CAT 2019 Slot 1 — DILR questions with answers

Q35MCQLinear Arrangement

1. A8
2. B4
3. C2
4. D1

Answer and solution

Answer: (A) 8

With 5 biscuits, 3 candies and 4 savouries on 16 shelves, 4 shelves are empty. Savouries come after all biscuits and candies, so K, on shelf 16, is the fourth savoury: D, E, F, K fill shelves 13 to 16, and shelf 12 is empty. C (a candy) has two empty shelves before it and L exactly one, so each starts a different group: L is a biscuit. J matches L, and I sits next to J with no gap, so both are biscuits. H differs from L, so H is a candy. A and B are adjacent, so they share a type; as candies they would make four, so they are biscuits. G is the third candy. Shelves 1 to 11 hold 8 items and 3 empty shelves. The biscuits run L, A, B, then I and J; the candies run C, then G and H. Biscuits first: shelf 1 empty; biscuits on 2 to 6; shelves 7, 8 empty; candies on 9 to 11. Candies first: shelves 1, 2 empty; candies on 3 to 5; shelf 6 empty; biscuits on 7 to 11. In each layout G and H can swap, and so can I and J: $2 \times 2 = 4$ arrangements, so $4 + 4 = 8$ in all. Option B (4) counts only one layout. Hence, option A (8).

Q36MCQLinear Arrangement

1. AL
2. BA
3. CB
4. DG

Answer and solution

Answer: (D) G

There are 5 biscuits, 3 candies and 4 savouries. D, E and F are savouries, and savouries come after all biscuits and candies, so K, on the last shelf (16), is the fourth savoury. L is therefore a biscuit or a candy. C, a candy, has two empty shelves before it, while L has exactly one. So each starts a group, and they are of different types: L is a biscuit. J is the same type as L, so J is a biscuit. I is on the shelf next to J with no empty shelf between them, so I is also a biscuit. H is of a different type from L, so H is a candy. A and B are on consecutive shelves, so they share a type. As candies they would make four (C, H, A, B), so A and B are biscuits. The biscuits are L, J, I, A and B, and the candies are C, H and G. Both possible layouts show these types: Options A, B and C (L, A and B) are all biscuits; G is a candy. Hence, option D (G).

Q37MCQLinear Arrangement

1. A1, 7, 11, 12
2. B1, 5, 6, 12
3. C1, 2, 6, 12
4. D1, 2, 8, 12

Answer and solution

Answer: (C) 1, 2, 6, 12

The savouries D, E, F and K fill shelves 13 to 16, so shelf 12 is empty. That leaves 8 items (5 biscuits, 3 candies) and 3 empty shelves on shelves 1 to 11. The candy group starts with C, which needs two empty shelves before it; the biscuit group starts with L, which needs exactly one. Candies first: two empty shelves, 3 candies, one empty shelf, 5 biscuits. This uses exactly 11 shelves, so the empty shelves are 1, 2, 6 and 12. Biscuits first: one empty shelf, 5 biscuits, two empty shelves, 3 candies. Again exactly 11 shelves, so the empty shelves are 1, 7, 8 and 12. Option C matches the candies-first layout. Option D (1, 2, 8, 12) is the closest trap: it would put the biscuits on 3 to 7, with two empty shelves before L, and the candies on 9 to 11, with only one before C. Options A and B would also leave only one empty shelf before C. Hence, option C (1, 2, 6, 12).

Q38MCQLinear Arrangement

1. AAll biscuits are kept before candies.
2. BThere are two empty shelves between the biscuits and the candies.
3. CAll candies are kept before biscuits.
4. DThere are at least four shelves between items B and C.

Answer and solution

Answer: (D) There are at least four shelves between items B and C.

