CAT 2025 Slot 2 — DILR questions with answers

Q25MCQTeam Selection & Scheduling

1. A2021
2. B2018
3. C2014
4. D2020

Answer and solution

Answer: (D) 2020

Write A, B, C, D for the four musicians, and R, S, M for Raghunath, Samiran and Meghnath. R teaches only in 2013–14 and 2019–20, so his span is 2 years, two students per window. In 2013 A is with M and B with S, so R has C and D in 2013–14 and A and B in 2019–20. That gives CD and AB their 2 Gurubhai years. S teaches one at a time, so AC and BD share 2 years under M. A starts with M in 2013 while C is with R until 2014, so a 3-year span would give them at most one year. Hence M's span is 4 and S's is 3: A is with M in 2013–16, joined by C in 2015–16 (AD is barred; B is with S), and C stays till 2018. B's only free 4-year stretch is 2021–24, his M span; D overlaps it by exactly 2 years with 2019–22. Under S: B 2013–15, D 2016–18. A is free for 3 years only from 2021, so A takes 2022–24 and C 2019–21. A and B are Gurubhai only under R in 2019–20. In 2014 A is with M and B with S; in 2018 neither trains; in 2021 B is with M but A is not training. Hence, option D (2020).

Q26MCQTeam Selection & Scheduling

1. A2021
2. B2015
3. C2016
4. D2017

Answer and solution

Answer: (B) 2015

Write A, B, C, D for the four musicians, and R, S, M for Raghunath, Samiran and Meghnath. R teaches only in 2013–14 and 2019–20, so his span is 2 years, two students per window. In 2013 A is with M and B with S, so R has C and D in 2013–14 and A and B in 2019–20. That gives CD and AB their 2 Gurubhai years. S teaches one at a time, so AC and BD share 2 years under M. A starts with M in 2013 while C is with R until 2014, so a 3-year span would give them at most one year. Hence M's span is 4 and S's is 3: A is with M in 2013–16, joined by C in 2015–16 (AD is barred; B is with S), and C stays till 2018. B's only free 4-year stretch is 2021–24, his M span; D overlaps it by exactly 2 years with 2019–22. Under S: B 2013–15, D 2016–18. A is free for 3 years only from 2021, so A takes 2022–24 and C 2019–21. Charu's training under M runs from 2015 to 2018, so it began in 2015. Option C (2016) is its second year, when A and C were Gurubhai. Hence, option B (2015).

Q27MCQTeam Selection & Scheduling

1. A2018
2. B2020
3. C2015
4. D2022

Answer and solution

Answer: (D) 2022

Write A, B, C, D for the four musicians, and R, S, M for Raghunath, Samiran and Meghnath. R teaches only in 2013–14 and 2019–20, so his span is 2 years, two students per window. In 2013 A is with M and B with S, so R has C and D in 2013–14 and A and B in 2019–20. That gives CD and AB their 2 Gurubhai years. S teaches one at a time, so AC and BD share 2 years under M. A starts with M in 2013 while C is with R until 2014, so a 3-year span would give them at most one year. Hence M's span is 4 and S's is 3: A is with M in 2013–16, joined by C in 2015–16 (AD is barred; B is with S), and C stays till 2018. B's only free 4-year stretch is 2021–24, his M span; D overlaps it by exactly 2 years with 2019–22. Under S: B 2013–15, D 2016–18. A is free for 3 years only from 2021, so A takes 2022–24 and C 2019–21. B and D are Gurubhai only under M in 2021–22. In 2020 D is with M but B is with R; in 2015 and 2018 only one of them is training. Hence, option D (2022).

Q28MCQTeam Selection & Scheduling

1. AAnanya was training under Ustad Samiran in 2018.
2. BCharu was training under Ustad Samiran in 2019.
3. CAnanya was training under Ustad Samiran in 2015.
4. DCharu was training under Ustad Samiran in 2018.

Answer and solution

Answer: (B) Charu was training under Ustad Samiran in 2019.

Write A, B, C, D for the four musicians, and R, S, M for Raghunath, Samiran and Meghnath. R teaches only in 2013–14 and 2019–20, so his span is 2 years, two students per window. In 2013 A is with M and B with S, so R has C and D in 2013–14 and A and B in 2019–20. That gives CD and AB their 2 Gurubhai years. S teaches one at a time, so AC and BD share 2 years under M. A starts with M in 2013 while C is with R until 2014, so a 3-year span would give them at most one year. Hence M's span is 4 and S's is 3: A is with M in 2013–16, joined by C in 2015–16 (AD is barred; B is with S), and C stays till 2018. B's only free 4-year stretch is 2021–24, his M span; D overlaps it by exactly 2 years with 2019–22. Under S: B 2013–15, D 2016–18. A is free for 3 years only from 2021, so A takes 2022–24 and C 2019–21. Charu is with S in 2019. Options A and C fail, as Ananya is with S only in 2022–24; D fails, as Charu is with M in 2018. Hence, option B (Charu was training under Ustad Samiran in 2019.).

