Page 1 Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 1 1. (B) Given + > â€“ × = I < L Given expression A | B × C A < B = C because option (B) follows the given expression here because C - B + A means C B > A, one of which meaning is A<B= C 2. (C)2 M C 3 K (2 M = N) (2 N 3 K) (2 N 3 K) 2 × 2 M 3 K (since N = 2 M) 4 M 3 K 4 M is not more than 3 K, then 2 M is less than 3 K. Hence 2 M C 3 K is correct. 3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8 Similarly, (64 + 53) ÷ 9 = 13 4. (B) 5. (C) 6. (D)54 â€“ 30 = 24 and 112 â€“ 42 = 70 Similarly, x â€“ 28 = 38 x = 38 + 28 = 66 7. (D)216 â€“ 7 = 209 209 â€“ 7 = 202 and 522 â€“ 7 = 515 515 â€“ 7 = 508 Similarly, 633 â€“ 7 = 626 626 â€“ 7 = 619 8. (B) 9. (B) 10. (A) 11. (A) 12. (A)a a b b/abba/a abb/a b ba 13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y 14. (C) 15. (D) (A) (B) (C) (D) 16. (B) 17. (A) 18. (B)(A) 7 × 11 â€“ 3 = 74 (B) 9 × 11 â€“ 3 = 96 (C) 4 × 11 â€“ 3 = 41 (D) 6 × 11 â€“ 6 = 63 19. (B) Given set = Similarly, in option (B) 20. (A) 21. (B) 22. (D) 23. (B) 24. (D) SSC MOCK TEST 29 (SOLUTION) Page 2 Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 1 1. (B) Given + > â€“ × = I < L Given expression A | B × C A < B = C because option (B) follows the given expression here because C - B + A means C B > A, one of which meaning is A<B= C 2. (C)2 M C 3 K (2 M = N) (2 N 3 K) (2 N 3 K) 2 × 2 M 3 K (since N = 2 M) 4 M 3 K 4 M is not more than 3 K, then 2 M is less than 3 K. Hence 2 M C 3 K is correct. 3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8 Similarly, (64 + 53) ÷ 9 = 13 4. (B) 5. (C) 6. (D)54 â€“ 30 = 24 and 112 â€“ 42 = 70 Similarly, x â€“ 28 = 38 x = 38 + 28 = 66 7. (D)216 â€“ 7 = 209 209 â€“ 7 = 202 and 522 â€“ 7 = 515 515 â€“ 7 = 508 Similarly, 633 â€“ 7 = 626 626 â€“ 7 = 619 8. (B) 9. (B) 10. (A) 11. (A) 12. (A)a a b b/abba/a abb/a b ba 13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y 14. (C) 15. (D) (A) (B) (C) (D) 16. (B) 17. (A) 18. (B)(A) 7 × 11 â€“ 3 = 74 (B) 9 × 11 â€“ 3 = 96 (C) 4 × 11 â€“ 3 = 41 (D) 6 × 11 â€“ 6 = 63 19. (B) Given set = Similarly, in option (B) 20. (A) 21. (B) 22. (D) 23. (B) 24. (D) SSC MOCK TEST 29 (SOLUTION) Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 2 25. (B) Total age of father and son = 22 × 2 = 44 years Father : Son 10 1 Son's age = 44 1 10 1 = 4 years 26. (B) Veni > Smith > Raju > Salim 27. (B) G E N E R A T E 28. (C) Q U A I N T 29. (B) According to question, the odd numerical value of HOTEL will be â€“ HOTEL = 15 + 29 + 39 + 9 + 23 = 115 30. (A) Similarly, 31. (A) 29 × 48 2 × 9 × 4 × 8 = 576 35 × 16 3 × 5 × 1 × 6 = 90, 22 × 46 2 × 2 × 4 × 6 = 96 and Similarly, 42 × 17 4 × 2 × 1 × 7 = 56 32. (C) 12 P 6 M 15 T 16 B 4 12 × 6 + 15 â€“ 16 ÷ 4 12 × 6 + 15 â€“ 4 72 + 15 â€“ 4 87 â€“ 4 = 83 33. (B) Meaningful word 34. (C) Similarly, 35. (A) (Here = DA = CB = 2 km) Required distance AE = DE â€“ DA = 3 â€“ 2 = 1 km 36. (C) According to given direction the faces of Rani and Sarita will be East and South direction with respect to x. 37. (D) 38. (B) Conclusion I â€” II â€” × 39. (B) 40. (B) 41. (A) 42. (C) 43. (B) 44. (A) 45. (A) 46. (B) 47. (C) 48. (A) 49. (D) 50. (D) The numerical groups of PLAY will be â€“ P â€“ 15, 43 L â€“ 36, 65 A â€“ 42, 46, 62 Y â€“ 45 51. (B)Straight line 4x + 3y = 12 passes through 1st, 2nd & 4th quadrant. 