CAT 2020 Slot 2 — DILR questions with answers

Q27MCQMissing Value Tables

1. ABengaluru and Delhi
2. BMumbai and Mumbai
3. CMumbai and Delhi
4. DDelhi and Delhi

Answer and solution

Answer: (D) Delhi and Delhi

Only the Home Décor row matters here. From the table, 2019 Home Décor sales total $100+72+80+54=306$ crore. Clue 5 says the total rose by 70 crore, so the 2018 total was $306-70=236$. Clue 6 says Delhi and Bengaluru each rose by 20, so in 2018 Delhi had $100-20=80$ and Bengaluru had $80-20=60$. By clue 3, Mumbai and Kolkata were equal in 2018, each $\frac{236-80-60}{2}=48$. So in 2018 Home Décor sales were Delhi 80, Mumbai 48, Bengaluru 60 and Kolkata 48, and Delhi was highest. In 2019, Delhi's 100 beats Mumbai's 72, Bengaluru's 80 and Kolkata's 54. Option C (Mumbai and Delhi) fails because Mumbai had only 48 in 2018, well below Delhi's 80. Option A fails for the same reason: Bengaluru had 60 in 2018. Hence, option D (Delhi and Delhi).

Q28MCQMissing Value Tables

1. A12
2. B8
3. C5
4. D10

Answer and solution

Answer: (A) 12

From the table, Electronics changed by $+20$ (Delhi), $+20$ (Mumbai), $-20$ (Bengaluru) and $+20$ (Kolkata), a net rise of 40. By clue 4, total Apparel sales also rose by 40. Let 2018 Apparel sales be $a$ for Delhi and Kolkata (clue 1) and $b$ for Mumbai and Bengaluru; Delhi's 2019 figure is also $b$ (clue 2). So Delhi rose by $b-a$, and by clue 7 Bengaluru also rose by $b-a$, reaching $2b-a$. Kolkata reached 54, a rise of $54-a$, and Mumbai rose by the same $54-a$. Clue 4 gives $2(b-a)+2(54-a)=40$, so $b=2a-34$. Clue 8: Delhi $b$, Kolkata 54 and Bengaluru $2b-a$ are in AP, so $b+(2b-a)=2\times54$, i.e. $3b-a=108$. Substituting, $3(2a-34)-a=108$, so $5a=210$, $a=42$ and $b=50$. Mumbai's Apparel sales rose by $54-a=12$ crore, from 50 to 62. Option D (10) would need $a=44$, giving $b=54$ and $3b-a=118$, which breaks clue 8. Hence, option A (12).

Q29MCQMissing Value Tables

1. A25
2. B28
3. C75
4. D50

Answer and solution

Answer: (D) 50

Home Décor: the 2019 total is $100+72+80+54=306$, so by clue 5 the 2018 total was $306-70=236$. Delhi and Bengaluru each rose by 20 (clue 6), so they had 80 and 60 in 2018; Mumbai and Kolkata had $\frac{236-140}{2}=48$ each (clue 3). Apparel: Electronics rose by $20+20-20+20=40$, so Apparel rose by 40 (clue 4). Let 2018 Apparel be $a$ (Delhi, Kolkata) and $b$ (Mumbai, Bengaluru), with Delhi's 2019 figure $b$ (clues 1 and 2). By clue 7, Delhi and Bengaluru rose by $b-a$, Mumbai and Kolkata by $54-a$. So $2(b-a)+2(54-a)=40$, giving $b=2a-34$, and the AP $b$, 54, $2b-a$ (clue 8) gives $3b-a=108$. Hence $a=42$ and $b=50$. Apparel increases: Delhi $\frac{8}{42}$ (about 19%), Mumbai $\frac{12}{50}$ (24%), Bengaluru $\frac{8}{50}$ (16%), Kolkata $\frac{12}{42}$ (about 28.6%). Electronics: at most Delhi's $\frac{20}{78}$ (about 25.6%); Bengaluru fell. Home Décor: Delhi $\frac{20}{80}$ (25%), Mumbai $\frac{24}{48}$ (50%), Bengaluru $\frac{20}{60}$ (about 33.3%), Kolkata $\frac{6}{48}$ (12.5%). The maximum is 50%, for Mumbai Home Décor. Option B (28) is roughly Kolkata Apparel's 28.6% rise, the largest outside Home Décor, but still well below 50%. Hence, option D (50).

