CAT 2023 Slot 1 — DILR questions with answers

Q25TITATeam Selection & Scheduling

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

Q26TITATeam Selection & Scheduling

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

Answer and solution

Answer: 200

Each slot has 5 US applications (half of 200, spread over 20 slots). Ira, Vijay and Nandini show at least 3 Schengen applications per slot, and Mahira and Osman at least 2 Others. That fills all 10 places, so each slot has 3 Schengen, 2 Others and no UK. The day has $2 \times 20 = 40$ Others applications. At 9:20 the UK counters have nothing to do. The 4 US counters are all busy, and both Schengen counters are either processing or about to take Ira and Vijay, who are waiting. These are the six active counters, so both Others counters were idle at 9:20. Let an Others application take $t$ minutes. The 9:15 pair arrived on time, could not start before 9:15, and had finished by 9:20. So $t \le 5$. The table shows this largest case, 5 minutes per Others application. The total time for all Others applications is therefore at most $40 \times 5 = 200$ minutes. The answer is 200.

Q27MCQTeam Selection & Scheduling

1. A9:50 am
2. B9:37 am
3. C9:35 am
4. D9:45 am

Answer and solution

Answer: (D) 9:45 am

Each slot has 3 Schengen applications (5 US, at least 3 Schengen and at least 2 Others fill the 10 places), handled at 2 counters in order. The 9:00 slot's three, S1, S2 and S3, come before Ira, Vijay and Nandini, who arrived at 9:20. Let a Schengen application take $t$ minutes. S1 and S2 start at 9:00, and S3 starts $t$ minutes past 9:00, when they finish. If $2t \le 20$, both counters are free by 9:20 and Vijay would be called at 9:20. If $t \ge 20$, S3 and Ira both start at or after 9:20, and Vijay waits until at least 9:40. So S3 is still at one counter at 9:20, Ira takes the other at 9:20, and Vijay is called when S3 ends, $2t$ minutes past 9:00. Since that is 9:25, $t = 12.5$. Ira runs 9:20 to 9:32:30 and Vijay 9:25 to 9:37:30. Nandini takes Ira's counter at 9:32:30 and finishes 12.5 minutes later, at 9:45. Option B (9:37) is when Vijay's processing ends, not Nandini's. Hence, option D (9:45 am).

Q28MCQTeam Selection & Scheduling

1. AThe application process of Osman was completed before 9:45 am.
2. BThe application process of Mahira started after Nandini’s.
3. CThe application process of Osman was completed before Vijay’s.
4. DThe application process of Mahira was completed before Nandini’s.

Answer and solution

Answer: (B) The application process of Mahira started after Nandini’s.

Each slot has 5 US, 3 Schengen and 2 Others applications (5 US from the 50% rule; the named applicants give at least 3 Schengen and 2 Others, which fills the 10 places). Schengen: Vijay, 5th in the Schengen line, was called at 9:25. That works only if a Schengen application takes 12.5 minutes, so that the third 9:00 applicant runs 9:12:30 to 9:25. So Ira runs 9:20 to 9:32:30, Vijay 9:25 to 9:37:30 and Nandini 9:32:30 to 9:45. Others: both Others counters were idle when Ira's group entered at 9:20, so the 9:15 pair had already finished. An Others application takes at most 5 minutes. Mahira and Osman arrived on time at 9:30 and found both Others counters free, so both started at 9:30 and finished by 9:35. A is true: Osman finished by 9:35, before 9:45. C is true: Osman finished by 9:35, before Vijay at 9:37:30. D is true: Mahira finished by 9:35, before Nandini at 9:45. B is false: Mahira started at 9:30 and Nandini at 9:32:30, so Mahira started first. Hence, option B (The application process of Mahira started after Nandini’s.).

Q29MCQTeam Selection & Scheduling

1. A2:05 pm
2. B2:25 pm
3. C2:00 pm
4. D3:40 pm

Answer and solution

Answer: (A) 2:05 pm

Each slot has 5 US applications (half of 200, spread over 20 slots), handled at 4 counters, 10 minutes each. 9:00 slot: four run 9:00 to 9:10, and the fifth runs 9:10 to 9:20. 9:15 slot: three counters are free at 9:15, so three run 9:15 to 9:25. The fourth starts at 9:20, when the 9:00 slot's fifth ends, and the fifth starts at 9:25 and ends at 9:35. Every later slot starting at time $T$ goes the same way: three start at $T$, the fourth at $T + 5$ and the fifth at $T + 10$, so the slot's last US application ends at $T + 20$. For the 9:30 slot that is 9:50, as the table shows. The last slot starts at 1:45 pm, so all US processing ends at 2:05 pm. Equivalently, the first slot takes 20 minutes and each of the other 19 adds 15: $20 + 19 \times 15 = 305$ minutes after 9:00 am. Option C (2:00 pm) is when the last slot's time runs out, but its fifth US applicant is at a counter until 2:05 pm. Hence, option A (2:05 pm).

