UGC NET Computer Science & Applications June 2025 Question PaperPaper 1 & Paper 2 with Answer Key

50 questions with the answer key and explanations.

1. Question 1

   What is the average of the number of students scoring 50 or more but less than 60 marks in Punjabi and in Hindi?

   Marks (out of 100)	Cumulative percentage of Hindi and Punjabi students	Cumulative number of Hindi students
   80 and above	15%	108
   70 and above	35%	264
   60 and above	65%	424
   50 and above	90%	554
   40 and above	100%	624

   Marks-wise distribution of students. Total students: 1200; no student failed; each student appeared in either Hindi or Punjabi.

   1. A180
   2. B150
   3. C200
   4. D140

   Answer: (B) 150

   Explanation

   The combined 50–60 group is 300. Hindi contributes 554 − 424 = 130, so Punjabi contributes 170; their average is (130 + 170) ÷ 2 = 150. Therefore, 150 is the correct choice. The Hindi and Punjabi student counts must be subtracted from the cumulative distribution before averaging.

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2. Question 2

   What is the difference between the number of Punjabi students scoring 70 or more but less than 80 marks and Hindi students scoring 40 or more but less than 50 marks?

   Marks (out of 100)	Cumulative percentage of Hindi and Punjabi students	Cumulative number of Hindi students
   80 and above	15%	108
   70 and above	35%	264
   60 and above	65%	424
   50 and above	90%	554
   40 and above	100%	624

   Marks-wise distribution of students. Total students: 1200; no student failed; each student appeared in either Hindi or Punjabi.

   1. A20
   2. B16
   3. C14
   4. D30

   Answer: (C) 14

   Explanation

   The combined 70–80 group is 240; Hindi has 156, leaving 84 Punjabi students. Hindi has 624 − 554 = 70 in 40–50, so the difference is 14. Therefore, 14 is the correct choice. The required Hindi and Punjabi groups are obtained by subtracting adjacent cumulative student counts.

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3. Question 3

   What is the ratio of Hindi students scoring 50 or more but less than 60 marks to Punjabi students scoring 40 or more but less than 50 marks?

   Marks (out of 100)	Cumulative percentage of Hindi and Punjabi students	Cumulative number of Hindi students
   80 and above	15%	108
   70 and above	35%	264
   60 and above	65%	424
   50 and above	90%	554
   40 and above	100%	624

   Marks-wise distribution of students. Total students: 1200; no student failed; each student appeared in either Hindi or Punjabi.

   1. A7:3
   2. B8:7
   3. C9:4
   4. D13:5

   Answer: (D) 13:5

   Explanation

   Hindi 50–60 is 554 − 424 = 130. The total 40–50 group is 120 and Hindi contributes 70, leaving 50 Punjabi students; 130:50 simplifies to 13:5. Therefore, 13:5 is the correct choice.

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4. Question 4

   What is the sum of Hindi students scoring 70 or more but less than 80 marks and Punjabi students scoring 80 or more marks?

   Marks (out of 100)	Cumulative percentage of Hindi and Punjabi students	Cumulative number of Hindi students
   80 and above	15%	108
   70 and above	35%	264
   60 and above	65%	424
   50 and above	90%	554
   40 and above	100%	624

   Marks-wise distribution of students. Total students: 1200; no student failed; each student appeared in either Hindi or Punjabi.

   1. A330
   2. B228
   3. C272
   4. D284

   Answer: (B) 228

   Explanation

   Hindi 70–80 is 264 − 108 = 156. The combined 80-and-above group is 180, of which 108 are Hindi, so Punjabi has 72; the sum is 228. Therefore, 228 is the correct choice. The cumulative Hindi values and combined Hindi-Punjabi percentages identify both required student groups.

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5. Question 5

   The number of Hindi students scoring 80 or more is what percentage less than the number of Punjabi students scoring 60 or more but less than 70?

   Marks (out of 100)	Cumulative percentage of Hindi and Punjabi students	Cumulative number of Hindi students
   80 and above	15%	108
   70 and above	35%	264
   60 and above	65%	424
   50 and above	90%	554
   40 and above	100%	624

   Marks-wise distribution of students. Total students: 1200; no student failed; each student appeared in either Hindi or Punjabi.

   1. A46%
   2. B54%
   3. C44%
   4. D52%

   Answer: (A) 46%

   Explanation

   Hindi 80-and-above is 108. Punjabi 60–70 is 360 combined minus 160 Hindi = 200; 108 is 92 less than 200, or 46% less. Therefore, 46% is the correct choice.

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6. Question 6

   Standard deviation (S) and coefficient of variation (CV, as a percentage) are given below. Compute each mean and arrange them in ascending order.

   A. S = 4, CV = 40/9
   B. S = 5, CV = 62.5
   C. S = 14, CV = 70/3
   D. S = 16, CV = 160/7

   Choose the correct answer.

   1. AB, A, C, D
   2. BC, D, B, A
   3. CA, D, B, C
   4. DD, B, C, A

   Answer: (A) B, A, C, D

   Explanation

   Using CV = S/mean × 100 gives means B = 8, A = 36, C = 60, and D = 70. Their ascending order is B, A, C, D. Therefore, B, A, C, D is the correct choice. The standard deviation and coefficient of variation determine every dataset mean through the stated formula.

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7. Question 7

   Which are the risks of plagiarism for a researcher who plagiarises?

   A. Losing the inspiration
   B. Losing obtained titles
   C. Destruction of obtained solutions
   D. Destroying the public image

   Choose the most appropriate answer.

   1. AA and C only
   2. BB and C only
   3. CB and D only
   4. DC and D only

   Answer: (C) B and D only

   Explanation

   Plagiarism can lead to withdrawal of academic titles and serious reputational damage. “Losing inspiration” and “destruction of solutions” are not standard consequences. Therefore, B and D only is the correct choice.

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8. Question 8

   The role of a teacher in the simulation model is that of

   1. AExplaining
   2. BDesigning
   3. CDescribing
   4. DChoosing activity

   Answer: (B) Designing

   Explanation

   In simulation-based teaching, the teacher designs the learning situation, rules, roles, and debriefing. Learners then participate in the simulated experience. Therefore, Designing is the correct choice.

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9. Question 9

   Match List-I with List-II.

   List-I: institution	List-II: function
   A. NIEPA	IV. Promoting research in planning and management
   B. RUSA	III. Infrastructure development for upgrading colleges into model colleges
   C. QAA	I. Ensuring quality
   D. HEFA	II. Providing loans to higher-education institutions

   Match List-I with List-II.

   1. AA-I, B-III, C-II, D-IV
   2. BA-I, B-II, C-III, D-IV
   3. CA-I, B-IV, C-III, D-II
   4. DA-IV, B-III, C-I, D-II

   Answer: (D) A-IV, B-III, C-I, D-II

   Explanation

   NIEPA promotes research in educational planning and management; RUSA supports infrastructure upgrading; QAA ensures quality; and HEFA provides higher-education loans. Therefore, A-IV, B-III, C-I, D-II is the correct choice.

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10. Question 10

    Learning for all implies:

    A. Inclusive Education
    B. Equitable Education
    C. Inclusive and Equitable Education
    D. Quality Education
    E. Categorized based Education

    1. AA, B and D only
    2. BA and E only
    3. CC only
    4. DD only

    Answer: (C) C only

    Explanation

    “Learning for all” combines inclusion with equity: all learners must participate and receive fair support. Quality education alone does not state both conditions. Therefore, C only is the correct choice.

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11. Question 11

    Match List-I with List-II.

    List-I: learning style	List-II: strength
    A. Convergent	II. Practical application of ideas
    B. Divergent	IV. Imaginative ability and generation of ideas
    C. Assimilation	I. Creating theoretical model
    D. Accommodative	III. Carrying out plans and tasks involving new experiences

    Match List-I with List-II.

    1. AA-I, B-II, C-III, D-IV
    2. BA-II, B-I, C-III, D-IV
    3. CA-II, B-IV, C-I, D-III
    4. DA-IV, B-I, C-III, D-II

    Answer: (C) A-II, B-IV, C-I, D-III

    Explanation

    Kolb’s convergent style applies ideas practically, divergent style generates imaginative ideas, assimilative style builds theoretical models, and accommodative style learns through active experience. Therefore, A-II, B-IV, C-I, D-III is the correct choice.

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12. Question 12

    Which is the correct sequence of increasing access time of storage/memory devices?

    A. Magnetic tape
    B. RAM
    C. Cache memory
    D. Hard disk
    E. Registers

    1. AA, B, C, D, E
    2. BB, E, C, A, D
    3. CE, C, B, D, A
    4. DC, E, B, A, D

    Answer: (C) E, C, B, D, A

    Explanation

    Registers are fastest, followed by cache, RAM, hard disk, and magnetic tape. Thus the increasing access-time sequence is E, C, B, D, A. Therefore, E, C, B, D, A is the correct choice.

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13. Question 13

    Who directed the first film made in India?

    1. ASatyajit Ray
    2. BDadasaheb Phalke
    3. CPrithviraj Kapoor
    4. DArdeshir Irani

    Answer: (B) Dadasaheb Phalke

    Explanation

    Dadasaheb Phalke directed Raja Harishchandra (1913), widely recognised as India’s first full-length feature film. Ardeshir Irani directed the first talkie, not the first film. Therefore, Dadasaheb Phalke is the correct choice.

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14. Question 14

    As per National Education Policy 1986, the major functions of the National Apex Body are:

    A. To allocate resources
    B. To evaluate the examination system
    C. To advise government on policy
    D. To encourage interdisciplinarity and interfaces among different areas
    E. To conduct research in higher education

    Choose the correct answer.

    1. AA, C and E only
    2. BB, D and E only
    3. CC, D and E only
    4. DA, C and D only

    Answer: (D) A, C and D only

    Explanation

    The apex body’s policy role includes resource allocation, advising government, and encouraging interdisciplinary interfaces. Examination evaluation and conducting research are not its stated major functions here. Therefore, A, C and D only is the correct choice.

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15. Question 15

    Arrange the following events in correct chronological order.

    A. Establishment of the University of Madras
    B. Introduction of the Indian Universities Act
    C. Establishment of UGC
    D. First National Education Policy of India

    1. AA, B, C, D
    2. BA, D, C, B
    3. CB, C, A, D
    4. DC, B, A, D

    Answer: (A) A, B, C, D

    Explanation

    The University of Madras was established in 1857, the Indian Universities Act came in 1904, UGC in 1956, and the first National Policy on Education in 1968. Therefore, A, B, C, D is the correct choice.

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16. Question 16

    According to classical Indian logic, which instrument of knowledge is knowledge of the relation between a word and its denotation, produced through knowledge of resemblance or similarity?

    1. APratyaksa (perception)
    2. BAnumana (inference)
    3. CUpamana (comparison)
    4. DSabda (verbal testimony)

    Answer: (C) Upamana (comparison)

    Explanation

    Upamana is knowledge through comparison or similarity. It helps one understand a word’s referent by comparing a known object with a described one. Therefore, Upamana (comparison) is the correct choice.

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17. Question 17

    Increasing the ratio of fissile uranium (U-235) in natural uranium is known as

    1. AEnhancement
    2. BConcentration
    3. CAccumulation
    4. DEnrichment

    Answer: (D) Enrichment

    Explanation

    Uranium enrichment increases the proportion of the fissile isotope U-235. The other terms are general words and not the technical nuclear-process term. Therefore, Enrichment is the correct choice.

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18. Question 18

    Which statements about coliform bacteria are true?

    A. They are used to test water contamination.
    B. Coliform bacteria are harmless and cause no disease.
    C. They are found naturally in the human intestinal tract.
    D. Their presence in water indicates faecal contamination.

    Choose the most appropriate answer.

    1. AA and D only
    2. BB and C only
    3. CA, C and D only
    4. DA, B, C and D

    Answer: (C) A, C and D only

    Explanation

    Coliforms are indicator organisms for water quality; many originate in intestinal tracts and their presence may indicate faecal contamination. Statement B is too absolute. Therefore, A, C and D only is the correct choice.