The savouries D, E, F and K fill shelves 13 to 16. The biscuits are L, A, B, I and J, and the candies C, G and H. C opens the candy group (two empty shelves before it), and L opens the biscuit group (one empty shelf before it), followed by A, B and then I, J. Only two layouts fit. Candies first: shelves 1, 2 empty; candies on 3 to 5 with C on 3; shelf 6 empty; biscuits on 7 to 11 with B on 9. Biscuits first: shelf 1 empty; biscuits on 2 to 6 with B on 4; shelves 7, 8 empty; candies on 9 to 11 with C on 9. Between B and C lie 5 shelves (4 to 8) in the first layout and 4 shelves (5 to 8) in the second, so there are always at least four. Option D is necessarily true. Options A and C each hold in only one layout. Option B fails in the candies-first layout, where only shelf 6 separates the two groups. Hence, option D (There are at least four shelves between items B and C.).

Q39MCQMissing Value Tables

1. A25
2. B21
3. C24
4. D23

Answer and solution

Answer: (A) 25

Each bull's eye in Rounds 1–3 earns one bonus round, so the table shows how many each player hit there: Xyla three, Umeza and Zeneca two, Tanzi and Yonita one, Wangdu none. Tanzi, Umeza and Yonita total $14+a$, $13+b$ and $13+c$, where $a,b,c \le 4$ are their non-5 scores in Rounds 1–3. Fact 1 makes these totals equal, so the common total is between 15 and 17. Fact 2 allows only one total that is not a multiple of 3, so all three are 15. Xyla scored 5 in each of Rounds 1–3, so she totals $15+1+5+x=21+x$, between 22 and 26. Zeneca totals $20+z$ with $z \le 4$, at most 24. Wangdu scored at most 4 in each of his three rounds, so at most 12, the lowest. By fact 3 the highest total is $2W+1$, with $W$ Wangdu's total: odd and at least 22, so 23 ($W=11$) or 25 ($W=12$). With 23 and 11, two totals would not be multiples of 3, breaking fact 2. So the highest total is 25. This rules out 24, which is even, and 23. Hence, option A (25).

Q40MCQMissing Value Tables

1. A21
2. B22
3. C23
4. D24

Answer and solution

Answer: (D) 24

Bonus rounds played count bull's eyes in Rounds 1–3: Xyla three, Umeza and Zeneca two, Tanzi and Yonita one, Wangdu none. Tanzi, Umeza and Yonita total $14+a$, $13+b$, $13+c$ (non-5 scores $a,b,c \le 4$). By fact 1 these are equal, hence 15 to 17; fact 2 allows only one non-multiple of 3, so all are 15 and $a=1$. Fact 5: a 1 for Tanzi in Round 1 would give Zeneca a 1 there and both a 5 in Round 3. So both scored 5 in Round 1, and Tanzi 1 in Round 3. Fact 4: Tanzi and Wangdu scored 4 in Round 2, so it has at most four bull's eyes and Round 3 one or two. With one (Xyla's), Umeza and Zeneca would both hit 5 in Round 2: three. So Round 2 has four, Zeneca's included, and her Round 3 score is neither 5 nor 1: she totals 22 to 24. Fact 3: the top total is $2W+1$, where $W \le 12$ is Wangdu's (lowest) total. Xyla has at least $15+1+5+1=22$, so the top is 23 or 25; 23 with $W=11$ breaks fact 2. So Xyla has 25, the one non-multiple of 3, and Zeneca, a multiple of 3, has 24. Option A (21) needs Zeneca's Round 3 to be 1, ruled out above. Hence, option D (24).

Q41MCQMissing Value Tables

1. AXyla’s score was 23.
2. BZeneca’s score was 23.
3. CZeneca was the highest scorer.
4. DXyla was the highest scorer.

Answer and solution

Answer: (D) Xyla was the highest scorer.

