Page 1 1 1 2 3 3 4 4 5 5 5 6 3 7 1 8 1 9 4 10 5 11 4 12 2 13 2 14 2 15 3 16 3 17 2 18 1 19 5 20 4 21 5 22 2,3 23 1 24 4 25 3 26 5 27 1 28 2 29 3 30 4 31 2 32 1 33 4 34 5 35 3 36 3 37 2 38 5 39 1 40 4 41 3 42 1 43 4 44 2 45 4 46 3 47 5 48 4 49 2 50 1 51 1 52 2 53 1 54 2 55 5 56 2 57 2 58 4 59 2 60 4 61 2 62 4 63 1 64 4 65 4 66 3 67 2 68 5 69 1 70 5 71 5 72 5 73 2 74 5 75 3 LRDI 1 to 25 25 EU + RC 26 to 50 25 QA 51 to 75 25 T otal 75 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number Page 2 1 1 2 3 3 4 4 5 5 5 6 3 7 1 8 1 9 4 10 5 11 4 12 2 13 2 14 2 15 3 16 3 17 2 18 1 19 5 20 4 21 5 22 2,3 23 1 24 4 25 3 26 5 27 1 28 2 29 3 30 4 31 2 32 1 33 4 34 5 35 3 36 3 37 2 38 5 39 1 40 4 41 3 42 1 43 4 44 2 45 4 46 3 47 5 48 4 49 2 50 1 51 1 52 2 53 1 54 2 55 5 56 2 57 2 58 4 59 2 60 4 61 2 62 4 63 1 64 4 65 4 66 3 67 2 68 5 69 1 70 5 71 5 72 5 73 2 74 5 75 3 LRDI 1 to 25 25 EU + RC 26 to 50 25 QA 51 to 75 25 T otal 75 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number For questions 1 to 5: From statement one, team would include exactly one among P, R, S ? P (or) R (or) S. From statement two, team would include either M, or Q ? M but not Q (or) Q but not M From statement three, if a team includes K, it will include L or vice versa. ? K, L always accompany each other. From statement four, if one of S, U, W is included, then the other two also have to be included. ? S, U, W are always together. From statement five, L and N cannot be included together ? L, N are never together. From statement six, L and U cannot be included together. ? L, U are never together. 1. 1 From statements one and two; one of P, R, S and one of M, Q are to be selected. We require one more member. But from statement three; (K, L) are always together. Hence 'L' cannot be included in a team of 3 members. 2. 3 Again, from statement one; one of P, R, S has to be selected. To make a team of '5' 'S' will be chosen (which leaves out P and R) ? If 'S' is chosen 'U' and ‘W’ have to be chosen (statement four) ? If 'U' is chosen 'L' cannot be chosen (statement five) ? K cannot be chosen (statement three) And from statement two; one of M (or) Q has to be chosen. 3. 4 From statements one and two Two members are to be selected. Of the remaining seven; To maximize the size of the team. We would chose S, ? U and W are included in the team (statement four) We cannot include K (or) L because we would then have to leave out N and U (from statements five and six) 4. 5 If 'K' is included, 'L' has to be included (statement (3)) If 'L' is chosen, neither N nor U can be chosen (statements (5) and (6)) ? S, W are also not included because S, U, W have to be always together. (Statement (4)) Hence one of P (or) R would be selected (statement (1)) and one of M (or) Q would be selected (statement (2)) (K, L) and two of the above five have to be included. 5. 5 If a team includes N, it cannot include 'L', and therefore, not even 'K'. (from statement five and three) According to statement (1), one of P or R or S has to be included. According to statement (2), one of M or Q has to be selected. So the following cases are possible P Q N, R Q N P M N, R M N If 'S' is selected, then S U W M N and S U W Q N are the only possible cases. Hence, in all 4 + 2 = 6 teams can be constituted. For questions 6 to 10: 6. 3 Let Dipan get x marks in paper II. Dipan's average in PCB group = 98 Maths group = 95 S.S. group = 95.5 Vernacular group = 95 English group = + ?? ?? ?? 96 x 2 Sum of all = 96 × 5 So 95.5 + 96 × 3 + + x 48 2 = 96 × 5 ?= × - - x 96 2 95.5 48 2 () =- =× = x 2 96.5 48 2 48.5 97 So (3) is the correct option. 7. 1 The only boy getting 95 in atleast one of the subjects of the group among all the groups is Dipan. So (1) is the correct option. 8. 1 A group score of 100 in Social Science would have increased the scores as follows: Page 3 1 1 2 3 3 4 4 5 5 5 6 3 7 1 8 1 9 4 10 5 11 4 12 2 13 2 14 2 15 3 16 3 17 2 18 1 19 5 20 4 21 5 22 2,3 23 1 24 4 25 3 26 5 27 1 28 2 29 3 30 4 31 2 32 1 33 4 34 5 35 3 36 3 37 2 38 5 39 1 40 4 41 3 42 1 43 4 44 2 45 4 46 3 47 5 48 4 49 2 50 1 51 1 52 2 53 1 54 2 55 5 56 2 57 2 58 4 59 2 60 4 61 2 62 4 63 1 64 4 65 4 66 3 67 2 68 5 69 1 70 5 71 5 72 5 73 2 74 5 75 3 LRDI 1 to 25 25 EU + RC 26 to 50 25 QA 51 to 75 25 T otal 75 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number For questions 1 to 5: From statement one, team would include exactly one among P, R, S ? P (or) R (or) S. From statement two, team would include either M, or Q ? M but not Q (or) Q but not M From statement three, if a team includes K, it will include L or vice versa. ? K, L always accompany each other. From statement four, if one of S, U, W is included, then the other two also have to be included. ? S, U, W are always together. From statement five, L and N cannot be included together ? L, N are never together. From statement six, L and U cannot be included together. ? L, U are never together. 