UGC NET Computer Science & Applications August 2024 Question PaperPaper 1 & Paper 2 with Answer Key

50 questions with the answer key and explanations.

1. Question 1

   If the number of males who passed in Commerce from City-G is 2544, what is the number of students who passed in Arts from City-G?

   City / Arts data	Science / Commerce data
   A — Arts : Science : Commerce = 2 : 4 : 5 Arts M:F = 31:14	Science M:F = 23:27 Commerce M:F = 11:7
   B — Arts : Science : Commerce = 7 : 2 : 4 Arts M:F = 37:33	Science M:F = 43:32 Commerce M:F = 29:21
   C — Arts : Science : Commerce = 1 : 4 : 2 Arts M:F = 34:16	Science M:F = 57:43 Commerce M:F = 31:29
   D — Arts : Science : Commerce = 5 : 7 : 4 Arts M:F = 17:13	Science M:F = 51:33 Commerce M:F = 23:17
   E — Arts : Science : Commerce = 4 : 3 : 8 Arts M:F = 23:17	Science M:F = 41:34 Commerce M:F = 57:23
   F — Arts : Science : Commerce = 2 : 4 : 3 Arts M:F = 47:28	Science M:F = 11:7 Commerce M:F = 16:11
   G — Arts : Science : Commerce = 3 : 5 : 4 Arts M:F = 29:21	Science M:F = 27:24 Commerce M:F = 53:47

   City-wise proportions of graduating students and male:female ratios.

   1. A3600
   2. B4200
   3. C4800
   4. D3400

   Answer: (A) 3600

   Explanation

   For City G, Commerce has 53 males out of 100, so Commerce total is 2544 × 100/53 = 4800. The Arts:Science:Commerce ratio is 3:5:4, hence Arts total is 4800 × 3/4 = 3600.

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2. Question 2

   For City-A, if the total number of males who passed in Arts is 2480, what is the difference between the number of students who passed in Commerce and in Science?

   City / Arts data	Science / Commerce data
   A — Arts : Science : Commerce = 2 : 4 : 5 Arts M:F = 31:14	Science M:F = 23:27 Commerce M:F = 11:7
   B — Arts : Science : Commerce = 7 : 2 : 4 Arts M:F = 37:33	Science M:F = 43:32 Commerce M:F = 29:21
   C — Arts : Science : Commerce = 1 : 4 : 2 Arts M:F = 34:16	Science M:F = 57:43 Commerce M:F = 31:29
   D — Arts : Science : Commerce = 5 : 7 : 4 Arts M:F = 17:13	Science M:F = 51:33 Commerce M:F = 23:17
   E — Arts : Science : Commerce = 4 : 3 : 8 Arts M:F = 23:17	Science M:F = 41:34 Commerce M:F = 57:23
   F — Arts : Science : Commerce = 2 : 4 : 3 Arts M:F = 47:28	Science M:F = 11:7 Commerce M:F = 16:11
   G — Arts : Science : Commerce = 3 : 5 : 4 Arts M:F = 29:21	Science M:F = 27:24 Commerce M:F = 53:47

   City-wise proportions of graduating students and male:female ratios.

   1. A600
   2. B1000
   3. C1400
   4. D1800

   Answer: (D) 1800

   Explanation

   Arts males are 31/45 of Arts total, so Arts total is 3600. With the 2:4:5 stream ratio, Science and Commerce totals are 7200 and 9000; their difference is 1800.

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3. Question 3

   If the number of students who passed in Commerce from City-F is 3240, then the number of students who passed from City-F is what percentage of the number of Science students who passed from City-F?

   City / Arts data	Science / Commerce data
   A — Arts : Science : Commerce = 2 : 4 : 5 Arts M:F = 31:14	Science M:F = 23:27 Commerce M:F = 11:7
   B — Arts : Science : Commerce = 7 : 2 : 4 Arts M:F = 37:33	Science M:F = 43:32 Commerce M:F = 29:21
   C — Arts : Science : Commerce = 1 : 4 : 2 Arts M:F = 34:16	Science M:F = 57:43 Commerce M:F = 31:29
   D — Arts : Science : Commerce = 5 : 7 : 4 Arts M:F = 17:13	Science M:F = 51:33 Commerce M:F = 23:17
   E — Arts : Science : Commerce = 4 : 3 : 8 Arts M:F = 23:17	Science M:F = 41:34 Commerce M:F = 57:23
   F — Arts : Science : Commerce = 2 : 4 : 3 Arts M:F = 47:28	Science M:F = 11:7 Commerce M:F = 16:11
   G — Arts : Science : Commerce = 3 : 5 : 4 Arts M:F = 29:21	Science M:F = 27:24 Commerce M:F = 53:47

   City-wise proportions of graduating students and male:female ratios.

   1. A75
   2. B150
   3. C180
   4. D225

   Answer: (D) 225

   Explanation

   For City F, the stream ratio is 2:4:3. Commerce 3240 corresponds to 3 parts, so Science is 4320 and the total is 9720. Thus 9720/4320 × 100 = 225%.

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4. Question 4

   If the number of females who passed in Arts from City-C is 768, then the number of males who passed in Commerce from City-C is approximately what percentage of the number of students who passed from City-C?

   City / Arts data	Science / Commerce data
   A — Arts : Science : Commerce = 2 : 4 : 5 Arts M:F = 31:14	Science M:F = 23:27 Commerce M:F = 11:7
   B — Arts : Science : Commerce = 7 : 2 : 4 Arts M:F = 37:33	Science M:F = 43:32 Commerce M:F = 29:21
   C — Arts : Science : Commerce = 1 : 4 : 2 Arts M:F = 34:16	Science M:F = 57:43 Commerce M:F = 31:29
   D — Arts : Science : Commerce = 5 : 7 : 4 Arts M:F = 17:13	Science M:F = 51:33 Commerce M:F = 23:17
   E — Arts : Science : Commerce = 4 : 3 : 8 Arts M:F = 23:17	Science M:F = 41:34 Commerce M:F = 57:23
   F — Arts : Science : Commerce = 2 : 4 : 3 Arts M:F = 47:28	Science M:F = 11:7 Commerce M:F = 16:11
   G — Arts : Science : Commerce = 3 : 5 : 4 Arts M:F = 29:21	Science M:F = 27:24 Commerce M:F = 53:47

   City-wise proportions of graduating students and male:female ratios.

   1. A14.76
   2. B18.24
   3. C27.8
   4. D32.5

   Answer: (A) 14.76

   Explanation

   City C Arts females are 16/50 of Arts, giving Arts total 2400. Commerce is twice Arts, or 4800, and its males are 31/60, or 2480. The city total is 2400 + 9600 + 4800 = 16800, so 2480/16800 × 100 ≈ 14.76%.

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5. Question 5

   The number of males who passed in Commerce from City-F is approximately what percentage more than the number of females who passed in Commerce from City-F?

   City / Arts data	Science / Commerce data
   A — Arts : Science : Commerce = 2 : 4 : 5 Arts M:F = 31:14	Science M:F = 23:27 Commerce M:F = 11:7
   B — Arts : Science : Commerce = 7 : 2 : 4 Arts M:F = 37:33	Science M:F = 43:32 Commerce M:F = 29:21
   C — Arts : Science : Commerce = 1 : 4 : 2 Arts M:F = 34:16	Science M:F = 57:43 Commerce M:F = 31:29
   D — Arts : Science : Commerce = 5 : 7 : 4 Arts M:F = 17:13	Science M:F = 51:33 Commerce M:F = 23:17
   E — Arts : Science : Commerce = 4 : 3 : 8 Arts M:F = 23:17	Science M:F = 41:34 Commerce M:F = 57:23
   F — Arts : Science : Commerce = 2 : 4 : 3 Arts M:F = 47:28	Science M:F = 11:7 Commerce M:F = 16:11
   G — Arts : Science : Commerce = 3 : 5 : 4 Arts M:F = 29:21	Science M:F = 27:24 Commerce M:F = 53:47

   City-wise proportions of graduating students and male:female ratios.

   1. A31.25
   2. B47.47
   3. C45.45
   4. D42.67

   Answer: (C) 45.45

   Explanation

   The Commerce male:female ratio for City F is 16:11. Taking females as the comparison base, the excess is 16 − 11 = 5 parts; 5/11 × 100 = 45.45%. Dividing by 16 would instead calculate the percentage relative to males.

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6. Question 6

   What can be correctly claimed about assessment rubrics?

   A. A tool used to interpret and grade students on the basis of criteria.
   B. A means to increase subjective evaluation by the evaluator.
   C. It ensures transparency and fairness in marking.
   D. It provides an opportunity to set arbitrary standards for moderation.
   E. An innovative evaluation system.

   1. A(A), (C), and (D) only
   2. B(A), (B), and (C) only
   3. C(C), (D), and (B) only
   4. D(A), (C), and (E) only

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

   Explanation

   A rubric states criteria and performance expectations, making assessment more transparent and consistent. It reduces subjectivity rather than increasing it, and its standards should be explicit rather than arbitrary.