52. (B) Page 3 Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 1 1. (B) Given + > â€“ × = I < L Given expression A | B × C A < B = C because option (B) follows the given expression here because C - B + A means C B > A, one of which meaning is A<B= C 2. (C)2 M C 3 K (2 M = N) (2 N 3 K) (2 N 3 K) 2 × 2 M 3 K (since N = 2 M) 4 M 3 K 4 M is not more than 3 K, then 2 M is less than 3 K. Hence 2 M C 3 K is correct. 3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8 Similarly, (64 + 53) ÷ 9 = 13 4. (B) 5. (C) 6. (D)54 â€“ 30 = 24 and 112 â€“ 42 = 70 Similarly, x â€“ 28 = 38 x = 38 + 28 = 66 7. (D)216 â€“ 7 = 209 209 â€“ 7 = 202 and 522 â€“ 7 = 515 515 â€“ 7 = 508 Similarly, 633 â€“ 7 = 626 626 â€“ 7 = 619 8. (B) 9. (B) 10. (A) 11. (A) 12. (A)a a b b/abba/a abb/a b ba 13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y 14. (C) 15. (D) (A) (B) (C) (D) 16. (B) 17. (A) 18. (B)(A) 7 × 11 â€“ 3 = 74 (B) 9 × 11 â€“ 3 = 96 (C) 4 × 11 â€“ 3 = 41 (D) 6 × 11 â€“ 6 = 63 19. (B) Given set = Similarly, in option (B) 20. (A) 21. (B) 22. (D) 23. (B) 24. (D) SSC MOCK TEST 29 (SOLUTION) Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 2 25. (B) Total age of father and son = 22 × 2 = 44 years Father : Son 10 1 Son's age = 44 1 10 1 = 4 years 26. (B) Veni > Smith > Raju > Salim 27. (B) G E N E R A T E 28. (C) Q U A I N T 29. (B) According to question, the odd numerical value of HOTEL will be â€“ HOTEL = 15 + 29 + 39 + 9 + 23 = 115 30. (A) Similarly, 31. (A) 29 × 48 2 × 9 × 4 × 8 = 576 35 × 16 3 × 5 × 1 × 6 = 90, 22 × 46 2 × 2 × 4 × 6 = 96 and Similarly, 42 × 17 4 × 2 × 1 × 7 = 56 32. (C) 12 P 6 M 15 T 16 B 4 12 × 6 + 15 â€“ 16 ÷ 4 12 × 6 + 15 â€“ 4 72 + 15 â€“ 4 87 â€“ 4 = 83 33. (B) Meaningful word 34. (C) Similarly, 35. (A) (Here = DA = CB = 2 km) Required distance AE = DE â€“ DA = 3 â€“ 2 = 1 km 36. (C) According to given direction the faces of Rani and Sarita will be East and South direction with respect to x. 37. (D) 38. (B) Conclusion I â€” II â€” × 39. (B) 40. (B) 41. (A) 42. (C) 43. (B) 44. (A) 45. (A) 46. (B) 47. (C) 48. (A) 49. (D) 50. (D) The numerical groups of PLAY will be â€“ P â€“ 15, 43 L â€“ 36, 65 A â€“ 42, 46, 62 Y â€“ 45 51. (B)Straight line 4x + 3y = 12 passes through 1st, 2nd & 4th quadrant. 