1. 1 From statements one and two; one of P, R, S and one of M, Q are to be selected. We require one more member. But from statement three; (K, L) are always together. Hence 'L' cannot be included in a team of 3 members. 2. 3 Again, from statement one; one of P, R, S has to be selected. To make a team of '5' 'S' will be chosen (which leaves out P and R) ? If 'S' is chosen 'U' and ‘W’ have to be chosen (statement four) ? If 'U' is chosen 'L' cannot be chosen (statement five) ? K cannot be chosen (statement three) And from statement two; one of M (or) Q has to be chosen. 3. 4 From statements one and two Two members are to be selected. Of the remaining seven; To maximize the size of the team. We would chose S, ? U and W are included in the team (statement four) We cannot include K (or) L because we would then have to leave out N and U (from statements five and six) 4. 5 If 'K' is included, 'L' has to be included (statement (3)) If 'L' is chosen, neither N nor U can be chosen (statements (5) and (6)) ? S, W are also not included because S, U, W have to be always together. (Statement (4)) Hence one of P (or) R would be selected (statement (1)) and one of M (or) Q would be selected (statement (2)) (K, L) and two of the above five have to be included. 5. 5 If a team includes N, it cannot include 'L', and therefore, not even 'K'. (from statement five and three) According to statement (1), one of P or R or S has to be included. According to statement (2), one of M or Q has to be selected. So the following cases are possible P Q N, R Q N P M N, R M N If 'S' is selected, then S U W M N and S U W Q N are the only possible cases. Hence, in all 4 + 2 = 6 teams can be constituted. For questions 6 to 10: 6. 3 Let Dipan get x marks in paper II. Dipan's average in PCB group = 98 Maths group = 95 S.S. group = 95.5 Vernacular group = 95 English group = + ?? ?? ?? 96 x 2 Sum of all = 96 × 5 So 95.5 + 96 × 3 + + x 48 2 = 96 × 5 ?= × - - x 96 2 95.5 48 2 () =- =× = x 2 96.5 48 2 48.5 97 So (3) is the correct option. 7. 1 The only boy getting 95 in atleast one of the subjects of the group among all the groups is Dipan. So (1) is the correct option. 8. 1 A group score of 100 in Social Science would have increased the scores as follows: Score Increase Group Score Final Score Increase Final group Score Pritam 22 11 = 11 2.2 5 96.1 Joseph 9 4.5 = 4.5 .9 5 95.9 Tirna 21 10.5 = 10.5 2.1 5 95.8 Agni 9 4.5 = 4.5 .9 5 95.2 So the order is Pritam > Joseph > Trina > Agni. So option (1) is the correct choice. 9. 4 The student having atleast 95 in every group is Dipan, so the answer is Dipan, option (4). 10. 5 Let us increase the score in one of the subjects of the following candidates Least Scores Contributio n in net Score Final Score Ram 94 in group of 2 3 in 5 groups 96.1 + .6 = 96.7 Agni 82 in group of 2 9 in 5 groups 94.3 + 1.8 = 96.1 Pritam 83 in group of 2 8.5 in 5 groups 93.9 + 1.7 = 95.6 Ayesha 93 in group of 2 3.5 in 5 groups 96.2 + .7 = 96.9 Dipan 95 in group of 1 5 in 5 groups 96 + 1 = 97.0 So, Dipan will end with a highest total. So the answer is option (5). For questions 11 to 15: As only Paul Erdös was having an Erdös number of zero, so the minimum Erdös number among A, B, C, D, E, F , G, H should be 1 or greater than one. At the end of the third day, F co- authored a paper with A and C. F had the minimum Erdös number among the 8 people. So if F's Erdös number is y, then A and C's Erdös number should change to (y + 1) after third day. As A and C decreased the average by maximum possible extent, it means C had the second-height Erdös number among all eight, as A had an Erdös number of infinity. Suppose Erdös numbers of A, B, C, D, E, F, G, H are y + 1, b, y + 1, c, d, e, y, g, h respectively at the end of third day. ? (y + 1 + b + y + 1 + c + d + e + y + g + h) = 24 = (3 × 8) ? 3y + 2 + b + d + e + g + h = 24 When E co-authored with F , the average Erdös number reduced again, it means, E's Erdös number was not the same with A & C initially. As at the end of third day, 5 people had same Erdös number, they should be A, C and any 3 out of B, D, G, H. Suppose those 3 people are B, D, G. Then (3y + 2 + y + 1 + y + 1 + y + 1 + e + h) = 24 ? 6y + h + e = 19 …(i) On the fifth day, E co-authored a paper with F and hence, Erdös number of E changed to (y + 1). Also the average decreased by 0.5 which means the total decreased by 4. Hence, e - (y + 1) = 4 ? e – y = 5 Putting the value of e in equation (i), we get 6y + h + (5 + y) = 19 ? 7y + h = 14 Only possible value of y = 1 as h cannot be zero. So after 3rd round Erdös number of A, C, E, F were 2, 2, 6, 1 respectively. 11. 4 Only A, C, E changed their Erdös number, rest 5 did not change their Erdös number. 12. 