CAT 2023 Slot 3 — DILR questions with answers

Q25TITATables & Caselets

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

Answer and solution

Answer: 14

Split inverter ACs are 36, which is 80% of inverter ACs, so inverter ACs $= 45$ and Window inverter $= 9$. Window ACs $= 9 + 6 = 15$, which is 25% of the total, so the total is $60$ and Split non-inverter $= 45 - 36 = 9$. Only D2 and D3 sold Window non-inverter ACs, D2 twice D3: D2 $= 4$, D3 $= 2$. Let D1 sell $x$ Window ACs (all inverter), D3 and D4 $y$ each, and D2 $3y$. Then $x + 5y = 15$ with $x, y \ge 2$, so $x = 5$, $y = 2$. D2's 6 Window ACs are 2 inverter and 4 non-inverter; D3's 2 are non-inverter; D4's 2 are inverter. D1: Split $= 2 \times 5 = 10$, of which inverter $13 - 5 = 8$ and non-inverter $2$; total $15$. Split inverter: D3 $=$ D4 $= z$ and D2 $= 2z$, so $8 + 4z = 36$ and $z = 7$. D3: its 5 non-inverter ACs are 2 Window and 3 Split, so its total is $7 + 3 + 2 = 12$. D2 sold $2z = 14$ Split inverter ACs. The answer is 14.

Q26MCQTables & Caselets

1. A33.33%
2. B75.00%
3. C25.00%
4. D20.00%

Answer and solution

Answer: (C) 25.00%

Let the total number of ACs be $T$ and the number of inverter ACs be $I$. Among inverter ACs, 20% are Window, so 80% are Split: $0.8I = 36$, which gives $I = 45$ and Window inverter $= 45 - 36 = 9$. Window ACs $= 9 + 6 = 15$ (inverter plus non-inverter). They are 25% of all ACs, so $T = 60$ and Split ACs $= 45$. Split non-inverter $= 45 - 36 = 9$. Non-inverter ACs $= 9 + 6 = 15$, which is $\frac{15}{60} = 25\%$ of all ACs. B (75%) is the share of inverter ACs, $\frac{45}{60}$, and D (20%) is the Window share among inverter ACs; neither is the non-inverter share. Hence, option C (25.00%).

Q27TITATables & Caselets

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

Answer and solution

Answer: 33

Split inverter ACs are 36, which is 80% of inverter ACs, so inverter ACs $= 45$ and Window inverter $= 9$. Window ACs $= 9 + 6 = 15$, which is 25% of the total, so the total is $60$ and Split non-inverter $= 45 - 36 = 9$. Only D2 and D3 sold Window non-inverter ACs, D2 twice D3: D2 $= 4$, D3 $= 2$. Let D1 sell $x$ Window ACs (all inverter), D3 and D4 $y$ each, and D2 $3y$. Then $x + 5y = 15$ with $x, y \ge 2$, so $x = 5$, $y = 2$. D2's 6 Window ACs are 2 inverter and 4 non-inverter; D3's 2 are non-inverter; D4's 2 are inverter. D1: Split $= 2 \times 5 = 10$, of which inverter $13 - 5 = 8$ and non-inverter $2$; total $15$. Split inverter: D3 $=$ D4 $= z$ and D2 $= 2z$, so $8 + 4z = 36$ and $z = 7$. D3: its 5 non-inverter ACs are 2 Window and 3 Split, so its total is $7 + 3 + 2 = 12$. D1 and D3 together sold $15 + 12 = 27$, so D2 and D4 together sold $60 - 27 = 33$. The answer is 33.

Q28MCQTables & Caselets

1. AD2 sold the highest number of ACs.
2. BD4 sold more Split ACs as compared to D3.
3. CD1 and D3 sold an equal number of Split ACs.
4. DD1 and D3 together sold more ACs as compared to D2 and D4 together.

Answer and solution

Answer: (D) D1 and D3 together sold more ACs as compared to D2 and D4 together.