52. (B) Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 3 In ADC , AC is the hypotenuse, which is the longest side of triangle. AC > AD Similarly, AC > CF AB > AD AB > BE BC > CF and BC > BE On adding above inequalities, we have 2(AB + BC + CA) > 2(AD + BE + CF) (AB + BC + CA) > (AD + BE + CF) 53. (B) ) ( ) ( C B B A = 65º + 140º B C B A ) ( = 205º 180º + B = 205º B = 205º â€“ 180º = 25º 54. (A) OT = 4 cm PO = 5 cm PO 2 = PT 2 + TO 2 5 2 = PT 2 + 4 2 PT = 3 cm 55. (B) x x 2 3 = x 3 3x â€“ 2 = x 3 3x â€“ x 3 = 2 x â€“ x 1 = 3 2 2 2 1 x x = 2 1 2 x x = 2 3 2 2 = 1 2 9 4 = 9 22 = 9 4 2 56. (A) 16a 2 â€“ 12a = (4a) 2 â€“ 2.4a × 2 2 3 2 3 = 2 2 3 4 a 4 9 must be added in 16a 2 â€“ 12a to make it a perfect square. 57. (C) Let AB & CD are two parallel chords of C(0, 10 cm). AB = 12 cm AM = 2 1 × 12 = 6 cm [perp. from centre to any chord, bisects the chord] Similarly CN = 2 1 × 16 = 8 cm AM 2 + OM 2 = OA 2 6 2 + OM 2 =10 2 OM = 36 100 = 64 = 8 cm Again, CN 2 + ON 2 = OC 2 8 2 + ON 2 = 10 2 ON = 36 100 = 36 = 6 cm The distance between the chords = MN = OM + ON = 8 + 6 = 14 cm 58. (D) BD = AD â€“ AD = 10 â€“ 4 = 6 cm BD : DA = BE : EC 6 : 4= BE : EC BE : CE = 3 : 2 59. (B) In AOB , OA = OB º 45 OBA OAB In AOC , OA = OC º 35 OCA OAC OAC OAB BAC º 80 º 35 º 45 Page 4 Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 1 1. (B) Given + > â€“ × = I < L Given expression A | B × C A < B = C because option (B) follows the given expression here because C - B + A means C B > A, one of which meaning is A<B= C 2. (C)2 M C 3 K (2 M = N) (2 N 3 K) (2 N 3 K) 2 × 2 M 3 K (since N = 2 M) 4 M 3 K 4 M is not more than 3 K, then 2 M is less than 3 K. Hence 2 M C 3 K is correct. 3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8 Similarly, (64 + 53) ÷ 9 = 13 4. (B) 5. (C) 6. (D)54 â€“ 30 = 24 and 112 â€“ 42 = 70 Similarly, x â€“ 28 = 38 x = 38 + 28 = 66 7. (D)216 â€“ 7 = 209 209 â€“ 7 = 202 and 522 â€“ 7 = 515 515 â€“ 7 = 508 Similarly, 633 â€“ 7 = 626 626 â€“ 7 = 619 8. (B) 9. (B) 10. (A) 11. (A) 12. (A)a a b b/abba/a abb/a b ba 13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y 14. (C) 15. (D) (A) (B) (C) (D) 16. (B) 17. (A) 18. (B)(A) 7 × 11 â€“ 3 = 74 (B) 9 × 11 â€“ 3 = 96 (C) 4 × 11 â€“ 3 = 41 (D) 6 × 11 â€“ 6 = 63 19. (B) Given set = Similarly, in option (B) 20. (A) 21. (B) 22. (D) 23. (B) 24. (D) SSC MOCK TEST 29 (SOLUTION) Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 2 25. (B) Total age of father and son = 22 × 2 = 44 years Father : Son 10 1 Son's age = 44 1 10 1 = 4 years 26. (B) Veni > Smith > Raju > Salim 27. (B) G E N E R A T E 28. (C) Q U A I N T 29. (B) According to question, the odd numerical value of HOTEL will be â€“ HOTEL = 15 + 29 + 39 + 9 + 23 = 115 30. (A) Similarly, 31. (A) 29 × 48 2 × 9 × 4 × 8 = 576 35 × 16 3 × 5 × 1 × 6 = 90, 22 × 46 2 × 2 × 4 × 6 = 96 and Similarly, 42 × 17 4 × 2 × 1 × 7 = 56 32. (C) 12 P 6 M 15 T 16 B 4 12 × 6 + 15 â€“ 16 ÷ 4 12 × 6 + 15 â€“ 4 72 + 15 â€“ 4 87 â€“ 4 = 83 33. (B) Meaningful word 34. (C) Similarly, 35. (A) (Here = DA = CB = 2 km) Required distance AE = DE â€“ DA = 3 â€“ 2 = 1 km 36. (C) According to given direction the faces of Rani and Sarita will be East and South direction with respect to x. 37. (D) 38. (B) Conclusion I â€” II â€” × 39. (B) 40. (B) 41. (A) 42. (C) 43. (B) 44. (A) 45. (A) 46. (B) 47. (C) 48. (A) 49. (D) 50. (D) The numerical groups of PLAY will be â€“ P â€“ 15, 43 L â€“ 36, 65 A â€“ 42, 46, 62 Y â€“ 45 51. (B)Straight line 4x + 3y = 12 passes through 1st, 2nd & 4th quadrant. 