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19. Question 19

    When a producer allows 36% commission on the retail price, the profit is 8.8%. What is the profit if commission is reduced by 24%?

    1. A36.6%
    2. B49.6%
    3. C59.6%
    4. D45.6%

    Answer: (B) 49.6%

    Explanation

    Take retail price as 100. Net receipt is 64 and equals 108.8% of cost, so cost is 58.82. At 12% commission the receipt is 88, giving (88−58.82)/58.82 = 49.6% profit. Therefore, 49.6% is the correct choice.

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20. Question 20

    With reference to cyber attacks, which statements are correct?

    A. Malware is malicious software designed to cause damage to a computer, server, or network.
    B. Malware includes viruses, worms, Trojan horses, ransomware, and spyware.
    C. Pegasus is ransomware.
    D. India is not a signatory to the Budapest Convention on Cybercrime.

    Choose the most appropriate answer.

    1. AA, B and C only
    2. BB, C and D only
    3. CA, B, C and D
    4. DA, B and D only

    Answer: (D) A, B and D only

    Explanation

    A and B correctly describe malware, and India has not signed the Budapest Convention. Pegasus is spyware, not ransomware, so C is false. Therefore, A, B and D only is the correct choice.

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21. Question 21

    Which international organisation coordinates the Education for All programme globally?

    1. AUNDP
    2. BUNESCO
    3. CIMF
    4. DUNEP

    Answer: (B) UNESCO

    Explanation

    UNESCO led global coordination of the Education for All initiative. UNDP, IMF, and UNEP have different development, finance, and environmental mandates. Therefore, UNESCO is the correct choice.

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22. Question 22

    In the ancient period, the Diksha Guru performed duty as a

    1. AGuru of secular subjects
    2. BSpiritual guide for both girls and boys
    3. CSpiritual guide of boys
    4. DTutor

    Answer: (C) Spiritual guide of boys

    Explanation

    The Diksha Guru initiated and spiritually guided boys in the traditional system. This role was distinct from general secular instruction. Therefore, Spiritual guide of boys is the correct choice.

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23. Question 23

    Find the number replacing the question mark: 7, 12, 32, 122, ?, 3602, 25202.

    1. A602
    2. B610
    3. C702
    4. D622

    Answer: (A) 602

    Explanation

    Each step multiplies by successive integers and subtracts successive even numbers: 7×2−2=12, 12×3−4=32, 32×4−6=122, so 122×5−8=602. Therefore, 602 is the correct choice.

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24. Question 24

    Arrange the following in increasing order of extension.

    A. Social Science
    B. Political Science
    C. Sciences
    D. Politics of International Relations

    1. AC, A, B, D
    2. BA, C, B, D
    3. CB, D, A, C
    4. DD, B, A, C

    Answer: (D) D, B, A, C

    Explanation

    A narrower concept has smaller extension. Politics of International Relations is within Political Science, which is within Social Science, which is within Sciences. Therefore, D, B, A, C is the correct choice.

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25. Question 25

    What is explanatory research?

    1. AIt studies a subject about which no or limited information is available.
    2. BIt describes social situations, events, systems, or structures.
    3. CIt explains the causes of social phenomena.
    4. DIt is based on accumulated common-sense beliefs.

    Answer: (C) It explains the causes of social phenomena.

    Explanation

    Explanatory research asks why a phenomenon occurs and investigates causal relationships. Option A describes exploratory research and B describes descriptive research. Therefore, It explains the causes of social phenomena. is the correct choice.

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26. Question 26

    Match List-I with List-II.

    List-I: operating system/program	List-II: task
    A. UNIX	IV. Multi-user operating system
    B. MS-DOS	III. Single-user operating system
    C. Android OS	II. Mobile device
    D. Utility programs	I. Assist users in system-maintenance tasks

    Match List-I with List-II.

    1. AA-III, B-IV, C-I, D-II
    2. BA-I, B-II, C-IV, D-III
    3. CA-IV, B-III, C-II, D-I
    4. DA-II, B-I, C-I, D-IV

    Answer: (C) A-IV, B-III, C-II, D-I

    Explanation

    UNIX is a multi-user operating system; MS-DOS is single-user; Android is for mobile devices; utility programs assist maintenance tasks. Therefore, A-IV, B-III, C-II, D-I is the correct choice.

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27. Question 27

    Match List-I with List-II.

    List-I: meaning	List-II: term
    A. Who	II. Communicator
    B. Says what	III. Message
    C. In which channel	IV. Medium
    D. To whom	I. Receiver

    Match List-I with List-II.

    1. AA-III, B-II, C-I, D-IV
    2. BA-I, B-IV, C-III, D-II
    3. CA-II, B-III, C-IV, D-I
    4. DA-IV, B-II, C-I, D-III

    Answer: (C) A-II, B-III, C-IV, D-I

    Explanation

    In Lasswell’s model, “Who” is the communicator, “says what” the message, “in which channel” the medium, and “to whom” the receiver. Therefore, A-II, B-III, C-IV, D-I is the correct choice.

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28. Question 28

    Why is the concentration of CO₂ higher in the Northern Hemisphere than in the Southern Hemisphere?

    A. Land mass is greater in the Northern Hemisphere.
    B. Ocean area is greater in the Southern Hemisphere.
    C. More fossil fuel is burnt in the Northern Hemisphere.
    D. The Northern Hemisphere is colder.

    Choose the most appropriate answer.

    1. AA and C only
    2. BB, C and D only
    3. CA, B and C only
    4. DB and D only

    Answer: (A) A and C only

    Explanation

    The Northern Hemisphere contains most land, population, and fossil-fuel emissions. Temperature difference is not the governing reason; B is complementary but not the direct causal statement tested. Therefore, A and C only is the correct choice.

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29. Question 29

    The CBCS (Choice Based Credit System) in higher education promotes

    1. ARigid curriculum structure
    2. BFlexibility in course selection
    3. CLimited student choices
    4. DOnly theoretical learning

    Answer: (B) Flexibility in course selection

    Explanation

    CBCS permits students to choose courses and credits across defined options. It is intended to increase flexibility rather than impose a rigid curriculum. Therefore, Flexibility in course selection is the correct choice.

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30. Question 30

    Which statements contradict each other?

    A. All ducks are birds.
    B. Some ducks are birds.
    C. Some ducks are not birds.
    D. No ducks are birds.

    Choose the most appropriate answer.

    1. AA and D only
    2. BB and C only
    3. CB and D only
    4. DC and D only

    Answer: (A) A and D only

    Explanation

    “All ducks are birds” and “No ducks are birds” cannot both be true. Statement B may follow from A if ducks exist, while C directly conflicts with A. Therefore, A and D only is the correct choice.

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31. Question 31

    Which is an example of IEEE 802.3?

    1. ALAN
    2. BMAN
    3. CWAN
    4. DPAN

    Answer: (A) LAN

    Explanation

    IEEE 802.3 specifies Ethernet, the standard technology used for local area networks. It is not a generic standard for MAN, WAN, or PAN. Therefore, LAN is the correct choice.

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32. Question 32

    A sum was put at simple interest for 3 years. At a rate 1% higher, it would have fetched ₹5,100 more. Find the sum.

    1. A₹170,000
    2. B₹150,000
    3. C₹175,000
    4. D₹200,000

    Answer: (A) ₹170,000

    Explanation

    An extra 1% simple interest for 3 years equals 3% of principal. Hence 0.03P = 5,100, giving P = ₹170,000. Therefore, ₹170,000 is the correct choice. The simple-interest increase applies to the principal for three years at one percentage point.

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33. Question 33

    Which technology was used in the second generation of computers?

    1. AVacuum tubes
    2. BTransistors
    3. CIntegrated circuits
    4. DVLSI

    Answer: (B) Transistors

    Explanation

    Second-generation computers replaced vacuum tubes with transistors. Integrated circuits belong to the third generation and VLSI to later generations. Therefore, Transistors is the correct choice.

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34. Question 34

    What is exploratory research?

    1. AIt describes social situations, events, systems, or structures.
    2. BIt studies a subject about which no or limited information is available.
    3. CIt explains causes of social phenomena.
    4. DIt is based on accumulated common-sense beliefs.

    Answer: (B) It studies a subject about which no or limited information is available.

    Explanation

    Exploratory research investigates an unfamiliar or insufficiently understood problem. Descriptive and explanatory research instead describe characteristics or test causes. Therefore, It studies a subject about which no or limited information is available. is the correct choice.

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35. Question 35

    Which statement(s) is/are wrong?

    A. Aman runs at 20 km/hr and covers 400 m in 8 minutes.
    B. A person travels equal distances at x km/hr and y km/hr; the average speed is 2xy/(x+y) km/hr.
    C. A cyclist covers 750 m in 2.5 minutes; the speed is 18 km/hr.
    D. x km/hr equals 5x/18 m/sec.

    Choose the most appropriate answer.

    1. AA only
    2. BA and B only
    3. CA and C only
    4. DB, C and D only

    Answer: (A) A only

    Explanation

    At 20 km/hr, 400 m takes 0.4/20 hour = 1.2 minutes, not 8. B is the harmonic mean for equal distances, and C and D are correct conversions. Therefore, A only is the correct choice.

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36. Question 36

    Match List-I with List-II.

    List-I: population mean (μ) and mean of x²	List-II: population standard deviation (σ)
    A. μ = 9, (1/N)Σxᵢ² = 117	III. 6
    B. μ = 10, (1/N)Σxᵢ² = 200	I. 10
    C. μ = 11, (1/N)Σxᵢ² = 202	IV. 9
    D. μ = 12, (1/N)Σxᵢ² = 160	II. 4

    Match List-I with List-II.

    1. AA-III, B-IV, C-I, D-II
    2. BA-II, B-I, C-IV, D-III
    3. CA-III, B-I, C-IV, D-II
    4. DA-I, B-IV, C-II, D-III

    Answer: (C) A-III, B-I, C-IV, D-II

    Explanation

    Population standard deviation is √(mean of squares − mean²): A = √36 = 6, B = √100 = 10, C = √81 = 9, and D = √16 = 4. Therefore, A-III, B-I, C-IV, D-II is the correct choice. Each population standard deviation follows from the population mean and the mean of squared values.

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37. Question 37

    In a code language, NOIDA is written as 1, 4, 9, 15, 14 and INDIA as 1, 9, 4, 14, 9. How will HELLO be written?

    1. A15, 12, 12, 5, 8
    2. B5, 12, 12, 15, 8
    3. C8, 5, 12, 12, 15
    4. D12, 5, 12, 8, 15

    Answer: (A) 15, 12, 12, 5, 8

    Explanation

    The code writes the alphabetical positions in reverse order. HELLO has positions 8, 5, 12, 12, 15, so the reversed sequence is 15, 12, 12, 5, 8. Therefore, 15, 12, 12, 5, 8 is the correct choice.

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38. Question 38

    What is the correct sequence, from initial to final, of surface wastewater treatment?

    A. Disinfection
    B. Sedimentation
    C. Filtration
    D. Screening

    1. AA, B, C, D
    2. BA, B, D, C
    3. CD, B, A, C
    4. DD, B, C, A

    Answer: (D) D, B, C, A

    Explanation

    Treatment starts with screening to remove large debris, then sedimentation, filtration, and finally disinfection. Disinfection must follow the removal stages. Therefore, D, B, C, A is the correct choice.

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39. Question 39

    Who directed the first talkie feature film made in India?

    1. AArdeshir Irani
    2. BRitwik Ghatak
    3. CDadasaheb Phalke
    4. DBimal Roy

    Answer: (A) Ardeshir Irani

    Explanation

    Ardeshir Irani directed Alam Ara (1931), India’s first talkie feature. Phalke directed the earlier silent landmark Raja Harishchandra. Therefore, Ardeshir Irani is the correct choice.