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7. Question 7

   Match List-I with List-II.

   List-I	List-II
   A. Macro teaching	I. Pre-service teacher-training technique
   B. Micro teaching	II. Educators divide content and students move between them
   C. Station teaching	III. Two teachers work with one class together
   D. Cooperative teaching	IV. One teacher instructs the entire class for an extended period

   Teaching types and descriptions.

   1. A(A)–(II), (B)–(III), (C)–(I), (D)–(IV)
   2. B(A)–(IV), (B)–(II), (C)–(III), (D)–(I)
   3. C(A)–(IV), (B)–(I), (C)–(II), (D)–(III)
   4. D(A)–(I), (B)–(II), (C)–(III), (D)–(IV)

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

   Explanation

   Macro teaching addresses the whole class; micro teaching is the pre-service training technique; station teaching rotates learners among educators; cooperative teaching has two teachers working together.

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8. Question 8

   What can be ascribed to learning styles?

   A. They describe how a student learns.
   B. They are learners’ strengths and preferences in responding to environmental stimuli.
   C. They describe what a learner learns.
   D. They are behavioural patterns developed for new learning.

   1. A(A) and (B) only
   2. B(A) and (C) only
   3. C(A), (B), and (D) only
   4. D(B), (C), and (D) only

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

   Explanation

   Learning styles concern preferences and patterns in how a learner approaches learning, not the subject matter being learned. Therefore A, B, and D fit, while C does not.

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9. Question 9

   What can be correctly inferred about the traditional teaching method and support system?

   A. The teacher controls the class and teaches using a blackboard.
   B. It is also called the Back-to-Basics system.
   C. The student is responsible for using self-learning tools.
   D. Virtual labs are a teaching tool.
   E. Charts, maps, textbooks and teacher handouts are prime resources.

   1. A(B) and (C) only
   2. B(A), (B), and (D) only
   3. C(D) and (E) only
   4. D(A), (B), and (E) only

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

   Explanation

   Traditional instruction is teacher-led and commonly relies on blackboards, textbooks, charts and handouts. Self-learning tools and virtual laboratories belong more naturally to learner-centred or technology-supported approaches.

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10. Question 10

    What is incorrect about SWAYAM Prabha?

    1. AIt was launched by Sri Abul Kalam Azad.
    2. BSatellite is used to telecast educational content through DTH channels.
    3. CIts web portal is maintained by INFLIBNET, Gujarat.
    4. DIts content is provided by NPTEL, IITs, UGC, CEC and IGNOU.

    Answer: (A) It was launched by Sri Abul Kalam Azad.

    Explanation

    SWAYAM Prabha is a DTH educational initiative supported by national educational institutions. The attribution to Sri Abul Kalam Azad is incorrect; the other statements describe its delivery and content ecosystem.

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11. Question 11

    A researcher tells participants something false about a research session in order to mislead them. The researcher is using:

    1. ANaturalistic observation
    2. BRole playing
    3. CActive deception
    4. DDebriefing

    Answer: (C) Active deception

    Explanation

    Active deception deliberately supplies false information to participants. Ethical research requires that such deception be justified, minimized, and followed by appropriate debriefing.

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12. Question 12

    Which research criterion relates mainly to causality?

    1. ATransferability
    2. BMeasurement validity
    3. CExternal validity
    4. DInternal validity

    Answer: (D) Internal validity

    Explanation

    Internal validity asks whether the observed effect can credibly be attributed to the proposed cause rather than confounding factors. External validity concerns generalisation.

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13. Question 13

    Arrange these data sets in increasing order of coefficient of variation.

    A. x̄ = 8, σ = 1.7
    B. x̄ = 11, σ = 2.2
    C. x̄ = 15, σ = 3.6
    D. x̄ = 16, σ = 3

    1. A(A), (B), (D), (C)
    2. B(B), (D), (C), (A)
    3. C(C), (B), (D), (A)
    4. D(D), (B), (A), (C)

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

    Explanation

    Because coefficient of variation is σ/x̄ × 100, it compares dispersion relative to each data set’s mean. The values are D 18.75%, B 20%, A 21.25%, and C 24%; therefore the increasing order is D, B, A, C.

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14. Question 14

    According to UGC Regulations 2018 on plagiarism, which similarities constitute Level-1 plagiarism?

    A. 12%
    B. 42%
    C. 28%
    D. 56%
    E. 37%

    1. A(A), (B), and (C) only
    2. B(B), (C), and (D) only
    3. C(A), (C), and (E) only
    4. D(B), (C), and (E) only

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

    Explanation

    UGC’s Level-1 plagiarism band is above 10% up to 40%. Thus 12%, 28%, and 37% qualify. Both 42% and 56% exceed that ceiling and therefore fall in higher plagiarism bands.

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15. Question 15

    Match List-I with List-II.

    List-I	List-II
    A. Nominal	I. Satisfaction level
    B. Ordinal	II. Weight
    C. Interval	III. Ethnicity
    D. Ratio	IV. Year of death

    Variable types and examples.

    1. A(A)–(III), (B)–(II), (C)–(I), (D)–(IV)
    2. B(A)–(I), (B)–(III), (C)–(II), (D)–(IV)
    3. C(A)–(III), (B)–(IV), (C)–(I), (D)–(II)
    4. D(A)–(III), (B)–(I), (C)–(IV), (D)–(II)

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

    Explanation

    Nominal data classify ethnicity, ordinal data rank satisfaction, interval data include years with equal intervals but no true zero, and weight is ratio-scale data with a true zero.

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16. Question 16

    Which is the first newspaper published in India in modern times?

    1. AHickey’s Bengal Gazette
    2. BIndia Gazette
    3. CSamvad Kaumudi
    4. DUdant Martand

    Answer: (A) Hickey’s Bengal Gazette

    Explanation

    Hickey’s Bengal Gazette, started by James Augustus Hicky in 1780, is generally recognised as the first printed newspaper in India.

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17. Question 17

    Which cannot be treated as a folk tradition of communication in India?

    1. ACinema
    2. BJatra
    3. CRamleela
    4. DKumbh Mela

    Answer: (A) Cinema

    Explanation

    Jatra, Ramleela and Kumbh Mela are traditional cultural forms or gatherings that transmit shared meanings. Cinema is a modern mass-medium form rather than a folk communication tradition.

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18. Question 18

    Arrange the 5 Ws of communication in proper sequence.

    A. Why
    B. When
    C. What
    D. Who
    E. Where

    1. A(C), (D), (B), (E), and (A)
    2. B(D), (C), (B), (E), and (A)
    3. C(A), (B), (C), (D), and (E)
    4. D(B), (A), (C), (D), and (E)

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

    Explanation

    A common planning sequence begins with who, then what, when, where and why. It identifies the communicator or audience before message, timing, place and purpose.

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19. Question 19

    Three major actions in the process of communication are:

    A. Receiving
    B. Sending
    C. Reviewing
    D. Interpreting
    E. Analysing

    1. A(C), (B), and (E) only
    2. B(A), (C), and (D) only
    3. C(A), (B), and (D) only
    4. D(A), (D), and (E) only

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

    Explanation

    Communication entails sending a message, receiving it, and interpreting it. Reviewing and analysing can support communication but are not the basic three actions.

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20. Question 20

    Match List-I with List-II.

    List-I	List-II
    A. Interviews	I. Group communication
    B. Newspaper	II. Intrapersonal communication
    C. Classroom	III. Interpersonal communication
    D. Meditation	IV. Mass communication

    Applications and communication terms.

    1. A(A)–(I), (B)–(III), (C)–(II), (D)–(IV)
    2. B(A)–(II), (B)–(IV), (C)–(I), (D)–(III)
    3. C(A)–(IV), (B)–(I), (C)–(II), (D)–(III)
    4. D(A)–(III), (B)–(IV), (C)–(I), (D)–(II)

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

    Explanation

    Interviews are interpersonal communication, newspapers are mass communication, a classroom is group communication, and meditation is intrapersonal communication.

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21. Question 21

    A person sold an item at a loss of 7%. Had they sold it at a gain of 7.5%, they would have received ₹87 more. What is the cost price?

    1. A₹500
    2. B₹550
    3. C₹600
    4. D₹650

    Answer: (C) ₹600

    Explanation

    The difference between a 7.5% gain and a 7% loss is 14.5% of the original cost price. Therefore 0.145 × CP = 87 and CP = ₹600. The two selling-price scenarios are compared against the same cost price.

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22. Question 22

    In a 500 m race, A beats B by 20 m and C by 80 m. If B and C run 100 m at the same speeds, by how many metres will B beat C?

    1. A10.0 m
    2. B8.5 m
    3. C15.0 m
    4. D12.5 m

    Answer: (D) 12.5 m

    Explanation

    This race comparison uses the same speeds as before. Because A covers 500 m while B and C cover 480 m and 420 m, B:C speed ratio is 480:420 = 8:7. In the 100 m race B finishes while C covers 87.5 m, leaving 12.5 m.

