CAT 2025 LRDI Slot 2 Paper
Q. 1 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.
In which of the following years were Ananya and Bhaskar Gurubhai?
Check Solution
Ans: A
To be published
Q. 2 In which year did Charu begin her training under Pandit Meghnath?
Check Solution
Ans: B
To be published
Q. 3 In which of the following years were Bhaskar and Devendra Gurubhai?
Check Solution
Ans: A
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Q. 4 Which of the following statements is TRUE?
Check Solution
Ans: D
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Q. 5 In how many of the years between 2013-24, were only two of these four musicians training under these three Gurus?
Check Solution
Ans: 4
To be published
Q. 6 Instructions
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.
What was the total number of two-author and three-author papers written by Brajen?
Check Solution
Ans: 4
To be published
Q. 7 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.
Check Solution
Ans: A
To be published
Q. 8 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.
Check Solution
Ans: B
To be published
Q. 9 If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?
Check Solution
Ans: 3
To be published
Q. 10 Instructions
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.
What was the total number of pings made by B1, B2, and B3?
Check Solution
Ans: 6
Let’s order the hoops and spheres by their diameter based on the provided details.
From the first piece of information, we learn that spheres B1 and B6 triggered a response from hoop H4, while sphere B5 did not. This allows us to conclude:
(B1, B6) < H4 < B5 --(1)
The second piece of information states that sphere B4 triggered a response from hoop H3, but sphere B1 did not. Therefore, we can establish:
B4 < H3 < B1 --(2)
According to the third piece of information, all spheres, with the exception of B3, triggered a response from hoop H1. This implies:
(B1, B2, B4, B5, B6) < H1 < B3 --(3)
The fourth piece of information tells us that only sphere B2 triggered a response from hoop H2, with no other spheres doing so. Consequently:
B2 < H2 < (B1, B3, B4, B5, B6) --(4)
By combining information (1) and (2), we can definitively arrange them as:
B4 < H3 < B1 < H4 < B5 --(5)
The position of sphere B6 remains uncertain; it could be smaller than H3 or larger than H3 but smaller than H4.
Integrating information (3), (4), and (5) yields the following arrangement:
B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3
Sphere B6 can occupy two potential positions, resulting in two possible sequences:
Sequence A: B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3
Sequence B: B2 < H2 < B4 < H3 < (B1, B6) < H4 < B5 < H1 < B3
Except for the exact placement of B6, the order of all other items is determined.
Now, let’s count the number of successful interactions (pings) for specific spheres:
Interactions by B1 = H4, H1 (Total: 2)
Interactions by B2 = H2, H3, H4, H1 (Total: 4)
Interactions by B3 = None (Total: 0)
The total count of interactions by spheres B1, B2, and B3 is the sum of their individual interactions: 2 + 4 + 0 = 6.
Therefore, the final answer is 6.
Q. 11 Which of the following statements about the relative sizes of the balls is NOT NECESSARILY true?
Check Solution
Ans: C
Let’s arrange the hoops and spheres by increasing diameter based on the provided details.
From the first piece of information, we learn that B1 and B6 triggered a signal at H4, while B5 did not. This allows us to establish:
(B1, B6) < H4 < B5 --(1)
The second detail states that B4 triggered a signal at H3, but B1 did not. Therefore, we can conclude:
B4 < H3 < B1 --(2)
According to the third piece of information, all balls except B3 triggered signals at H1. This leads to the conclusion:
(B1, B2, B4, B5, B6) < H1 < B3 --(3)
The fourth detail indicates that only B2 triggered a signal at H2, with no other balls doing so. This gives us:
B2 < H2 < (B1, B3, B4, B5, B6) --(4)
By combining insights (1) and (2), we can definitively state:
B4 < H3 < B1 < H4 < B5 --(5)
The exact placement of B6 remains uncertain; it could be smaller than H3 or larger than H3 but smaller than H4.
When we integrate (3), (4), and (5), we arrive at:
B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3
There are two potential arrangements for B6, resulting in these two possible orderings:
B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3
B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3
The position of B6 is not fully determined, but all other positions are fixed.
Let’s evaluate the options:
Option A) B4 < B5 < B3. This statement is always true.
Option B) B2 < B1 < B5. This statement is always true.
Option C) B1 < B6 < B3. This statement is not necessarily true, as the relative diameters of B1 and B6 are unknown.