52. (B) Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 3 In ADC , AC is the hypotenuse, which is the longest side of triangle. AC > AD Similarly, AC > CF AB > AD AB > BE BC > CF and BC > BE On adding above inequalities, we have 2(AB + BC + CA) > 2(AD + BE + CF) (AB + BC + CA) > (AD + BE + CF) 53. (B) ) ( ) ( C B B A = 65º + 140º B C B A ) ( = 205º 180º + B = 205º B = 205º â€“ 180º = 25º 54. (A) OT = 4 cm PO = 5 cm PO 2 = PT 2 + TO 2 5 2 = PT 2 + 4 2 PT = 3 cm 55. (B) x x 2 3 = x 3 3x â€“ 2 = x 3 3x â€“ x 3 = 2 x â€“ x 1 = 3 2 2 2 1 x x = 2 1 2 x x = 2 3 2 2 = 1 2 9 4 = 9 22 = 9 4 2 56. (A) 16a 2 â€“ 12a = (4a) 2 â€“ 2.4a × 2 2 3 2 3 = 2 2 3 4 a 4 9 must be added in 16a 2 â€“ 12a to make it a perfect square. 57. (C) Let AB & CD are two parallel chords of C(0, 10 cm). AB = 12 cm AM = 2 1 × 12 = 6 cm [perp. from centre to any chord, bisects the chord] Similarly CN = 2 1 × 16 = 8 cm AM 2 + OM 2 = OA 2 6 2 + OM 2 =10 2 OM = 36 100 = 64 = 8 cm Again, CN 2 + ON 2 = OC 2 8 2 + ON 2 = 10 2 ON = 36 100 = 36 = 6 cm The distance between the chords = MN = OM + ON = 8 + 6 = 14 cm 58. (D) BD = AD â€“ AD = 10 â€“ 4 = 6 cm BD : DA = BE : EC 6 : 4= BE : EC BE : CE = 3 : 2 59. (B) In AOB , OA = OB º 45 OBA OAB In AOC , OA = OC º 35 OCA OAC OAC OAB BAC º 80 º 35 º 45 Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 4 60. (A) Let h m be the height of the ballon. In CBD , tan60º = BD CB 3 = BD h BD = 3 h cm In CBA tan30º = BA CB 3 1 = DA BD h = 1000 3 h h 1000 3 h = 3 h 3 3 h h = 1000 3 1 3 h = 1000 h = 2 3 1000 = 500 3 = 1000 3 500 km = 3 2 1 km 61. (C) CBD CAD (Angles in the same segment of a circle) CBD Now, COB AOB (by RHS) CBO ABO (by CPCT) CBD ABO ABC = 2 CBO ABO 62. (C) Let ABC is equilateral & AD is its height. Let 'a' unit is the side of the ABC . AB = a & BD = 2 a height AD = 2 2 BD AB 15 = 4 2 2 a a 15 = a 2 3 a = 3 2 15 cm Hence BC = 3 30 = 3 3 3 10 = 3 10 cm Area of ABC = 2 4 3 a = 3 10 3 10 4 3 = 3 75 cm 2 . 63. (A)Work done by Ram & Shyam in 1 day = 12 1 Work done by Shyam & Hari in 1 day = 15 1 Work done by Hari & Ram in 1 day = 20 1 Work done by 2[Ram + Shyam + Hari] = 20 1 15 1 12 1 = 60 3 4 5 = 60 12 = 5 1 Work done by (Ram + Shyam + Hari) in 1 day = 10 1 Work done by Ram alone in 1 day = 10 1 5 1 = 30 2 3 = 30 1 Ram can do the whole work in 30 days. Page 5 Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 1 1. (B) Given + > â€“ × = I < L Given expression A | B × C A < B = C because option (B) follows the given expression here because C - B + A means C B > A, one of which meaning is A<B= C 2. (C)2 M C 3 K (2 M = N) (2 N 3 K) (2 N 3 K) 2 × 2 M 3 K (since N = 2 M) 4 M 3 K 4 M is not more than 3 K, then 2 M is less than 3 K. Hence 2 M C 3 K is correct. 