I have posted these solution keys here because all the books, available in the market, have lots of wrong solutions. Most of the students, like me, find themselves in dilemma and waste a lot of time in finding the correct keys. These solution keys are those which I had corrected while preparing for GATE Computer Science 2012 Exam. You will find most of these solution keys to be correct, except few for which I could not find the correct solution. Still I suggest you that please keep your mind open while using these keys.
GATE 2005 CS Solution Keys
1. C
2. C 3. C 4. B 5. A 6. C 7. D 8. A 9. B 10. B 11. A 12. C 13. C 14. B 15. A 16. A 17. A 18. A |
19. B
20. A 21. B 22. C 23. B 24. D 25. B 26. B 27. D 28. D 29. C 30. C 31. D 32. D 33. A 34. D 35. B 36. C |
37. C
38. D 39. A 40. B 41. B 42. C 43. B 44. C 45. C 46. A 47. A 48. B 49. B 50. B 51. A 52. D 53. B 54. D |
55. A
56. B 57. B 58. C 59. D 60. B 61. C 62. A 63. B 64. D 65. B 66. C 67. A 68. A 69. B 70. B 71. C 72. B |
73. D
74. D 75. B 76. C 77. D 78. D 79. B 80. B 81A. B 81B. B 82A. A 82B. B 83A. C 83B. A 84A. D 84B. A 85A. D 85B. D |
Comments 3
I guess answer to Q19 should be C.
The configuration given is similar to ‘daisy chain’.
Q 74 – C,
2* prop delay = Round trip propagation delay
min. frame size = 2* prop delay* transmission speed
= 46.4*10^-6 * 10 * 10^6 = 464 bits
Please explain solution for question no. 53 in GATE 2005 CS paper.
Thanks in advance.