Q30TITALogical Puzzles

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

Answer and solution

Answer: 3

Quoted price $= \text{base} + 5a + 3b$, where $a$ is the road adjacency value, $b$ is the neighbour count, and the base is 12 with parking or 10 without. A house has at most 3 neighbours, so $b \le 3$. The highest quoted price in Block XX is 24. With parking: $12 + 5a + 3b = 24$ gives $5a + 3b = 12$, whose only whole-number solution is $a = 0, b = 4$. That is impossible. Without parking: $10 + 5a + 3b = 24$ gives $5a + 3b = 14$, so $a = 1, b = 3$. In Block XX only B1 and B2 can have 3 neighbours, and only B2 touches a road. So B2 is vacant, and its neighbours B1, A2 and C2 are occupied. Row-1 has only one occupied house in each block. In Block XX that is B1, so A1 and C1 are vacant. (In the figure, U marks a vacant house and O an occupied one.) Check: A1 is quoted at most $12 + 0 + 3 \times 2 = 18$ and C1 at most $12 + 5 + 3 \times 2 = 23$, both below 24. The vacant houses in Block XX are A1, C1 and B2. The answer is 3.

Q32MCQLogical Puzzles

1. AExactly 3
2. BEither 3 or 4
3. CExactly 2
4. DEither 2 or 3

Answer and solution

Answer: (D) Either 2 or 3

Quoted price $= \text{base} + 5a + 3b$ ($a$ = road adjacency, $b$ = neighbour count; base 12 with parking, 10 without). Block XX: a price of 24 needs $5a + 3b = 12$ with parking (only $b = 4$, impossible) or $5a + 3b = 14$ without, so $a = 1, b = 3$. Only B2 fits: B2 is vacant and B1, A2, C2 are occupied. Block YY: E1 and E2 are vacant, and exactly one of D1, F1 is occupied (Row-1 has one per block). E2 at 15 would need no parking and $b = 0$ (D2, F2 vacant); Columns D and F would then need D1 and F1 both occupied, which Row-1 forbids. So E1 is at 15: with $a = 0$ and $b = 1$, only $12 + 3 = 15$ works, so E1 has parking. If D1 were occupied, F1 (no parking, $a = 0$, $b \le 1$) would be quoted at most 13, below 15. So F1 is occupied, D1 vacant, and D2 occupied (Column-D). F2 may be either. (U = vacant, O = occupied.) Vacant in Row-2: B2, E2 and possibly F2, so 2 or 3. Option B fails, as A2, C2 and D2 are occupied; options A and C wrongly fix F2. Hence, option D (Either 2 or 3).

Q33TITALogical Puzzles

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

Answer and solution

Answer: 21

Quoted price $= \text{base} + 5a + 3b$, where $a$ is road adjacency and $b$ is neighbour count. Block XX: a price of 24 needs $5a + 3b = 12$ (with parking) or $14$ (without). Only $a = 1, b = 3$ works, and only B2 can have both. So B2 is vacant and B1, A2 and C2 are occupied. Block YY: if E2 ($a = 1$) cost 15, it would need no parking and $b = 0$. Then D2 and F2 would be vacant, forcing D1 and F1 to be occupied: two occupied Row-1 houses in YY, against fact 2. So E1 ($a = 0$) is the 15-lakh house. $10 + 3b = 15$ has no whole-number solution, so E1 has YY's only parking space, and $12 + 3b = 15$ gives $b = 1$: exactly one of D1, F1 is occupied. If that were D1, F2 would have to be occupied (fact 3) and F1 would cost only $10 + 3 = 13 < 15$. So F1 is occupied, D1 vacant and D2 occupied (O is occupied, U vacant). E2 has no parking, $a = 1$, and neighbours D2 (occupied), E1 (vacant) and F2. Its price is highest when F2 is occupied: $10 + 5 + 3 \times 2 = 21$. The answer is 21.