Q30MCQMissing Value Tables

1. A600
2. B900
3. C150
4. D750

Answer and solution

Answer: (B) 900

The 2019 Electronics and Home Décor figures are in the table: Electronics $98+102+70+100=370$ and Home Décor $100+72+80+54=306$. For Apparel: Electronics rose by $20+20-20+20=40$, so Apparel also rose by 40 (clue 4). Let 2018 Apparel be $a$ for Delhi and Kolkata and $b$ for Mumbai and Bengaluru, with Delhi's 2019 figure $b$ (clues 1 and 2). By clue 7, Delhi and Bengaluru each rose by $b-a$, and Mumbai and Kolkata each by $54-a$. So $2(b-a)+2(54-a)=40$, giving $b=2a-34$. Clue 8 (Delhi $b$, Kolkata 54, Bengaluru $2b-a$ in AP) gives $3b-a=108$. So $a=42$ and $b=50$. Apparel in 2019: Delhi 50, Mumbai $50+12=62$, Bengaluru $50+8=58$ and Kolkata 54, a total of 224. Total 2019 sales: $224+370+306=900$ crore. Option D (750) would leave only $750-370-306=74$ for Apparel, yet Delhi and Kolkata alone had $50+54=104$. Hence, option B (900).

Q31TITAMissing Value Tables

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

Q32TITAMissing Value Tables

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

Q33MCQMissing Value Tables

1. A1,40,010
2. Bbetween 1,40,005 and 1,40,010
3. Cless than 2,00,010
4. D1,40,006

Q34MCQMissing Value Tables

1. A1,25,000
2. B1,50,000
3. C1,75,000
4. D62,500

Q35MCQMissing Value Tables

1. AD, B, C, A
2. BB, D, C, A
3. CB, C, D, A
4. DD, C, B, A

Q36MCQMissing Value Tables

1. A38.25%
2. B23.54%
3. C23.91%
4. D32.00%

Q37TITALogical Puzzles

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

Answer and solution

Answer: 2

Red cannot be one of the two colours. With only Red and Green (or only Red and Blue), rule 1 forces every row to alternate, so a row of five has two Reds with only one bead between them, which breaks rule 3. So the colours must be Blue and Green. Rule 1 then forces every row and every column to alternate Blue and Green, which is a checkerboard pattern. Rule 2 holds, because any two Blues in a line have a Green between them. A checkerboard is fixed completely by the colour of the top-left bead, which can be Green or Blue. These are the only two configurations: The answer is 2.

Q38TITALogical Puzzles

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

Answer and solution

Answer: 9

Two Reds in the same line need a Blue and a Green between them, so they must be at least three places apart. In a line of five, two Reds can sit only at positions (1, 4), (1, 5) or (2, 5), and three Reds never fit. So each row and each column has at most two Reds, and a Red in the middle position must be the only Red in its line. Ten Reds would need two in every row, and no allowed pair uses column 3, so all ten would lie in columns 1, 2, 4 and 5. Those columns hold at most $4\times2=8$ Reds. So there are at most 9 Reds. Nine is achievable, with the Reds placed as below: A full configuration with these Reds, rows from top to bottom: RGBRG / GRGBR / BGRGB / RBGRG / GRBGR. Every row and column satisfies all three rules. The answer is 9.

Q39TITALogical Puzzles

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

Answer and solution

Answer: 6

Every line of five needs a Blue: with only Red and Green, rule 1 forces alternation, which puts two Reds with only a Green between them and breaks rule 3. So each row and each column has at least one Blue, giving at least 5 Blues. If a line has exactly one Blue, that Blue cannot be at an end: the other four beads would alternate Red and Green, again putting two Reds with only a Green between them. With exactly 5 Blues, every row and column would have exactly one Blue. The Blue in column 1 would then be the only Blue in its row and sit at that row's end, which is impossible. So at least 6 Blues are needed. Six is achievable. Rows from top to bottom: RGBRG / GRGBR / BGRGB / RBGRG / GRBGR. This has six Blues, and every row and column satisfies all three rules. The answer is 6.