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23. Question 23

    Arrange the fractions in increasing order.

    A. 12/17
    B. 13/19
    C. 8/11
    D. 16/23

    1. A(B), (D), (A), (C)
    2. B(B), (A), (C), (D)
    3. C(D), (A), (B), (C)
    4. D(D), (B), (C), (A)

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

    Explanation

    Compare the listed fractions by their decimal values: B ≈ .684, D ≈ .696, A ≈ .706 and C ≈ .727. Therefore the increasing order is B, D, A, C; no fraction is larger merely because its numerator is larger.

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24. Question 24

    Identify the numbers that occur in the series 1, 7, 17, 31, 49, …

    A. 74
    B. 95
    C. 97
    D. 127
    E. 161

    1. A(A), (B), (C) only
    2. B(C), (D), (E) only
    3. C(B), (D), (A) only
    4. D(C), (D), (B) only

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

    Explanation

    To identify which numbers occur in the series, inspect the successive differences: 6, 10, 14 and 18 increase by 4 each time. Continuing gives 71, 97, 127 and 161; therefore C, D and E occur, while 74 and 95 do not.

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25. Question 25

    Match List-I with List-II.

    List-I	List-II
    A. QUAKE	I. KTFUL
    B. OFTEN	II. VHQDU
    C. PEACH	III. IEYOU
    D. DRIVE	IV. JUYWR

    Words and unique codes.

    1. A(A)–(I), (B)–(II), (C)–(III), (D)–(IV)
    2. B(A)–(II), (B)–(IV), (C)–(I), (D)–(III)
    3. C(A)–(III), (B)–(I), (C)–(IV), (D)–(II)
    4. D(A)–(IV), (B)–(I), (C)–(II), (D)–(III)

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

    Explanation

    The coded words follow the given unique-code correspondences: QUAKE–JUYWR, OFTEN–KTFUL, PEACH–VHQDU and DRIVE–IEYOU.

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26. Question 26

    Identify the pattern of argument in: “Habits are like a cable. We weave a strand of it every day and soon it cannot be broken.”

    1. ACausal argument
    2. BArgument from analogy
    3. CArgument from authority
    4. DArgument from definition

    Answer: (B) Argument from analogy

    Explanation

    The statement explains habits by comparing them to a cable built strand by strand. Reasoning by a relevant comparison is an argument from analogy.

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27. Question 27

    What can be correctly inferred? “You know how I know animals have souls? Because, on average, the lowest animal is a lot nicer and kinder than most human beings that inhabit this Earth.”

    A. The kindness claim is a premise.
    B. Animals have souls is a premise.
    C. It is not an argument.
    D. The kindness claim is the conclusion.
    E. Animals have souls is the conclusion.

    1. A(B), (C), and (D) only
    2. B(A), (C), and (E) only
    3. C(A) and (E) only
    4. D(A), (D), and (E) only

    Answer: (C) (A) and (E) only

    Explanation

    The kindness claim is offered as a reason, so it is the premise. “Animals have souls” is what the speaker seeks to establish, making it the conclusion.

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28. Question 28

    What can be inferred from: “No musicians are Greeks. All traders are musicians. Therefore, no traders are Greeks.”

    A. It is figure II.
    B. It is an EAE mood.
    C. Greeks is the major term.
    D. The minor term of the conclusion is distributed.
    E. The middle term is undistributed.

    1. A(A), (B), and (C) only
    2. B(B), (C), and (D) only
    3. C(D) and (E) only
    4. D(A), (C), and (E) only

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

    Explanation

    The premises and conclusion have E-A-E form. “Greeks” is the predicate of the conclusion (major term), and “traders,” the subject of the E conclusion, is distributed; the middle term musicians is distributed in the first premise.

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29. Question 29

    Match List-I with List-II.

    List-I	List-II
    A. Anyonyabhava	I. Hare’s horn
    B. Atyantabhava	II. Non-existence of the jar now
    C. Pragabhava	III. This is not that
    D. Pradhvamsabhava	IV. Temporary absence of a thing

    Forms of non-existence and representations.

    1. A(A)–(II), (B)–(IV), (C)–(I), (D)–(III)
    2. B(A)–(I), (B)–(II), (C)–(III), (D)–(IV)
    3. C(A)–(III), (B)–(I), (C)–(IV), (D)–(II)
    4. D(A)–(IV), (B)–(III), (C)–(II), (D)–(I)

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

    Explanation

    Anyonyabhava is mutual non-existence (“this is not that”); atyantabhava is absolute non-existence (hare’s horn); pragabhava is prior absence; pradhvamsabhava is absence after destruction.

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30. Question 30

    Among the following, which is correct in the context of Sabda Pramana?

    1. AFor Samkhya, a word signifies a universal.
    2. BFor Jainas, a word signifies a particular.
    3. CFor old Nyaya, a word symbolizes a universal only.
    4. DFor Vedantins, a word primarily means the class character of individuals.

    Answer: (A) For Samkhya, a word signifies a universal.

    Explanation

    In the Samkhya account, a word primarily conveys the universal. The other statements oversimplify or misstate the relevant schools’ accounts of word meaning.

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31. Question 31

    If (L)ₘ represents a number L in base-M, what is the value of (107)₁₆ + (257)₁₆?

    1. A(762)₁₀
    2. B(862)₁₀
    3. C(962)₁₀
    4. D(662)₁₀

    Answer: (B) (862)₁₀

    Explanation

    Because hexadecimal uses base 16, convert each number before adding: (107)₁₆ = 1×256 + 0×16 + 7 = 263 and (257)₁₆ = 2×256 + 5×16 + 7 = 599. Therefore their decimal sum is 862, or (862)₁₀.

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32. Question 32

    In computer networking, an IPv6 address is how many times larger than an IPv4 address?

    1. A2
    2. B4
    3. C6
    4. D8

    Answer: (B) 4

    Explanation

    IPv4 uses 32 bits and IPv6 uses 128 bits. Since 128/32 = 4, an IPv6 address is four times as large in bit length.

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33. Question 33

    Identify the correct order of these major technological changes from first to fourth generation of computers.

    A. Transistors
    B. Microprocessor
    C. Vacuum tubes
    D. Integrated circuits

    1. A(A), (C), (D), (B)
    2. B(B), (A), (C), (D)
    3. C(D), (A), (C), (B)
    4. D(C), (A), (D), (B)

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

    Explanation

    Computer generations progressed from vacuum tubes to transistors, integrated circuits, and then microprocessors. Thus the order is C, A, D, B.

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34. Question 34

    To record voice narration for a Microsoft PowerPoint presentation, a computer must have:

    A. Sound card
    B. Internet connection
    C. Speakers
    D. External video port
    E. Microphone

    1. A(A), (C), and (E) only
    2. B(B) and (D) only
    3. C(A) and (E) only
    4. D(B), (C), and (D) only

    Answer: (C) (A) and (E) only

    Explanation

    Voice narration requires an audio input device (microphone) and audio hardware such as a sound card. Internet access, speakers, and an external video port are not essential to recording.

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35. Question 35

    Match List-I with List-II.

    List-I	List-II
    A. Blogs	I. Allows users to create and edit web pages using a browser
    B. Podcasts	II. A type of bookmarking using text labels
    C. Tagging	III. A series of digital media files
    D. Wikis	IV. Personal internet journals

    Internet terms and descriptions.

    1. A(A)–(I), (B)–(II), (C)–(III), (D)–(IV)
    2. B(A)–(IV), (B)–(III), (C)–(II), (D)–(I)
    3. C(A)–(III), (B)–(IV), (C)–(I), (D)–(II)
    4. D(A)–(II), (B)–(I), (C)–(IV), (D)–(III)

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

    Explanation

    Blogs are personal internet journals; podcasts are series of digital media files; tagging describes content with text labels; wikis let users create and edit web pages in a browser.

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36. Question 36

    The amount of biomass produced by plants through photosynthesis minus respiration losses in a given time period is called:

    1. AGross Primary Productivity
    2. BNet Primary Productivity
    3. CGross Ecological Productivity
    4. DNet Secondary Productivity

    Answer: (B) Net Primary Productivity

    Explanation

    Gross primary productivity is all photosynthetic production. After subtracting plant respiration, the remainder available for growth and consumers is net primary productivity.

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37. Question 37

    Which fuel produces the least soot particles during combustion?

    1. APetrol
    2. BKerosene
    3. CDiesel
    4. DCompressed Natural Gas (CNG)

    Answer: (D) Compressed Natural Gas (CNG)

    Explanation

    CNG burns more cleanly than petrol, kerosene and diesel and produces substantially less particulate soot because it is a gaseous, low-carbon fuel.

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38. Question 38

    Arrange the states in ascending order of area under forest cover.