2 At the end of conference 6 people including E were having an Erdös number of 2 and F was having 1 as Erdös number. So 8th person was having an Erdös number of [20 – (2 × 6 + 1)] = 7 13. 2 At the end of 3rd round, 5 people were having same Erdös number. A and C changed their Erdös number after coauthoring with F. So, the other 3 will have same Erdös number in the beginning. 14. 2 2 15. 3 After co-authoring with F, E was having Erdös number of 2, which was 4 less than initial Erdös number of E. So answer is 2 + 4 = 6. For questions 16 to 20: The MCS share price at the beginning of first day is Rs.100 and at the close of day 5 is Rs.110. The following cases of the closing prices can be derived. At the end of Day 1Day 2Day 3Day 4Day 5 1 90 80 90 100 110 2 90 100 90 100 110 3 90 100 110 120 110 4 90 100 110 100 110 5 110 100 90 100 110 6 110 100 110 100 110 7 110 120 110 100 110 8 110 120 110 120 110 9 110 120 130 120 110 10 110 120 110 100 110 16. 3 As Chetan sold 10 shares on three consecutive days, therefore, of the five days, there must be an increase for three of the five days and a decrease for the remaining two days. It is given that Michael sold 10 shares only once. Hence, the price is more than 110 Page 4 1 1 2 3 3 4 4 5 5 5 6 3 7 1 8 1 9 4 10 5 11 4 12 2 13 2 14 2 15 3 16 3 17 2 18 1 19 5 20 4 21 5 22 2,3 23 1 24 4 25 3 26 5 27 1 28 2 29 3 30 4 31 2 32 1 33 4 34 5 35 3 36 3 37 2 38 5 39 1 40 4 41 3 42 1 43 4 44 2 45 4 46 3 47 5 48 4 49 2 50 1 51 1 52 2 53 1 54 2 55 5 56 2 57 2 58 4 59 2 60 4 61 2 62 4 63 1 64 4 65 4 66 3 67 2 68 5 69 1 70 5 71 5 72 5 73 2 74 5 75 3 LRDI 1 to 25 25 EU + RC 26 to 50 25 QA 51 to 75 25 T otal 75 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number For questions 1 to 5: From statement one, team would include exactly one among P, R, S ? P (or) R (or) S. From statement two, team would include either M, or Q ? M but not Q (or) Q but not M From statement three, if a team includes K, it will include L or vice versa. ? K, L always accompany each other. From statement four, if one of S, U, W is included, then the other two also have to be included. ? S, U, W are always together. From statement five, L and N cannot be included together ? L, N are never together. From statement six, L and U cannot be included together. ? L, U are never together. 1. 1 From statements one and two; one of P, R, S and one of M, Q are to be selected. We require one more member. But from statement three; (K, L) are always together. Hence 'L' cannot be included in a team of 3 members. 2. 3 Again, from statement one; one of P, R, S has to be selected. To make a team of '5' 'S' will be chosen (which leaves out P and R) ? If 'S' is chosen 'U' and ‘W’ have to be chosen (statement four) ? If 'U' is chosen 'L' cannot be chosen (statement five) ? K cannot be chosen (statement three) And from statement two; one of M (or) Q has to be chosen. 3. 4 From statements one and two Two members are to be selected. Of the remaining seven; To maximize the size of the team. We would chose S, ? U and W are included in the team (statement four) We cannot include K (or) L because we would then have to leave out N and U (from statements five and six) 4. 5 If 'K' is included, 'L' has to be included (statement (3)) If 'L' is chosen, neither N nor U can be chosen (statements (5) and (6)) ? S, W are also not included because S, U, W have to be always together. (Statement (4)) Hence one of P (or) R would be selected (statement (1)) and one of M (or) Q would be selected (statement (2)) (K, L) and two of the above five have to be included. 5. 5 If a team includes N, it cannot include 'L', and therefore, not even 'K'. (from statement five and three) According to statement (1), one of P or R or S has to be included. According to statement (2), one of M or Q has to be selected. So the following cases are possible P Q N, R Q N P M N, R M N If 'S' is selected, then S U W M N and S U W Q N are the only possible cases. Hence, in all 4 + 2 = 6 teams can be constituted. For questions 6 to 10: 6. 3 Let Dipan get x marks in paper II. Dipan's average in PCB group = 98 Maths group = 95 S.S. group = 95.5 Vernacular group = 95 English group = + ?? ?? ?? 96 x 2 Sum of all = 96 × 5 So 95.5 + 96 × 3 + + x 48 2 = 96 × 5 ?= × - - x 96 2 95.5 48 2 () =- =× = x 2 96.5 48 2 48.5 97 So (3) is the correct option. 7. 1 The only boy getting 95 in atleast one of the subjects of the group among all the groups is Dipan. So (1) is the correct option. 8. 1 A group score of 100 in Social Science would have increased the scores as follows: Score Increase Group Score Final Score Increase Final group Score Pritam 22 11 = 11 2.2 5 96.1 Joseph 9 4.5 = 4.5 .9 5 95.9 Tirna 21 10.5 = 10.5 2.1 5 95.8 Agni 9 4.5 = 4.5 .9 5 95.2 So the order is Pritam > Joseph > Trina > Agni. So option (1) is the correct choice. 9. 4 The student having atleast 95 in every group is Dipan, so the answer is Dipan, option (4). 