Q34MCQLogical Puzzles

1. AE1
2. BF2
3. CE2
4. DF1

Answer and solution

Answer: (A) E1

Quoted price $= \text{base} + 5a + 3b$, where $a$ is the road adjacency value, $b$ the neighbour count, and the base is 12 with parking or 10 without. Both Column-E houses are vacant. Row-1 has exactly one occupied house in Block YY, so exactly one of D1 and F1 is occupied. The lowest quoted price in Block YY is 15, and a Column-E house has it. E2 touches the front road, so $a = 1$. With parking it costs at least 17. Without parking, $10 + 5 + 3b = 15$ needs $b = 0$: D2 and F2 vacant. Then Columns D and F could have occupied houses only at D1 and F1, and Row-1 allows only one of them. So E2 is not quoted at 15. So E1 is. It has $a = 0$, and its neighbours are D1, F1 and the vacant E2, so $b = 1$. Without parking, $10 + 3 = 13$; with parking, $12 + 3 = 15$. So E1 has parking. (In the figure, U marks a vacant house and O an occupied one.) Block YY has only one house with parking, so F2, E2 and F1 (options B, C and D) have none. Hence, option A (E1).

Q35TITAMissing Value Tables

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

Answer and solution

Answer: 0

Totals are 5 × mean: U 11, V 19, W 17, X 18, Y 13; R1 17, R2 11, R3 19, R4 14, R5 17. The summary table, given 1s and 5s and totals fix each worker's ratings: U 1, 2, 2, 2, 4 (a 1 and range 3 cap it at 4); V 2, 4, 4, 4, 5 (a 5 and range 3 set the lowest at 2); W 1, 2, 4, 5, 5; X 1, 3, 4, 5, 5 (range 4 below the 5s gives a 1); Y 1, 1, 3, 4, 4 (modes 1 and 4, median 3). R3: U + V = 8, each 2 or 4, so both give 4; U's other ratings are all 2. R2: V = 11 − 9 = 2, so V gives 4 in R1 and R4. R5: U 2 and V 5 leave W + X + Y = 10: (W, X, Y) is (2, 4, 4) or (4, 3, 3). R1 (U 1, V 4, W 5) needs X + Y = 7, which needs X's or Y's 3; (4, 3, 3) would use both, so R5 is (2, 4, 4). Then R1 is X 3, Y 4, and R4 is W 4, X 1, Y 3. All ratings are fixed. The answer is 0.

Q36TITAMissing Value Tables

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

Answer and solution

Answer: 0

Totals are 5 × mean: R2 gives 11 in all and R3 gives 19; U receives 11 and V receives 19. Given ratings: in R2, W 1, X 5, Y 1; in R3, W 5, X 5, Y 1. U has a 1 (from R1) and range 3, so the top is 4. With median and mode 2 and sum 11, U's ratings are 1, 2, 2, 2, 4. V has a 5 (from R5) and range 3, so the lowest is 2. With median and mode 4 and sum 19, V's ratings are 2, 4, 4, 4, 5. In R3, U + V = 19 − (5 + 5 + 1) = 8. Each of them has only 2 or 4 left, so both give 4. U's one 4 is now used, so U's R2 rating is 2. In R2, V = 11 − (2 + 1 + 5 + 1) = 2. R2's ratings are U 2, V 2, W 1, X 5, Y 1. None of them is 4. The answer is 0.

Q37TITAMissing Value Tables

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

Answer and solution

Answer: 3

Totals are 5 × mean: U 11, V 19, W 17, X 18, Y 13; R1 17, R2 11, R3 19, R5 17. The summary table, given 1s and 5s and totals fix each worker's ratings: U 1, 2, 2, 2, 4 (a 1 and range 3 cap it at 4); V 2, 4, 4, 4, 5 (a 5 and range 3 set the lowest at 2); W 1, 2, 4, 5, 5; X 1, 3, 4, 5, 5 (range 4 below the 5s gives a 1); Y 1, 1, 3, 4, 4 (modes 1 and 4, median 3). R3: U + V = 8, each 2 or 4, so both give 4; U's other ratings are all 2. R2: V = 11 − 9 = 2, so V gives 4 in R1. R1: U 1, V 4, W 5 leave X + Y = 7, which needs a 3 from X or Y. R5: U 2 and V 5 leave W + X + Y = 10, so (W, X, Y) is (2, 4, 4) or (4, 3, 3). The latter leaves no 3 for R1, so R5 is (2, 4, 4). X is left with 1 and 3, so X = 3 and Y = 4. The answer is 3.