Q29TITATeam Selection & Scheduling

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 4

Write A, B, C, D for the four musicians, and R, S, M for Raghunath, Samiran and Meghnath. R teaches only in 2013–14 and 2019–20, so his span is 2 years, two students per window. In 2013 A is with M and B with S, so R has C and D in 2013–14 and A and B in 2019–20. That gives CD and AB their 2 Gurubhai years. S teaches one at a time, so AC and BD share 2 years under M. A starts with M in 2013 while C is with R until 2014, so a 3-year span would give them at most one year. Hence M's span is 4 and S's is 3: A is with M in 2013–16, joined by C in 2015–16 (AD is barred; B is with S), and C stays till 2018. B's only free 4-year stretch is 2021–24, his M span; D overlaps it by exactly 2 years with 2019–22. Under S: B 2013–15, D 2016–18. A is free for 3 years only from 2021, so A takes 2022–24 and C 2019–21. All four train in 2013–14 and 2019–20, and three in 2015–16 and 2021–22. Only C and D train in 2017–18, and only A and B in 2023–24. The answer is 4.

Q30TITAPie & Scatter Charts

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Answer and solution

Answer: 60

In the plot, the x-coordinate is the % change in SI from 2016 to 2020 and the y-coordinate the % change from 2020 to 2024. E is at $(25\%, 20\%)$. By fact 3, E's SI in 2024 is 90. A 20% rise from 2020 to 2024 means $E_{2020} \times 1.2 = 90$, so $E_{2020} = 75$. A 25% rise from 2016 to 2020 means $E_{2016} \times 1.25 = 75$, so $E_{2016} = 60$. This fits fact 1, where E ranks third in 2016, below B and C. The table shows E's values alongside those of B and F, which follow from facts 2 and 4: The answer is 60.

Q31TITAPie & Scatter Charts

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Answer and solution

Answer: 40

In the plot, x is the % change from 2016 to 2020 and y the % change from 2020 to 2024: A $(25\%, 50\%)$, B $(-25\%, -25\%)$, C $(-20\%, 40\%)$, D $(100\%, 20\%)$, E $(25\%, 20\%)$, F $(100\%, -25\%)$. E: $E_{2024} = 90$ (fact 3), so $E_{2020} = 90/1.2 = 75$ and $E_{2016} = 75/1.25 = 60$. No country is above 90 in 2024. Every SI is an integer. C: $C_{2024} = 1.12c = \frac{28c}{25}$, so $c$ is a multiple of 25 between 60 and 100: $c = 75$, $C_{2024} = 84$. A: $A_{2024} = \frac{15a}{8}$ with $a < 60$ a multiple of 8. $a = 56$ gives 105 (over 100) and $a = 48$ gives 90 (only E has 90), so $a \le 40$ and $A_{2024} \le 75$. D: $D_{2024} = 2.4d$ with $d < a \le 40$ and $d$ a multiple of 5, so $d \le 35$ and $D_{2024} \le 84$. B falls in both periods, to at most 56. So the 2024 maximum is 90, and with a range of 60 the lowest, F, has 30. Then $F_{2020} = \frac{30}{0.75} = 40$. The answer is 40.

Q32TITAPie & Scatter Charts

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Q33MCQPie & Scatter Charts

1. A54
2. B60
3. C45
4. D80

Answer and solution

Answer: (C) 45

In the plot, x is the % change from 2016 to 2020 and y the % change from 2020 to 2024. B is at $(-25\%, -25\%)$ and F at $(100\%, -25\%)$. E is at $(25\%, 20\%)$ with $E_{2024} = 90$ (fact 3), so $E_{2020} = 75$ and $E_{2016} = 60$. No country tops 90 in 2024 (every SI is an integer). C at $(-20\%, 40\%)$ gives $C_{2024} = \frac{28c}{25}$, so $c$ is a multiple of 25 between 60 and 100: $c = 75$ and $C_{2024} = 84$. A at $(25\%, 50\%)$ gives $\frac{15a}{8}$ with $a < 60$ a multiple of 8; $a = 56$ gives 105 and $a = 48$ gives 90, so A is at most 75. D at $(100\%, 20\%)$ gives $2.4d$ with $d < a \le 40$ a multiple of 5, so at most 84. B falls. With a 2024 range of 60, F has $90 - 60 = 30$, so $F_{2020} = 40$ and $F_{2016} = 20$. The 2016 range of 60 then gives the top country $B_{2016} = 80$, so $B_{2020} = 60$ and $B_{2024} = 45$. Option B (60) is B's 2020 value and option D (80) its 2016 value. Hence, option C (45).