Option D) B1 < B5 < B3. This statement is always true.
Therefore, the correct answer is option C.
Q. 12 Which of the following statements about the relative sizes of the hoops is true?
Check Solution
Ans: A
Let’s arrange the hoops and spheres in ascending order of their diameter based on the provided details.
From the first piece of information, we know that B1 and B6 registered on H4, while B5 did not. This establishes that:
(B1, B6) < H4 < B5 --(1)
The second piece of information tells us B4 registered on H3, but B1 did not. This means:
B4 < H3 < B1 --(2)
The third piece of information states that all spheres, except B3, registered on H1. Therefore:
(B1, B2, B4, B5, B6) < H1 < B3 --(3)
The fourth piece of information indicates that only B2 registered on H2. This implies:
B2 < H2 < (B1, B3, B4, B5, B6) --(4)
By merging information (1) and (2), we can confidently deduce:
B4 < H3 < B1 < H4 < B5 --(5)
The only sphere whose position relative to H3 is not fully determined is B6. It could be smaller than H3 or larger than H3 but smaller than H4.
Combining details (3), (4), and (5) leads to:
B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3
Sphere B6 can occupy two possible positions, resulting in two potential orderings:
B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3
B2 < H2 < B4 < H3 < (B1, B6) < H4 < B5 < H1 < B3
The precise placement of B6 remains uncertain, but the positions of all other items are fixed.
We have determined the definitive order of hoop diameters as H2 < H3 < H4 < H1.
Therefore, the correct selection is option A.
Q. 13 What BEST can be said about the total number of pings from all the tests undertaken?
Check Solution
Ans: A
Let’s establish an ordering of the hoops and spheres based on their diameters, using the provided details.
From the first piece of information, we learn that both B1 and B6 triggered a signal at H4, whereas B5 did not. This leads to the deduction:
(B1, B6) < H4 < B5 --(1)
The second detail states that B4 triggered a signal at H3, but B1 did not. Therefore, we can conclude:
B4 < H3 < B1 --(2)
The third piece of information indicates that all balls, with the exception of B3, triggered signals at H1. This means:
(B1, B2, B4, B5, B6) < H1 < B3 --(3)
According to the fourth detail, only B2 triggered a signal at H2; none of the other balls did. This implies:
B2 < H2 < (B1, B3, B4, B5, B6) --(4)
Combining the insights from (1) and (2) allows us to establish:
B4 < H3 < B1 < H4 < B5 --(5)
At this point, the exact placement of B6 is uncertain. It could be smaller than H3, or it could be larger than H3 but still smaller than H4.
Now, integrating (3), (4), and (5), we arrive at a primary ordering:
B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3
Considering the ambiguity with B6, there are two possible positions, leading to two potential overall orderings:
Scenario 1:
B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3
In this scenario, the possible number of signals is: B2 (4), B4 (3), B6 (3), B1 (2), B5 (1), and B3 (0). The total count is 4 + 3 + 3 + 2 + 1 = 13.
Scenario 2:
B2 < H2 < B4 < H3 < (B1, B6) < H4 < B5 < H1 < B3
In this scenario, the possible number of signals is: B2 (4), B4 (3), B6 (2), B1 (2), B5 (1), and B3 (0). The total count is 4 + 3 + 2 + 2 + 1 = 12.
Therefore, the total number of possible signals is either 12 or 13.
This leads to the conclusion that the correct option is A.
Q. 14 Instructions
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.
What was the SI of E in 2016?
Check Solution
Ans: 60
To be published
Q. 15 What was the SI of F in 2020?
Check Solution
Ans: 40
To be published
Q. 16 What was the SI of C in 2024?
Check Solution
Ans: 84
To be published
Q. 17 What was the SI of B in 2024?
Check Solution
Ans: D
To be published
Q. 18 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.
What is the PI of Whimshire?
Check Solution
Ans: 45
To be published
Q. 19 What is the PI of Fogglia?
Check Solution
Ans: 35
To be published
Q. 20 What is the PI of Humbleset?
Check Solution
Ans: 50
To be published
Q. 21 Which pair of cities definitely belong to the same state?
Check Solution
Ans: D
To be published
Q. 22 For how many of the cities and NURs is it possible to identify their PM and the state they belong to?
Check Solution
Ans: 9
To be published