Each bull's eye in Rounds 1–3 earns one bonus round, so the table shows how many each player hit there: Xyla three, Umeza and Zeneca two, Tanzi and Yonita one, Wangdu none. Tanzi, Umeza and Yonita total $14+a$, $13+b$ and $13+c$, where $a,b,c \le 4$ are their non-5 scores in Rounds 1–3. Fact 1 makes these totals equal, so the common total is between 15 and 17. Fact 2 allows only one total that is not a multiple of 3, so all three are 15. Xyla scored 5 in each of Rounds 1–3, so she totals $15+1+5+x=21+x$, between 22 and 26. Zeneca totals $20+z$ with $z \le 4$, at most 24. Wangdu scored at most 4 in each of his three rounds, so at most 12, the lowest. By fact 3 the highest total is $2W+1$, with $W$ Wangdu's total: odd and at least 22, so 23 ($W=11$) or 25 ($W=12$). With 23 and 11, two totals would not be multiples of 3, breaking fact 2. So the highest total is 25. Only Xyla can reach 25, so she, not Zeneca, was the highest scorer. As 25 is the one total that is not a multiple of 3, no one scored 23. Hence, option D (Xyla was the highest scorer.).

Q42MCQMissing Value Tables

1. A4
2. B5
3. C3
4. D1

Answer and solution

Answer: (D) 1

Each bull's eye in Rounds 1–3 earns one bonus round. Tanzi's only bonus round was Round 4, so she hit exactly one bull's eye in Rounds 1–3; Umeza and Zeneca hit two each, and Yonita one. Tanzi scored 4 in Round 2, so her Rounds 1 and 3 are a 5 and some $a \le 4$, and she totals $14+a$. Umeza totals $13+b$ and Yonita $13+c$, where $b, c \le 4$ are their non-5 scores. Fact 1 gives $b = c = a+1$, so the common total is between 15 and 17. Fact 2 allows only one total that is not a multiple of 3, so these three equal totals are 15 and $a = 1$. Fact 5: Tanzi and Zeneca match in Round 1 but differ in Round 3. If Tanzi's Round 1 score were 1, Zeneca's would be too, so Zeneca's two 5s would fall in Rounds 2 and 3, and both would score 5 in Round 3. So Tanzi scored 5 in Round 1 and 1 in Round 3. A 5 in Round 3 would be a second bull's eye, and 3 or 4 would take her total past 15. Hence, option D (1).

Q43TITA

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

Answer and solution

Answer: 1

Label the places from the right: units, tens, hundreds, thousands, ten-thousands, lakhs. Units: $F+F$ ends in $F$, which happens only for $F = 0$. So $F = 0$, with no carry. The sum has seven digits while each number added has six, so the sum's leftmost digit, $A$, is the carry out of the lakhs column. That column adds two digits and at most a carry of 1, so its total is at most $9+9+1 = 19$ and the carry is 1. So $A = 1$ (it cannot be 0, which is $F$). This fits the rest of the sum: the tens column $G+K$ must then end in 1, and since 0 and 1 are already used, $G+K = 11$. The answer is 1.

Q44TITA

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

Answer and solution

Answer: 9

Label the places from the right: units, tens, hundreds, thousands, ten-thousands, lakhs. Units: $F+F$ ends in $F$, so $F = 0$. The sum has a seventh digit, $A$, which can only be a carry of 1, so $A = 1$. Ten-thousands: $H+H$ plus a carry of 0 or 1 ends in $F = 0$. As $H \ne 0$, this total is 10. Since $2H$ is even, the carry is 0, so $H = 5$ and 1 is carried into the lakhs. Lakhs: $B+A+1 = B+2$ must end in $A = 1$ and carry 1 into the seventh place. So $B+2 = 11$ and $B = 9$. Completing the other columns gives three possible sums, and $B = 9$ in each: The answer is 9.