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40. Question 40

    “This is the richest fraternity on campus; therefore Mr. X, a member of it, must be one of the richest young men on campus.” Which fallacy is involved?

    1. AFallacy of division
    2. BFallacy of irrelevant conclusion
    3. CRed herring
    4. DSlippery slope

    Answer: (A) Fallacy of division

    Explanation

    The argument wrongly transfers a property of the whole group to every individual member. That is the fallacy of division. Therefore, Fallacy of division is the correct choice.

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41. Question 41

    What is the correct chronological order of these Doordarshan programmes, from earliest to latest?

    A. Mahabharat
    B. Chandrakanta
    C. Hum Log
    D. Ramayan
    E. Buniyaad

    1. AA, B, C, D, E
    2. BC, E, D, A, B
    3. CB, C, A, D, E
    4. DB, A, D, E, C

    Answer: (B) C, E, D, A, B

    Explanation

    Hum Log preceded Buniyaad; Ramayan and then Mahabharat followed, while Chandrakanta came later. Thus the sequence is C, E, D, A, B. Therefore, C, E, D, A, B is the correct choice.

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42. Question 42

    Which are differences between the Modified Mercalli Scale (MMS) and Richter Scale (RS)?

    A. RS measures released energy while MMS measures both energy and devastation.
    B. RS uses a linear scale while MMS uses a logarithmic scale.
    C. RS value is constant for an earthquake while MMS varies with distance.
    D. RS has more levels than MMS.

    Choose the most appropriate answer.

    1. AA and B only
    2. BB and C only
    3. CC and D only
    4. DA and C only

    Answer: (C) C and D only

    Explanation

    Richter magnitude is a single logarithmic magnitude for an event, whereas Modified Mercalli intensity varies by location and has twelve levels. A and B reverse or overstate the distinctions. Therefore, C and D only is the correct choice.

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43. Question 43

    If “No birds are animals” is false, which statements can be inferred true?

    A. Some animals are not birds.
    B. Some birds are not animals.
    C. Some animals are birds.
    D. Some birds are animals.

    Choose the most appropriate answer.

    1. AA and B only
    2. BC and D only
    3. CB only
    4. DA, B, C and D

    Answer: (B) C and D only

    Explanation

    Negating “No birds are animals” yields “Some birds are animals.” This is equivalent to “Some animals are birds,” so C and D are both true. Therefore, C and D only is the correct choice.

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44. Question 44

    Which films were directed by Satyajit Ray?

    A. Madhumati
    B. Apur Sansar
    C. Sujata
    D. Aparajito
    E. Pather Panchali

    Choose the most appropriate answer.

    1. AB, D and E only
    2. BA and C only
    3. CA, B, C and D only
    4. DA, B and C only

    Answer: (A) B, D and E only

    Explanation

    Ray directed Apur Sansar, Aparajito, and Pather Panchali. Madhumati and Sujata were directed by Bimal Roy. Therefore, B, D and E only is the correct choice.

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45. Question 45

    Arrange the problem-solving strategy.

    A. Generate many possible solutions
    B. Determine how to implement the solution and evaluate its success
    C. Make a decision
    D. Evaluate each solution
    E. Define the problem

    Choose the correct answer.

    1. AB, E, A, D, C
    2. BE, A, D, C, B
    3. CA, B, C, D, E
    4. DC, A, B, E, D

    Answer: (B) E, A, D, C, B

    Explanation

    A sound strategy defines the problem, generates alternatives, evaluates them, makes a decision, and then implements and evaluates the chosen solution. Therefore, E, A, D, C, B is the correct choice.

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46. Question 46

    Which Act/section gives parents the right to seek relief from any kind of abuse?

    Read the following passage carefully and answer the questions.

    Forced by circumstances and in search of success and status, educated young men and women settle far away from their parents. Some go abroad. The very education that the parents manage to give to their children turns against them in the end. Their children go away in search of green pastures leaving them alone. International community as well as our government is aware of their problems. The UNO proclaimed the year 1999 as ‘International Year for the Aged’. In our own country too the year 2000 was proclaimed ‘The Year for the Aged’.

    Article 41 of our constitution clearly makes it incumbent on the state to look after the aged. There is a social welfare ministry responsible for the welfare of the aged. Laws are made from time-to-time to protect the rights of senior citizens. For example under law parents cannot be evicted from a house without due process of law under section 105 of the Criminal Procedure Code. A magistrate can order a person to maintain his/her old parents under the Maintenance of Parents Act. The Hindu Adoption and Maintenance Act says that the aged parents can demand maintenance from their children. The Domestic Violence Act also provides parents with the rights to seek relief from any kind of abuse.

    1. AHindu Adoption and Maintenance Act
    2. BSection 105 of CPC
    3. CThe Domestic Violence Act
    4. DSection 98 of CPC

    Answer: (C) The Domestic Violence Act

    Explanation

    The passage expressly states that the Domestic Violence Act provides parents the right to seek relief from abuse. The other provisions concern maintenance or eviction. Therefore, The Domestic Violence Act is the correct choice.

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47. Question 47

    Who is empowered to order the concerned person to maintain their parents?

    Read the following passage carefully and answer the questions.

    Forced by circumstances and in search of success and status, educated young men and women settle far away from their parents. Some go abroad. The very education that the parents manage to give to their children turns against them in the end. Their children go away in search of green pastures leaving them alone. International community as well as our government is aware of their problems. The UNO proclaimed the year 1999 as ‘International Year for the Aged’. In our own country too the year 2000 was proclaimed ‘The Year for the Aged’.

    Article 41 of our constitution clearly makes it incumbent on the state to look after the aged. There is a social welfare ministry responsible for the welfare of the aged. Laws are made from time-to-time to protect the rights of senior citizens. For example under law parents cannot be evicted from a house without due process of law under section 105 of the Criminal Procedure Code. A magistrate can order a person to maintain his/her old parents under the Maintenance of Parents Act. The Hindu Adoption and Maintenance Act says that the aged parents can demand maintenance from their children. The Domestic Violence Act also provides parents with the rights to seek relief from any kind of abuse.

    1. AA retired judge
    2. BA High Court judge
    3. CLocal Sarpanch
    4. DA magistrate

    Answer: (D) A magistrate

    Explanation

    The passage states that a magistrate can order a person to maintain their aged parents. It does not assign that power to the other authorities. Therefore, A magistrate is the correct choice.

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48. Question 48

    The year 1999 was proclaimed as

    Read the following passage carefully and answer the questions.

    Forced by circumstances and in search of success and status, educated young men and women settle far away from their parents. Some go abroad. The very education that the parents manage to give to their children turns against them in the end. Their children go away in search of green pastures leaving them alone. International community as well as our government is aware of their problems. The UNO proclaimed the year 1999 as ‘International Year for the Aged’. In our own country too the year 2000 was proclaimed ‘The Year for the Aged’.

    Article 41 of our constitution clearly makes it incumbent on the state to look after the aged. There is a social welfare ministry responsible for the welfare of the aged. Laws are made from time-to-time to protect the rights of senior citizens. For example under law parents cannot be evicted from a house without due process of law under section 105 of the Criminal Procedure Code. A magistrate can order a person to maintain his/her old parents under the Maintenance of Parents Act. The Hindu Adoption and Maintenance Act says that the aged parents can demand maintenance from their children. The Domestic Violence Act also provides parents with the rights to seek relief from any kind of abuse.

    1. ANational Year for the Aged
    2. BUniversal Year for the Aged
    3. CInternational Year for the Aged
    4. DSAARC Year for the Aged

    Answer: (C) International Year for the Aged

    Explanation

    The passage explicitly identifies 1999 as the UNO’s “International Year for the Aged.” The other labels are not used in the passage. Therefore, International Year for the Aged is the correct choice.

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49. Question 49

    Under section 105 of the criminal procedure code, parents cannot be evicted from a house

    Read the following passage carefully and answer the questions.

    Forced by circumstances and in search of success and status, educated young men and women settle far away from their parents. Some go abroad. The very education that the parents manage to give to their children turns against them in the end. Their children go away in search of green pastures leaving them alone. International community as well as our government is aware of their problems. The UNO proclaimed the year 1999 as ‘International Year for the Aged’. In our own country too the year 2000 was proclaimed ‘The Year for the Aged’.

    Article 41 of our constitution clearly makes it incumbent on the state to look after the aged. There is a social welfare ministry responsible for the welfare of the aged. Laws are made from time-to-time to protect the rights of senior citizens. For example under law parents cannot be evicted from a house without due process of law under section 105 of the Criminal Procedure Code. A magistrate can order a person to maintain his/her old parents under the Maintenance of Parents Act. The Hindu Adoption and Maintenance Act says that the aged parents can demand maintenance from their children. The Domestic Violence Act also provides parents with the rights to seek relief from any kind of abuse.

    1. AWithout due process of law
    2. BIn any situation
    3. CWithout submitting their will
    4. DUnless property is transferred to their legal heir

    Answer: (A) Without due process of law

    Explanation

    According to the passage, eviction is prohibited without due process of law. It is not an unconditional prohibition or linked to a will or property transfer. Therefore, Without due process of law is the correct choice.

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50. Question 50

    According to the passage, who usually leaves their parents in the lurch?

    Read the following passage carefully and answer the questions.

    Forced by circumstances and in search of success and status, educated young men and women settle far away from their parents. Some go abroad. The very education that the parents manage to give to their children turns against them in the end. Their children go away in search of green pastures leaving them alone. International community as well as our government is aware of their problems. The UNO proclaimed the year 1999 as ‘International Year for the Aged’. In our own country too the year 2000 was proclaimed ‘The Year for the Aged’.

    Article 41 of our constitution clearly makes it incumbent on the state to look after the aged. There is a social welfare ministry responsible for the welfare of the aged. Laws are made from time-to-time to protect the rights of senior citizens. For example under law parents cannot be evicted from a house without due process of law under section 105 of the Criminal Procedure Code. A magistrate can order a person to maintain his/her old parents under the Maintenance of Parents Act. The Hindu Adoption and Maintenance Act says that the aged parents can demand maintenance from their children. The Domestic Violence Act also provides parents with the rights to seek relief from any kind of abuse.

    1. ADisobedient wards
    2. BFrustrated youth
    3. CCareer-seeking children settled in distant places
    4. DThose with job insecurity

    Answer: (C) Career-seeking children settled in distant places

    Explanation

    The passage describes educated children moving far away, often in search of success, status, and greener pastures, leaving their parents alone. Therefore, Career-seeking children settled in distant places is the correct choice.

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100 questions with the answer key and explanations.

1. Question 1

   In a group of 120 people, 65 eat rice, 45 eat bread, 42 eat curd, 20 eat both rice and bread, 25 eat both rice and curd, 15 eat both bread and curd, and 8 eat all three. How many eat at least one of the three items?

   1. A56
   2. B100
   3. C92
   4. D65

   Answer: (B) 100

   Explanation

   Apply inclusion-exclusion: 65 + 45 + 42 - 20 - 25 - 15 + 8 = 100. The pairwise overlaps were subtracted once too often, so the triple overlap is added back once.

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2. Question 2

   In a pack of 42 cards, 3 cards are chosen one after another without replacement. How many ordered selections are possible?

   1. A1722
   2. B1752
   3. C68880
   4. D6880

   Answer: (C) 68880

   Explanation

   Because the cards are chosen successively, order matters. There are 42 choices, then 41, then 40: 42 × 41 × 40 = 68,880.

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3. Question 3

   A positive integer is selected uniformly from the integers not exceeding 200. What is the probability that it is divisible by either 2 or 5?

   1. A2/5
   2. B3/5
   3. C4/5
   4. D1/5

   Answer: (B) 3/5

   Explanation

   There are 100 multiples of 2 and 40 multiples of 5 up to 200. The 20 multiples of 10 were counted twice, so the favourable count is 100 + 40 - 20 = 120. Thus the probability is 120/200 = 3/5.

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4. Question 4

   Consider the Boolean expression A(x, y, z) = x(y'z)'. Which is its complete sum-of-products form?

