CAT 2025 Slot 2 DILR Paper with Solutions

Set 1CAT 2025 · Slot 2 · DILR Q1–Q55 questionsScheduling & Timelines

Musicians training under three Gurus

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.

Set solution: how to crack this set

Raghunath can only teach in 2013–14 and 2019–20, so his span is 2 years. Samiran and Meghnath have spans 3 and 4.

Samiran teaches one student at a time, so Gurubhai pairs can form only under Meghnath or Raghunath.

In 2013 Ananya is with Meghnath and Bhaskar with Samiran, so Raghunath's 2013–14 block must be Charu and Devendra (their 2 Gurubhai years). His 2019–20 block is Ananya and Bhaskar (their 2 years).

Ananya–Charu and Bhaskar–Devendra must each overlap for exactly 2 years under Meghnath. Charu can't start before 2015, so Ananya's Meghnath block (from 2013) must run to 2016: Meghnath's span is 4 and Samiran's is 3. Charu: Meghnath 2015–18.

Samiran's four 3-year blocks fill 2013–24 exactly: 2013–15 Bhaskar, 2016–18 Devendra, 2019–21 Charu, 2022–24 Ananya.

Bhaskar's Meghnath years: 2021–24. Devendra must overlap him by 2 years: Meghnath 2019–22.

Student	Meghnath (4)	Samiran (3)	Raghunath (2)
Ananya	2013–16	2022–24	2019–20
Bhaskar	2021–24	2013–15	2019–20
Charu	2015–18	2019–21	2013–14
Devendra	2019–22	2016–18	2013–14

Q1CAT 2025 · Slot 2 · DILR Q1MCQ

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

Q2CAT 2025 · Slot 2 · DILR Q2MCQ

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

Q3CAT 2025 · Slot 2 · DILR Q3MCQ

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

Q4CAT 2025 · Slot 2 · DILR Q4MCQ

Which of the following statements is TRUE?

Q5CAT 2025 · Slot 2 · DILR Q5TITA

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

Set 2CAT 2025 · Slot 2 · DILR Q6–Q94 questionsCharts & Graphs

Research papers by four authors

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.

Data for this set (chart, table or figure)

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.

Set solution: how to crack this set

Papers: 10 single, 4 two-author, 3 three-author, 2 four-author. Author totals: Arman 5, Brajen 8, Chintan 12, Devon 10.

Every author is on both four-author papers. Brajen: singles $=$ two-author, so $2s + (\text{three}) = 6$ ; three-author must be even, so Brajen has 2 of each type.

Chintan and Devon wrote more three-author papers than Brajen, so 3 each, and Arman 1 (total incidences 9).

Three-author papers: $\{A, C, D\}$ , $\{B, C, D\}$ , $\{B, C, D\}$ .

Singles are distinct and sum to 10: $\{1, 2, 3, 4\}$ . Arman: $5 = s + t + 1 + 2$ , so 1 single and 1 two-author paper. Brajen has 2 singles, so Chintan and Devon have 3 and 4.

Two-author incidences $= 8$ : $1 + 2 + t_C + t_D = 8$ .

Case 1: Chintan 3 singles, 4 two-author; Devon 4 singles, 1 two-author.

Case 2: Chintan 4 singles, 3 two-author; Devon 3 singles, 2 two-author.

Q6CAT 2025 · Slot 2 · DILR Q6TITA

What was the total number of two-author and three-author papers written by Brajen?

Q7CAT 2025 · Slot 2 · DILR Q7MCQ

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.

Q8CAT 2025 · Slot 2 · DILR Q8MCQ

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.

Q9CAT 2025 · Slot 2 · DILR Q9TITA

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

Set 3CAT 2025 · Slot 2 · DILR Q10–Q134 questionsLogic Puzzles

Balls passing through hoops

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.

Set solution: how to crack this set

A ping means ball $\leq$ hoop.

H2: only B2 passes, so B2 is the smallest ball and H2 is smaller than every other ball. H1: everyone except B3 passes, so B3 is the largest ball.