Q45TITA

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

Answer and solution

Answer: 7

Label the places from the right: units, tens, hundreds, thousands, ten-thousands, lakhs. Units: $F+F$ ends in $F$, so $F = 0$. The sum's seventh digit $A$ is a carry, so $A = 1$. Tens: $G+K$ ends in 1 and cannot equal 1 (0 and 1 are taken), so $G+K = 11$, carrying 1. Hundreds: $1+0+1 = 2$, so $C = 2$. Thousands: $1+J = G$ (a carry would make $G = 0$). Ten-thousands: $2H$ ends in 0, so $H = 5$; the lakhs then give $B = 9$. The unused digits are 3, 4, 6, 7 and 8. With $G+K = 11$ and $G = J+1$: $G = 8$, $K = 3$, $J = 7$ leaves 4 and 6 for $D$ and $E$; $G = 4$, $K = 7$, $J = 3$ leaves 6 and 8; $G = 7$, $K = 4$, $J = 6$ leaves 3 and 8; $G = 3$ fails, as $J$ would be 2, which is $C$; $G = 6$ fails, as $K$ would be 5, which is $H$. So $D$ can be 3, 4, 6 or 8, but never 7, which is always $J$, $K$ or $G$. The answer is 7.

Q46TITA

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

Answer and solution

Answer: 6

The sum of two six-digit numbers is less than 2,000,000, so its extra leading digit is $A = 1$. Units column: $F + F$ ends in $F$, so $F = 0$. Hundred-thousands column: $B + A + \text{carry} = A + 10$, so $B + \text{carry} = 10$. This forces $B = 9$, with a carry of 1 from the H column. H column: it shows $F = 0$ and carries 1, so $2H + \text{carry} = 10$. As $2H$ is even, that carry is 0 and $H = 5$. Hundreds column: $1 + 0 + \text{carry} = C$ and $C \ne 1$, so the carry is 1, $C = 2$, and nothing carries onward. Thousands column: $A + J = G$, so $G = J + 1$. Tens column: $G + K = 11$, giving digit $A$ and the carry into the hundreds. The unused digits are 3, 4, 6, 7 and 8; pairs adding to 11 are 3 and 8, and 4 and 7. $G = 3$ needs $J = 2 = C$, impossible. $G = 8$, 7 and 4 all work, with $J = G - 1$ and $K = 11 - G$: $G = 6$ would need $K = 5 = H$, so G can never be 6. The answer is 6.

Q47MCQRadar & Special Graphs

1. ACustomer Service
2. BCost
3. CReliability
4. DQuality

Q48MCQRadar & Special Graphs

1. AVendor 4
2. BVendor 2
3. CVendor 1
4. DVendor 3

Q49MCQRadar & Special Graphs

1. AVendor 2, Vendor 3 and Vendor 4
2. BVendor 1 and Vendor 5
3. CVendor 2 and Vendor 5
4. DVendor 1 and Vendor 2

Answer and solution

Answer: (B) Vendor 1 and Vendor 5

The radar chart's scores, read approximately, are in this table: The top two vendors on each aspect: Reliability: Vendor 3 (75), Vendor 5 (60). Reach: Vendor 1 (80), Vendor 5 (70). Quality: Vendor 1 (72), Vendor 2 (69). Features: Vendor 4 (90), Vendor 5 (75). Customer Service: Vendor 4 (70), Vendor 1 (55). Cost: Vendor 3 (90), Vendor 2 (81). Counting appearances: Vendor 1 three times (Reach, Quality, Customer Service), Vendor 5 three times (Reliability, Reach, Features), and Vendors 2, 3 and 4 twice each. The maximum, three, is reached only by Vendors 1 and 5. Options C and D fail because Vendor 2 is in the top two only twice, and option A lists three vendors who appear only twice each. Hence, option B (Vendor 1 and Vendor 5).