   1. Axyz + x'yz' + x'y'z'
   2. Bxyz' + x'yz' + x'y'z'
   3. Cxyz' + xy'z + x'y'z'
   4. Dxyz + xyz' + xy'z'

   Answer: (D) xyz + xyz' + xy'z'

   Explanation

   By De Morgan's law, (y'z)' = y + z'. Hence A = x(y + z') = xy + xz'. Expanding xy as xyz + xyz' and xz' as xyz' + xy'z', then removing the duplicate xyz', gives xyz + xyz' + xy'z'. The other choices introduce minterms with x' or omit one required minterm.

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5. Question 5

   Choose the correct statement for a group G.

   1. AIf (xy)^2 = x^2y^2 for all x, y in G, then G is commutative.
   2. BIf x^3 = 1 for all x in G, then G is commutative.
   3. CIf x^5 = 1 for all x in G, then G is commutative.
   4. DIf G is commutative, a subgroup of G need not be commutative.

   Answer: (A) If (xy)^2 = x^2y^2 for all x, y in G, then G is commutative.

   Explanation

   The stated identity gives xyxy = xxyy. Multiplying by x⁻¹ on the left and y⁻¹ on the right yields yx = xy, so G is commutative. The exponent conditions alone do not force commutativity, and every subgroup of an abelian group is abelian.

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6. Question 6

   What is the binary equivalent of (A0F)₁₆?

   1. A111000111
   2. B101001111
   3. C101000001111
   4. D111100001010

   Answer: (C) 101000001111

   Explanation

   Convert each hexadecimal digit to four bits: A = 1010, 0 = 0000, and F = 1111. Concatenating them gives 101000001111.

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7. Question 7

   In an I/O operation, what does a write operation do?

   1. ATransfers data from an I/O device to memory
   2. BTransfers data from memory to an I/O device
   3. CTransfers data from a CPU register to memory
   4. DTransfers data from a CPU register to an I/O device

   Answer: (B) Transfers data from memory to an I/O device

   Explanation

   A write operation sends data outward from the computer's memory to a peripheral or I/O device. The opposite direction, from a device into memory, is a read operation.

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8. Question 8

   The transformation of data from main memory to cache memory is called:

   1. AData exchange
   2. BData transformation
   3. CMapping
   4. DMatching

   Answer: (C) Mapping

   Explanation

   A cache uses a mapping scheme—such as direct, associative, or set-associative mapping—to determine where a main-memory block is placed in cache.

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9. Question 9

   What is the Gray-code equivalent of decimal 8?

   1. A1000
   2. B1100
   3. C1010
   4. D1110

   Answer: (B) 1100

   Explanation

   Decimal 8 is binary 1000. Gray code is found by XORing the binary number with itself shifted right one bit: 1000 XOR 0100 = 1100.

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10. Question 10

    What is printed by this C program?

    int x = 128, y = 110;
    do {
      if (x > y) x = x - y;
      else y = y - x;
    } while (x != y);
    printf("%d", x);
    

    1. A18
    2. B2
    3. C92
    4. D74

    Answer: (B) 2

    Explanation

    The loop repeatedly replaces the larger value by the difference, which is Euclid's subtraction algorithm. It ends when x and y both equal gcd(128, 110) = 2, so it prints 2.

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11. Question 11

    Which technique is used for clipping?

    1. AStack-based seed
    2. BScan-line seed
    3. CSutherland-Cohen
    4. DInverse scaling

    Answer: (C) Sutherland-Cohen

    Explanation

    Cohen-Sutherland is the standard line-clipping algorithm; the option reverses its two names. Seed-fill and scan-line methods are filling techniques, while inverse scaling is a transformation idea.

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12. Question 12

    What is printed by this C program?

    int i, j;
    for (i = 1; i < 5; i += 2)
      for (j = 1; j < i; j += 2)
        printf("%d", j);
    

    1. A1
    2. B1 2
    3. C1 3
    4. D1 1 3

    Answer: (A) 1

    Explanation

    The outer loop uses i = 1 and then i = 3. At i = 1, j starts at 1 and fails j < i, so nothing prints. At i = 3, j = 1 satisfies the test and prints once; incrementing j to 3 ends that inner loop. Hence the only printed value is 1, without spaces because the format string contains none.

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13. Question 13

    Which pointer expression refers to the array element ar[m][n][o]?

    1. A((*(ar) + m + n) + o)
    2. B(((*ar + m) + n) + o)
    3. C(((ar + m) + n) + o)
    4. D((*(ar + m) + n) + o)

    Answer: (D) ((*(ar + m) + n) + o)

    Explanation

    For a multidimensional array, ar + m reaches the mth two-dimensional slice; *(ar + m) + n reaches the nth row; and *(*(ar + m) + n) + o reaches the oth element. Dereferencing that final address gives the element.

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14. Question 14

    Consider relation R with attributes A, B, C, D, and E, and dependencies C → F, E → A, EC → D, and A → B. Which option is a key for R?

    1. ACD
    2. BEC
    3. CAE
    4. DAC

    Answer: (B) EC

    Explanation

    EC is a key: E determines A and then B, while EC determines D. Thus EC determines every attribute listed in R. The printed dependency C → F names an attribute omitted from the displayed schema, but it does not change which listed attributes EC determines. None of the other choices contains both E and C, so none can derive all of R's attributes.

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15. Question 15

    Let T1 have attributes A, B, C, D, and E with dependencies {EB → C, D → E, EA → B}; T2 has attributes A, B, C, and D with dependencies {C → A, A → B, A → D}. Which statement is true?

    1. AT1 is in 3NF
    2. BT2 is in 3NF
    3. CT1 is not in 3NF
    4. DT1 is in 2NF

    Answer: (C) T1 is not in 3NF

    Explanation

    For T1, AD is a candidate key: D gives E, then EA gives B, and EB gives C. In the dependency EB → C, EB is not a superkey and C is non-prime, so T1 violates 3NF. T1 is therefore not in 3NF.

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16. Question 16

    In a relational database, which statement is correct?

    1. AA relation with only two attributes is always in BCNF.
    2. BIf all attributes of a relation are prime attributes, then the relation is in BCNF.
    3. CEvery relation has at least one non-prime attribute.
    4. DBCNF decomposition preserves functional dependencies.

    Answer: (A) A relation with only two attributes is always in BCNF.

    Explanation

    For a two-attribute relation, any nontrivial functional dependency has one attribute determining the other, making that determinant a key. Therefore every nontrivial dependency has a superkey on the left, satisfying BCNF. BCNF decomposition need not preserve every dependency.

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17. Question 17

    Consider Employee(eid, eName), Comp(cid, cName), and Own(eid, cid). Which relational-algebra expression returns eid values for people who own every brand?

    1. Aπ_eid(π_eid,cid(Own) ÷ π_cid(Comp))
    2. Bπ_eid(π_eid(Own) × π_cid(Comp))
    3. Cπ_eid(π_eid,cid(Own) × π_cid(Comp))
    4. Dπ_eid(π_eid(Own) × (π_cid,cName(Own) ÷ π_cid(Comp)))

    Answer: (A) π_eid(π_eid,cid(Own) ÷ π_cid(Comp))

    Explanation

    Division returns the eid values paired in Own with every cid in Comp. Projecting eid and cid from Own before dividing supplies exactly the required dividend relation, and the final projection returns only eids.

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18. Question 18

    Using non-preemptive SJF and Round Robin with time quantum 10, which process finishes second last in each schedule?

    Process	Burst time	Arrival time
    P1	9	0
    P2	30	0
    P3	4	0
    P4	8	2
    P5	11	6

    1. ASJF: P4; RR: P4
    2. BSJF: P5; RR: P4
    3. CSJF: P4; RR: P3
    4. DSJF: P5; RR: P5

    Answer: (D) SJF: P5; RR: P5

    Explanation

    For non-preemptive SJF the finishing order is P1, P3, P4, P5, P2, so P5 is second last. With RR(10), the completion order is P1, P3, P4, P5, P2; again P5 is second last.

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19. Question 19

    Which statement about Pthreads is correct?

    1. AIt refers to POSIX standard IEEE 1002.1c.
    2. BThe standard defines an API only for thread creation.
    3. CIt specifies thread behaviour but not an implementation.
    4. DSolaris, Linux, and macOS, except Tru64 UNIX, implement Pthreads.

    Answer: (C) It specifies thread behaviour but not an implementation.

    Explanation

    Pthreads is an interface specification: it defines the observable thread API and behaviour, while operating systems may implement it differently. It covers much more than creation, including synchronization and thread management. The standard number in option A is not the POSIX Threads designation.

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20. Question 20

    On average, 10 processes arrive each second and 20 processes are normally in the queue. What is the average waiting time per process?

    1. A3 seconds
    2. B2 seconds
    3. C18 seconds
    4. D9 seconds

    Answer: (B) 2 seconds

    Explanation

    Little's law gives L = λW. With average queue length L = 20 and arrival rate λ = 10 processes per second, W = 20/10 = 2 seconds.

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21. Question 21

    What is the total swap time (swap-in and swap-out) for a 15 MB process at a transfer rate of 30 MB/s, with average latency 12 ms and no head seeks?

    1. A1.024 s
    2. B1.00 s
    3. C0.512 s
    4. D12 s

    Answer: (A) 1.024 s

    Explanation

    Each transfer takes 15/30 = 0.5 s. Swapping in and out takes 1.0 s of transfer time, and two 12 ms latencies add 0.024 s, for 1.024 s total.

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22. Question 22

    Which is not a requirements-elicitation technique?

    1. AInterviews
    2. BThe use-case approach
    3. CFacilitated Application Specification Technique (FAST)
    4. DData-flow diagram

    Answer: (D) Data-flow diagram

    Explanation

    Interviews, use cases, and FAST are techniques for discovering and negotiating requirements. A data-flow diagram models how data moves through a system; it may document requirements but is not itself an elicitation technique.

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23. Question 23

    In basic COCOMO, the effort estimate for organic software is:

    1. AE = 2.4(KLOC)^1.05 person-months
    2. BE = 3.4(KLOC)^1.06 person-months
    3. CE = 2.0(KLOC)^1.05 person-months
    4. DE = 2.4(KLOC)^1.07 person-months

    Answer: (A) E = 2.4(KLOC)^1.05 person-months

    Explanation

    The basic COCOMO organic-mode effort equation is E = 2.4 × (KLOC)^1.05 person-months. The coefficients and exponents for semi-detached and embedded modes differ.

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24. Question 24

    The process used to discover difficult, unknown, and hidden information about an existing software system is called:

    1. ASoftware engineering
    2. BSoftware re-engineering
    3. CReverse engineering
    4. DInverse engineering

    Answer: (C) Reverse engineering

    Explanation

    Reverse engineering analyzes an existing system to recover its design, specifications, or other knowledge that is not readily visible. Re-engineering is broader and includes modifying or improving the system after such understanding is obtained.

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25. Question 25

    Which is not a software characteristic?

    1. ASoftware does not wear out.
    2. BSoftware is flexible.
    3. CSoftware is not manufactured.
    4. DSoftware is always correct.

    Answer: (D) Software is always correct.

    Explanation

    Software can contain defects, so it is not always correct. Unlike physical hardware it does not wear out through use, it can be changed, and it is developed rather than manufactured in the usual physical sense.

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26. Question 26

    For infix expression ((A+B)*D)↑(E-F), what is its equivalent postfix expression?

    1. AA B + D * E F ↑
    2. BA B + D * - ↑ E F
    3. CA B + D E * F - ↑
    4. DA B + D * E F - ↑

    Answer: (D) A B + D * E F - ↑

    Explanation

    Postfix for the left operand ((A+B)*D) is A B + D *; postfix for (E-F) is E F -. Appending the exponent operator after the two complete operands gives A B + D * E F - ↑.

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27. Question 27

    What is the tight asymptotic bound for T(n) = 2T(n/4) + √n?