10. 5 Let us increase the score in one of the subjects of the following candidates Least Scores Contributio n in net Score Final Score Ram 94 in group of 2 3 in 5 groups 96.1 + .6 = 96.7 Agni 82 in group of 2 9 in 5 groups 94.3 + 1.8 = 96.1 Pritam 83 in group of 2 8.5 in 5 groups 93.9 + 1.7 = 95.6 Ayesha 93 in group of 2 3.5 in 5 groups 96.2 + .7 = 96.9 Dipan 95 in group of 1 5 in 5 groups 96 + 1 = 97.0 So, Dipan will end with a highest total. So the answer is option (5). For questions 11 to 15: As only Paul Erdös was having an Erdös number of zero, so the minimum Erdös number among A, B, C, D, E, F , G, H should be 1 or greater than one. At the end of the third day, F co- authored a paper with A and C. F had the minimum Erdös number among the 8 people. So if F's Erdös number is y, then A and C's Erdös number should change to (y + 1) after third day. As A and C decreased the average by maximum possible extent, it means C had the second-height Erdös number among all eight, as A had an Erdös number of infinity. Suppose Erdös numbers of A, B, C, D, E, F, G, H are y + 1, b, y + 1, c, d, e, y, g, h respectively at the end of third day. ? (y + 1 + b + y + 1 + c + d + e + y + g + h) = 24 = (3 × 8) ? 3y + 2 + b + d + e + g + h = 24 When E co-authored with F , the average Erdös number reduced again, it means, E's Erdös number was not the same with A & C initially. As at the end of third day, 5 people had same Erdös number, they should be A, C and any 3 out of B, D, G, H. Suppose those 3 people are B, D, G. Then (3y + 2 + y + 1 + y + 1 + y + 1 + e + h) = 24 ? 6y + h + e = 19 …(i) On the fifth day, E co-authored a paper with F and hence, Erdös number of E changed to (y + 1). Also the average decreased by 0.5 which means the total decreased by 4. Hence, e - (y + 1) = 4 ? e – y = 5 Putting the value of e in equation (i), we get 6y + h + (5 + y) = 19 ? 7y + h = 14 Only possible value of y = 1 as h cannot be zero. So after 3rd round Erdös number of A, C, E, F were 2, 2, 6, 1 respectively. 11. 4 Only A, C, E changed their Erdös number, rest 5 did not change their Erdös number. 12. 2 At the end of conference 6 people including E were having an Erdös number of 2 and F was having 1 as Erdös number. So 8th person was having an Erdös number of [20 – (2 × 6 + 1)] = 7 13. 2 At the end of 3rd round, 5 people were having same Erdös number. A and C changed their Erdös number after coauthoring with F. So, the other 3 will have same Erdös number in the beginning. 14. 2 2 15. 3 After co-authoring with F, E was having Erdös number of 2, which was 4 less than initial Erdös number of E. So answer is 2 + 4 = 6. For questions 16 to 20: The MCS share price at the beginning of first day is Rs.100 and at the close of day 5 is Rs.110. The following cases of the closing prices can be derived. At the end of Day 1Day 2Day 3Day 4Day 5 1 90 80 90 100 110 2 90 100 90 100 110 3 90 100 110 120 110 4 90 100 110 100 110 5 110 100 90 100 110 6 110 100 110 100 110 7 110 120 110 100 110 8 110 120 110 120 110 9 110 120 130 120 110 10 110 120 110 100 110 16. 3 As Chetan sold 10 shares on three consecutive days, therefore, of the five days, there must be an increase for three of the five days and a decrease for the remaining two days. It is given that Michael sold 10 shares only once. Hence, the price is more than 110 for only one day and on all the remaining days, it cannot exceed 110. The only satisfying case is (3). Hence, the price at the end of Day 3 is Rs.110. 17. 2 The satisfying cases are (1), (2), (4), (5), (6). Hence, the price at the end of Day 4 is Rs.100. 18. 1 Let Chetan and Michael start with x number of shares initially. From case (1), we get that the number of shares with Michael = x + 10 and number of shares with Chetan = x + 10 + 10 – 10 – 10 – 10 = x – 10. So Michael has 20 more shares than Chetan. This is the only satisfying case. Hence, the share price at the end of Day 3 is Rs.90. 19. 5 Consider cases (3) and (7). Only these two satisfies the condition that Michael had Rs.100 less than Chetan at the end of day 5. For case (3), Number of shares with Chetan = x + 10 – 10 – 10 – 10 + 10 = x – 10 And with Michael = x – 10 For case (7), Number of shares with Chetan = x – 10 – 10 + 10 + 10 – 10 = x – 10 And with Michael = x – 10 In either case, number of shares with Michael and Chetan are the same. 20. 4 To maximise the amount gathered by both of them, we need to look into those cases wherein we have maximum number of 110 excess figures. It is only then that Michael and Chetan both will make money. So we check for case (9). For case (9), Extra cash with Chetan by the end of day 5 = 1100 + 1200 + 1300 – 1200 – 1100 = Rs.1300 And that with Michael = 1200 + 1300 + 1200 = Rs.3700 Total extra cash with both of them = 1300 + 3700 = Rs.5000 For questions 21 to 25: In this set, the fuel cost for each of the path is given. In addition, there are four toll collection junctions. 