Q50MCQRadar & Special Graphs

1. AVendor 1 and Vendor 3
2. BNone of the Vendors
3. CVendor 3
4. DVendor 1

Answer and solution

Answer: (C) Vendor 3

The radar chart's scores, read approximately, are in this table: The top three vendors on each aspect: Reliability: Vendors 3 (75), 5 (60), 1 (52). Reach: Vendors 1 (80), 5 (70), 3 (63). Quality: Vendors 1 (72), 2 (69), 3 (62). Features: Vendors 4 (90), 5 (75), 3 (56). Customer Service: Vendors 4 (70), 1 (55), 3 (50). Cost: Vendors 3 (90), 2 (81), 1 (77). Vendor 3 is in all six lists. Vendor 1 is in five but misses Features, where its score of about 40 is the lowest of all five vendors, so options A and D fail. Vendor 5 misses three aspects and Vendors 2 and 4 miss four each, so option B fails too. Hence, option C (Vendor 3).

Q51TITABar & Line Charts

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

Answer and solution

Answer: 5

The chart's counts, read approximately, are in this table: IPC cases: Delhi 64, Goa 26, Karnataka 16, Maharashtra 16, Kerala 8, Telangana 4, and fewer for Haryana, Tamil Nadu, Puducherry and West Bengal. The rank table gives Telangana rank 6 in IPC crimes, so exactly five states/UTs have more IPC cases than Telangana. Delhi, Goa, Karnataka, Maharashtra and Kerala all do, so they are those five, and no state outside the chart is above Telangana. Hence exactly four states (Delhi, Goa, Karnataka, Maharashtra) have more IPC cases than Kerala, and Kerala's rank is $4 + 1 = 5$. This fits the table: Karnataka and Maharashtra share rank 3 behind Delhi and Goa, so rank 4 is skipped. The answer is 5.

Q52MCQBar & Line Charts

1. A3 : 2
2. B19 : 20
3. C11 : 10
4. D1 : 9

Answer and solution

Answer: (D) 1 : 9

The chart's counts, read approximately, are in this table: The two states with the most cases are West Bengal (520, all of them SLL crimes) and Delhi (about 142); the next, Karnataka, has only about 91. IPC cases: West Bengal 0 + Delhi about 64 = about 64. SLL cases: West Bengal 520 + Delhi about 35 = about 555. The ratio is about $\frac{64}{555} \approx 0.115$, and $1 : 9 \approx 0.111$ is the closest option. The other options, 3 : 2, 19 : 20 and 11 : 10, are all near or above 1, which would need about as many IPC cases as SLL cases. Here SLL cases outnumber IPC cases about nine to one. Hence, option D (1 : 9).

Q53MCQBar & Line Charts

1. Aboth i) and ii)
2. Bonly ii)
3. Cneither i) , nor ii)
4. Donly i)

Answer and solution

Answer: (A) both i) and ii)

The chart's counts, read approximately, are in this table: 'Other crimes' counts: Delhi 43, Tamil Nadu 35, Puducherry 30, Karnataka 26, Goa 19, and fewer for the rest of the ten. Every state outside the top ten has no more cases in total than Telangana, about 25, so none of them can have more than about 25 other-crime cases. So only Delhi has more other-crime cases than Tamil Nadu, giving Tamil Nadu rank 2, and only Delhi and Tamil Nadu have more than Puducherry, giving Puducherry rank 3. Both statements are definitely true. Option D (only i) would need some state to overtake Puducherry's 30, but every state not already above it has at most about 26. Hence, option A (both i) and ii)).

Q54TITABar & Line Charts

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

Answer and solution

Answer: 5

The chart's counts, read approximately, are in this table: Delhi has about 64 IPC, 35 SLL and 43 other-crime cases. States outside the top ten have no more cases in total than Telangana (about 25). IPC: no other state comes close to Delhi's 64 (Goa is next with about 26), so Delhi's rank is 1. Other crimes: Delhi's 43 is again the most (Tamil Nadu is next with about 35), so its rank is 1. SLL: West Bengal (520) and Karnataka (rank 2 in the table) are ahead of Delhi. Goa and Maharashtra, with about the same count as Delhi, have rank 4, so exactly three states have more SLL cases than they do: West Bengal, Karnataka and one more. Every other state has at most about 26, so the third must be Delhi, whose rank is 3. Sum of ranks $= 1 + 3 + 1 = 5$. The answer is 5.