Q34TITALogical Puzzles

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Q35TITALogical Puzzles

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Q36TITALogical Puzzles

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Q37MCQLogical Puzzles

1. ABlusterburg, Mumpypore
2. BMumpypore, Zingaloo
3. CNoodleton, Quackford
4. DSplutterville, Quackford

Q38TITALogical Puzzles

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Q39TITA

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Answer and solution

Answer: 4

From the charts: Arman wrote 5 papers, Brajen 8, Chintan 12 and Devon 10; there were 10 single-, 4 two-, 3 three- and 2 four-author papers. Every four-author paper includes all four authors, so each author wrote 2 of them. Arman's other $5 - 2 = 3$ papers cover the three remaining types, at least one each, so Arman wrote exactly 1 three-author paper. The 3 three-author papers have $3 \times 3 = 9$ author places, so Brajen, Chintan and Devon fill 8. Nobody can be on more than 3, and Chintan and Devon each exceed Brajen (fact 3). Brajen cannot have 3, and with 1 the other two would need 7 places, more than $3 + 3$. So Brajen wrote 2 (Chintan and Devon 3 each). Brajen's single- and two-author papers total $8 - 2 - 2 = 4$ and are equal (fact 4), so he wrote 2 of each. Brajen's two-author plus three-author papers: $2 + 2 = 4$. The answer is 4.

Q40MCQ

1. AOnly ii
2. BBoth i and ii
3. CNeither i nor ii
4. DOnly i

Answer and solution

Answer: (C) Neither i nor ii

From the charts: Arman wrote 5 papers, Brajen 8, Chintan 12, Devon 10; there were 10 single-, 4 two-, 3 three- and 2 four-author papers. Every author is on both four-author papers, so Arman's other 3 papers are one of each other type. The 9 three-author places then split Arman 1, Brajen 2, Chintan 3, Devon 3 (Chintan and Devon above Brajen, nobody above 3). Brajen's other 4 papers are 2 single and 2 two-author (fact 4). The four single-author counts are distinct and total 10, so they are 1, 2, 3, 4; Chintan and Devon have 3 and 4 in some order. Chintan's single plus two-author papers total $12 - 3 - 2 = 7$, Devon's $10 - 3 - 2 = 5$. Both orders work (two-author places total 8 either way): Case 1: Chintan 3 single, 4 two-author; Devon 4 single, 1 two-author. Case 2: Chintan 4 single, 3 two-author; Devon 3 single, 2 two-author. Statement i (Chintan wrote three two-author papers) and statement ii (Chintan wrote more single-author papers than Devon) both hold only in Case 2. Case 1 breaks both, so neither is necessarily true; options A, B and D each need at least one of them. Hence, option C (Neither i nor ii).

Q41MCQ

1. AOnly I
2. BNeither i or ii
3. COnly ii
4. DBoth i and ii

Answer and solution

Answer: (D) Both i and ii

From the charts: Arman wrote 5 papers, Brajen 8, Chintan 12, Devon 10; there were 3 three-author and 2 four-author papers. Every author is on both four-author papers. Arman's other 3 papers must then be one single, one two-author and one three-author paper, since each type needs at least one. The 3 three-author papers have $3 \times 3 = 9$ author places. Arman takes 1, leaving 8 for Brajen, Chintan and Devon, with nobody above 3 and Chintan and Devon each above Brajen (fact 3). Brajen cannot have 3, and with 1 the other two would need 7 places, more than $3 + 3$. So Brajen 2, Chintan 3, Devon 3. The two possible full tables differ only in single- and two-author counts; the three-author column is 1, 2, 3, 3 in both. Chintan and Devon are on all 3 three-author papers, so each such paper is Chintan, Devon and one more author: Arman once, Brajen twice. The papers are {Arman, Chintan, Devon} and {Brajen, Chintan, Devon} twice. So both statements are true. Options A and C each leave out a statement that holds. Hence, option D (Both i and ii).

Q42TITA

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 3

From the charts: Arman wrote 5 papers, Brajen 8, Chintan 12, Devon 10; there were 10 single-, 4 two-, 3 three- and 2 four-author papers. Every author is on both four-author papers, so Arman's other 3 papers are one of each other type. The 9 three-author places then split Arman 1, Brajen 2, Chintan 3, Devon 3 (Chintan and Devon above Brajen, nobody above 3). Brajen's other 4 papers are 2 single and 2 two-author (fact 4). The four single-author counts are distinct and total 10, so they are 1, 2, 3, 4, and Chintan and Devon have 3 and 4. Devon's single plus two-author papers total $10 - 3 - 2 = 5$, and Chintan's total $12 - 3 - 2 = 7$. If Devon wrote more than one two-author paper, Devon cannot have 4 singles, which would leave only 1 two-author paper. So Devon has 3 singles and 2 two-author papers, and Chintan has 4 singles and $7 - 4 = 3$ two-author papers. Check: two-author places $1 + 2 + 3 + 2 = 8 = 4 \times 2$. The answer is 3.