21. 5 No traffic flows on the street from D to T. Now, we have fuel cost on different paths as SAT : 9 + 5 = Rs. 14 + toll at junction A SBAT : 2 + 2 + 5 = Rs. 9 + toll at junction B and A SBCT : 2 3 + 2 = Rs. 7 + toll at junction B and C SDCT : 7 + 1 + 2 = Rs. 10 + toll at junction D and C Now, checking the options we find that toll at junction A is 0 or 1. When toll is 0, fuel cost on SAT = 14 + 0 = Rs. 14 When toll is 1, fuel cost on SAT = 14 + 1 = Rs. 15 The fuel cost on all the paths should be equal. Options (1), (2), (3) can be ruled out as in all these options toll at C and D add up to more than Rs. 5. As fuel cost on SDCT is Rs. 10 without toll, so with toll it cannot exceed Rs. 15 (i.e. toll of path SAT). Option (4) is ruled out as in this option SAT comes out to be Rs. 14 and SDCT sums up to Rs.15. So correct answer is option (5). 22. 2 & 3 Note: Both the options b and c are correct. Available routes are: SAT ? Rs. 14 SBAT ? Rs. 9 SDCT ? Rs. 10 SDT ? Rs. 13 Now, fuel cost of SAT - fuel of SDT = 14 – 13 = Rs. 1. Hence toll at junction D should be 1 more than the toll at A. So option (a), (d) and (e) are ruled out. Now, fuel cost of SAT - fuel cost of SBAT = 14 – 9 = Rs. 5. So toll at junction B should be Rs. 5. So answer could be either (2) or option (3). 23. 1 Available paths considering no toll are SAT ? Rs. 14 SBCT ? Rs. 7 SBAT ? Rs. 9 SDCT ? Rs. 10 SDT ? Rs. 13 It is very likely that option (4) is selected. But, if all the five routes have the same cost, then there will be an equal flow on all the five routes i.e., 20% on each route. But, then the percentage of traffic. on S – A ? 20% S – B ? 40% (As there are two routes involving S – B.) S – D ? 40% (As there are two routes involving S – D.) But, it is given that traffic on S – A = traffic on S – B = traffic on S – D. 24. 4 Available routes are SAT ? Rs. 14 SBAT ? Rs. 9 SBCT ? Rs. 7 SDCT ? Rs. 10 SDT ? Rs. 13 Fuel cost on path SAT - fuel cost on path SDT = 14 - 13 = Rs. 1. So the toll at junction D should be 1 more than toll at junction A. So option a and c are ruled out. Fuel cost on path SAT - fuel cost on path SBCT = 14 - 7 = Rs. 7. So sum of toll at junction B and C should be 7 more than the toll at A. Hence, only option (d) matches. 25. 3 We have to find a path on which minimum cost is incurred and such that total traffic through B does not exceed 70%. So option (5) is ruled out because we can send all the traffic through SDCT or SDT and meet all conditions. Option (1) is also ruled out as in that case all traffic will Page 5 1 1 2 3 3 4 4 5 5 5 6 3 7 1 8 1 9 4 10 5 11 4 12 2 13 2 14 2 15 3 16 3 17 2 18 1 19 5 20 4 21 5 22 2,3 23 1 24 4 25 3 26 5 27 1 28 2 29 3 30 4 31 2 32 1 33 4 34 5 35 3 36 3 37 2 38 5 39 1 40 4 41 3 42 1 43 4 44 2 45 4 46 3 47 5 48 4 49 2 50 1 51 1 52 2 53 1 54 2 55 5 56 2 57 2 58 4 59 2 60 4 61 2 62 4 63 1 64 4 65 4 66 3 67 2 68 5 69 1 70 5 71 5 72 5 73 2 74 5 75 3 LRDI 1 to 25 25 EU + RC 26 to 50 25 QA 51 to 75 25 T otal 75 T otal questions T otal attempted T otal correct T otal wrong Net Score Time T aken Question number For questions 1 to 5: From statement one, team would include exactly one among P, R, S ? P (or) R (or) S. From statement two, team would include either M, or Q ? M but not Q (or) Q but not M From statement three, if a team includes K, it will include L or vice versa. ? K, L always accompany each other. From statement four, if one of S, U, W is included, then the other two also have to be included. ? S, U, W are always together. From statement five, L and N cannot be included together ? L, N are never together. From statement six, L and U cannot be included together. ? L, U are never together. 1. 1 From statements one and two; one of P, R, S and one of M, Q are to be selected. We require one more member. But from statement three; (K, L) are always together. Hence 'L' cannot be included in a team of 3 members. 2. 3 Again, from statement one; one of P, R, S has to be selected. To make a team of '5' 'S' will be chosen (which leaves out P and R) ? If 'S' is chosen 'U' and ‘W’ have to be chosen (statement four) ? If 'U' is chosen 'L' cannot be chosen (statement five) ? K cannot be chosen (statement three) And from statement two; one of M (or) Q has to be chosen. 3. 4 From statements one and two Two members are to be selected. Of the remaining seven; To maximize the size of the team. We would chose S, ? U and W are included in the team (statement four) We cannot include K (or) L because we would then have to leave out N and U (from statements five and six) 4. 