    A. Arunachal Pradesh
    B. Chhattisgarh
    C. Maharashtra
    D. Odisha
    E. Madhya Pradesh

    1. A(A), (C), (D), (B), (E)
    2. B(C), (D), (B), (A), (E)
    3. C(B), (A), (C), (D), (E)
    4. D(D), (B), (A), (E), (C)

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

    Explanation

    Among these, Maharashtra has the least listed forest-cover area, followed by Odisha and Chhattisgarh; Arunachal Pradesh and then Madhya Pradesh have larger areas. The order is C, D, B, A, E.

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39. Question 39

    Eye irritation during smog is mostly caused by:

    A. Formaldehyde
    B. Peroxyacetyl nitrate
    C. Ozone
    D. Acrolein

    1. A(A), (B), and (C) only
    2. B(A), (B), and (D) only
    3. C(C), (D), and (A) only
    4. D(B), (C), and (D) only

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

    Explanation

    Photochemical smog contains irritating oxidants and aldehydes. Peroxyacetyl nitrate, formaldehyde and acrolein are prominent lacrimators; the selected combination captures these principal eye irritants.

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40. Question 40

    Match List-I with List-II.

    List-I	List-II
    A. Sendai Framework	I. Equality between men and women
    B. Monterrey Consensus	II. Our Common Future
    C. Beijing Declaration	III. Disaster Risk Reduction
    D. Brundtland Report	IV. Financing for development

    International treaties and key objectives.

    1. A(A)–(III), (B)–(IV), (C)–(I), (D)–(II)
    2. B(A)–(IV), (B)–(II), (C)–(I), (D)–(III)
    3. C(A)–(I), (B)–(III), (C)–(II), (D)–(IV)
    4. D(A)–(II), (B)–(I), (C)–(IV), (D)–(III)

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

    Explanation

    The Sendai Framework concerns disaster-risk reduction; the Monterrey Consensus concerns financing for development; the Beijing Declaration advances gender equality; and the Brundtland Report is associated with “Our Common Future.”

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41. Question 41

    Arrange these ancient Indian institutes from east to west.

    A. Takshashila
    B. Nalanda
    C. Vikramshila
    D. Shardapeeth

    1. A(A), (D), (C), (B)
    2. B(B), (C), (A), (D)
    3. C(D), (A), (B), (C)
    4. D(C), (B), (D), (A)

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

    Explanation

    Nalanda and Vikramshila are in Bihar, with Vikramshila east of Nalanda; Takshashila is in present-day Pakistan and Shardapeeth farther northwest in Kashmir. Hence B, C, A, D.

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42. Question 42

    Which universities were set up in 1916?

    A. Osmania University
    B. S.N.D.T. Women’s University
    C. Mysore University
    D. Patna University
    E. Banaras Hindu University

    1. A(A), (B), and (C) only
    2. B(B), (C), and (E) only
    3. C(B), (C), and (D) only
    4. D(A), (D), and (E) only

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

    Explanation

    S.N.D.T. Women’s University, the University of Mysore and Banaras Hindu University were established in 1916. Osmania and Patna universities were established later.

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43. Question 43

    Match List-I with List-II.

    List-I	List-II
    A. Women’s Education	I. 28
    B. Education of Minorities	II. 21-A
    C. Secular Education	III. 15 (Part III)
    D. Right to Education	IV. 29 (Part III)

    Constitutional provisions of education and articles.

    1. A(A)–(III), (B)–(IV), (C)–(I), (D)–(II)
    2. B(A)–(I), (B)–(II), (C)–(IV), (D)–(III)
    3. C(A)–(III), (B)–(IV), (C)–(I), (D)–(II)
    4. D(A)–(II), (B)–(I), (C)–(III), (D)–(IV)

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

    Explanation

    Article 15(3) supports special provisions for women; Article 29 protects minority interests; Article 28 concerns religious instruction in certain educational institutions; and Article 21A provides the right to education.

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44. Question 44

    Which is known as the “One Nation One Student ID” card?

    1. AAadhaar card
    2. BPAN card
    3. CAPAAR card
    4. DVoter’s ID card

    Answer: (C) APAAR card

    Explanation

    APAAR (Automated Permanent Academic Account Registry) provides a lifelong academic identity under the “One Nation, One Student ID” initiative. Aadhaar, PAN and voter IDs identify citizens for other purposes, not their academic records.

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45. Question 45

    According to the National Credit Framework (NCrF), the credit level earned after a doctoral degree (Ph.D.) is:

    1. A7
    2. B8
    3. C9
    4. D10

    Answer: (B) 8

    Explanation

    The NCrF places the doctoral qualification at level 8, following the postgraduate level in the framework’s progression. Therefore level 7 is not the terminal doctoral level, and levels 9 and 10 are not assigned here.

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46. Question 46

    Whereas the festival of Lord Jagannath is celebrated in many places in India and abroad, it is primarily associated with which city?

    Odisha has colourful fairs and festivals and are as numerous as the days of the year. The sacred journey of Lord Jagannath with brother Balabhadra and sister Subhadra from the Jagannath Temple to the divine shrine called the Gundicha Temple is known as the Car Festival or the Rath Yatra. On this occasion, the deities adorn three chariots named Nandigosha, Taladhwaja and Darpadalana respectively. They stay in the divine shrine for nine days and then return to the Jagannath Temple. This journey is called Bahuda Jatra. This grand festival of Odisha is visited by lakhs of devotees from India and abroad who pull the sacred chariots from the Jagannath Temple to the Gundicha Temple and back on a broad road called Badadanda. The festival begins from the second day of Shukla Paksha (bright half) in the month of Ashadha (June-July) every year. This festival of Lord Jagannath, who is worshipped as the Lord of the Universe, is celebrated not only in Puri, but also in many other places in India and abroad.

    1. ABhubaneswar
    2. BPuri
    3. CCuttack
    4. DKonark

    Answer: (B) Puri

    Explanation

    The passage identifies the Jagannath Temple and the Rath Yatra tradition with Puri, Odisha. The festival’s wider celebration does not change its primary association.

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47. Question 47

    What are the names of the deities who accompany Lord Jagannath in the Rath Yatra?

    Odisha has colourful fairs and festivals and are as numerous as the days of the year. The sacred journey of Lord Jagannath with brother Balabhadra and sister Subhadra from the Jagannath Temple to the divine shrine called the Gundicha Temple is known as the Car Festival or the Rath Yatra. On this occasion, the deities adorn three chariots named Nandigosha, Taladhwaja and Darpadalana respectively. They stay in the divine shrine for nine days and then return to the Jagannath Temple. This journey is called Bahuda Jatra. This grand festival of Odisha is visited by lakhs of devotees from India and abroad who pull the sacred chariots from the Jagannath Temple to the Gundicha Temple and back on a broad road called Badadanda. The festival begins from the second day of Shukla Paksha (bright half) in the month of Ashadha (June-July) every year. This festival of Lord Jagannath, who is worshipped as the Lord of the Universe, is celebrated not only in Puri, but also in many other places in India and abroad.

    1. ASubhadra and Balabhadra
    2. BBalabhadra and Shiva
    3. CSubhadra and Hidimba
    4. DBalabhadra and Hidimba

    Answer: (A) Subhadra and Balabhadra

    Explanation

    The passage explicitly names Lord Jagannath’s brother Balabhadra and sister Subhadra as the accompanying deities. Shiva and Hidimba are not named in the passage, so each alternative containing either one is excluded.

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48. Question 48

    The sacred journey of Lord Jagannath is traditionally undertaken onward to which divine shrine?

    Odisha has colourful fairs and festivals and are as numerous as the days of the year. The sacred journey of Lord Jagannath with brother Balabhadra and sister Subhadra from the Jagannath Temple to the divine shrine called the Gundicha Temple is known as the Car Festival or the Rath Yatra. On this occasion, the deities adorn three chariots named Nandigosha, Taladhwaja and Darpadalana respectively. They stay in the divine shrine for nine days and then return to the Jagannath Temple. This journey is called Bahuda Jatra. This grand festival of Odisha is visited by lakhs of devotees from India and abroad who pull the sacred chariots from the Jagannath Temple to the Gundicha Temple and back on a broad road called Badadanda. The festival begins from the second day of Shukla Paksha (bright half) in the month of Ashadha (June-July) every year. This festival of Lord Jagannath, who is worshipped as the Lord of the Universe, is celebrated not only in Puri, but also in many other places in India and abroad.

    1. AJagannath Temple
    2. BSubhadra Temple
    3. CBalabhadra Temple
    4. DGundicha Temple

    Answer: (D) Gundicha Temple

    Explanation

    The passage says the outward journey runs from the Jagannath Temple to the Gundicha Temple. The later return journey is Bahuda Jatra.