The 36 Split inverter ACs are 80% of inverter ACs, so inverter ACs $= 45$ and Window inverter $9$. Window ACs ($9 + 6 = 15$) are 25% of all ACs, so the total is $60$ and Split non-inverter $9$. Window non-inverter: only D2 and D3, D2 twice D3, so 4 and 2. With D1 selling $x$ Window ACs (all inverter), D3 and D4 $y$ each, D2 $3y$: $x + 5y = 15$ and $x, y \ge 2$, so $x = 5$, $y = 2$. D1: Split $= 2 \times 5 = 10$, inverter $13 - 5 = 8$, non-inverter $2$, total $15$. Split inverter, with D3 $=$ D4 $= z$ and D2 $= 2z$: $8 + 4z = 36$, so $z = 7$ and D2 has $14$. D3: non-inverter $5 = 2 + 3$, total $12$. The 4 Split non-inverter ACs left go $a$ to D2, $4 - a$ to D4: D2 sold $20 + a$, D4 sold $13 - a$. D1 and D3 sold $27$, D2 and D4 $33$: D is always false. A always holds (D2 at least 20, D4 at most 13). B holds when $a = 0$ (D4's Split 11, D3's 10). C holds (10 each). Hence, option D (D1 and D3 together sold more ACs as compared to D2 and D4 together.).

Q29MCQTables & Caselets

1. A4
2. B5
3. C7
4. D6

Answer and solution

Answer: (B) 5

Split inverter ACs are 36, which is 80% of inverter ACs, so inverter ACs $= 45$ and Window inverter $= 9$. Window ACs $= 9 + 6 = 15$, which is 25% of the total, so the total is $60$ and Split non-inverter $= 9$. Window non-inverter: D2 $= 4$, D3 $= 2$ (only they sold these, D2 twice D3). With D1 selling $x$ Window ACs (all inverter), D3 and D4 $y$ each and D2 $3y$: $x + 5y = 15$ and $x, y \ge 2$, so $x = 5$, $y = 2$. D4's 2 Window ACs are inverter. D1: Split $= 10$, of which inverter $13 - 5 = 8$ and non-inverter $2$. Split inverter: $8 + 2z + z + z = 36$, so $z = 7$ for D3 and D4. D3: non-inverter $5 = 2 + 3$, total $7 + 3 + 2 = 12$. The 4 Split non-inverter ACs left go $a$ to D2 and $4 - a$ to D4, so D4 sold $7 + (4 - a) + 2 = 13 - a$. If D4 equals D3's 12, then $a = 1$. D2's non-inverter ACs $= 1$ Split $+ 4$ Window $= 5$. A (4) counts only the Window non-inverter ACs. Hence, option B (5).

Q30MCQNetworks & Routes

1. AStation C
2. BStation E
3. CStation D
4. DStation F

Answer and solution

Answer: (B) Station E

Four streets leave A (to B, C, E, F), and no street holds two teams at once, so at 9:00 each team takes a different one. Only Teams 1 and 4 use A–E, so Team 2 (at E at 10:00) went A→F→E and Team 3 (at D) went A→C→D. Team 4 avoids B, so Team 4 took A→E and Team 1 took A→B. At 10:00 Team 1 is back at A; Team 4 must leave E (only Team 2 is there) and avoids D and F, so it is at A. At 10:30 Teams 1 and 3 reach E (A→E, D→E). Team 2 cannot use E–A or the busy E–D, so it goes to F; Team 4 cannot use the busy A–E, so it goes to C. To be at B and E at 11:30, Team 1 runs E→A→B→A and Team 4 runs C→A→E→A. Team 3 can return from E in three steps without A–E only by E→D→C→A. Team 2 ends F→A→F→A or F→E→F→A (F→A→C→A would share C–A with Team 3 after 11:30). Visits: E gets 5 or 6 (Team 4 twice, Teams 1 and 3 once, Team 2 once or twice); C gets 3, F gets 3 and D gets 2. Hence, option B (Station E).