    1. AΘ(√n)
    2. BΘ(n log n)
    3. CΘ(√n log n)
    4. DΘ(n log √n)

    Answer: (C) Θ(√n log n)

    Explanation

    Here a = 2 and b = 4, so n^(log₄2) = √n. Since f(n) is also √n, this is the equal-order case of the Master Theorem, giving Θ(√n log n).

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28. Question 28

    Which sequence is a longest common subsequence of {1, 2, 3, 2, 4, 1, 2} and {2, 4, 3, 1, 2, 1}?

    1. A2, 1, 2, 3
    2. B1, 3, 2, 1
    3. C2, 3, 2, 1
    4. D2, 3, 1, 2, 1

    Answer: (C) 2, 3, 2, 1

    Explanation

    2, 3, 2, 1 appears in order in the first sequence at positions 2, 3, 4, 6 and in the second at positions 1, 3, 5, 6. Option D cannot occur in the first sequence because after 2, 3, 1, 2 there is no later 1.

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29. Question 29

    Maintaining a graph in memory using its adjacency matrix is called:

    1. AComplete representation
    2. BLinked representation
    3. CCircular representation
    4. DSequential representation

    Answer: (D) Sequential representation

    Explanation

    An adjacency matrix is a two-dimensional array whose entries are stored in contiguous, indexed locations. That array-based form is the sequential representation; linked representation instead uses adjacency lists.

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30. Question 30

    What is printed by this C code?

    int *i, a = 12, b = 2, c;
    c = (a = a + b, b = a / b, a = a * b, b = a - b);
    i = &c;
    printf("%d", --(*i));
    

    1. A91
    2. B90
    3. C98
    4. D92

    Answer: (B) 90

    Explanation

    The displayed code supplies the source's evidently omitted terminating semicolon. The comma expressions execute left to right: a becomes 14, b becomes 7, a becomes 98, and b becomes 91. The assignment sets c to the final expression value 91; --(*i) decrements c through its pointer before printing, giving 90. It does not print 91 because the pre-decrement happens before printf, and 98 is only the intermediate value of a.

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31. Question 31

    Which is the best description of the language accepted by the DFA?

    

    DFA over the alphabet {a, b, c}.

    1. AL = (a* + b* + c*)+
    2. BL = (a + b + c)(abc)+(a + b + c)
    3. CThe set of strings that start with a, b, or c and end with c
    4. DThe set of strings having an even count, including zero, of the substring abc

    Answer: (D) The set of strings having an even count, including zero, of the substring abc

    Explanation

    The automaton's states track progress through a possible abc and whether a completed occurrence has changed the parity. The initial accepting state represents zero completed occurrences; completing abc moves to the opposite-parity group, and completing another returns to the accepting group. Thus it accepts exactly strings with an even number of abc occurrences.

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32. Question 32

    Which pictured NFA is valid for the given DFA?

    

    Given DFA and four candidate NFA diagrams.

    1. ADiagram 1
    2. BDiagram 2
    3. CDiagram 3
    4. DDiagram 4

    Answer: (D) Diagram 4

    Explanation

    A valid NFA must preserve the DFA's accepted language while allowing nondeterministic state choices. Comparing the transition alternatives and accepting states shows that diagram 4 retains the required accepting paths without accepting a string that the DFA rejects; the other diagrams alter an accepting state or omit a required transition alternative.

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33. Question 33

    In a stop-and-wait system, line bandwidth is 1 Mbps and round-trip delay is 30 ms. What is the bandwidth-delay product?

    1. A15,000 bits
    2. B20,000 bits
    3. C30,000 bits
    4. D60,000 bits

    Answer: (C) 30,000 bits

    Explanation

    The bandwidth-delay product is the amount of data that can be in flight: 1,000,000 bits/s × 0.030 s = 30,000 bits.

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34. Question 34

    CB84000D001C001C is a UDP header in hexadecimal. What is the source-port number?

    1. A52100
    2. B13
    3. C28
    4. D52000

    Answer: (A) 52100

    Explanation

    The first 16 bits of a UDP header are the source port. The first four hexadecimal digits are CB84, and 0xCB84 = 12×4096 + 11×256 + 8×16 + 4 = 52,100.

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35. Question 35

    A packet sent by a router or node to the source to inform it of congestion is called a:

    1. AExplicit packet
    2. BBackpressure packet
    3. CChoke packet
    4. DRetransmission packet

    Answer: (C) Choke packet

    Explanation

    A choke packet explicitly tells a source to reduce its sending rate because congestion has been detected. Backpressure instead propagates congestion control hop by hop toward the source.

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36. Question 36

    Which is not a characteristic of packet switching?

    1. AData is broken into packets.
    2. BEach packet may take a different route.
    3. CIt requires a dedicated path.
    4. DIt supports dynamic routing.

    Answer: (C) It requires a dedicated path.

    Explanation

    Packet switching shares network links and routes packets independently, often dynamically. Requiring an exclusive end-to-end path is the defining property of circuit switching, not packet switching.

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37. Question 37

    Which is not a valid property of λ-cuts of fuzzy relations R̃ and S̃?

    1. A(R̃ ∪ S̃)λ = Rλ ∪ Sλ
    2. B(R̃ ∩ S̃)λ = Rλ ∩ Sλ
    3. CThe λ-cut of the complement of R̃ differs from the complement of Rλ except when λ = 1.
    4. DFor λ ≤ β with 0 ≤ β ≤ 1, Rβ ⊆ Rλ.

    Answer: (C) The λ-cut of the complement of R̃ differs from the complement of Rλ except when λ = 1.

    Explanation

    Union and intersection commute with λ-cuts, and increasing the threshold can only shrink a cut, so Rβ ⊆ Rλ for β ≥ λ. Complement does not have the stated exceptional equality at λ = 1: its cut is governed by the complementary membership threshold, not simply by taking the ordinary complement of Rλ.

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38. Question 38

    Which is not a parent-selection technique in genetic-algorithm implementations?

    1. ARadial
    2. BTournament
    3. CBoltzmann
    4. DRank

    Answer: (A) Radial

    Explanation

    Tournament, rank-based, and Boltzmann selection are ways to choose parents according to fitness. “Radial” is not a standard parent-selection method; it is more commonly associated with radial-basis functions.

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39. Question 39

    With η as learning rate and xi as the ith input, which update rule best describes a Kohonen self-organizing map weight wij?

    1. Awij(new) = (1 − η)wij(old) + ηxi
    2. Bwij(new) = ηwij(old) + (1 − η)xi
    3. Cwij(new) = (1 − η)wij(old) + xi
    4. Dwij(new) = wij(old) + ηxi

    Answer: (A) wij(new) = (1 − η)wij(old) + ηxi

    Explanation

    A SOM moves a winning neuron's weight vector a fraction η toward the input: w(new) = w(old) + η(x − w(old)). Rearranging gives (1 − η)w(old) + ηx, so the new weight is a convex combination of the old weight and input.

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40. Question 40

    After applying samples S1 through S4 in order to a perceptron with initial weights w1 = w2 = 0, learning rate 0.1, no bias, and output 0 when yin = 0, what are the updated weights?

    Sample	x1	x2	Target
    S1	0	0	0
    S2	0	1	1
    S3	1	0	1
    S4	1	1	1

    1. Aw1 = 0.1, w2 = 0.1
    2. Bw1 = 0.0, w2 = 0.2
    3. Cw1 = 0.0, w2 = 0.1
    4. Dw1 = 0.2, w2 = 0.2

    Answer: (A) w1 = 0.1, w2 = 0.1

    Explanation

    S1 has zero inputs and target 0, so it changes nothing. S2 has input (0,1), target 1, and adds 0.1 to w2; S3 similarly adds 0.1 to w1. For S4, yin = 0.2 and the observed output is already 1, matching its target, so no further update occurs.

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41. Question 41

    A computer processes an instruction by fetching it, decoding it, calculating its effective address, fetching operands, and executing it. Which sequence is correct?

    1. AA, B, C, D, E
    2. BB, A, D, A, E
    3. CB, C, A, D, E
    4. DC, B, A, D, E

    Answer: (C) B, C, A, D, E

    Explanation

    The processor must first fetch the instruction and decode it to know its addressing mode. It can then calculate the effective address, fetch the needed operands, and execute the operation. That order is B, C, A, D, E.

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42. Question 42

    In an arithmetic pipeline, the operations are compare exponents, align mantissas, add or subtract mantissas, and normalize the result. Which sequence is correct?

    1. AD, A, B, C
    2. BD, B, A, C
    3. CB, C, A, D
    4. DB, C, D, A

    Answer: (A) D, A, B, C

    Explanation

    Floating-point operands must first have their exponents compared so the smaller mantissa can be aligned. After alignment, the mantissas are added or subtracted, and the resulting normalized form is produced last.

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43. Question 43

    To input data through a CPU-controlled I/O program, the actions are reading the status register, checking its flag bit, reading the data register, and transferring data to memory. Which sequence is correct?

    1. AA, B, C, D
    2. BB, A, C, D
    3. CC, B, A, D
    4. DA, C, B, D

    Answer: (A) A, B, C, D

    Explanation

    The program first reads the device status, then tests the readiness flag. Only after the flag indicates data is available should it read the data register and transfer that value into memory. Thus the sequence is A, B, C, D.

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44. Question 44

    For C operators &&, +=, >>, >=, and ?:, what is the precedence order from higher to lower?

    1. A

       , >=, ?:, +=, &&

    2. B

       , &&, >=, ?:, +=

    3. C

       , >=, &&, ?:, +=

    4. D

       =, >>, &&, +=, ?:

    Answer: (C)

    , >=, &&, ?:, +=

    Explanation

    C precedence places shift operators above relational operators, relational above logical AND, logical AND above conditional, and conditional above assignment. Therefore the order is >>, >=, &&, ?:, +=.

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45. Question 45

    Arrange Linux interrupt-protection levels from increasing priority: user-mode programs, preemptible kernel system-service routines, bottom-half interrupt handlers, and top-half interrupt handlers.

    1. AB → A → C → D
    2. BC → B → A → D
    3. CC → A → B → D
    4. DA → D → B → C

    Answer: (D) A → D → B → C

    Explanation

    User-mode work is the most preemptible and therefore lowest priority. Preemptible kernel service routines rank above it, bottom halves above ordinary kernel work, and top-half interrupt handlers are highest because they run immediately in interrupt context.

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46. Question 46

    Which sequence gives the hierarchy of a layered file system from application level down to I/O control?

    1. AA → B → D → C → E
    2. BD → E → C → A → B
    3. CE → A → B → C → D
    4. DE → C → B → A → D

    Answer: (C) E → A → B → C → D

    Explanation

    Application programs invoke the logical file system, which manages directories and metadata. It calls the file-organization module, then the basic file system, which finally relies on I/O control for device operations. The sequence is E, A, B, C, D.

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47. Question 47

    Arrange stamp coupling, control coupling, external coupling, common coupling, and content coupling from lowest to highest coupling.

    1. AE, A, C, B, D
    2. BD, B, A, E, C
    3. CB, C, D, A, E
    4. DC, A, B, D, E

    Answer: (C) B, C, D, A, E

    Explanation

    Stamp coupling passes a composite data structure and is relatively weak. Control coupling passes control information; external coupling shares externally imposed formats; common coupling shares global data; and content coupling, where one module relies on another's internals, is strongest. Thus B, C, D, A, E is increasing order.

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48. Question 48

    Arrange unit, integration, system, and acceptance testing in their usual software-development order.

    1. AB, C, A, D
    2. BB, A, C, D
    3. CC, B, A, D
    4. DC, B, D, A

    Answer: (B) B, A, C, D

    Explanation

    Testing starts with individual units, then verifies interactions between integrated units. The complete system is tested next, and acceptance testing follows to confirm that stakeholder requirements are met. The order is B, A, C, D.

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49. Question 49

    To construct a Huffman tree, first create leaf nodes, build a priority queue, repeatedly combine the lowest-frequency nodes, and stop when the root is formed. Which sequence is correct?