Q43TITALogical Puzzles

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Answer and solution

Answer: 6

A ball pings on a hoop exactly when it is not larger than the hoop. Fact 3: every ball but B3 passes H1, so B3 is larger than every other ball. Fact 4: only B2 passes H2, so B2 is the smallest. On H3, B4 passes and B1 does not, so $B1 > B4$. On H4, B1 and B6 pass but B5 does not, so $B5 > B1$ and $B5 > B6$. So $B3 > B5 > B1 > B4 > B2$ and $B2 < B6 < B5$. This fills the table: B3 fails H3 and H4 (it is larger than B1 and B5, which fail there); B2 passes H3 and H4 (it is smaller than B4 and B1, which pass); B4 passes H4 (smaller than B1); B5 fails H3 (larger than B1). Only B6 on H3 stays open. B1 pings on H1 and H4 (2), B2 on all four hoops (4), and B3 on none (0). Total $= 2 + 4 + 0 = 6$. The answer is 6.

Q44MCQLogical Puzzles

1. AB1 < B6 < B3
2. BB2 < B1 < B5
3. CB4 < B5 < B3
4. DB1 < B5 < B3

Answer and solution

Answer: (A) B1 < B6 < B3

A ball pings on a hoop exactly when it is not larger than the hoop. Fact 3: every ball but B3 passes H1, so B3 is larger than every other ball. Fact 4: only B2 passes H2, so B2 is the smallest. On H3, B4 passes and B1 does not, so $B1 > B4$. On H4, B1 and B6 pass but B5 does not, so $B5 > B1$ and $B5 > B6$. So $B3 > B5 > B1 > B4 > B2$ and $B2 < B6 < B5$, with B6 never compared with B1 or B4. Options B, C and D follow from the chain: $B2 < B1 < B5$, $B4 < B5 < B3$ and $B1 < B5 < B3$ are all certain. Option A needs $B1 < B6$, but B6 is only known to be below B5. It could be smaller than B4 (then it passes H3) or lie between B1 and B5 (then it does not), and every fact still holds. So $B1 < B6 < B3$ is not necessarily true. Hence, option A (B1 < B6 < B3).

Q45MCQLogical Puzzles

1. AH2 < H4 < H3 < H1
2. BH2 < H3 < H4 < H1
3. CH1 < H3 < H4 < H2
4. DH1 < H4 < H3 < H2

Answer and solution

Answer: (B) H2 < H3 < H4 < H1

A ball pings on a hoop exactly when it is not larger than the hoop. So if a ball passes one hoop but gets stuck on another, the first hoop is the larger. B4 passes H3 (fact 2) but not H2 (fact 4): $H2 < B4 \le H3$. B1 passes H4 (fact 1) but not H3 (fact 2): $H3 < B1 \le H4$. B5 passes H1 (fact 3) but not H4 (fact 1): $H4 < B5 \le H1$. So $H2 < H3 < H4 < H1$. The completed table agrees: H2 gets 1 ping (B2), H3 gets 2 or 3 (B2, B4 and perhaps B6), H4 gets 4 (B1, B2, B4, B6) and H1 gets 5 (all but B3). A larger hoop lets through every ball a smaller one does. Option A swaps H3 and H4, but B1 passes H4 and not H3, so H4 is the larger. Options C and D put H1 below H2, but H2 is the smallest hoop and H1 the largest. Hence, option B (H2 < H3 < H4 < H1).

Q46MCQLogical Puzzles

1. A12 or 13 or 14
2. B13 or 14
3. CAt least 9
4. D12 or 13

Answer and solution

Answer: (D) 12 or 13

A ball pings on a hoop exactly when it is not larger than the hoop. Fact 3: every ball but B3 passes H1, so B3 is the largest. Fact 4: only B2 passes H2, so B2 is the smallest. On H3, B4 passes and B1 does not, so $B1 > B4$. On H4, B1 and B6 pass but B5 does not, so $B5 > B1$ and $B5 > B6$. So B3 fails H3 and H4 (larger than B1 and B5), B2 passes H3 and H4 (smaller than B4 and B1), B4 passes H4 (smaller than B1), and B5 fails H3 (larger than B1). Pings: H1 5, H2 1, H4 4 (B1, B2, B4, B6), H3 2 (B2, B4) plus B6 if it fits. B6 only has to lie between B2 and B5, so it can be smaller than B4 (a ping on H3) or larger than B1 (no ping). Both are possible. Total $= 5 + 1 + 4 + 2 = 12$, or 13 with B6's ping. Option A adds 14, but B6 on H3 is the only open test. Option C (at least 9) is true but not the best. Hence, option D (12 or 13).