3. (B) (15 +12) ÷ 9 = 3 and (44 + 28) ÷ 9 = 8 Similarly, (64 + 53) ÷ 9 = 13 4. (B) 5. (C) 6. (D)54 â€“ 30 = 24 and 112 â€“ 42 = 70 Similarly, x â€“ 28 = 38 x = 38 + 28 = 66 7. (D)216 â€“ 7 = 209 209 â€“ 7 = 202 and 522 â€“ 7 = 515 515 â€“ 7 = 508 Similarly, 633 â€“ 7 = 626 626 â€“ 7 = 619 8. (B) 9. (B) 10. (A) 11. (A) 12. (A)a a b b/abba/a abb/a b ba 13. (A) AZ B Y , A Z B Y, A Z B Y, A Z B Y 14. (C) 15. (D) (A) (B) (C) (D) 16. (B) 17. (A) 18. (B)(A) 7 × 11 â€“ 3 = 74 (B) 9 × 11 â€“ 3 = 96 (C) 4 × 11 â€“ 3 = 41 (D) 6 × 11 â€“ 6 = 63 19. (B) Given set = Similarly, in option (B) 20. (A) 21. (B) 22. (D) 23. (B) 24. (D) SSC MOCK TEST 29 (SOLUTION) Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 2 25. (B) Total age of father and son = 22 × 2 = 44 years Father : Son 10 1 Son's age = 44 1 10 1 = 4 years 26. (B) Veni > Smith > Raju > Salim 27. (B) G E N E R A T E 28. (C) Q U A I N T 29. (B) According to question, the odd numerical value of HOTEL will be â€“ HOTEL = 15 + 29 + 39 + 9 + 23 = 115 30. (A) Similarly, 31. (A) 29 × 48 2 × 9 × 4 × 8 = 576 35 × 16 3 × 5 × 1 × 6 = 90, 22 × 46 2 × 2 × 4 × 6 = 96 and Similarly, 42 × 17 4 × 2 × 1 × 7 = 56 32. (C) 12 P 6 M 15 T 16 B 4 12 × 6 + 15 â€“ 16 ÷ 4 12 × 6 + 15 â€“ 4 72 + 15 â€“ 4 87 â€“ 4 = 83 33. (B) Meaningful word 34. (C) Similarly, 35. (A) (Here = DA = CB = 2 km) Required distance AE = DE â€“ DA = 3 â€“ 2 = 1 km 36. (C) According to given direction the faces of Rani and Sarita will be East and South direction with respect to x. 37. (D) 38. (B) Conclusion I â€” II â€” × 39. (B) 40. (B) 41. (A) 42. (C) 43. (B) 44. (A) 45. (A) 46. (B) 47. (C) 48. (A) 49. (D) 50. (D) The numerical groups of PLAY will be â€“ P â€“ 15, 43 L â€“ 36, 65 A â€“ 42, 46, 62 Y â€“ 45 51. (B)Straight line 4x + 3y = 12 passes through 1st, 2nd & 4th quadrant. 52. (B) Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 3 In ADC , AC is the hypotenuse, which is the longest side of triangle. AC > AD Similarly, AC > CF AB > AD AB > BE BC > CF and BC > BE On adding above inequalities, we have 2(AB + BC + CA) > 2(AD + BE + CF) (AB + BC + CA) > (AD + BE + CF) 53. (B) ) ( ) ( C B B A = 65º + 140º B C B A ) ( = 205º 180º + B = 205º B = 205º â€“ 180º = 25º 54. (A) OT = 4 cm PO = 5 cm PO 2 = PT 2 + TO 2 5 2 = PT 2 + 4 2 PT = 3 cm 55. (B) x x 2 3 = x 3 3x â€“ 2 = x 3 3x â€“ x 3 = 2 x â€“ x 1 = 3 2 2 2 1 x x = 2 1 2 x x = 2 3 2 2 = 1 2 9 4 = 9 22 = 9 4 2 56. (A) 16a 2 â€“ 12a = (4a) 2 â€“ 2.4a × 2 2 3 2 3 = 2 2 3 4 a 4 9 must be added in 16a 2 â€“ 12a to make it a perfect square. 