    1. AB, C, A, D
    2. BD, B, A, C
    3. CB, C, D, A
    4. DC, A, B, D

    Answer: (C) B, C, D, A

    Explanation

    Huffman coding begins with a leaf for every symbol and places those leaves in a min-priority queue. It repeatedly removes the two least-frequent nodes and combines them until one root remains, giving B, C, D, A.

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50. Question 50

    In dynamic programming, the steps are characterizing an optimal solution, recursively defining its value, computing that value bottom-up, and constructing an optimal solution. Which sequence is correct?

    1. AB, C, A, D
    2. BB, A, C, D
    3. CC, B, A, D
    4. DC, B, D, A

    Answer: (D) C, B, D, A

    Explanation

    First identify the optimal substructure, then write a recurrence for the optimal value. Compute the recurrence bottom-up and finally trace the stored choices to construct an optimal solution. This is C, B, D, A.

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51. Question 51

    Arrange LL(0), LR(0), SLR, LALR(1), and LR(1) parsers from least to greatest grammar-handling power.

    1. ALL(0) → LR(0) → SLR → LR(1) → LALR(1)
    2. BSLR → LR(0) → LL(0) → LR(1) → LALR(1)
    3. CLL(0) → LR(0) → SLR → LALR(1) → LR(1)
    4. DLR(0) → LL(0) → SLR → LR(1) → LALR(1)

    Answer: (C) LL(0) → LR(0) → SLR → LALR(1) → LR(1)

    Explanation

    LL(0) has no lookahead and is most restricted. LR(0) adds bottom-up item sets, SLR uses FOLLOW sets to reduce conflicts, LALR(1) merges compatible LR(1) states, and canonical LR(1) retains the most lookahead information. Hence C gives the increasing order.

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52. Question 52

    Arrange the steps of accessing a website after a user enters a URL: DNS resolution, HTTP request, URL entry, IP-address acquisition, and webpage display.

    1. AB, A, C, E, D
    2. BC, B, A, E, D
    3. CB, C, D, A, E
    4. DC, A, D, B, E

    Answer: (D) C, A, D, B, E

    Explanation

    The user first enters the URL. The browser resolves the domain name, obtains an IP address from that resolution, sends the HTTP request to the server, and finally renders the received webpage. The order is C, A, D, B, E.

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53. Question 53

    Arrange virtualization steps in a cloud environment: install a hypervisor, create virtual machines, allocate their resources, and run isolated workloads.

    1. AB, C, A, D
    2. BA, B, C, D
    3. CA, C, B, D
    4. DB, C, D, A

    Answer: (B) A, B, C, D

    Explanation

    A hypervisor must be installed on the physical server before it can create VMs. Resources are then allocated to those VMs, which can subsequently run isolated workloads. The order is A, B, C, D.

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54. Question 54

    For knowledge-base design, arrange choosing a task domain, selecting atoms for propositions, axiomatizing the domain, and asking questions about the intended interpretation.

    1. AC → A → D → B
    2. BC → A → B → D
    3. CB → C → A → D
    4. DA → C → B → D

    Answer: (A) C → A → D → B

    Explanation

    The task domain sets the scope. The designer then chooses the propositional vocabulary, writes axioms in that vocabulary, and can ask whether the resulting representation has the intended interpretation. This produces C, A, D, B.

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55. Question 55

    Arrange these genetic-algorithm actions: calculate population fitness, select parents, apply crossover, and mutate each locus.

    1. AA → C → B → D
    2. BC → A → D → B
    3. CC → A → B → D
    4. DA → D → B → C

    Answer: (B) C → A → D → B

    Explanation

    Fitness must be computed before fitness-informed parent selection. Selected parents undergo crossover to create offspring, and mutation is then applied probabilistically at loci. The standard order is C, A, D, B.

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56. Question 56

    Let P mean “she is intelligent” and Q mean “she is happy.” Which set correctly formalizes: intelligent implies unhappy; neither intelligent nor happy; being not intelligent is necessary for happiness; and not intelligent implies unhappy?

    1. AP → ¬Q; P ∧ ¬Q; Q → ¬P; P → ¬Q
    2. BP → Q; ¬P ∧ ¬Q; ¬Q → P; ¬P → ¬Q
    3. CP → ¬Q; ¬P ∧ ¬Q; Q → ¬P; ¬P → ¬Q
    4. DP → ¬Q; P ∧ ¬Q; Q → P; ¬P → ¬Q

    Answer: (C) P → ¬Q; ¬P ∧ ¬Q; Q → ¬P; ¬P → ¬Q

    Explanation

    “Neither” means ¬P ∧ ¬Q. “Not intelligent is necessary for happy” makes happiness imply not intelligent, Q → ¬P. The first and fourth implications are P → ¬Q and ¬P → ¬Q respectively, so only option C has all four forms.

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57. Question 57

    For positive integers m and n, which statements are correct: n ≠ 1 implies m < mn; a composite k can be written k = mn with 1 < m,n > k; mn = 1 implies m = n = 1; and a composite k can be written k = mn with 1 < m,n < k?

    1. AA, C, D only
    2. BB, C, D only
    3. CA, B only
    4. DA, B, C

    Answer: (A) A, C, D only

    Explanation

    If n ≠ 1, positivity makes n at least 2, hence m < mn. A product of positive integers can equal 1 only when both are 1. A composite number has proper factors strictly between 1 and k, so D is true; B is impossible because two factors greater than k cannot multiply to k.

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58. Question 58

    Which statements are true: write-through updates only cache; write-back updates cache and main memory together; private writable caches create coherence problems; and cache coherence can be solved by a hardware-only scheme?

    1. AA and B only
    2. BB and C only
    3. CC and D only
    4. DA, B, and C only

    Answer: (C) C and D only

    Explanation

    Write-through updates cache and main memory on each write, so A is false. Write-back normally updates main memory later, on eviction, so B is false. Private writable caches do create coherence issues, and hardware protocols such as snooping or directory mechanisms can solve them; C and D are true.

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59. Question 59

    Which gate does not output 1 when both inputs are 0?

    1. ANAND and NOR only
    2. BNOR and XOR only
    3. CXOR and XNOR only
    4. DXOR only

    Answer: (D) XOR only

    Explanation

    At inputs 0 and 0, NAND outputs 1, NOR outputs 1, and XNOR outputs 1 because the inputs match. XOR alone outputs 0, so it is the only listed gate that does not output 1.

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60. Question 60

    Editorially repaired version of an ambiguous NTA item. Consider these XML statements:

    A. XML has no predefined tag set, and users can define their own tags.
    B. XML is used primarily for presentation.
    C. XML is case-sensitive.
    D. XML supports dynamic data representation.

    Which statements are true?

    1. AB, C, and D only
    2. BA, B, and D only
    3. CA, C, and D only
    4. DB and D only

    Answer: (C) A, C, and D only

    Explanation

    XML has no fixed, predefined vocabulary, so applications may define their own element names. XML names are case-sensitive, and XML can represent data whose values or structure change over time. XML itself is not a presentation language; presentation is normally handled by CSS, XSLT, or another rendering layer. Thus A, C, and D are true.

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61. Question 61

    Which pointer operations are legal in C?

    1. AA, B, C only
    2. BA, B, D only
    3. CA and B only
    4. DA and E only

    Answer: (D) A and E only

    Explanation

    Adding an integer to a pointer yields another pointer. Subtracting two pointers into the same array yields an integer difference (ptrdiff_t). Pointer plus pointer is invalid; pointer minus integer yields a pointer, not a number; and pointer minus pointer does not yield a pointer. Therefore A and E only are legal as stated.

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62. Question 62

    Once a process is executing, which events could not occur: issuing I/O then joining the ready queue; creating a child and waiting for it; time-slice expiry then joining the waiting queue; or interrupt-driven removal then joining the waiting queue?

    1. AB only
    2. BD only
    3. CA and C only
    4. DA, C, and D only

    Answer: (D) A, C, and D only

    Explanation

    An I/O request blocks the process, sending it to waiting rather than ready. Expiry of a time slice preempts it to ready, not waiting. An interrupt can also preempt it to ready; it does not by itself make the process wait. Waiting for a child is a valid blocking event, so A, C, and D are the impossible descriptions.

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63. Question 63

    Which statement about spinlock semaphores is correct?

    1. AThey are busy-waiting semaphores.
    2. BThey are not useful when locks are held briefly.
    3. CWaiting for one may require multiple context switches.
    4. DThey are often employed on uniprocessor systems.

    Answer: (A) They are busy-waiting semaphores.

    Explanation

    A spinlock repeatedly tests a lock while busy waiting. It is useful for very short critical sections because it can avoid sleep-and-wakeup overhead; waiting does not itself cause context switches, and spinning is generally unsuitable on a uniprocessor because the lock holder needs CPU time to run.

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64. Question 64

    Which cohesions are better than procedural cohesion?

    1. AFunctional and communicational only
    2. BFunctional, sequential, and communicational only
    3. CTemporal, communicational, and logical only
    4. DFunctional, communicational, and logical only

    Answer: (B) Functional, sequential, and communicational only

    Explanation

    Functional cohesion is strongest, followed by sequential and communicational cohesion; each is tighter than procedural cohesion. Temporal and logical cohesion are weaker because they group work by time or category rather than by direct data-flow or a single focused purpose.

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65. Question 65

    Which are McCall quality factors: maintainability, usability, integrity, and functionality?

    1. AA and D only
    2. BA, B, and D only
    3. CC and D only
    4. DA, B, and C only

    Answer: (D) A, B, and C only

    Explanation

    McCall's quality model includes maintainability, usability, and integrity among its factors. “Functionality” is prominent in later models such as ISO/IEC 25010, but is not one of McCall's named factors.

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66. Question 66

    Which algorithms use a greedy strategy: Dijkstra's algorithm, Kruskal's algorithm, Huffman coding, and Bellman-Ford?

    1. AA, B, and D only
    2. BA, B, and C only
    3. CC and D only
    4. DA and D only

    Answer: (B) A, B, and C only

    Explanation

    Dijkstra greedily fixes the smallest tentative distance, Kruskal greedily adds the cheapest edge that avoids a cycle, and Huffman greedily combines the two least-frequent nodes. Bellman-Ford repeatedly relaxes edges and is not a greedy algorithm.

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67. Question 67

    Which trees are height balanced: binary search tree, AVL tree, red-black tree, and B-tree?

    1. AA and D only
    2. BA, B, and D only
    3. CC and D only
    4. DB and C only

    Accepted answers: (C) C and D only, (D) B and C only

    Explanation

    An ordinary binary-search tree has no balancing guarantee. AVL trees enforce a strict height-balance condition, while red-black trees guarantee logarithmic height through black-height constraints. The term “height balanced” is also used broadly for B-trees because all leaves are at the same level; the supplied options therefore reflect two defensible conventions.

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68. Question 68

    Which are controlled-access protocols: reservation, polling, TDMA, token passing, and CSMA/CA?

    1. AA and D only
    2. BA, B, and D only
    3. CC, D, and E only
    4. DA, D, and E only

    Answer: (B) A, B, and D only

    Explanation

    Controlled access coordinates stations so they take turns. Reservation, polling, and token passing are the standard controlled-access methods. TDMA is a channelization technique, while CSMA/CA is contention based.

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69. Question 69

    Which are data-encoding schemes: NRZ, Manchester encoding, amplitude modulation, Hamming code, and bipolar AMI?

    1. AA, C, and D only
    2. BA, B, and D only
    3. CC, D, and E only
    4. DA, B, and E only

    Answer: (D) A, B, and E only

    Explanation

    NRZ, Manchester, and bipolar AMI are line-encoding schemes. Amplitude modulation is a modulation technique and Hamming code is an error-control code, so neither belongs in the list of data-encoding schemes here.

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70. Question 70

    Which statements about agent systems are correct: an agent system comprises an agent and its environment; an agent controller receives percepts and sends commands; actuators are always reliable; and actuators convert stimuli into percepts?