5 If 'K' is included, 'L' has to be included (statement (3)) If 'L' is chosen, neither N nor U can be chosen (statements (5) and (6)) ? S, W are also not included because S, U, W have to be always together. (Statement (4)) Hence one of P (or) R would be selected (statement (1)) and one of M (or) Q would be selected (statement (2)) (K, L) and two of the above five have to be included. 5. 5 If a team includes N, it cannot include 'L', and therefore, not even 'K'. (from statement five and three) According to statement (1), one of P or R or S has to be included. According to statement (2), one of M or Q has to be selected. So the following cases are possible P Q N, R Q N P M N, R M N If 'S' is selected, then S U W M N and S U W Q N are the only possible cases. Hence, in all 4 + 2 = 6 teams can be constituted. For questions 6 to 10: 6. 3 Let Dipan get x marks in paper II. Dipan's average in PCB group = 98 Maths group = 95 S.S. group = 95.5 Vernacular group = 95 English group = + ?? ?? ?? 96 x 2 Sum of all = 96 × 5 So 95.5 + 96 × 3 + + x 48 2 = 96 × 5 ?= × - - x 96 2 95.5 48 2 () =- =× = x 2 96.5 48 2 48.5 97 So (3) is the correct option. 7. 1 The only boy getting 95 in atleast one of the subjects of the group among all the groups is Dipan. So (1) is the correct option. 8. 1 A group score of 100 in Social Science would have increased the scores as follows: Score Increase Group Score Final Score Increase Final group Score Pritam 22 11 = 11 2.2 5 96.1 Joseph 9 4.5 = 4.5 .9 5 95.9 Tirna 21 10.5 = 10.5 2.1 5 95.8 Agni 9 4.5 = 4.5 .9 5 95.2 So the order is Pritam > Joseph > Trina > Agni. So option (1) is the correct choice. 9. 4 The student having atleast 95 in every group is Dipan, so the answer is Dipan, option (4). 10. 5 Let us increase the score in one of the subjects of the following candidates Least Scores Contributio n in net Score Final Score Ram 94 in group of 2 3 in 5 groups 96.1 + .6 = 96.7 Agni 82 in group of 2 9 in 5 groups 94.3 + 1.8 = 96.1 Pritam 83 in group of 2 8.5 in 5 groups 93.9 + 1.7 = 95.6 Ayesha 93 in group of 2 3.5 in 5 groups 96.2 + .7 = 96.9 Dipan 95 in group of 1 5 in 5 groups 96 + 1 = 97.0 So, Dipan will end with a highest total. So the answer is option (5). For questions 11 to 15: As only Paul Erdös was having an Erdös number of zero, so the minimum Erdös number among A, B, C, D, E, F , G, H should be 1 or greater than one. At the end of the third day, F co- authored a paper with A and C. F had the minimum Erdös number among the 8 people. So if F's Erdös number is y, then A and C's Erdös number should change to (y + 1) after third day. As A and C decreased the average by maximum possible extent, it means C had the second-height Erdös number among all eight, as A had an Erdös number of infinity. Suppose Erdös numbers of A, B, C, D, E, F, G, H are y + 1, b, y + 1, c, d, e, y, g, h respectively at the end of third day. ? (y + 1 + b + y + 1 + c + d + e + y + g + h) = 24 = (3 × 8) ? 3y + 2 + b + d + e + g + h = 24 When E co-authored with F , the average Erdös number reduced again, it means, E's Erdös number was not the same with A & C initially. As at the end of third day, 5 people had same Erdös number, they should be A, C and any 3 out of B, D, G, H. Suppose those 3 people are B, D, G. Then (3y + 2 + y + 1 + y + 1 + y + 1 + e + h) = 24 ? 6y + h + e = 19 …(i) On the fifth day, E co-authored a paper with F and hence, Erdös number of E changed to (y + 1). Also the average decreased by 0.5 which means the total decreased by 4. Hence, e - (y + 1) = 4 ? e – y = 5 Putting the value of e in equation (i), we get 6y + h + (5 + y) = 19 ? 7y + h = 14 Only possible value of y = 1 as h cannot be zero. So after 3rd round Erdös number of A, C, E, F were 2, 2, 6, 1 respectively. 11. 4 Only A, C, E changed their Erdös number, rest 5 did not change their Erdös number. 12. 2 At the end of conference 6 people including E were having an Erdös number of 2 and F was having 1 as Erdös number. So 8th person was having an Erdös number of [20 – (2 × 6 + 1)] = 7 13. 2 At the end of 3rd round, 5 people were having same Erdös number. A and C changed their Erdös number after coauthoring with F. So, the other 3 will have same Erdös number in the beginning. 14. 2 2 15. 3 After co-authoring with F, E was having Erdös number of 2, which was 4 less than initial Erdös number of E. So answer is 2 + 4 = 6. For questions 16 to 20: The MCS share price at the beginning of first day is Rs.100 and at the close of day 5 is Rs.110. The following cases of the closing prices can be derived. At the end of Day 1Day 2Day 3Day 4Day 5 1 90 80 90 100 110 2 90 100 90 100 110 3 90 100 110 120 110 4 90 100 110 100 110 5 110 100 90 100 110 6 110 100 110 100 110 7 110 120 110 100 110 8 110 120 110 120 110 9 110 120 130 120 110 10 110 120 110 100 110 16. 3 As Chetan sold 10 shares on three consecutive days, therefore, of the five days, there must be an increase for three of the five days and a decrease for the remaining two days. It is given that Michael sold 10 shares only once. Hence, the price is more than 110 for only one day and on all the remaining days, it cannot exceed 110. The only satisfying case is (3). Hence, the price at the end of Day 3 is Rs.110. 