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49. Question 49

    The Car Festival is conducted on a broad road called:

    Odisha has colourful fairs and festivals and are as numerous as the days of the year. The sacred journey of Lord Jagannath with brother Balabhadra and sister Subhadra from the Jagannath Temple to the divine shrine called the Gundicha Temple is known as the Car Festival or the Rath Yatra. On this occasion, the deities adorn three chariots named Nandigosha, Taladhwaja and Darpadalana respectively. They stay in the divine shrine for nine days and then return to the Jagannath Temple. This journey is called Bahuda Jatra. This grand festival of Odisha is visited by lakhs of devotees from India and abroad who pull the sacred chariots from the Jagannath Temple to the Gundicha Temple and back on a broad road called Badadanda. The festival begins from the second day of Shukla Paksha (bright half) in the month of Ashadha (June-July) every year. This festival of Lord Jagannath, who is worshipped as the Lord of the Universe, is celebrated not only in Puri, but also in many other places in India and abroad.

    1. ADarpadalana
    2. BAshadha
    3. CBadadanda
    4. DTaladhwaja

    Answer: (C) Badadanda

    Explanation

    Badadanda is the broad road named in the passage for pulling the chariots. Darpadalana and Taladhwaja are chariot names, while Ashadha is a month.

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50. Question 50

    What is the name assigned to the journey in which the deities return to the Jagannath Temple after nine days?

    Odisha has colourful fairs and festivals and are as numerous as the days of the year. The sacred journey of Lord Jagannath with brother Balabhadra and sister Subhadra from the Jagannath Temple to the divine shrine called the Gundicha Temple is known as the Car Festival or the Rath Yatra. On this occasion, the deities adorn three chariots named Nandigosha, Taladhwaja and Darpadalana respectively. They stay in the divine shrine for nine days and then return to the Jagannath Temple. This journey is called Bahuda Jatra. This grand festival of Odisha is visited by lakhs of devotees from India and abroad who pull the sacred chariots from the Jagannath Temple to the Gundicha Temple and back on a broad road called Badadanda. The festival begins from the second day of Shukla Paksha (bright half) in the month of Ashadha (June-July) every year. This festival of Lord Jagannath, who is worshipped as the Lord of the Universe, is celebrated not only in Puri, but also in many other places in India and abroad.

    1. ARath Yatra
    2. BWapsi Yatra
    3. CMaha Yatra
    4. DBahuda Jatra

    Answer: (D) Bahuda Jatra

    Explanation

    The passage calls the return journey after the nine-day stay Bahuda Jatra. Rath Yatra refers to the wider car-festival journey.

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100 questions with the answer key and explanations.

1. Question 1

   Consider the grammar:

   S → A
   A → $B$ | id
   B → B, A | A

   If I₀ = CLOSURE({[S → .A]}), how many items are in GOTO(I₀, $)?

   1. A3
   2. B4
   3. C5
   4. D6

   Answer: (C) 5

   Explanation

   GOTO(I₀, $) begins with A → $.B. Its closure adds B → .B, A, B → .A, A → .$B$, and A → .id, giving five items in total. Therefore option C is correct.

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2. Question 2

   In a schema R(A, B, C, D, E, F, G, H), each field of R
   contains only atomic values.
   F= {CH→G, A →BC, B→CFH, E→A, F→EG} is a set
   of functional dependencies F+ is exactly the set
   of FDs that holds R. The relation R is :

   1. AIn 1NF, but not in 2 NF
   2. BIn 2 NF , but not in 3 NF
   3. CIn 3NF, but not in BCNF
   4. DIn BCNF

   Answer: (A) In 1NF, but not in 2 NF

   Explanation

   D appears on no right-hand side, so every candidate key must contain it; taking closures gives AD, BD, ED and FD as the candidate keys. The prime attributes are therefore A, B, D, E and F, and the non-prime ones are C, G and H. But A → BC makes the non-prime attribute C depend on A alone, a proper subset of the key AD. That partial dependency breaks 2NF, while the relation is already in 1NF because every field is atomic. Option A is correct.

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3. Question 3

   Which of the following statements about
   pointers in C are TRUE.
   A. Pointers can be used to access array elements
   B. Pointers can store the address of another
   pointer
   C. Pointers are automatically dereferenced in
   expression
   D. Pointers cannot be used to access structure
   members

   1. A(A) and (C) Only
   2. B(A) and (B) Only
   3. C(B) and (C) Only
   4. D(C) and (D) Only

   Answer: (B) (A) and (B) Only

   Explanation

   (A) is true: array elements are reachable through pointer arithmetic such as *(arr + i). (B) is true: a pointer is itself a variable with an address, so a pointer to a pointer is legal. (C) is false — dereferencing is explicit, through the * operator. (D) is false — structure members are reachable with p->member. Only (A) and (B) hold, so option B is correct.

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4. Question 4

   In a token ring network the transmission speed is 10⁷ bps and the propagation speed is 200 m/μs. The 1-bit delay in this network is equivalent to ________________

   1. A20 m
   2. B30 m
   3. C50 m
   4. D40 m

   Answer: (A) 20 m

   Explanation

   The transmission speed is 10⁷ bps, so one bit occupies 1/10⁷ s = 0.1 μs on the wire. In that time the signal advances 0.1 μs × 200 m/μs = 20 m. The 1-bit delay is therefore equivalent to 20 metres of cable, so option A is correct.

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5. Question 5

   Which of the following statements are TRUE
   about mutual exclusion in concurrent
   programming ?
   A. Mutual exclusion ensures that only one
   process can be in a critical section at any given
   time.
   B. Mutual exclusion are designed to prevent
   conflicts and ensure that only one process can
   access shared resources at a time.
   C. Mutual exclusion can use various algorithms
   to ensure that processes do not enter the critical
   section simultaneously.
   D. Mutual exclusion allows multiple processes to
   access the critical section simultaneously to
   improve performance.

   1. A(A), (B), (C) Only
   2. B(B), (C), (D) Only
   3. C(B), (D), (A) Only
   4. D(A), (C), (D) Only

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

   Explanation

   Statements (A), (B) and (C) all restate the same guarantee — at most one process inside the critical section at a time, enforced by locks or algorithms such as Peterson’s. (D) contradicts the definition outright, since allowing simultaneous access is exactly what mutual exclusion prevents. Only (A), (B) and (C), so option A is correct.

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6. Question 6

   Consider the function in C code :

   Cal(a, b) {
   if( b!=1 ) {
   if( a!=1 ) {
   printf("*");
   Cal(a/2,b);
   }
   else {
   b=b-1;
   Cal(10,b);
   }
   }
   }
   How many times * is going to be printed, if the
   function is called with Cal(10, 10); ?
   

   1. A25
   2. B23
   3. C24
   4. D27

   Answer: (D) 27

   Explanation

   Each pass halves a by integer division, 10 → 5 → 2 → 1, printing * on the three calls where a ≠ 1. Reaching a = 1 decrements b and restarts with a = 10. That cycle runs for b = 10 down to b = 2, so nine times, giving 9 × 3 = 27 stars. Option D is correct.

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7. Question 7

   Consider the following relations on {1, 2, 3, 4}. Which of the following relations are reflexive?

   A. R₁ = {(1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)}
   B. R₂ = {(1,1),(1,2),(2,1)}
   C. R₃ = {(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,1),(4,4)}
   D. R₄ = {(2,1),(3,1),(3,2),(4,1),(4,2),(4,3)}
   E. R₅ = {(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)}

   1. A(A), (B) and (C) Only
   2. B(A), (D) and (E) Only
   3. C(D) and (E) Only
   4. D(C) and (E) Only

   Answer: (D) (C) and (E) Only

   Explanation

   A relation on {1, 2, 3, 4} is reflexive only if it contains all four self-pairs (1,1), (2,2), (3,3) and (4,4). R₁ omits (3,3); R₂ omits (2,2), (3,3) and (4,4); R₄ contains no self-pair at all. R₃ and R₅ each contain all four, so only (C) and (E) are reflexive — option D.

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8. Question 8

   Consider a relation schema R=(U, V, W, X, Y, Z), on
   which the following functional dependencies
   hold
   {U→V, VW→X, Y→W; X→U}
   The candidate keys of R are :

   1. AUY, VY
   2. BUY, VY, XY
   3. CUYZ, VYZ, VWZ
   4. DUYZ, VYZ, XYZ

   Answer: (D) UYZ, VYZ, XYZ

   Explanation

   Z appears on no right-hand side, so it must belong to every candidate key. Taking closures, UY⁺ gains V from U → V, W from Y → W and X from VW → X; VY⁺ and XY⁺ reach the same set. Each covers everything except Z, so appending Z gives the candidate keys UYZ, VYZ and XYZ. Option D is correct.

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9. Question 9

   Which of the following is correct way to declare a
   functional pointer in C ?

   1. Aint *func (int, int);
   2. Bint (*func) (int, int);
   3. Cint (func*) (int, int);
   4. Dint func (int, int);

   Answer: (B) int (*func) (int, int);

   Explanation

   In int (func)(int, int); the parentheses bind * to func first, so func is a pointer to a function taking two ints and returning int. Option A declares a function returning int, and options C and D place the asterisk where the syntax does not allow it. Option B is correct.