Q55MCQNetworks & Routes

1. AW
2. BY
3. CNo one
4. DV

Answer and solution

Answer: (C) No one

By fact 1, X, U and Z fill triangle b–f–g or b–c–g; by fact 2, no one else stands where X can see. Fact 4 needs an intersection just beyond Z on line UZ, other than the barred d, leaving five placements: 1. U=b, Z=f, V=j, X=g: W, seeing neither f nor j, is at a; Y, seeing a and b, is at c (seen by X) or the barred d. Fails. 2. U=c, Z=b, V=a, X=g, and 3. U=c, Z=g, V=k, X=b: apart from d, every free spot that sees c is in X's view, so Y cannot see U. Fails. 4. U=f, Z=g, V=h, X=b: W, seeing neither g nor h, is at i; Y, seeing f and i, is at e (also sees Z) or j (seen by X). Fails. 5. U=g, Z=f, V=e, X=b: Y sees g from a spot hidden from X, so h or k; h also sees Z, so Y=k. W is seen by Y: c and j are seen by X and i sees V, so W=l. Only placement 5 works: No one stands at a: X at b sees a, so it stays empty. V was at a only in rejected placement 2, so option D fails. Hence, option C (No one).

Q56MCQNetworks & Routes

1. AZ only
2. BU, W and Z only
3. CU and Z only
4. DU only

Answer and solution

Answer: (C) U and Z only

By fact 1, X, U and Z fill triangle b–f–g or b–c–g; by fact 2, no one else stands where X can see. Fact 4 needs an intersection just beyond Z on line UZ, other than the barred d, leaving five placements: 1. U=b, Z=f, V=j, X=g: W, seeing neither f nor j, is at a; Y, seeing a and b, is at c (seen by X) or the barred d. Fails. 2. U=c, Z=b, V=a, X=g, and 3. U=c, Z=g, V=k, X=b: apart from d, every free spot that sees c is in X's view, so Y cannot see U. Fails. 4. U=f, Z=g, V=h, X=b: W, seeing neither g nor h, is at i; Y, seeing f and i, is at e (also sees Z) or j (seen by X). Fails. 5. U=g, Z=f, V=e, X=b: Y sees g from a spot hidden from X, so h or k; h also sees Z, so Y=k. W is seen by Y: c and j are seen by X and i sees V, so W=l. Only placement 5 works: V at e sees along row a–e–i (both empty) and column e–f–g–h: Z at f and U at g. W at l is on neither street, so option B fails. Hence, option C (U and Z only).

Q57MCQNetworks & Routes

1. A1
2. B4
3. C2
4. D3

Answer and solution

Answer: (C) 2

By fact 1, X, U and Z fill triangle b–f–g or b–c–g; by fact 2, no one else stands where X can see. Fact 4 needs an intersection just beyond Z on line UZ, other than the barred d, leaving five placements: 1. U=b, Z=f, V=j, X=g: W, seeing neither f nor j, is at a; Y, seeing a and b, is at c (seen by X) or the barred d. Fails. 2. U=c, Z=b, V=a, X=g, and 3. U=c, Z=g, V=k, X=b: apart from d, every free spot that sees c is in X's view, so Y cannot see U. Fails. 4. U=f, Z=g, V=h, X=b: W, seeing neither g nor h, is at i; Y, seeing f and i, is at e (also sees Z) or j (seen by X). Fails. 5. U=g, Z=f, V=e, X=b: Y sees g from a spot hidden from X, so h or k; h also sees Z, so Y=k. W is seen by Y: c and j are seen by X and i sees V, so W=l. Only placement 5 works: X at b and Y at k share no street, so one segment (option A) is not enough. The route b–g along the diagonal, then g–k along row c–g–k, uses 2. Hence, option C (2).