Q31TITANetworks & Routes

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

Answer and solution

Answer: 2

Four streets leave A (to B, C, E, F), and no street holds two teams at once, so at 9:00 each team takes a different one. Only Teams 1 and 4 use A–E, so Team 2 (at E at 10:00) went A→F→E and Team 3 (at D) went A→C→D. Team 4 avoids B, so Team 4 took A→E and Team 1 took A→B. At 10:00 Team 1 is back at A; Team 4 must leave E (only Team 2 is there) and avoids D and F, so it is at A. At 10:30 Teams 1 and 3 reach E (A→E, D→E). Team 2 cannot use E–A or the busy E–D, so it goes to F; Team 4 cannot use the busy A–E, so it goes to C. To be at B and E at 11:30, Team 1 runs E→A→B→A and Team 4 runs C→A→E→A. Team 3 can return from E in three steps without A–E only by E→D→C→A. Team 2 ends F→A→F→A or F→E→F→A (F→A→B→A would share A–B with Team 1). Only Team 1 reaches B, at 9:30 and at 11:30. The answer is 2.

Q32MCQNetworks & Routes

1. ATeam 4
2. BTeam 1
3. CTeam 2
4. DTeam 3

Answer and solution

Answer: (D) Team 3

Four streets leave A (to B, C, E, F), and no street holds two teams at once, so at 9:00 each team takes a different one. Only Teams 1 and 4 use A–E, so Team 2 (at E at 10:00) went A→F→E and Team 3 (at D) went A→C→D. Team 4 avoids B, so Team 4 took A→E and Team 1 took A→B. At 10:00 Team 1 is back at A; Team 4 must leave E (only Team 2 is there) and avoids D and F, so it is at A. At 10:30 Teams 1 and 3 reach E (A→E, D→E). Team 2 cannot use E–A or the busy E–D, so it goes to F; Team 4 cannot use the busy A–E, so it goes to C. To be at B and E at 11:30, Team 1 runs E→A→B→A and Team 4 runs C→A→E→A; Team 3 returns by E→D→C→A. Between 10:00 and 10:30 Team 3 moves from D to E, so it is on street D–E at 10:15. At that time Team 1 is on A–E, Team 2 on E–F and Team 4 on A–C. Hence, option D (Team 3).

Q33TITANetworks & Routes

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

Answer and solution

Answer: 2

Four streets leave A (to B, C, E, F), and no street holds two teams at once, so at 9:00 each team takes a different one. Only Teams 1 and 4 use A–E, so Team 2 (at E at 10:00) went A→F→E and Team 3 (at D) went A→C→D. Team 4 avoids B, so Team 4 took A→E and Team 1 took A→B. At 10:00 Team 4 must leave E (only Team 2 is there); it avoids D and F, so it returns to A. At 10:30 Teams 1 and 3 are the only ones at E, and Team 1 is travelling A→E, so Team 4 cannot use that street; it goes to C. Team 4 must be at E at 11:30 and never enters B, D or F, so it goes C→A at 11:00, A→E at 11:30 and E→A at 12:00. Team 4's route is A→E→A→C→A→E→A, so it is at E at 9:30 and at 11:30. The answer is 2.

Q34MCQNetworks & Routes

1. A4
2. B3
3. C1
4. D2

Answer and solution

Answer: (D) 2

Four streets leave A (to B, C, E, F), and no street holds two teams at once, so at 9:00 each team takes a different one. Only Teams 1 and 4 use A–E, so Team 2 (at E at 10:00) went A→F→E and Team 3 (at D) went A→C→D. Team 4 avoids B, so Team 4 took A→E and Team 1 took A→B. At 10:00 Team 1 is back at A; Team 4 must leave E (only Team 2 is there) and avoids D and F, so it is at A. At 10:30 Teams 1 and 3 reach E (A→E, D→E). Team 2 cannot use E–A or the busy E–D, so it goes to F; Team 4 cannot use the busy A–E, so it goes to C. To be at B and E at 11:30, Team 1 runs E→A→B→A and Team 4 runs C→A→E→A. Team 3 can return from E in three steps without A–E only by E→D→C→A. Teams 3 and 4 pass through C. Team 1 never does, and Team 2's only route via C, F→A→C→A, would put it on C–A with Team 3 after 11:30. So B (3) is impossible. Hence, option D (2).