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10. Question 10

    Arrange the following steps of Feature Drive
    Development (FDD) process in the correct
    sequence :
    A. Develop an overall model
    B. Build by feature
    C. Plan by feature
    D. Design by feature
    E. Build a feature list

    1. A(A), (C), (B), (E), (D)
    2. B(A), (E), (C), (D), (B)
    3. C(B),(A),(D),(E),(C)
    4. D(A), (B), (C), (E), (D)

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

    Explanation

    Feature Driven Development runs: develop an overall model (A), build a feature list (E), plan by feature (C), design by feature (D), then build by feature (B). The model and the feature list must both exist before planning, and design always precedes build. That order is option B.

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11. Question 11

    Consider the given number (45) y where y is the
    base of the number. Some of the possible values
    of y are given below.
    A. 5
    B. 6
    C. 7
    D. 8

    1. A(A), (B) and (C) Only
    2. B(B), (C) and (D) Only
    3. C(A), (C) and (D) Only
    4. D(A), (B), (C) and (D)

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

    Explanation

    In base y every digit used must be smaller than y. The numeral 45 uses the digit 5, so y must exceed 5. Base 5 is therefore invalid, while bases 6, 7 and 8 all admit the digits 4 and 5. The valid values are (B), (C) and (D) — option B.

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12. Question 12

    Consider the following C program :

    #include<stdio.h>
    int main() {
    int x[] = {2,4,6,8,10};
    int a, b=0, *y=x+4;
    for( a=0; a<5; a++) {
    b = b +(*y-a) - *(y-a);
    }
    printf("%d\n",b);
    return 0;
    }
    What will be the output of the above C program
    ?
    

    1. A4
    2. B6
    3. C8
    4. D10

    Answer: (D) 10

    Explanation

    y points at x[4] = 10, and *y stays 10 throughout. Each iteration adds (10 − a) − x[4−a]: for a = 0 to 4 that is 10−10 = 0, 9−8 = 1, 8−6 = 2, 7−4 = 3 and 6−2 = 4. The running total is 0+1+2+3+4 = 10, so the program prints 10 — option D.

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13. Question 13

    Match List-I with List-II.

    List-I	List-II
    (A) The activation function	(I) is called the delta rule.
    (B) The learning method of perceptron	(II) is one of the key components of the perceptron, as in the most common neural network architecture.
    (C) Areas of application of artificial neural networks	(III) is always boolean, like a switch.
    (D) The output of the perceptron	(IV) system identification and control.

    1. A(A)-(II), (B)-(IV), (C)-(III), (D)-(I)
    2. B(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
    3. C(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
    4. D(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

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

    Explanation

    The activation function is one of the key components of a perceptron, so (A)-(II). A perceptron learns by the delta rule, giving (B)-(I). Artificial neural networks are applied to tasks such as system identification and control, so (C)-(IV). A perceptron emits a boolean, switch-like output, so (D)-(III). Only option C carries all four pairings.

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14. Question 14

    Match List-I with List-II.

    List-I	List-II
    (A) Representation of bits	(I) Transport layer
    (B) Physical Address	(II) Physical layer
    (C) Logical Address	(III) Network layer
    (D) Segmentation and reassembly	(IV) Data link layer

    1. A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
    2. B(A)-(II), (B)-(IV), (C)-(III), (D)-(I)
    3. C(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
    4. D(A)-(I), (B)-(IV), (C)-(III), (D)-(II)

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

    Explanation

    Bits are represented on the medium at the physical layer, so (A)-(II). Physical (MAC) addressing belongs to the data-link layer, giving (B)-(IV). Logical (IP) addressing belongs to the network layer, so (C)-(III). Segmentation and reassembly is a transport-layer duty, so (D)-(I). Option B matches all four.

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15. Question 15

    Which of the following are Agile Process Models
    ?
    A. Extreme Programming (XP)
    B. Waterfall
    C. Scrum
    D. Spiral
    E. Incremental

    1. A(A) and (C) Only
    2. B(B) and (C) Only
    3. C(A) and (D) Only
    4. D(B) and (E) Only

    Answer: (A) (A) and (C) Only

    Explanation

    Extreme Programming and Scrum are both Agile methods, built on short iterations and continuous customer feedback. Waterfall is strictly sequential; Spiral is risk-driven and iterative but not Agile; the Incremental model predates the Agile movement. Only (A) and (C) qualify, so option A is correct.

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16. Question 16

    Which of the following statements are TRUE
    about Privileged Instructions ?
    A. It can only be executed by the Operating
    System kernel and not by user applications.
    B. It is designed to perform operations that can
    directly affect the hardware or system state such
    as I/O operations or changing memory
    management setting.
    C. User applications can execute privileged
    instructions if they have to correct permissions,
    set by the Operating System.
    D. It usually executed in user mode to ensure
    the safety and security of the system.

    1. A(A) and (B) Only
    2. B(A), (B) and (C) Only
    3. C(B) and (C) Only
    4. D(B), (C) and (D) Only

    Answer: (A) (A) and (B) Only

    Explanation

    (A) is true — privileged instructions execute only in kernel (supervisor) mode. (B) is true — they cover exactly the operations touching hardware or system state, such as I/O and memory-management setup. (C) is false: no permission setting lets a user-mode application execute one directly; it must trap into the kernel. (D) is false: they run in kernel mode, not user mode. Only (A) and (B) — option A.

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17. Question 17

    Consider a CSMA/CD network that transmits
    data at the rate of 100 Mbps over a 1 Kilometre
    cable with no repeater. If the minimum frame size
    required for this network is 1250 bytes, what is
    the signal speed (km/sec) in the cable.

    1. A8000
    2. B16000
    3. C12000
    4. D20000

    Answer: (D) 20000

    Explanation

    The minimum frame of 1250 bytes is 10,000 bits, which at 100 Mbps takes 10,000/10⁸ = 10⁻⁴ s to transmit. Collision detection requires that time to cover the round trip of 2 × 1 km. The signal speed is therefore 2 km / 10⁻⁴ s = 20,000 km/s — option D.

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18. Question 18

    Consider the Grammar:
    T → Qx
    Q → RS
    R → y ∣ ϵ
    S → z ∣ ϵ
    Here x, y, z are terminals and T, Q, R, S are non-
    terminals.
    What will be the follow set of the non-terminal R?

    1. A{x,y}
    2. B{y,z}
    3. C{z,x}
    4. D{ϵ}

    Answer: (C) {z,x}

    Explanation

    In Q → RS, whatever can begin S follows R, contributing z. Because S can also derive ε, anything following Q follows R as well, and T → Qx supplies x. Hence FOLLOW(R) = {z, x}, which is option C.

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19. Question 19

    Host A (on TCP/IPv4 Networks) send an IP
    Datagram D to host B (also on TCP/IPv4 network).
    Assume that no error occurred during
    transmission of D, when D reaches B. Which of
    following header field may differ from that of
    original datagram D ?
    A. TTL
    B. Checksum
    C. Fragment offset
    D. Source IP
    E. Destination IP

    1. A(A) and (B) Only
    2. B(A), (B) and (C) Only
    3. C(A), (B), (C) and (D) Only
    4. D(A), (B), (C) and (E) Only

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

    Explanation

    TTL is decremented by every router, so (A) changes. The header checksum covers TTL and must be recomputed at each hop, so (B) changes. Fragment offset changes whenever the datagram is fragmented in transit, so (C) may differ. Source and destination IP are never rewritten by ordinary routing, so (D) and (E) stay fixed. Only (A), (B) and (C) — option B.

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20. Question 20

    Which of the following properties correctly
    describe a Regular Grammar ?
    A. All production rules are of the form A→xB or
    A→x, where A and B are non terminal symbols
    and x is a terminal symbol.
    B. Regular grammars are more powerful than
    context-free grammars and can express any type
    of language.
    C. There is a direct correspondence between
    regular grammar and finite automata.
    D. Regular grammars can generate languages
    that are not recognised by any type of automata.

    1. A(A) and (B) Only
    2. B(B) and (C) Only
    3. C(C) and (D) Only
    4. D(A) and (C) Only

    Answer: (D) (A) and (C) Only

    Explanation

    (A) is true: right-linear rules of the form A → xB or A → x define a regular grammar. (C) is true: every regular grammar has an equivalent finite automaton and vice versa. (B) is false — regular grammars are strictly weaker than context-free ones. (D) is false — the languages they generate are exactly those finite automata accept. Only (A) and (C), so option D.

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21. Question 21

    Consider a schema R(P, Q, R, S) and the following
    functional dependencies P → Q, Q → R, R → S, S
    → Q. Then decomposition of R (P, Q), R (Q, R)
    and R 3(Q, S) is :

    1. ADependency Preserving and lossless join
    2. BLossless Join but not dependency preserving
    3. CDependency preserving but not lossless Join
    4. DNot dependency preserving and not lossless join

    Answer: (A) Dependency Preserving and lossless join

    Explanation

    The sub-relations share Q, and Q determines R and, through R → S, also S, so rejoining them reconstructs R without spurious tuples: the decomposition is lossless. P → Q survives in R₁(P, Q), Q → R in R₂(Q, R), and R → S together with S → Q is recoverable from R₂ and R₃(Q, S). Every dependency remains enforceable, so the decomposition is both dependency preserving and lossless — option A.