Q58MCQNetworks & Routes

1. AW and X only
2. BU and W only
3. CU and Z only
4. DV and X only

Answer and solution

Answer: (A) W and X only

By fact 1, X, U and Z fill triangle b–f–g or b–c–g; by fact 2, no one else stands where X can see. Fact 4 needs an intersection just beyond Z on line UZ, other than the barred d, leaving five placements: 1. U=b, Z=f, V=j, X=g: W, seeing neither f nor j, is at a; Y, seeing a and b, is at c (seen by X) or the barred d. Fails. 2. U=c, Z=b, V=a, X=g, and 3. U=c, Z=g, V=k, X=b: apart from d, every free spot that sees c is in X's view, so Y cannot see U. Fails. 4. U=f, Z=g, V=h, X=b: W, seeing neither g nor h, is at i; Y, seeing f and i, is at e (also sees Z) or j (seen by X). Fails. 5. U=g, Z=f, V=e, X=b: Y sees g from a spot hidden from X, so h or k; h also sees Z, so Y=k. W is seen by Y: c and j are seen by X and i sees V, so W=l. Only placement 5 works: From d one sees column a–b–c–d, holding only X (at b), and row d–h–l, holding only W (at l). U at g is on neither, so options B and C fail. Hence, option A (W and X only).

Q59MCQTeam Selection & Scheduling

1. AThe second performance was composed by Dyu.
2. BThe third performance was composed by Dyu.
3. CThe third performance was composed by Ashman.
4. DThe second performance was composed by Gagan.

Answer and solution

Answer: (A) The second performance was composed by Dyu.

All first items came before any second item, and second items kept the same order, so performances 1–4 and 5–8 follow the same dancer order: Princess (1st, Badal), ?, ?, Rani, Princess, ?, ?, Rani (8th, Gagan). By (i) and (ii), each composer works with one of Princess and Queen and one of Rani and Samragni: two items each. The first items have different composers, so each composer has one item in 1–4 and one in 5–8, and Ashman's consecutive pair is the 4th and 5th. Badal has the 1st and the 6th or 7th, Gagan the 8th and the 2nd or 3rd, and Dyu the rest; the three gaps must be equal. Badal at 6th (gap 5) needs Gagan 3rd, leaving Dyu 2nd and 7th, also gap 5. Badal at 7th (gap 6) needs Gagan 2nd, leaving Dyu 3rd and 6th, gap 3. So the composers run Badal, Dyu, Gagan, Ashman, Ashman, Badal, Dyu, Gagan. Princess has Badal and Ashman, so by (i) Queen has Gagan and Dyu (3rd and 7th), and Samragni the 2nd and 6th (Dyu, Badal). The 2nd was Dyu's and the 3rd Gagan's, so options B, C and D fail. Hence, option A (The second performance was composed by Dyu.).

Q60MCQTeam Selection & Scheduling

1. ASamragni did not perform in any item composed by Ashman.
2. BPrincess did not perform in any item composed by Dyu.
3. CRani did not perform in any item composed by Badal.
4. DQueen did not perform in any item composed by Gagan.

Answer and solution

Answer: (D) Queen did not perform in any item composed by Gagan.

All first items came before any second item, and second items kept the same order, so performances 1–4 and 5–8 follow the same dancer order: Princess (1st, Badal), ?, ?, Rani, Princess, ?, ?, Rani (8th, Gagan). By (i) and (ii), each composer works with one of Princess and Queen and one of Rani and Samragni: two items each. The first items have different composers, so each composer has one item in 1–4 and one in 5–8, and Ashman's consecutive pair is the 4th and 5th. Badal has the 1st and the 6th or 7th, Gagan the 8th and the 2nd or 3rd, and Dyu the rest; the three gaps must be equal. Badal at 6th (gap 5) needs Gagan 3rd, leaving Dyu 2nd and 7th, also gap 5. Badal at 7th (gap 6) needs Gagan 2nd, leaving Dyu 3rd and 6th, gap 3. So the composers run Badal, Dyu, Gagan, Ashman, Ashman, Badal, Dyu, Gagan. Princess has Badal and Ashman, so by (i) Queen has Gagan and Dyu (3rd and 7th), and Samragni the 2nd and 6th (Dyu, Badal). Samragni (Dyu, Badal), Princess (Badal, Ashman) and Rani (Ashman, Gagan) make A, B and C true, but Queen danced Gagan's item 3rd. Hence, option D (Queen did not perform in any item composed by Gagan.).