57. (C) Let AB & CD are two parallel chords of C(0, 10 cm). AB = 12 cm AM = 2 1 × 12 = 6 cm [perp. from centre to any chord, bisects the chord] Similarly CN = 2 1 × 16 = 8 cm AM 2 + OM 2 = OA 2 6 2 + OM 2 =10 2 OM = 36 100 = 64 = 8 cm Again, CN 2 + ON 2 = OC 2 8 2 + ON 2 = 10 2 ON = 36 100 = 36 = 6 cm The distance between the chords = MN = OM + ON = 8 + 6 = 14 cm 58. (D) BD = AD â€“ AD = 10 â€“ 4 = 6 cm BD : DA = BE : EC 6 : 4= BE : EC BE : CE = 3 : 2 59. (B) In AOB , OA = OB º 45 OBA OAB In AOC , OA = OC º 35 OCA OAC OAC OAB BAC º 80 º 35 º 45 Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 4 60. (A) Let h m be the height of the ballon. In CBD , tan60º = BD CB 3 = BD h BD = 3 h cm In CBA tan30º = BA CB 3 1 = DA BD h = 1000 3 h h 1000 3 h = 3 h 3 3 h h = 1000 3 1 3 h = 1000 h = 2 3 1000 = 500 3 = 1000 3 500 km = 3 2 1 km 61. (C) CBD CAD (Angles in the same segment of a circle) CBD Now, COB AOB (by RHS) CBO ABO (by CPCT) CBD ABO ABC = 2 CBO ABO 62. (C) Let ABC is equilateral & AD is its height. Let 'a' unit is the side of the ABC . AB = a & BD = 2 a height AD = 2 2 BD AB 15 = 4 2 2 a a 15 = a 2 3 a = 3 2 15 cm Hence BC = 3 30 = 3 3 3 10 = 3 10 cm Area of ABC = 2 4 3 a = 3 10 3 10 4 3 = 3 75 cm 2 . 63. (A)Work done by Ram & Shyam in 1 day = 12 1 Work done by Shyam & Hari in 1 day = 15 1 Work done by Hari & Ram in 1 day = 20 1 Work done by 2[Ram + Shyam + Hari] = 20 1 15 1 12 1 = 60 3 4 5 = 60 12 = 5 1 Work done by (Ram + Shyam + Hari) in 1 day = 10 1 Work done by Ram alone in 1 day = 10 1 5 1 = 30 2 3 = 30 1 Ram can do the whole work in 30 days. Centres at: MUKHERJEE NAGAR MUNIRKA UTTAM NAGAR DILSHAD GARDEN ROHINI BADARPUR BORDER ============================================================= 5 64. (A) 3 men = 5 women 1 man = 3 5 women 6 men = 3 5 × 6 = 10 women 6 men + 5 women = (10 + 5) women 5 women complete a work in 15 days. 1 woman completes a work in 15 × 5 15 women complete a work in 15 5 15 = 5 days 65. (C) Work done by (A + B) in 1 day = 12 1 Work done by (B + C) in 1 day = 15 1 Work done by (C + A) in 1 day = 20 1 Work done by 2(A + B + C) in 1 day = 20 1 15 1 12 1 = 60 3 4 5 = 60 12 = 5 1 Work done by (A + B + C) in 1 day = 2 5 1 = 10 1 Work done by A alone in 1 day = 15 10 1 = 30 2 3 = 30 1 Hence A can complete the work alone in 30 days. 66. (A)Perimeter â€“ Diameter = X r r 2 2 = X ) 1 ( 2 r = X 2r = 1 X 67. (D) r 2 = a r = 2 a Volume = V h r 2 = V h a ² 4 ² = V h = 2 4 a V 68. (D) 4 th root of 24010000 = 4 1 ) 24010000 ( = 4 1 ) 10 10 10 10 7 7 7 7 ( = 7 × 10 = 70 69. (C) Greatest 4-digit number = 9999 Greatest 4-digit perfect square number = 9999 â€“ 198 = 9801 70. (A)MP = Rs. x (say) SP = 90% of x = Rs. 10 9x % profit = 12% CP = % 12 100 100 SP = 112 100 10 9 x = 112 90x Now, MP CP = x x 112 90 = 56 45 71. (D) MP = Rs. 160 SP after two successive discounts = Rs. 122.40 Ist discount = 10% ATQ, 100 100 100 10 100 160 y = Rs. 122.40 100 100 100 90 160 y = Rs. 122.40 (100 â€“ y) = 160 90 100 100 40 . 122 = 85 y = 100 â€“ 85 = 15 2nd discount = 15% 72. (B)Let 'g' stands for the no. of girls & 'b' stands for the no. of boys 10% of g = 20 1 of b 100 10 × g = 20 1 × b g b = 100 20 10 = 1 2 b : g = 2 : 1Read More

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