Q35TITAMissing Value Tables

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

Answer and solution

Answer: 120

All numbers are multiples of 10, and every total is between 110 and 130. January: offline is twice online, so the total is 3 times online, a multiple of 30: 120. January is 40 online, 80 offline. April likewise: 80 online, 40 offline, total 120. May has the largest online number, 100. With a total of at most 130, its offline is at most 30, the minimum: May is 100 and 30, total 130. Let February be $y$ online, $x$ offline; March online is also $x$, March offline $z$. The online median 80 (values 40, 80, 100, $x$, $y$) needs $x$ or $y$ to be at least 80. If $x = 80$, the offline median 50 forces $z = 50$, March totals 130, and February must be the minimum 110, so $y = 30$, below the online minimum. So $y \ge 80$, and $x + y \le 130$ gives $x \le 50$. Offline values 30, 40, 80, $x$, $z$ then have median 50 only if $x = 50$. So $y = 80$, and March must be the minimum: $50 + z = 110$, $z = 60$. April's total is $80 + 40 = 120$. The answer is 120.

Q36TITAMissing Value Tables

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

Answer and solution

Answer: 40

All numbers are multiples of 10, and every total is between 110 and 130. January: offline is twice online, so the total is 3 times online. It must be a multiple of 30 between 110 and 130, so it is 120, and January's online figure is $120 \div 3 = 40$, with 80 offline. April likewise: 80 online, 40 offline, total 120. May has the largest online number, 100; with a total of at most 130, its offline is the minimum 30. Let February be $y$ online, $x$ offline; March online is also $x$, March offline $z$. The online median 80 needs $x$ or $y$ to be at least 80. If $x = 80$, the offline median forces $z = 50$, March totals 130, and February must be the minimum 110, giving $y = 30$, below the online minimum. So $y \ge 80$ and $x \le 50$; the offline median 50 then forces $x = 50$, so $y = 80$, and March gives the minimum: $z = 60$. January had 40 online registrations. The answer is 40.

Q37MCQMissing Value Tables

1. ABoth I and II
2. BOnly II
3. CNeither I nor II
4. DOnly I

Answer and solution

Answer: (D) Only I

All numbers are multiples of 10, and every total is between 110 and 130. January: offline is twice online, so the total is 3 times online, a multiple of 30: 120. January is 40 online, 80 offline. April likewise: 80 online, 40 offline, total 120. May has the largest online number, 100. With a total of at most 130, its offline is at most 30, the minimum: May is 100 and 30, total 130. Let February be $y$ online, $x$ offline; March online is also $x$, March offline $z$. The online median 80 (values 40, 80, 100, $x$, $y$) needs $x$ or $y$ to be at least 80. If $x = 80$, the offline median 50 forces $z = 50$, March totals 130, and February must be the minimum 110, so $y = 30$, below the online minimum. So $y \ge 80$, and $x + y \le 130$ gives $x \le 50$. Offline values 30, 40, 80, $x$, $z$ then have median 50 only if $x = 50$. So $y = 80$, and March must be the minimum: $50 + z = 110$, $z = 60$. I is true: May's 30 offline is the smallest. II is false: February totals 130, March 110. Hence, option D (Only I).

Q38MCQMissing Value Tables

1. A80
2. B50 or 80
3. C30 or 50 or 80
4. D50

Answer and solution

Answer: (D) 50

All numbers are multiples of 10, and every total is between 110 and 130. January: offline is twice online, so the total is 3 times online, a multiple of 30: 120. January is 40 online, 80 offline. April likewise: 80 online, 40 offline, total 120. May has the largest online number, 100. With a total of at most 130, its offline is at most 30, the minimum: May is 100 and 30, total 130. Let February be $y$ online, $x$ offline; March online is also $x$, March offline $z$. The online median 80 (values 40, 80, 100, $x$, $y$) needs $x$ or $y$ to be at least 80. If $x = 80$, the offline median 50 forces $z = 50$, March totals 130, and February must be the minimum 110, so $y = 30$, below the online minimum. So $y \ge 80$, and $x + y \le 130$ gives $x \le 50$. Offline values 30, 40, 80, $x$, $z$ then have median 50 only if $x = 50$ (30 or 40 would make the median 40 or less). So February's offline figure is exactly 50; 80 and 30 are both ruled out, so B and C fail. Hence, option D (50).