Q38TITAMissing Value Tables

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

Answer and solution

Answer: 4

R3's ratings total $5 \times 3.8 = 19$. The given ones are W 5, X 5 and Y 1, so U + V = 8. U has a 1 (from R1) and range 3, so the top is 4. Mean 2.2 gives a total of 11, and with median and mode 2 the ratings are 1, 2, 2, 2, 4. So U's R3 rating is 2 or 4. V has a 5 (from R5) and range 3, so the lowest is 2. Mean 3.8 gives a total of 19, and with median and mode 4 the ratings are 2, 4, 4, 4, 5. So V's R3 rating is 2 or 4. U + V = 8 then forces both to be 4. R3's ratings in order are 1, 4, 4, 5, 5, and the middle one is 4. The answer is 4.

Q39MCQMissing Value Tables

1. AR2
2. BR5
3. CR4
4. DR3

Answer and solution

Answer: (C) R4

Totals are 5 × mean: workers U 11, V 19, W 17, X 18, Y 13; restaurants R1 17, R2 11, R3 19, R4 14, R5 17. Range, sum, median and mode fix each worker's ratings: U 1, 2, 2, 2, 4; V 2, 4, 4, 4, 5; W 1, 2, 4, 5, 5; X 1, 3, 4, 5, 5; Y 1, 1, 3, 4, 4. R3: U + V = 8, each 2 or 4, so both give 4, and U gives 2 elsewhere. R2: V = 11 − 9 = 2, so V gives 4 in R1 and R4. R5: W + X + Y = 10, so (W, X, Y) is (2, 4, 4) or (4, 3, 3). R1 needs X + Y = 7, which takes a 3 from X or Y, so R5 cannot use both 3s: R5 is (2, 4, 4). Then R1 is X 3, Y 4, and R4 is W 4, X 1, Y 3. Medians: R2 (1, 1, 2, 2, 5) is 2, given to U and V; R3 (1, 4, 4, 5, 5) is 4, to U and V; R5 (2, 2, 4, 4, 5) is 4, to X and Y; R4 (1, 2, 3, 4, 4) is 3, to Y only. Hence, option C (R4).

Q40MCQLogical Puzzles

1. AProf. Pakrasi and Prof. Qureshi
2. BProf. Pakrasi and Prof. Samuel
3. CProf. Qureshi and Prof. Ramaswamy
4. DProf. Ramaswamy and Prof. Samuel

Answer and solution

Answer: (A) Prof. Pakrasi and Prof. Qureshi

Each candidate got one vote from another candidate, so non-candidates gave P $3 - 1 = 2$ votes, Q $14 - 1 = 13$, R $6 - 1 = 5$ and S $1 - 1 = 0$. Each department's non-candidates vote as one block: O&Q's is 4 (one candidate), F&A's at least 7, M&S's at least 5, B&H's at most 3. P's 2 can only be B&H's block, so B&H has 1 candidate. R's 5 cannot use a block of 7 or more, or 4 + 1 (no block of 1 is left), so it is M&S's block: M&S has 2 candidates, leaving none for F&A. Q's 13 is F&A's 9 plus O&Q's 4. M&S voted for R, so R is not in M&S. S voted for P and Q voted for S, so S shares a department with neither. M&S's two candidates are therefore P and Q, and R and S are in O&Q and B&H, one each, in either order. Options B and C pair candidates who must be in different departments. In option D, R and S are in different departments in both arrangements. Hence, option A (Prof. Pakrasi and Prof. Qureshi).

Q41MCQLogical Puzzles

1. A7
2. B6
3. C8
4. D9

Answer and solution

Answer: (D) 9

Each candidate got one vote from another candidate, so non-candidates gave P $3 - 1 = 2$ votes, Q $14 - 1 = 13$, R $6 - 1 = 5$ and S $1 - 1 = 0$. Each department's non-candidates vote as one block: O&Q's is 4 (one candidate), F&A's at least 7, M&S's at least 5, B&H's at most 3. P's 2 can only be B&H's block, so B&H has 1 candidate. R's 5 cannot use a block of 7 or more, or 4 + 1 (no block of 1 is left), so it is M&S's block: M&S has 2 candidates, leaving none for F&A. Q's 13 is F&A's 9 plus O&Q's 4. R is not in M&S (M&S voted for him), and S cannot share a department with P or Q (S voted for P, Q voted for S). So P and Q are in M&S, and R and S split O&Q and B&H. Q's votes by department: F&A 9; O&Q 4, or 5 if R is there (R voted for Q); B&H 1 if R is there, else 0; M&S 0, as it is his own department. So one department can give Q 9, 5, 4, 1 or 0 votes, never 6, 7 or 8. Hence, option D (9).