H3: B4 passes, B1 doesn't, so $B4 \leq H3 < B1$ . H4: B1, B6 pass, B5 doesn't, so $B1, B6 \leq H4 < B5$ . And $B5 \leq H1 < B3$ .

Hoops: $H2 < H3 < B1 \leq H4 < B5 \leq H1$ , so $H2 < H3 < H4 < H1$ .

Pings: B2 passes all 4; B3 none; B1 on H1, H4 (2); B4 on H1, H3, H4 (3); B5 on H1 only (1); B6 on H1, H4 and maybe H3 (2 or 3).

Q10CAT 2025 · Slot 2 · DILR Q10TITA

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

Q11CAT 2025 · Slot 2 · DILR Q11MCQ

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

Q12CAT 2025 · Slot 2 · DILR Q12MCQ

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

Q13CAT 2025 · Slot 2 · DILR Q13MCQ

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

Set 4CAT 2025 · Slot 2 · DILR Q14–Q174 questionsCharts & Graphs

Sustainability Index of six countries

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.

Set solution: how to crack this set

From the plot (2016→2020, 2020→2024): A (+25%, +50%), B (−25%, −25%), C (−20%, +40%), D (+100%, +20%), E (+25%, +20%), F (+100%, −25%).

E: $SI_{2016} \times 1.25 \times 1.2 = 90 \Rightarrow 60$ .

All SIs are integers. B needs to be a multiple of 16, and C a multiple of 25. C (rank 2) lies between 60 and B, so C = 75 and B is 80 or 96.

2016 range 60: F = B − 60 = 20 or 36. F's 2024 value $= 1.5F$ , and 2024's range is 60 with a maximum of at most 100, so F = 20 (2024: 30, max 90 = E).

So B = 80, C = 75, F = 20 in 2016.

Q14CAT 2025 · Slot 2 · DILR Q14TITA

What was the SI of E in 2016?

Q15CAT 2025 · Slot 2 · DILR Q15TITA

What was the SI of F in 2020?

Q16CAT 2025 · Slot 2 · DILR Q16TITA

What was the SI of C in 2024?

Q17CAT 2025 · Slot 2 · DILR Q17MCQ

What was the SI of B in 2024?

Set 5CAT 2025 · Slot 2 · DILR Q18–Q225 questionsQuant-based Sets

Pollution measures of cities and non-urban regions

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.

Set solution: how to crack this set

Only one NUR beats only one city, and both are in Humbleset. Any NUR above a city is above all smaller cities too, so that city must be the smallest, Blusterburg, and Humbleset's NUR lies between Blusterburg and Noodleton. The other two NURs are below all cities.

So: NURs 10, 20; Blusterburg 30; Humbleset NUR 40; Noodleton 50, Splutterville 60, Quackford 70, Mumpypore 80, Zingaloo 90.

Humbleset PI $= 20 + 7.5 + \dfrac{x}{4}$ (x = its other city) must be an integer, so x is 50, 70 or 90. For the other states, the two city PMs must add to a multiple of 20.

x = 50 or 70 makes another state's PI at least Humbleset's. x = 90 gives Humbleset 50, and the remaining cities pair as {50, 70} and {60, 80}.

NUR 10 + {50, 70}: 35 (Fogglia, lowest). NUR 20 + {60, 80}: 45 (Whimshire). (The other pairing gives a tie at 40.)

Q18CAT 2025 · Slot 2 · DILR Q18TITA

What is the PI of Whimshire?

Q19CAT 2025 · Slot 2 · DILR Q19TITA

What is the PI of Fogglia?

Q20CAT 2025 · Slot 2 · DILR Q20TITA

What is the PI of Humbleset?

Q21CAT 2025 · Slot 2 · DILR Q21MCQ

Which pair of cities definitely belong to the same state?

Q22CAT 2025 · Slot 2 · DILR Q22TITA

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

CAT 2025 Slot 2 DILR paper with solutions

Musicians training under three Gurus

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Research papers by four authors

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Balls passing through hoops

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Sustainability Index of six countries

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Pollution measures of cities and non-urban regions

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