Q39MCQMissing Value Tables

1. ABoth I and II
2. BOnly II
3. COnly I
4. DNeither I nor II

Answer and solution

Answer: (A) Both I and II

All numbers are multiples of 10, and every total is between 110 and 130. January: offline is twice online, so the total is 3 times online, a multiple of 30: 120. January is 40 online, 80 offline. April likewise: 80 online, 40 offline, total 120. May has the largest online number, 100. With a total of at most 130, its offline is at most 30, the minimum: May is 100 and 30, total 130. Let February be $y$ online, $x$ offline; March online is also $x$, March offline $z$. The online median 80 (values 40, 80, 100, $x$, $y$) needs $x$ or $y$ to be at least 80. If $x = 80$, the offline median 50 forces $z = 50$, March totals 130, and February must be the minimum 110, so $y = 30$, below the online minimum. So $y \ge 80$, and $x + y \le 130$ gives $x \le 50$. Offline values 30, 40, 80, $x$, $z$ then have median 50 only if $x = 50$. So $y = 80$ (February totals 130), and March gives the minimum: $z = 60$. The table is unique: January and April total 120, February and May 130. Hence, option A (Both I and II).

Q40TITALogical Puzzles

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

Answer and solution

Answer: 60

Project: the three pair scores have minimum 40, maximum 80 and average 60, so they are 40, 60 and 80. Amala's is double Koli's: Amala 80, Koli 40, so Rini 60. Test: the scores total $6 \times 60 = 360$; two are 60, and the four distinct ones include 40 and 80, so the other two sum to 120: 50 and 70. Shyamal is second in the test: 70. Koli is 20 above Amala, so (Amala, Koli) is (40, 60) or (60, 80). Shyamal is 2 below Amala and 2 above Koli, so Amala exceeds Koli by 4. Let $w$ be the project weight. (40, 60): $40 + 40w = 64 - 20w$ gives $w = 0.4$ and Amala 56, but Shyamal gets at least $0.6(70) + 0.4(40) = 58$. Rejected. (60, 80): $60 + 20w = 84 - 40w$ gives $w = 0.4$: Amala 68, Koli 64, Shyamal 66, so Shyamal's project is $(66 - 42) \div 0.4 = 60$, with Rini. Mathew beats Rini in the project: Mathew 80 (with Amala), Biman 40 (with Koli). Biman is second lowest in the test (50); Mathew is below Rini: Mathew 40, Rini 60. Rini's project score is 60. The answer is 60.

Q41MCQLogical Puzzles

1. A0.60
2. B0.50
3. C0.75
4. D0.40

Answer and solution

Answer: (A) 0.60

Project: the three pair scores have minimum 40, maximum 80 and average 60, so they are 40, 60 and 80. Amala's is double Koli's: Amala 80, Koli 40, so Rini 60. Test: the scores total $6 \times 60 = 360$; two are 60, and the four distinct ones include 40 and 80, so the other two sum to 120: 50 and 70. Shyamal is second in the test: 70. Koli is 20 above Amala, so (Amala, Koli) is (40, 60) or (60, 80). Shyamal is 2 below Amala and 2 above Koli, so Amala exceeds Koli by 4. Let $w$ be the project weight. (40, 60): $40 + 40w = 64 - 20w$ gives $w = 0.4$ and Amala 56, but Shyamal gets at least $0.6(70) + 0.4(40) = 58$, beating Amala. Rejected. (60, 80): $60 + 20w = 84 - 40w$ gives $60w = 24$, so $w = 0.4$: Amala 68, Koli 64, Shyamal 66, and the rest of the table follows. The project weight is 0.4, so the test weight is $1 - 0.4 = 0.6$. D (0.40) is the project weight, not the test weight. Hence, option A (0.60).