    1. AA and B only
    2. BB and D only
    3. CC and D only
    4. DB, C, and D only

    Answer: (A) A and B only

    Explanation

    An agent acts in an environment and its controller maps percepts to actions. Sensors, not actuators, convert environmental stimuli to percepts; actuators execute actions and can be noisy or unreliable. Thus only A and B are correct.

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71. Question 71

    Which STRIPS statements are correct: it is feature-centric; world-state features are primitive and derived; an action has preconditions and effects; and it directly defines conditional effects?

    1. AA and C only
    2. BB and C only
    3. CA and D only
    4. DB, C, and D only

    Answer: (B) B and C only

    Explanation

    STRIPS represents a state using features or propositions, commonly distinguished as primitive and derived, and describes actions through preconditions and effects. Basic STRIPS effects are unconditional, so it does not directly express conditional effects. Therefore B and C are the correct pair.

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72. Question 72

    Editorially repaired version of an ambiguous NTA item. Which grammars are context-free but not regular?

    Grammar	Productions
    A	S → Ab; aS → aA; A → a
    B	S → Ab; A → Sa; A → a
    C	S → AS; S → aA; A → a
    D	S → Ab; S → aA; A → a
    E	S → Sb; A → Aa; A → ε

    1. AA and B only
    2. BC and D only
    3. CC only
    4. DA, B, D, and E only

    Answer: (C) C only

    Explanation

    A context-free grammar must have one nonterminal on the left of every production. A is not context-free because it has aS → aA. B, D, and E are regular because their productions are consistently linear. C is context-free but not regular because S → AS has two nonterminals on its right side. Thus only C qualifies. The source's answer set was repaired to include that result explicitly.

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73. Question 73

    The source prints 196 bits for Twofish; that typo is corrected to 256 bits below. Which encryption statements are true: symmetric-key cryptography requires secrecy of the key-generating function above the decrypting function; DES uses a 64-bit block and 56-bit key; Twofish uses a 128-bit block with variable keys up to 256 bits; and RC4 is a stream cipher operating on bits or bytes rather than blocks?

    1. AB and C only
    2. BA and D only
    3. CA and C only
    4. DB, C, and D only

    Answer: (D) B, C, and D only

    Explanation

    Symmetric encryption depends on keeping the key secret, not the algorithm or a key-generating function. DES uses a 64-bit block and a 56-bit key; Twofish has a 128-bit block and supports keys up to 256 bits; and RC4 is a stream cipher. Therefore B, C, and D are true. The source's printed 196-bit maximum has been explicitly corrected to 256 bits in this learner-facing version.

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74. Question 74

    Editorially repaired version of an ambiguous NTA item. In the broad Knowledge Discovery in Databases (KDD) process, consider these statements:

    A. Data transformation is involved.
    B. An online database can serve as a data source for a data warehouse.
    C. Classification is a measure of accuracy.
    D. K-means clustering minimizes within-cluster variance.
    E. Pattern evaluation identifies knowledge using interestingness measures.

    Which statements are true?

    1. AA, B, D, and E only
    2. BB, D, and E only
    3. CA, C, and E only
    4. DA, C, and D only

    Answer: (A) A, B, D, and E only

    Explanation

    Data transformation is a standard KDD step, and online operational databases can provide source data to a warehouse. K-means minimizes within-cluster variance, while pattern evaluation uses interestingness measures. Classification is a predictive task; accuracy is one measure used to evaluate a classifier. Therefore A, B, D, and E are true.

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75. Question 75

    Which serializability statements are true: conflict serializability means conflict equivalence to a serial schedule; view serializability means view equivalence to a serial schedule; serial schedules interleave participating transactions; and every conflict-serializable schedule is view-serializable but not conversely?

    1. AA, B, C, and D
    2. BA, B, and C only
    3. CA, B, and D only
    4. DB, C, and D only

    Answer: (C) A, B, and D only

    Explanation

    Conflict serializability and view serializability have the stated equivalence definitions. A serial schedule does not interleave transactions; it completes one before another starts. Conflict equivalence implies view equivalence, but some view-serializable schedules are not conflict-serializable, so A, B, and D are true.

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76. Question 76

    Match the statements about the Indian team with inverse, converse, contrapositive, and conditional forms.

    List I	List II
    A. If the Indian team wins, then it is raining	I. Inverse
    B. If the Indian team does not win, then it is not raining	II. Converse
    C. If it is raining, then the Indian team wins	III. Contrapositive
    D. If it is not raining, then the Indian team does not win	IV. Conditional

    1. AA-II, B-I, C-IV, D-III
    2. BA-III, B-I, C-II, D-IV
    3. CA-II, B-III, C-IV, D-I
    4. DA-III, B-II, C-IV, D-I

    Answer: (C) A-II, B-III, C-IV, D-I

    Explanation

    Take “if it is raining, the team wins” as the conditional, C-IV. Its converse is “if the team wins, it is raining,” A-II; its contrapositive is “if the team does not win, it is not raining,” B-III; and its inverse is D-I. Hence option C.

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77. Question 77

    Match Types 3, 2, 1, and 0 grammars to their production restrictions.

    List I	List II
    A. Type 3 Grammar	I. V → (V ∪ Σ)*
    B. Type 2 Grammar	II. λVβ → λ(V ∪ Σ)*β, where (λ, β) ∈ (V ∪ Σ)*
    C. Type 1 Grammar	III. (V ∪ Σ)* → (V ∪ Σ)*
    D. Type 0 Grammar	IV. V → VΣ | Σ

    1. AA-II, B-III, C-IV, D-I
    2. BA-III, B-I, C-IV, D-II
    3. CA-IV, B-I, C-II, D-III
    4. DA-III, B-II, C-I, D-IV

    Answer: (C) A-IV, B-I, C-II, D-III

    Explanation

    Type 3 grammars use the regular production form V → VΣ or Σ. Type 2 allows a single variable on the left, V → (V ∪ Σ)*. Type 1 uses noncontracting contextual productions λVβ → λ(V ∪ Σ)β, and Type 0 has unrestricted productions over (V ∪ Σ). This gives A-IV, B-I, C-II, D-III.

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78. Question 78

    Match address space, memory space, cache memory, and segmented program with associative mapping, logical address, physical address, and virtual address.

    List I	List II
    A. Address Space	I. Associative Mapping
    B. Memory Space	II. Logical address
    C. Cache Memory	III. Physical address
    D. Segmented Program	IV. Virtual address

    1. AA-IV, B-III, C-I, D-II
    2. BA-I, B-II, C-IV, D-III
    3. CA-III, B-IV, C-I, D-II
    4. DA-II, B-IV, C-I, D-III

    Answer: (A) A-IV, B-III, C-I, D-II

    Explanation

    Address space is the virtual address range visible to a process, whereas memory space is physical memory. Cache memory uses associative mapping, and a segmented program's segment-plus-offset form is a logical address. The matching A-IV, B-III, C-I, D-II is option A.

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79. Question 79

    Editorially repaired version of an NTA-dropped item. Treat the following descriptions as a one-to-one matching. Which mapping is correct?

    List I	List II
    A. Overloading	I. An overridden method call is selected at runtime
    B. Early binding	II. A function call is resolved at compile time
    C. Method overriding	III. Supports compile-time polymorphism
    D. Late binding	IV. Supports runtime polymorphism

    1. AA-III, B-II, C-I, D-IV
    2. BA-IV, B-II, C-III, D-I
    3. CA-III, B-I, C-IV, D-II
    4. DA-IV, B-I, C-III, D-II

    Answer: (A) A-III, B-II, C-I, D-IV

    Explanation

    Overloading is compile-time polymorphism, so A maps to III. Early binding resolves a call at compile time, so B maps to II. Method overriding uses runtime dispatch, so C maps to I; late binding enables runtime polymorphism, so D maps to IV. The resulting mapping is A-III, B-II, C-I, D-IV.

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80. Question 80

    Match oblique, cavalier, cabinet, and orthographic projection with their defining projection properties.

    List I	List II
    A. Oblique projection	I. Perpendicular lines are foreshortened
    B. Cavalier projection	II. Perpendicular lines are not foreshortened
    C. Cabinet projection	III. Projectors are perpendicular to the plane
    D. Orthographic projection	IV. Projector angle to plane is not 90°

    1. AA-III, B-II, C-IV, D-I
    2. BA-IV, B-II, C-I, D-III
    3. CA-III, B-I, C-IV, D-II
    4. DA-IV, B-I, C-III, D-II

    Answer: (B) A-IV, B-II, C-I, D-III

    Explanation

    Oblique projectors meet the plane at an angle other than 90 degrees. Cavalier projection keeps receding lengths unforeshortened, cabinet projection foreshortens them, and orthographic projection uses projectors perpendicular to the plane. The mapping is A-IV, B-II, C-I, D-III.

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81. Question 81

    Match clustered page table, hierarchical page table, segmentation, and inverted page table with their characteristics.

    List I	List II
    A. Clustered Page Table	I. Generally inappropriate for 64-bit architectures
    B. Hierarchical Page Table	II. One entry for each real page/frame
    C. Segmentation	III. Useful for sparse address spaces
    D. Inverted Page Table	IV. Supports a user view of the system

    1. AA-III, B-IV, C-II, D-I
    2. BA-III, B-I, C-IV, D-II
    3. CA-II, B-III, C-I, D-IV
    4. DA-IV, B-II, C-I, D-III

    Answer: (B) A-III, B-I, C-IV, D-II

    Explanation

    Clustered page tables are useful for sparse address spaces. Large hierarchical page tables are generally inappropriate for 64-bit address spaces, segmentation supports the user's logical view, and an inverted page table has one entry per physical frame. The mapping is A-III, B-I, C-IV, D-II.

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82. Question 82

    Match RAID levels 2, 3, 5, and 6 with their organizations.

    List I	List II
    A. RAID Level 2	I. Block-interleaved distributed parity
    B. RAID Level 3	II. P+Q redundancy scheme
    C. RAID Level 5	III. Bit-interleaved parity
    D. RAID Level 6	IV. Memory-style error-correcting code organization

    1. AA-IV, B-III, C-I, D-II
    2. BA-III, B-IV, C-I, D-II
    3. CA-I, B-III, C-IV, D-II
    4. DA-II, B-IV, C-I, D-III

    Answer: (A) A-IV, B-III, C-I, D-II

    Explanation

    RAID 2 uses memory-style error-correcting code organization, RAID 3 uses bit-interleaved parity, RAID 5 uses distributed block-interleaved parity, and RAID 6 is P+Q dual-parity redundancy. Thus A-IV, B-III, C-I, D-II.

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83. Question 83

    Match reliability, efficiency, maintainability, and portability with their software-quality descriptions.

    Characteristic	Description
    A. Reliability	I. Transfer between environments
    B. Efficiency	II. Sustained performance over time
    C. Maintainability	III. Performance relative to resources
    D. Portability	IV. Effort needed for modification

    1. AA-II, B-III, C-I, D-IV
    2. BA-II, B-III, C-IV, D-I
    3. CA-III, B-II, C-I, D-IV
    4. DA-IV, B-II, C-I, D-III

    Answer: (B) A-II, B-III, C-IV, D-I

    Explanation

    Reliability is sustained performance under stated conditions and time; efficiency relates performance to resources; maintainability is the effort to modify software; and portability is transfer between environments. The mapping is A-II, B-III, C-IV, D-I.

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84. Question 84

    Match equivalence partitioning, boundary-value analysis, cyclomatic complexity, and decision-table testing with their testing focus.

    Technique	Focus
    A. Equivalence Partitioning	I. Independent code paths
    B. Boundary Value Analysis	II. Valid/invalid input sets
    C. Cyclomatic Complexity	III. Input-range limits
    D. Decision Table Testing	IV. Action combinations and conditions

    1. AA-II, B-III, C-I, D-IV
    2. BA-II, B-III, C-IV, D-I
    3. CA-III, B-II, C-I, D-IV
    4. DA-IV, B-II, C-I, D-III

    Answer: (A) A-II, B-III, C-I, D-IV

    Explanation

    Equivalence partitioning divides inputs into valid and invalid classes; boundary-value analysis tests input limits; cyclomatic complexity measures independent code paths; and decision tables test action combinations under varying conditions. Therefore A-II, B-III, C-I, D-IV.