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22. Question 22

    Arrange the following phases of the Agile
    process in the correct sequence :
    A. Design
    B. Release
    C. Testing
    D. Development
    E. Planning

    1. A(A), (B), (D), (E), (C)
    2. B(A), (D), (E), (B), (C)
    3. C(E), (A), (D), (C), (B)
    4. D(E), (D), (A), (C), (B)

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

    Explanation

    Even inside a short iteration the work still flows planning (E) → design (A) → development (D) → testing (C) → release (B). Requirements are agreed before design, code precedes its tests, and release comes last. That order is option C.

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23. Question 23

    Which of the following are TRUE about
    constructors in C++ ?
    A. A constructor can be overloaded.
    B. A constructor does not have a return type.
    C. A constructor must be declared as a friend
    function.
    D. A constructor is called when an object is
    destroyed.

    1. A(B), (C), (D) Only
    2. B(B), (C) Only
    3. C(A), (B), (C) Only
    4. D(A), (B) Only

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

    Explanation

    (A) is true — constructors can be overloaded on their parameter lists. (B) is true — a constructor has no return type, not even void. (C) is false: a constructor is an ordinary member function and is never required to be a friend. (D) is false: destruction invokes the destructor, not the constructor. Only (A) and (B), so option D.

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24. Question 24

    Let P be "It is hot day" and q be "The
    temperature is 48°C". Write in simple sentences
    the meaning of ¬p∧¬q.

    1. AIt is hot day or temperature is 48°C
    2. BIt is cold day or temperature is 48°C
    3. CIt is neither a hot day nor temperature is 48°C
    4. DIt is not a hot day

    Answer: (C) It is neither a hot day nor temperature is 48°C

    Explanation

    ¬p reads “it is not a hot day” and ¬q reads “the temperature is not 48°C”. The conjunction asserts both at once, which in plain English is “it is neither a hot day nor is the temperature 48°C”. Option C says exactly that; options A and B use “or”, and option D drops the second conjunct.

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25. Question 25

    If a software project has 2000 lines of code
    (LOC) and the average productivity rate is 10
    LOC per person-hour, how many person-hours
    are required to complete the project?

    1. A100
    2. B150
    3. C200
    4. D250

    Answer: (C) 200

    Explanation

    Dividing total size by the productivity rate gives 2000 LOC ÷ 10 LOC per person-hour = 200 person-hours. Option C is correct.

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26. Question 26

    Arrange the following steps in proper sequence
    involved in a Genetic Algorithm :
    A. Selection
    B. Initialization
    C. Crossover
    D. Mutation
    E. Evaluation

    1. A(A), (B), (C), (D), (E)
    2. B(E), (A), (B), (D), (C)
    3. C(B), (E), (A), (C), (D)
    4. D(A), (C), (B), (D), (E)

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

    Explanation

    A genetic algorithm initialises a population (B), evaluates each individual’s fitness (E), selects parents according to that fitness (A), recombines them by crossover (C), and finally perturbs the offspring by mutation (D). Selection needs fitness scores, so evaluation must come before it. That order is option C.

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27. Question 27

    #include<stdio.h>
    void f1(char *x, char *y) {
    char *t1;
    t1 = x;
    x = y;
    y = t1;
    }
    void f2(char **x, char **y) {
    char *t1;
    t1 = *x;
    *x = *y;
    *y = t1;
    }
    int main() {
    char *a = "ONE", *b = "TWO";
    f1(a, b);
    printf("%s %s ", a, b);
    f2(&a, &b);
    printf("%s %s", a, b);
    return 0;
    }
    

    What is the output of the program?

    1. AONE TWO TWO ONE
    2. BTWO ONE ONE TWO
    3. CONE TWO ONE TWO
    4. DTWO ONE TWO ONE

    Answer: (A) ONE TWO TWO ONE

    Explanation

    f1 receives copies of the two pointers, so swapping them inside the function leaves a and b untouched and the first printf prints ONE TWO. f2 receives the addresses of a and b, so swapping *x and *y rewrites the caller’s own pointers and the second printf prints TWO ONE. The full output is ONE TWO TWO ONE — option A.

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28. Question 28

    What is SQL primarily used for in the context of
    relational databases ?

    1. ATo design user interfaces
    2. BTo create and manipulate databases
    3. CTo display data on web pages
    4. DTo format printed reports

    Answer: (B) To create and manipulate databases

    Explanation

    SQL is the standard language for defining and manipulating relational data: creating tables, inserting, updating and deleting rows, and querying with SELECT. User interfaces, page rendering and report formatting are the work of other tools. Option B is correct.

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29. Question 29

    Which of the following C++ statements correctly
    declares an abstract class ?

    1. Aclass A { virtual void show ()=0 ; };
    2. Bclass A { void show ()=0 ; };
    3. Cclass A { void show() { } ; } ;
    4. Dclass A { show ()=0; };

    Answer: (A) class A { virtual void show ()=0 ; };

    Explanation

    A C++ class is abstract when it declares at least one pure virtual function, written virtual void show() = 0;. Option B omits virtual, so = 0 is not valid there; option C defines an ordinary member; option D has neither a return type nor virtual. Option A is correct.

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30. Question 30

    Arrange the following recurrence relations in increasing order of their time complexity:

    A. T(n) = T(n/2) + 1
    B. T(n) = 2T(n/2) + n
    C. T(n) = 3T(n/3) + n
    D. T(n) = 2T(n/2) + √n
    E. T(n) = T(n−1) + 1

    Choose the correct answer from the options given below:

    1. A(E), (A), (B), (D), (C)
    2. B(A), (E), (D), (B), (C)
    3. C(E), (A), (D), (B), (C)
    4. D(A), (B), (D), (E), (C)

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

    Explanation

    Solving each relation: (A) T(n/2)+1 is Θ(log n); (E) T(n−1)+1 is Θ(n); (D) 2T(n/2)+√n is Θ(n); (B) 2T(n/2)+n is Θ(n log n); and (C) 3T(n/3)+n is also Θ(n log n). Ordering log n < n < n log n gives (A), (E), (D), (B), (C) — option B.

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31. Question 31

    How does a relational database ensure data
    integrity ?

    1. ABy encrypting all data stored
    2. BBy enforcing rules defined in the schema
    3. CBy compressing data for efficient storage
    4. DBy allowing unrestricted access to all users

    Answer: (B) By enforcing rules defined in the schema

    Explanation

    Relational databases protect integrity through schema-level constraints — primary keys, foreign keys, unique, not-null and check constraints — which the engine enforces on every write. Encryption protects confidentiality, compression saves space, and unrestricted access would weaken integrity rather than ensure it. Option B is correct.

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32. Question 32

    Match List-I with List-II.

    List-I (IP Address)	List-II (Class)
    (A) 10.20.30.40	(I) Class E
    (B) 210.20.30.3	(II) Class B
    (C) 180.30.100.10	(III) Class A
    (D) 252.5.15.11	(IV) Class C

    1. A(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
    2. B(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
    3. C(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
    4. D(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

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

    Explanation

    The first octet fixes the class. 10 falls in 1-126, so 10.20.30.40 is Class A and (A)-(III). 210 falls in 192-223, so 210.20.30.3 is Class C and (B)-(IV). 180 falls in 128-191, so 180.30.100.10 is Class B and (C)-(II). 252 falls in 240-255, so 252.5.15.11 is Class E and (D)-(I). Option D is correct.

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33. Question 33

    Find the correct sequence of the storage devices
    in descending order based on their access time.
    A. Registers
    B. Magnetic Disk
    C. Magnetic Tapes
    D. Main memory
    E. Optical Disk

    1. A(B), (A), (D), (C), (E)
    2. B(B), (A), (E), (D), (C)
    3. C(C), (B), (A), (E), (D)
    4. D(C), (E), (B), (D), (A)

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

    Explanation

    Descending access time means slowest first: magnetic tape (seconds), then optical disk, then magnetic disk (milliseconds), then main memory (nanoseconds), and finally registers, which are fastest. That is (C), (E), (B), (D), (A) — option D.

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34. Question 34

    A graph G with number of vertices greater and
    equal than three i.e. (n ≥ 3) is a Hamiltonian
    graph, if the degree of each vertex is greater and
    equal to . . . .

    1. AEqual to number of vertices
    2. BDouble of number of vertices
    3. CHalf of number of vertices
    4. DFour times of number of vertices

    Answer: (C) Half of number of vertices

    Explanation

    Dirac’s theorem states that a simple graph on n ≥ 3 vertices in which every vertex has degree at least n/2 is Hamiltonian. The threshold is therefore half the number of vertices, so option C is correct.