Q40TITALogical Puzzles

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

Answer and solution

Answer: 6

Two Reds in a line need a Blue and a Green between them, so they must be at least three places apart. A pair of Reds can sit only at positions (1, 4), (1, 5) or (2, 5), and a Red in position 3 is alone in its line. The Red at row 2, column 3 is in the middle of row 2, so it is the only Red in row 2. The Red at row 3, column 2 is in the middle of column 2, so it is the only Red in that column. Row 3 can add a second Red only at column 5. Rows 1, 4 and 5 cannot use column 2, so any pair of Reds in them must be (1, 4) or (1, 5), both using column 1. All three rows having a pair would put three Reds in column 1, which is impossible. So rows 1, 4 and 5 hold at most $2+2+1=5$ Reds, and the total is at most $1+2+5=8$. Eight is achievable, with the Reds placed as below: A full configuration, rows from top to bottom: RGBRG / GBRGB / BRGBR / RGBRG / GBRGB. That is 6 Reds added to the two given. The answer is 6.

Q41MCQTeam Selection & Scheduling

1. ATwo seminars are scheduled in the first slot.
2. BOnly one seminar is scheduled in the second slot.
3. CThree seminars are scheduled in the first slot.
4. DThree seminars are scheduled in the last slot.

Answer and solution

Answer: (A) Two seminars are scheduled in the first slot.

A is the only seminar at 10:00 (clue 2) and F is at 10:30 (clue 3). By clue 7, G is next to A, so at 9:30 or 10:30. By clue 5, B, G and an Economics student guided by T share a slot; at 10:30 that slot would hold four seminars with F, over the limit of three. So B, G and the T student are at 9:30, which is then full. P guides B and C and has students in the first two slots (clue 6), so C is at 9:00. T's two students must be in consecutive slots, and 10:00 holds only A, so T's other student is at 9:00. R guides A, and R's second student must be next to A; 9:30 is full, so it is at 10:30. The schedule is: 9:00 C and a T student; 9:30 B, G and a T student; 10:00 A; 10:30 F and an R student. So the first slot has two seminars. Option C fails because 9:00 has only two, option B fails because 9:30 has three, and option D fails because 10:30 has two. Hence, option A (Two seminars are scheduled in the first slot.).

Q42MCQTeam Selection & Scheduling

1. AQ, R and S
2. BP, Q and R
3. CP, R and S
4. DP, Q and S

Answer and solution

Answer: (D) P, Q and S

P guides B and C (clue 6), and B is a Sociology student (clue 5), so P is a Sociology guide and C is also Sociology. With G (clue 5), the three Sociology students are B, C and G. F is the only Anthropology student, so the four Economics students are A, D, E and H. R guides A, an Economics student (clue 2), and T guides an Economics student (clue 5). Each student's guide is from the student's department, so R and T are Economics guides. With two students each, they guide exactly the four Economics students. That leaves G and F for Q and S, one each. So P, Q and S guide no Economics student. Options A, B and C all include R, and they fail because R guides A, an Economics student. Hence, option D (P, Q and S).

Q43MCQTeam Selection & Scheduling

1. AQ is guiding G.
2. BH is an Economics student.
3. CS is guiding F.
4. DB is scheduled in the first slot.

Answer and solution

Answer: (B) H is an Economics student.

P guides B and C (clue 6), and B is a Sociology student (clue 5), so C is also Sociology. With G (clue 5), the three Sociology students are B, C and G. F is the only Anthropology student (clue 3), so the four Economics students are A, D, E and H. So H is necessarily an Economics student. R (guiding A) and T are Economics guides with two students each, so they guide all four Economics students. G and F are then guided by Q and S, one each, but nothing fixes which way round. So options A (Q is guiding G) and C (S is guiding F) are possible but not necessary. For option D: A is alone at 10:00 and F is at 10:30. By clue 7, G is at 9:30 or 10:30. B, G and a T-guided Economics student share a slot (clue 5); at 10:30 they would make four seminars with F, so they are at 9:30. B is in the second slot, so option D is false. Hence, option B (H is an Economics student.).