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85. Question 85

    Match circular queue, priority queue, deque, and simple queue with their common applications.

    List I	List II
    A. Circular Queue	I. Print Queue
    B. Priority Queue	II. CPU Scheduling
    C. Double Ended Queue	III. Dijkstra algorithm
    D. Simple Queue	IV. Palindrome checking

    1. AA-II, B-III, C-I, D-IV
    2. BA-II, B-III, C-IV, D-I
    3. CA-III, B-II, C-I, D-IV
    4. DA-IV, B-II, C-I, D-III

    Answer: (B) A-II, B-III, C-IV, D-I

    Explanation

    A circular queue supports round-robin CPU scheduling, a priority queue supports Dijkstra's algorithm, a deque can test palindromes by removing from both ends, and a simple FIFO queue serves a print queue. The mapping is A-II, B-III, C-IV, D-I.

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86. Question 86

    Match chaining, linear probing, quadratic probing, and double hashing with their collision-handling strategies.

    Method	Strategy
    A. Chaining	I. Check next slot
    B. Linear Probing	II. Use second hash function
    C. Quadratic Probing	III. Linked list at index
    D. Double Hashing	IV. Quadratic step

    1. AA-II, B-III, C-I, D-IV
    2. BA-II, B-III, C-IV, D-I
    3. CA-III, B-I, C-IV, D-II
    4. DA-IV, B-II, C-I, D-III

    Answer: (C) A-III, B-I, C-IV, D-II

    Explanation

    Chaining keeps a linked list at an index; linear probing checks the next slot; quadratic probing skips by a quadratic step; and double hashing uses a second hash function. The mapping is A-III, B-I, C-IV, D-II.

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87. Question 87

    Match count-to-infinity, IEEE 802.16, IEEE 802.15, and AODV with their networking associations.

    List I	List II
    A. Count-to-infinity problem	I. Mobile Adhoc Networks
    B. IEEE 802.16	II. Distance-vector routing
    C. IEEE 802.15	III. WiMAX
    D. AODV	IV. Bluetooth

    1. AA-II, B-III, C-I, D-IV
    2. BA-II, B-III, C-IV, D-I
    3. CA-II, B-IV, C-III, D-I
    4. DA-IV, B-III, C-I, D-II

    Answer: (B) A-II, B-III, C-IV, D-I

    Explanation

    Count-to-infinity is a distance-vector-routing problem. IEEE 802.16 is WiMAX, IEEE 802.15 includes Bluetooth personal-area networking, and AODV is used in mobile ad hoc networks. The mapping is A-II, B-III, C-IV, D-I.

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88. Question 88

    Match digital signature, hash function, AES, and RSA with their security roles.

    List I	List II
    A. Digital Signature	I. Asymmetric encryption algorithm
    B. Hash Function	II. Confirms authenticity and integrity
    C. AES	III. Produces a fixed-size digest
    D. RSA	IV. Symmetric encryption algorithm

    1. AA-II, B-III, C-I, D-IV
    2. BA-II, B-III, C-IV, D-I
    3. CA-III, B-II, C-I, D-IV
    4. DA-IV, B-II, C-I, D-III

    Answer: (B) A-II, B-III, C-IV, D-I

    Explanation

    A digital signature confirms authenticity and integrity; a hash function produces a fixed-size digest; AES is a symmetric encryption algorithm; and RSA is asymmetric. The matching is A-II, B-III, C-IV, D-I.

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89. Question 89

    Match fuzzy-set intersection, bounded sum, bounded difference, and algebraic sum with their membership functions.

    Operation	Membership function
    A. Intersection	I. μA(x) + μB(x) − μA(x)μB(x)
    B. Bounded Sum	II. max(0, μA(x) − μB(x))
    C. Bounded Difference	III. min(1, μA(x) + μB(x))
    D. Algebraic Sum	IV. min(μA(x), μB(x))

    1. AA-III, B-I, C-II, D-IV
    2. BA-III, B-II, C-IV, D-I
    3. CA-IV, B-III, C-II, D-I
    4. DA-II, B-I, C-III, D-IV

    Answer: (C) A-IV, B-III, C-II, D-I

    Explanation

    Intersection takes the smaller membership, min(μA, μB). Bounded sum caps ordinary addition at 1, while bounded difference floors subtraction at 0. Algebraic sum is the probabilistic-style formula μA + μB − μAμB, not simple capped addition. Therefore A-IV, B-III, C-II, D-I.

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90. Question 90

    Match worst individual, best individual, sum of fitness, and median fitness termination criteria with their convergence descriptions.

    Criterion	Description
    A. Worst Individual	I. At least half the individuals are at or above convergence value
    B. Best Individual	II. Virtually all individuals are within a fitness range
    C. Sum of Fitness	III. Entire population meets a minimum standard
    D. Median Fitness	IV. At least one best solution / faster convergence

    1. AA-III, B-IV, C-II, D-I
    2. BA-IV, B-III, C-I, D-II
    3. CA-II, B-I, C-III, D-IV
    4. DA-II, B-III, C-IV, D-I

    Answer: (A) A-III, B-IV, C-II, D-I

    Explanation

    A worst-individual threshold ensures the entire population meets a minimum standard. A best-individual criterion guarantees at least one best solution, sum of fitness constrains the population to a fitness range, and median fitness means at least half the population is at or above the convergence value. Thus A-III, B-IV, C-II, D-I.

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91. Question 91

    Which high-level program corresponds to the displayed three-address code?

    

    Three-address code and its four high-level-code alternatives.

    1. ASource code option 1
    2. BSource code option 2
    3. CSource code option 3
    4. Dinit = x/(y+2); limit = 10; while (i > j && num > b) { num = limit-1; b = y+2; i = i-num; } j = i-num; k = j+b;

    Answer: (D) init = x/(y+2); limit = 10; while (i > j && num > b) { num = limit-1; b = y+2; i = i-num; } j = i-num; k = j+b;

    Explanation

    The two conditional jumps send execution to the loop body only when both i > j and num > b hold. The loop body recomputes num, b, and i, then jumps back to the first condition; on either failed condition it computes j and k. That control flow is the while form in option D.

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92. Question 92

    What are the leaders for this three-address code?

    100: t1 = y + 2
    101: initial = x / t1
    102: limit = 10
    103: if i > j goto 105
    104: goto 111
    105: if num > b goto 107
    106: goto 111
    107: num = limit - 1
    108: b = y + 2
    109: i = i - num
    110: goto 103
    111: j = i - num
    112: k = j + b
    

    1. A100, 102, 103, 104, 105, 107, 110
    2. B100, 103, 105, 107, 111, 112
    3. C100, 103, 105, 106, 107, 111
    4. D100, 103, 104, 105, 106, 107, 111

    Answer: (D) 100, 103, 104, 105, 106, 107, 111

    Explanation

    The first statement is a leader. Every jump target—103, 105, 107, and 111—is a leader, and each statement immediately following a conditional jump is also a leader: 104 and 106. This gives 100, 103, 104, 105, 106, 107, and 111.

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93. Question 93

    Which displayed flow graph is correct for the given three-address code?

    

    Four adjacency-matrix flow-graph alternatives for the three-address code.

    1. AAdjacency-matrix option 1
    2. BAdjacency-matrix option 2
    3. CAdjacency-matrix option 3
    4. DAdjacency-matrix option 4

    Answer: (A) Adjacency-matrix option 1

    Explanation

    Partitioning at the leaders yields blocks for initialization, both tests and fall-throughs, the loop body, and the exit. The graph must have the two test branches, the loop-back edge from the body to the first test, and both failed-condition edges to the exit; only option A has that structure.

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94. Question 94

    In the flow graph for the given three-address code, which blocks are dominated by Block 3 and Block 5?

    1. ABlock 3: 3; Block 5: 5
    2. BBlock 3: 3, 4, 5, 6; Block 5: 5, 6
    3. CBlock 3: 3, 7; Block 5: 5, 7
    4. DBlock 3: 3, 4, 7; Block 5: 5, 6, 7

    Answer: (A) Block 3: 3; Block 5: 5

    Explanation

    A block dominates another only when every path from the entry to that block passes through it. The branch structure allows the exit to be reached through paths that bypass each of the named decision blocks, so neither dominates another distinct block. Each block dominates itself, giving option A.

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95. Question 95

    Editorially repaired version of an NTA-dropped item. Consider this three-address code. Assume y is not reassigned between lines 100 and 108. Which single change is a semantics-preserving common-subexpression elimination?

    100: t1 = y + 2
    101: initial = x / t1
    102: limit = 10
    103: if i > j goto 105
    104: goto 111
    105: if num > b goto 107
    106: goto 111
    107: num = limit - 1
    108: b = y + 2
    109: i = i - num
    110: goto 103
    111: j = i - num
    112: k = j + b
    

    1. AReplace only line 108 with b = t1.
    2. BMove num = limit - 1 before the test at line 105.
    3. CReplace lines 111–112 with k = i + b.
    4. DReplace line 109 with i = i - limit - 1.

    Answer: (A) Replace only line 108 with b = t1.

    Explanation

    Line 100 already computes y + 2 into t1, and the stated assumption ensures that value is still valid at line 108. Replacing b = y + 2 with b = t1 therefore eliminates a repeated computation without changing the program. Moving the assignment to num changes the value tested at line 105; k = i + b omits the subtraction of num; and i - limit - 1 is not equal to i - (limit - 1).

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96. Question 96

    Using FCFS disk scheduling, what is the total head movement for request queue 45, 81, 185, 33, 175, 99, 150, 77 with initial head location 58?

    1. A536
    2. B594
    3. C647
    4. D700

    Answer: (C) 647

    Explanation

    FCFS visits requests in the given order rather than choosing a nearby request. The movements are 13, 36, 104, 152, 142, 76, 51, and 73 cylinders. Their sum is 647 cylinders; the smaller alternatives would require reordering the queue, which FCFS never does.

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97. Question 97

    Using SSTF disk scheduling, what is the total head movement for request queue 45, 81, 185, 33, 175, 99, 150, 77 with initial head location 58?

    1. A160
    2. B124
    3. C177
    4. D240

    Answer: (C) 177

    Explanation

    SSTF chooses the nearest pending request: 58 → 45 → 33 → 77 → 81 → 99 → 150 → 175 → 185. The movements are 13 + 12 + 44 + 4 + 18 + 51 + 25 + 10 = 177.

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98. Question 98

    For SCAN moving left from initial head location 58, what is the second request processed for queue 45, 81, 185, 33, 175, 99, 150, 77?

    1. A81
    2. B45
    3. C33
    4. D175

    Answer: (C) 33

    Explanation

    Moving left, SCAN serves the requests below 58 in descending proximity: 45 first and then 33. Thus the second processed request is 33.

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99. Question 99

    For C-SCAN moving right from initial head location 58, what is the total head movement for queue 45, 81, 185, 33, 175, 99, 150, 77?

    1. A327
    2. B307
    3. C525
    4. D385

    Answer: (D) 385

    Explanation

    The head services 77, 81, 99, 150, 175, and 185 while moving right, continues to the end cylinder 199, wraps to 0, then services 33 and 45. The movement is 127 to cylinder 185, then 14 to 199, 199 for the wrap, and 45 after wrapping: 385 cylinders.

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100. Question 100

     For C-LOOK moving right from initial head location 58, what is the second request processed for queue 45, 81, 185, 33, 175, 99, 150, 77?

     1. A81
     2. B45
     3. C77
     4. D33

     Answer: (A) 81

     Explanation

     C-LOOK services requests in the current direction only as far as the last requested cylinder. Moving right from 58, it processes 77 first and 81 second, so the second request is 81.

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