17. 2 The satisfying cases are (1), (2), (4), (5), (6). Hence, the price at the end of Day 4 is Rs.100. 18. 1 Let Chetan and Michael start with x number of shares initially. From case (1), we get that the number of shares with Michael = x + 10 and number of shares with Chetan = x + 10 + 10 – 10 – 10 – 10 = x – 10. So Michael has 20 more shares than Chetan. This is the only satisfying case. Hence, the share price at the end of Day 3 is Rs.90. 19. 5 Consider cases (3) and (7). Only these two satisfies the condition that Michael had Rs.100 less than Chetan at the end of day 5. For case (3), Number of shares with Chetan = x + 10 – 10 – 10 – 10 + 10 = x – 10 And with Michael = x – 10 For case (7), Number of shares with Chetan = x – 10 – 10 + 10 + 10 – 10 = x – 10 And with Michael = x – 10 In either case, number of shares with Michael and Chetan are the same. 20. 4 To maximise the amount gathered by both of them, we need to look into those cases wherein we have maximum number of 110 excess figures. It is only then that Michael and Chetan both will make money. So we check for case (9). For case (9), Extra cash with Chetan by the end of day 5 = 1100 + 1200 + 1300 – 1200 – 1100 = Rs.1300 And that with Michael = 1200 + 1300 + 1200 = Rs.3700 Total extra cash with both of them = 1300 + 3700 = Rs.5000 For questions 21 to 25: In this set, the fuel cost for each of the path is given. In addition, there are four toll collection junctions. 21. 5 No traffic flows on the street from D to T. Now, we have fuel cost on different paths as SAT : 9 + 5 = Rs. 14 + toll at junction A SBAT : 2 + 2 + 5 = Rs. 9 + toll at junction B and A SBCT : 2 3 + 2 = Rs. 7 + toll at junction B and C SDCT : 7 + 1 + 2 = Rs. 10 + toll at junction D and C Now, checking the options we find that toll at junction A is 0 or 1. When toll is 0, fuel cost on SAT = 14 + 0 = Rs. 14 When toll is 1, fuel cost on SAT = 14 + 1 = Rs. 15 The fuel cost on all the paths should be equal. Options (1), (2), (3) can be ruled out as in all these options toll at C and D add up to more than Rs. 5. As fuel cost on SDCT is Rs. 10 without toll, so with toll it cannot exceed Rs. 15 (i.e. toll of path SAT). Option (4) is ruled out as in this option SAT comes out to be Rs. 14 and SDCT sums up to Rs.15. So correct answer is option (5). 22. 2 & 3 Note: Both the options b and c are correct. Available routes are: SAT ? Rs. 14 SBAT ? Rs. 9 SDCT ? Rs. 10 SDT ? Rs. 13 Now, fuel cost of SAT - fuel of SDT = 14 – 13 = Rs. 1. Hence toll at junction D should be 1 more than the toll at A. So option (a), (d) and (e) are ruled out. Now, fuel cost of SAT - fuel cost of SBAT = 14 – 9 = Rs. 5. So toll at junction B should be Rs. 5. So answer could be either (2) or option (3). 23. 1 Available paths considering no toll are SAT ? Rs. 14 SBCT ? Rs. 7 SBAT ? Rs. 9 SDCT ? Rs. 10 SDT ? Rs. 13 It is very likely that option (4) is selected. But, if all the five routes have the same cost, then there will be an equal flow on all the five routes i.e., 20% on each route. But, then the percentage of traffic. on S – A ? 20% S – B ? 40% (As there are two routes involving S – B.) S – D ? 40% (As there are two routes involving S – D.) But, it is given that traffic on S – A = traffic on S – B = traffic on S – D. 24. 4 Available routes are SAT ? Rs. 14 SBAT ? Rs. 9 SBCT ? Rs. 7 SDCT ? Rs. 10 SDT ? Rs. 13 Fuel cost on path SAT - fuel cost on path SDT = 14 - 13 = Rs. 1. So the toll at junction D should be 1 more than toll at junction A. So option a and c are ruled out. Fuel cost on path SAT - fuel cost on path SBCT = 14 - 7 = Rs. 7. So sum of toll at junction B and C should be 7 more than the toll at A. Hence, only option (d) matches. 25. 3 We have to find a path on which minimum cost is incurred and such that total traffic through B does not exceed 70%. So option (5) is ruled out because we can send all the traffic through SDCT or SDT and meet all conditions. Option (1) is also ruled out as in that case all traffic will be passed through SBCT [not possible as traffic at B can't be more than 70%] Option (2) is also ruled out as it is possible only when toll at junction C is 2. In that case also all traffic will pass through B. Option (3) can be the answer, when toll at junction B is 4 and toll at junction C is 0. Then SDCT will have toll equal to Rs. 10. As Rs. 10 is less than Rs. 13, so option (4) is also ruled out. Hence, option (3) is the correct choice. 26. 