5}
But 12.5 and 17.5 will not satisfy the condition that Humblest has the highest PI. So the second city in Humbleset has to be Zingaloo with a weighted PI of 22.5.
PI(Humbleset) = 7.5 + 20 + 22.5 = 50
For the PIs of the states to be integers, the cities with PIs 12.5 and 17.5 have to be in the same state. This will make
the sum 12.5 + 17.5 = 30 and the sum of the other pair of cities = 15 + 20 = 35 Note that Fogglia has to have the lowest PI. The only possible combination is:
PI (Fogglia) = 12.5 + 17.5 + 5 = 35
PI (Whimshire) = 15 + 20 + 10 = 45
So we have the final table as follows:

The PI of Whimshire = 45

Q. 19 What is the PI of Fogglia?

Correct Answer

35

Explanation

Step 1:
According to the given information the nine PIs are all distinct multiples of 10, ranging from 10 to 90. The values will be {10, 20, 30, …, 80, 90}
Order of PIs of the 6 cities: Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo.
There is only one pair of an NUR and a city (considering all cities and all NURs) where the PI of the NUR is greater than that of the city. This has to be Blusterburg with a PI of 30 and the respective NUR has to have a PI of 40. Any other combination would contradict the condition. Also these two (NUR and Blusterburg) belong to Humbleset.
The remaining cities in increasing order will have PIs from 50 to 90 and the remaining two NURs will have PIs 10 and 20.

Step 2:
Given that, the PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.
For Humbleset to have an integer PI, the remaining city can have a weighted PI of (12.5, 17.5, 22.5}
But 12.5 and 17.5 will not satisfy the condition that Humblest has the highest PI. So the second city in Humbleset has to be Zingaloo with a weighted PI of 22.5.
PI(Humbleset) = 7.5 + 20 + 22.5 = 50
For the PIs of the states to be integers, the cities with PIs 12.5 and 17.5 have to be in the same state. This will make
the sum 12.5 + 17.5 = 30 and the sum of the other pair of cities = 15 + 20 = 35 Note that Fogglia has to have the lowest PI. The only possible combination is:
PI (Fogglia) = 12.5 + 17.5 + 5 = 35
PI (Whimshire) = 15 + 20 + 10 = 45
So we have the final table as follows:

The PI of Fogglia = 35

Q. 20 What is the PI of Humbleset?

Correct Answer

50

Explanation

Step 1:
According to the given information the nine PIs are all distinct multiples of 10, ranging from 10 to 90. The values will be {10, 20, 30, …, 80, 90}
Order of PIs of the 6 cities: Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo.
There is only one pair of an NUR and a city (considering all cities and all NURs) where the PI of the NUR is greater than that of the city. This has to be Blusterburg with a PI of 30 and the respective NUR has to have a PI of 40. Any other combination would contradict the condition. Also these two (NUR and Blusterburg) belong to Humbleset.
The remaining cities in increasing order will have PIs from 50 to 90 and the remaining two NURs will have PIs 10 and 20.

Step 2:
Given that, the PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.
For Humbleset to have an integer PI, the remaining city can have a weighted PI of (12.5, 17.5, 22.5}
But 12.5 and 17.5 will not satisfy the condition that Humblest has the highest PI. So the second city in Humbleset has to be Zingaloo with a weighted PI of 22.5.
PI(Humbleset) = 7.5 + 20 + 22.5 = 50
For the PIs of the states to be integers, the cities with PIs 12.5 and 17.5 have to be in the same state. This will make
the sum 12.5 + 17.5 = 30 and the sum of the other pair of cities = 15 + 20 = 35 Note that Fogglia has to have the lowest PI. The only possible combination is:
PI (Fogglia) = 12.5 + 17.5 + 5 = 35
PI (Whimshire) = 15 + 20 + 10 = 45
So we have the final table as follows:

The PI of Humbleset = 50

Q. 21 Which pair of cities definitely belong to the same state?

1) Splutterville, Quackford

2) Mumpypore, Zingaloo

3) Noodleton, Quackford

4) Blusterburg, Mumpypore

Correct Answer

3

Explanation

Step 1:
According to the given information the nine PIs are all distinct multiples of 10, ranging from 10 to 90. The values will be {10, 20, 30, …, 80, 90}
Order of PIs of the 6 cities: Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo.
There is only one pair of an NUR and a city (considering all cities and all NURs) where the PI of the NUR is greater than that of the city. This has to be Blusterburg with a PI of 30 and the respective NUR has to have a PI of 40. Any other combination would contradict the condition. Also these two (NUR and Blusterburg) belong to Humbleset.
The remaining cities in increasing order will have PIs from 50 to 90 and the remaining two NURs will have PIs 10 and 20.

Step 2:
Given that, the PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.
For Humbleset to have an integer PI, the remaining city can have a weighted PI of (12.5, 17.5, 22.5}
But 12.5 and 17.5 will not satisfy the condition that Humblest has the highest PI. So the second city in Humbleset has to be Zingaloo with a weighted PI of 22.5.
PI(Humbleset) = 7.5 + 20 + 22.5 = 50
For the PIs of the states to be integers, the cities with PIs 12.5 and 17.5 have to be in the same state. This will make the sum 12.5 + 17.5 = 30 and the sum of the other pair of cities = 15 + 20 = 35 Note that Fogglia has to have the lowest PI. The only possible combination is:
PI (Fogglia) = 12.5 + 17.5 + 5 = 35
PI (Whimshire) = 15 + 20 + 10 = 45
So we have the final table as follows:

The pair of cities that definitely belong to the same states Noodleton and Quackford.

Q. 22 For how many of the cities and NURs is it possible to identify their PM and the state they belong to?

Correct Answer

9

Explanation

Step 1:
According to the given information the nine PIs are all distinct multiples of 10, ranging from 10 to 90. The values will be {10, 20, 30, …, 80, 90}
Order of PIs of the 6 cities: Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo.
There is only one pair of an NUR and a city (considering all cities and all NURs) where the PI of the NUR is greater than that of the city. This has to be Blusterburg with a PI of 30 and the respective NUR has to have a PI of 40. Any other combination would contradict the condition. Also these two (NUR and Blusterburg) belong to Humbleset.
The remaining cities in increasing order will have PIs from 50 to 90 and the remaining two NURs will have PIs 10 and 20.

Step 2:
Given that, the PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.
For Humbleset to have an integer PI, the remaining city can have a weighted PI of (12.5, 17.5, 22.5}
But 12.5 and 17.5 will not satisfy the condition that Humblest has the highest PI. So the second city in Humbleset has to be Zingaloo with a weighted PI of 22.5.
PI(Humbleset) = 7.5 + 20 + 22.5 = 50
For the PIs of the states to be integers, the cities with PIs 12.5 and 17.5 have to be in the same state. This will make the sum 12.5 + 17.5 = 30 and the sum of the other pair of cities = 15 + 20 = 35 Note that Fogglia has to have the lowest PI. The only possible combination is:
PI (Fogglia) = 12.5 + 17.5 + 5 = 35
PI (Whimshire) = 15 + 20 + 10 = 45
So we have the final table as follows:

We can identify the pIs of all cities and NURs and also identify the state they belong to.