Q42MCQLogical Puzzles

1. ANeither statement A nor statement B
2. BOnly statement B
3. COnly statement A
4. DBoth statements A and B

Answer and solution

Answer: (D) Both statements A and B

Each candidate got one vote from another candidate, so non-candidates gave P $3 - 1 = 2$ votes, Q $14 - 1 = 13$, R $6 - 1 = 5$ and S $1 - 1 = 0$. Each department's non-candidates vote as one block: O&Q's is 4 (one candidate), F&A's at least 7, M&S's at least 5, B&H's at most 3. P's 2 can only be B&H's block, so B&H has 1 candidate. R's 5 cannot use a block of 7 or more, or 4 + 1 (no block of 1 is left), so it is M&S's block: M&S has 2 candidates, leaving none for F&A. Q's 13 is F&A's 9 plus O&Q's 4. R is not in M&S, since M&S voted for him. S voted for P and Q voted for S, so S is in neither P's nor Q's department. Hence M&S's two candidates are P and Q, and R and S fill O&Q and B&H. If S is in B&H, then R is the O&Q candidate (case 1 in the figure). P is in M&S, so Statement A is true; R is in O&Q, so Statement B is true. Options A, B and C each reject a statement that holds. Hence, option D (Both statements A and B).

Q43MCQLogical Puzzles

1. AIt was Prof. Samuel.
2. BIt was either Prof. Ramaswamy or Prof. Samuel.
3. CIt was Prof. Ramaswamy.
4. DIt was either Prof. Pakrasi or Prof. Qureshi.

Answer and solution

Answer: (B) It was either Prof. Ramaswamy or Prof. Samuel.

Each candidate got one vote from another candidate, so non-candidates gave P $3 - 1 = 2$ votes, Q $14 - 1 = 13$, R $6 - 1 = 5$ and S $1 - 1 = 0$. Each department's non-candidates vote as one block: O&Q's is 4 (one candidate), F&A's at least 7, M&S's at least 5, B&H's at most 3. P's 2 can only be B&H's block, so B&H has 1 candidate. R's 5 cannot use a block of 7 or more, or 4 + 1 (no block of 1 is left), so it is M&S's block: M&S has 2 candidates, leaving none for F&A. Q's 13 is F&A's 9 plus O&Q's 4. R is not in M&S (M&S voted for him), and S shares a department with neither P nor Q (S voted for P; Q voted for S). So P and Q are in M&S, and R and S fill O&Q and B&H. Both arrangements obey every rule: no one votes within his own department. So the O&Q candidate is R or S, and the data cannot say which. Options A and C name only one; option D is impossible, as P and Q are in M&S. Hence, option B (It was either Prof. Ramaswamy or Prof. Samuel.).

Q44MCQLogical Puzzles

1. ABoth statements A and B
2. BOnly statement B
3. COnly statement A
4. DNeither statement A nor statement B

Answer and solution

Answer: (B) Only statement B

Each candidate got one vote from another candidate, so non-candidates gave P $3 - 1 = 2$ votes, Q $14 - 1 = 13$, R $6 - 1 = 5$ and S $1 - 1 = 0$. Each department's non-candidates vote as one block: O&Q's is 4 (one candidate), F&A's at least 7, M&S's at least 5, B&H's at most 3. P's 2 can only be B&H's block, so B&H has 1 candidate. R's 5 cannot use a block of 7 or more, or 4 + 1 (no block of 1 is left), so it is M&S's block: M&S has 2 candidates, leaving none for F&A. Q's 13 is F&A's 9 plus O&Q's 4. R is not in M&S, since M&S voted for him. S voted for P and Q voted for S, so S shares a department with neither; hence P and Q are the two M&S candidates. Statement A is false: the M&S non-candidates voted for R, and they could not have voted for Q, who is in their own department. Statement B is true: the 9 F&A non-candidates voted for Q. Options A and C accept the false statement A, and option D rejects the true statement B. Hence, option B (Only statement B).