Q44MCQTeam Selection & Scheduling

1. AOnly (ii)
2. BNeither (i) nor (ii)
3. COnly (i)
4. DBoth (i) and (ii)

Answer and solution

Answer: (D) Both (i) and (ii)

P guides B and C, so B, C and G are Sociology and F is Anthropology; the Economics students are A, D, E and H. R (guiding A) and T are Economics guides, so R's second student and T's two students are D, E and H, and Q and S guide G and F, one each. A alone is at 10:00 and F at 10:30. By clue 7, G is at 9:30 or 10:30; at 10:30, F, B, G and the T student of clue 5 would make four, so B, G and a T student are at 9:30. T's other student must be in an adjacent slot, and 10:00 holds only A, so it is at 9:00. R's second student must be next to A, and 9:30 is full, so it is at 10:30. Q's student is G (9:30) or F (10:30). If Q guided F, D would need a slot after 10:30, which is impossible. So Q guides G, and (ii) holds. D is then after 9:30; 10:00 holds only A, so D is at 10:30 as R's second student. So E and H are T's students, and (i) holds. Both statements hold, so options A and C, which keep only one, fail. Hence, option D (Both (i) and (ii)).

Q45MCQTeam Selection & Scheduling

1. AExactly one of D and H is guided by T.
2. BBoth D and H are guided by T.
3. CAt least one of D and H is guided by T.
4. DNeither D nor H is guided by T.

Answer and solution

Answer: (C) At least one of D and H is guided by T.

P guides B and C, so B, C and G are the Sociology students, F is the Anthropology student, and A, D, E and H are the Economics students, guided by R and T. Q and S guide G and F in some order. A is alone at 10:00, so G is at 9:30 or 10:30 (clue 7). B shares G's slot and is in the first two slots (clue 6), so B, G and a T-guided Economics student (clue 5) are at 9:30, and C is at 9:00. T's other student must be next to 9:30 but not at 10:00, so it is at 9:00. Clue 1 needs an Economics student at 10:30, so R's second student is there, with F. If Q guides G, E is T's student at 9:30; D and H are then T's 9:00 student and R's 10:30 student, so exactly one of them is with T. If Q guides F, E is at 10:30 with R, and D and H are both with T. Either way, at least one of D and H is with T. Option A fails in the second case, option B in the first, and option D in both. Hence, option C (At least one of D and H is guided by T.).

Q46MCQTeam Selection & Scheduling

1. AG is guided by Q.
2. BE is guided by R.
3. CF is guided by S.
4. DD is guided by T.

Answer and solution

Answer: (B) E is guided by R.

P guides B and C, so B, C and G are the Sociology students, F is the Anthropology student, and A, D, E and H are the Economics students, guided by R and T. Q and S guide G and F in some order. A is alone at 10:00, so G is at 9:30 or 10:30 (clue 7). B shares G's slot and is in the first two slots (clue 6), so B, G and a T-guided Economics student (clue 5) are at 9:30, and C is at 9:00. T's other student must be next to 9:30 but not at 10:00, so it is at 9:00. Clue 1 needs an Economics student at 10:30, so R's second student is there, with F. Q's student is G (9:30) or F (10:30). The slot just before 10:30 is 10:00, which holds only A, so D cannot precede F. Hence Q guides G, D is at 9:00 as T's student there, and S guides F. E and H are T's 9:30 student and R's 10:30 student, in either order. So options A, C and D must be true, while E could be guided by T instead of R. Hence, option B (E is guided by R.).

Q47MCQLogical Puzzles

1. A8
2. B5
3. C6
4. D7

Q48MCQLogical Puzzles

1. A8, 2, 3, V, 6, 5, 7
2. BV, 2, 3, 7, 5, 6, 8
3. C8, 2, 3, V, 5, 7, 6
4. D8, 2, 3, V, 5, 6, 7

Q49MCQLogical Puzzles

1. ACar 4 is a compact.
2. BCar 1 is an SUV.
3. CCar 3 is an SUV
4. DCar 5 is a compact.

Q50MCQLogical Puzzles

1. ACar 2 is a compact.
2. BCar 7 is a compact.
3. CCar 4 is an SUV.
4. DCar 6 is a compact.