5 The paragraph stresses on the relationships between the factories, dealers and the consumers. Every entity has certain short-term expectations from each other. This makes these relationships strenuous. This strain leads to feelings of mistrust and lack of commitment. So the longer this continues, the more the chances of everyone succumbing to this vicious trap and they would soon realize that they have sacrificed long- term stability and gain for short-term benefits. Hence Option (5). Option (4) is too specific to industry (at the cost of the other players – dealers and customers), option (2) suffers from the same short-comings together with throwing the technical (unexplained) jargon ‘supply chain’ to us. Option (1) takes into account only 2 players and repeats what is stated in the passage about “dealers adjusting prices and making deals” in the term ‘Deal making’; option (3) seems close but can be eliminated as the word ‘adversary’ is too strong. The passage implies that everyone tries to maximize his benefits, not that they ‘oppose’ one another. 27. 1 The passage heads towards describing the functions that bad / good maps (and therefore theories) serve. Just as a ‘Bad theory’ does not help us understand a problem, a ‘good theory’ is invaluable to us, though it may be simplified. ‘Simplified’ here implies that less valuable information is left out. According to this logic, option (2), (3), (4), get eliminated. Option (5) is close but more negative in tone than required. The word ‘limitation’ here indicates a short coming whereas the passage implies that it is a simplification as it would not be of practical use otherwise. 28. 2 Going with the direction of the passage, the last line is stating ‘now all players “profess” to seek only peace’. Profess means to mask or to pretend. Thus option (2) which talks about the veil being lifted is the most logical statement that completes the passage. More so this also follows from the source of the text. 29. 3 The answer is very direct. With every statement of his, the author seeks to show how foolish those people are who call his advice ‘rules’. After his first statement he has posed the rhetorical question “Call that a rule?” The same should follow after his second “scarcely a rule!” 30. 4 In the first part of the passage, the author seeks to explain why one who is young would exploit an entrepreneurial opportunity. Thus, in the second part of the passage once the “however” is established, evidence will seek to show how older people will be reluctant to exploit entrepreneurial opportunity. Option (2) seems correct but it only gives a general statement that with age, people become reluctant to new ideas. Between option (2) and (4), option (4) goes in continuation with the text as it states that at a mature age, people are unwilling to utilize entrepreneurial opportunities. So option (4) is correct. 31. 2 According to the passage, "A critical attitude needs for its raw material, as it were, theories or beliefs which are held more or less dogmatically". Therefore, our critical attitude is the tool by which we shape our dogmatic beliefs. Thus, the relationship of dogmatic beliefs and critical attitude is equivalent to that of a chisel and that of a marble stone. 32. 1 Option (3), (4) and (5) are ruled out because they are not supported by the passage. (negative, neutral, inferior) - Option (1) and (2) are close but (1) is better because the question is about the role of dogmatic behaviour with respect to the development of science. In the third paragraph, 8 th line, it is mentioned that dogmatic attitude is pseudo/pre-scientific attitude. Science needs dogmatic beliefs for their critical revision. Beginning of fourth paragraph states that science begins with myths and criticism of myths. Thus, dogmatic behavior is required to develop science because the former serves as the base on which science is made. 33. 4 Refer to the last sentence of the second paragraph. It is clear from the context (especially from the words - 'experience', 'maturity') that time has a direct effect on the evolution of thinking. Option (4) is the only option which takes into account the element of time (the word - 'stages'). 34. 5 Option (5) is correct because this statement suggests that critical attitude is a process of questioning which leads to tentative hypothesis. A critical attitude by itself is not opposed to conviction, but it tries to modify the conviction according to reason. 35. 3 Refer to the third last paragraph of the passage; dogmatic attitude is pseudo-scientific because its aim is only to verify its laws and schemata even if it has to neglect the refutations. Whereas critical attitude is flexible enough to change, refute or falsify its tenets and therefore has a questioning attitude. 36. 3 We refer to the tenth line of the third paragraph. Here Mr. Goran Lindblad admits that communism did commit brutalities but it also had positive consequences like rapid industrialization. Hence option (3) is the best answer.Read More

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