Q61MCQTeam Selection & Scheduling

1. ABadal
2. BDyu
3. CAshman
4. DGagan

Answer and solution

Answer: (A) Badal

All first items came before any second item, and second items kept the same order, so performances 1–4 and 5–8 follow the same dancer order: Princess (1st, Badal), ?, ?, Rani, Princess, ?, ?, Rani (8th, Gagan). By (i) and (ii), each composer works with one of Princess and Queen and one of Rani and Samragni: two items each. The first items have different composers, so each composer has one item in 1–4 and one in 5–8, and Ashman's consecutive pair is the 4th and 5th. Badal has the 1st and the 6th or 7th, Gagan the 8th and the 2nd or 3rd, and Dyu the rest; the three gaps must be equal. Badal at 6th (gap 5) needs Gagan 3rd, leaving Dyu 2nd and 7th, also gap 5. Badal at 7th (gap 6) needs Gagan 2nd, leaving Dyu 3rd and 6th, gap 3. So the composers run Badal, Dyu, Gagan, Ashman, Ashman, Badal, Dyu, Gagan. Princess has Badal and Ashman, so by (i) Queen has Gagan and Dyu (3rd and 7th), and Samragni the 2nd and 6th (Dyu, Badal). The 6th was Samragni's second item, by Badal; Ashman's items were the 4th and 5th, so option C fails. Hence, option A (Badal).

Q62MCQTeam Selection & Scheduling

1. AThe first and the seventh
2. BThe third and the seventh
3. CThe second and the sixth
4. DThe first and the sixth

Answer and solution

Answer: (D) The first and the sixth

All first items came before any second item, and second items kept the same order, so performances 1–4 and 5–8 follow the same dancer order: Princess (1st, Badal), ?, ?, Rani, Princess, ?, ?, Rani (8th, Gagan). By (i) and (ii), each composer works with one of Princess and Queen and one of Rani and Samragni: two items each. The first items have different composers, so each composer has one item in 1–4 and one in 5–8, and Ashman's consecutive pair is the 4th and 5th. Badal has the 1st and the 6th or 7th, Gagan the 8th and the 2nd or 3rd, and Dyu the rest; the three gaps must be equal. Badal at 6th (gap 5) needs Gagan 3rd, leaving Dyu 2nd and 7th, also gap 5. Badal at 7th (gap 6) needs Gagan 2nd, leaving Dyu 3rd and 6th, gap 3. So the composers run Badal, Dyu, Gagan, Ashman, Ashman, Badal, Dyu, Gagan. Princess has Badal and Ashman, so by (i) Queen has Gagan and Dyu (3rd and 7th), and Samragni the 2nd and 6th (Dyu, Badal). Badal composed the 1st and 6th. The 7th was Dyu's, so options A and B fail, and the 2nd (Dyu) and 6th (Badal) differ, so option C fails. Hence, option D (The first and the sixth).

Q63TITALogical Puzzles

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

Q64TITALogical Puzzles

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

Q65MCQLogical Puzzles

1. AThere are exactly 75 items of type e.
2. BThere are exactly 30 items of type b.
3. CThere are exactly 45 items of type c.
4. DThere are exactly 60 items of type d.

Q66MCQLogical Puzzles

1. A5
2. B6
3. C4
4. D3