Q42MCQLogical Puzzles

1. A68
2. B80
3. C62
4. D66

Answer and solution

Answer: (A) 68

Project: the three pair scores have minimum 40, maximum 80 and average 60, so they are 40, 60 and 80. Amala's is double Koli's: Amala 80, Koli 40, so Rini 60. Test: the scores total $6 \times 60 = 360$; two are 60, and the four distinct ones include 40 and 80, so the other two sum to 120: 50 and 70. Shyamal is second in the test: 70. Koli is 20 above Amala, so (Amala, Koli) is (40, 60) or (60, 80). Shyamal is 2 below Amala and 2 above Koli, so Amala exceeds Koli by 4. Let $w$ be the project weight. (40, 60): $40 + 40w = 64 - 20w$ gives $w = 0.4$ and Amala 56, but Shyamal gets at least $0.6(70) + 0.4(40) = 58$. Rejected. (60, 80): $60 + 20w = 84 - 40w$ gives $w = 0.4$: Amala 68, Koli 64, Shyamal 66 (project 60, with Rini). Mathew beats Rini in the project, so Mathew 80, Biman 40; Biman's test is 50, Mathew's 40, Rini's 60. Aggregates: Amala 68, Shyamal 66, Koli 64, Rini 60, Mathew 56, Biman 46. The maximum is Amala's 68; 66 is second, and 80 is only a top component score. Hence, option A (68).

Q43TITALogical Puzzles

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

Answer and solution

Answer: 40

Project: the pairs share scores with minimum 40, maximum 80 and average 60, so the three pair scores are 40, 60 and 80. Amala's is double Koli's, so Amala has 80, Koli 40 and Rini 60. Test: the scores are multiples of 10 adding to $6 \times 60 = 360$. Two are 60, and the four distinct ones include 40 and 80, so the other two add to 120 and must be 50 and 70. Shyamal (second highest) has 70 and Biman (second lowest) has 50. Let the project weight be $w$. Koli's test score is Amala's plus 20, and Amala's aggregate is 4 more than Koli's, since Shyamal is 2 above Koli and 2 below Amala. So $40w - 20(1 - w) = 4$, giving $w = 0.4$. If Amala had scored 40 in the test, her aggregate would be $0.6 \times 40 + 0.4 \times 80 = 56$ and Shyamal's $54 = 0.6 \times 70 + 0.4p$, giving $p = 30$, which is not a project score. So Amala scored 60 and Koli 80. That leaves test scores 40 and 60 for Rini and Mathew. Mathew scored less than Rini, so Mathew scored 40. The answer is 40.

Q44MCQLogical Puzzles

1. AOnly ii)
2. BOnly i)
3. CNeither i) nor ii)
4. DBoth i) and ii)

Answer and solution

Answer: (C) Neither i) nor ii)

Project: the three pair scores have minimum 40, maximum 80 and average 60, so they are 40, 60 and 80. Amala's is double Koli's: Amala 80, Koli 40, so Rini 60. Test: the scores total $6 \times 60 = 360$; two are 60, and the four distinct ones include 40 and 80, so the other two sum to 120: 50 and 70. Shyamal is second in the test: 70. Koli is 20 above Amala, so (Amala, Koli) is (40, 60) or (60, 80). Shyamal is 2 below Amala and 2 above Koli, so Amala exceeds Koli by 4. Let $w$ be the project weight. (40, 60): $40 + 40w = 64 - 20w$ gives $w = 0.4$ and Amala 56, but Shyamal gets at least $0.6(70) + 0.4(40) = 58$. Rejected. (60, 80): $60 + 20w = 84 - 40w$ gives $w = 0.4$: Amala 68, Shyamal 66, so Shyamal's project is $(66 - 42) \div 0.4 = 60$, with Rini. Mathew beats Rini in the project, so Mathew has 80 (with Amala) and Biman 40 (with Koli). The teams are Amala–Mathew, Koli–Biman and Rini–Shyamal. Neither Amala–Biman nor Koli–Mathew is a team; D swaps the male partners. Hence, option C (Neither i) nor ii)).