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35. Question 35

    Fifth normal form is concerned with :

    1. AJoin Dependency
    2. BDomain-Key
    3. CMultivalued dependency
    4. DFunctional dependency

    Answer: (A) Join Dependency

    Explanation

    Fifth normal form, also called project-join normal form, is concerned with join dependency: a relation is in 5NF when it cannot be decomposed further without losing information. Multivalued dependency belongs to 4NF, and functional dependency to 2NF, 3NF and BCNF. Option A is correct.

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36. Question 36

    Consider the transactions T₁, T₂, T₃, and the
    schedules S₁ and S₂ given below.
    T₁: r₁(x); r₁(z); w₁(z)
    T₂: r₂(y); r₂(z); w₂(z)
    T₃: r₃(y); r₃(x); w₃(y)
    S₁ : r₁(x); r₃(y); r₃(x); r (y); r₂(z); w (y); w (z);
    r (z); w (x); w (z)
    S₂ : r₁(x); r₃(y); r (y), r₃(x); r₁(z); r₂(z); w₃(y);
    w₁(x); w₂(z); w₁(z)
    Which one of the following statements about the
    schedules is TRUE?

    1. AOnly S₁ is conflict-serializable
    2. BOnly S₂ is conflict-serializable
    3. CBoth S₁ and S₂ are conflict-serializable
    4. DNeither S₁ nor S₂ is conflict-serializable

    Answer: (A) Only S₁ is conflict-serializable

    Explanation

    Constructing the precedence graph for S₁ from its read-write, write-read and write-write conflicts produces no cycle, so S₁ is conflict-serializable. The graph for S₂ does contain a cycle, so no serial order is conflict-equivalent to it. Only S₁ qualifies — option A.

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37. Question 37

    A CPU has a 5-stage pipeline with the following
    stages Fetch (F), Decode (D), Execute (E), Memory
    (M) and Write-back (W). Each stage takes one
    clock cycle to complete. Assume there are no
    pipeline stalls and the pipeline is initially empty.
    How many clock cycles are required to complete
    the execution of 10 instructions?

    1. A10
    2. B14
    3. C15
    4. D19

    Answer: (B) 14

    Explanation

    A k-stage pipeline completes n instructions in k + n − 1 cycles once it has been filled. With k = 5 and n = 10 that is 5 + 10 − 1 = 14 cycles, so option B is correct.

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38. Question 38

    Match List-I with List-II.

    List-I (computing systems)	List-II (working/output)
    (A) Half Adder	(I) Has n input and 2ⁿ output
    (B) Decoder	(II) CPU Storage Unit
    (C) Register	(III) Used to store program at runtime
    (D) Main memory	(IV) 2-bit addition circuit

    1. A(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
    2. B(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
    3. C(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
    4. D(A)-(II), (B)-(IV), (C)-(III), (D)-(I)

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

    Explanation

    A half adder adds two bits, so (A)-(IV). A decoder turns n inputs into 2ⁿ outputs, giving (B)-(I). A register is storage inside the CPU, so (C)-(II). Main memory holds the program while it runs, so (D)-(III). Option C is correct.

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39. Question 39

    Arrange the following steps in a proper sequence
    for the process of training a neural network.
    A. Weight initialization
    B. Feed forward
    C. Back Propagation
    D. Loss Calculation
    E. Weight Update

    1. A(A), (B), (D), (C), (E)
    2. B(D), (B), (A), (C), (E)
    3. C(A), (C), (D), (B), (E)
    4. D(E), (C), (B), (D), (A)

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

    Explanation

    Training runs: initialise the weights (A), feed the input forward to obtain predictions (B), compute the loss against the targets (D), back-propagate the gradients (C), then update the weights (E). The loss must exist before gradients can be taken, so (D) precedes (C). That order is option A.

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40. Question 40

    Consider the following statements regarding
    combinational and sequential circuits.
    A. Output of combinational circuits depends on
    the only current input.
    B. Output of combinational circuit depends on
    the both current input and previous output.
    C. Output of sequential circuit depends on the
    current input.
    D. Output of sequential circuit depends on both
    current input and previous output.

    1. A(A) and (C) Only
    2. B(A) and (D) Only
    3. C(B) and (D) Only
    4. D(B) and (C) Only

    Answer: (B) (A) and (D) Only

    Explanation

    A combinational circuit’s output is a pure function of its present inputs, which makes (A) true and (B) false. A sequential circuit adds memory, so its output depends on both the present input and the stored previous state, making (D) true and (C) false. Only (A) and (D) — option B.

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41. Question 41

    Which of the following relations can not be
    decomposed in to BCNF with a lossless join and
    dependency-preserving decomposition?

    1. AR (V,W,X,Y,Z) {VW→X, WX→V, VX→W, W→Y, Y→Z}
    2. BR (V,W,X,Y,Z) {VW→W, X→Y}
    3. CR (V,W,X,Y,Z) {VW→X,X→VY}
    4. DR (V,W,X,Y,Z) {VW→X, X→Y, Y→Z, Z→V}

    Answer: (A) R (V,W,X,Y,Z) {VW→X, WX→V, VX→W, W→Y, Y→Z}

    Explanation

    For R(V, W, X, Y, Z) with {VW→X, WX→V, VX→W, W→Y, Y→Z}, the closure of VW reaches every attribute, so VW is a candidate key. But W → Y and Y → Z both have non-superkey left-hand sides, violating BCNF. Any decomposition that removes them separates W from Y or Y from Z, and the lost dependency cannot be re-derived from the fragments. This relation therefore admits no lossless, dependency-preserving BCNF decomposition — option A.

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42. Question 42

    #include<iostream.h>
    using namespace std;
    void swap(int &x, int &y)
    {
    int temp=x;
    x=y;
    y=temp;
    }
    int main()
    {
    int a=5, b=10;
    swap(a,b);
    swap(a,b);
    cout<<a<<" "<<b;
    return 0;
    }
    What will be the output of the above code?
    

    1. A5, 10
    2. B10, 5
    3. C5, 5
    4. D10, 10

    Answer: (A) 5, 10

    Explanation

    swap takes its parameters by reference, so the first call genuinely exchanges a and b, giving a = 10 and b = 5. The second call exchanges them straight back, restoring a = 5 and b = 10. The program prints 5 10, so option A is correct.

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43. Question 43

    Which of the following are typical activities in the
    software process lifecycle?
    A. Requirement Analysis
    B. System Design
    C. Code Refactoring
    D. Deployment
    E. Substructure

    1. A(A), (B), (C) Only
    2. B(B), (C), (D) Only
    3. C(A), (B), (D) Only
    4. D(A), (D), (E) Only

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

    Explanation

    Requirement analysis (A), system design (B) and deployment (D) are standard phases of the software process lifecycle. Code refactoring is a development practice rather than a lifecycle phase, and “substructure” is not a recognised activity at all. Only (A), (B) and (D) — option C.

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44. Question 44

    Which of the following best describes the
    structure of a relational database ?

    1. AData organized into tables with rows and columns
    2. BData organized into files and folders
    3. CData organized into a hierarchical tree structure
    4. DData organized into a network of interconnected nodes

    Answer: (A) Data organized into tables with rows and columns

    Explanation

    A relational database stores data in tables of rows and columns, where a row is a record and a column an attribute, with relationships expressed through keys. Files and folders, tree hierarchies and node networks describe other data models. Option A is correct.

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45. Question 45

    A coin is tossed successively three times. Find the
    Probability (P), Event (E), Sample space (S) of
    getting exactly one head or two heads, where n
    is number of occurrence.
    A. n(S) = 8 and n(E) = 4
    B. n(E) = 6 and n(S) = 8
    C. P(E) = 3/4
    D. P(E) = 1/2

    1. A(A), (B), (C) and (D)
    2. B(B) and (C) Only
    3. C(A) and (D) Only
    4. D(C) and (D) Only

    Answer: (B) (B) and (C) Only

    Explanation

    Three tosses give n(S) = 2³ = 8. “Exactly one head or exactly two heads” covers every outcome except HHH and TTT, so n(E) = 6 and P(E) = 6/8 = 3/4. Statements (B) and (C) hold while (A) and (D) do not, so option B is correct.

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46. Question 46

    Out of the following steps in the proper
    sequence for simplifying a Boolean function
    using a Karnaugh map (K-map).
    A. Identify and group the largest possible
    cluster of 1's
    B. Draw the K-map for the given Boolean
    function
    C. Write the simplified Boolean expression from
    the grouped clusters
    D. Transfer the truth table values to the K-map

    1. A(B), (D), (A), (C)
    2. B(D), (B), (A), (C)
    3. C(B), (A), (D), (C)
    4. D(A), (B), (C), (D)

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

    Explanation

    Simplifying with a Karnaugh map goes: draw the map for the function (B), transfer the truth-table values into its cells (D), group the largest possible clusters of 1s (A), then read the simplified expression off those groups (C). That order is option A.

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47. Question 47

    Consider the following, where p and q are two statements.

    A. ¬(p ∧ q) ≡ ¬p ∨ ¬q
    B. ¬(p ∨ q) ≡ ¬p ∧ ¬q
    C. p ∧ ¬p ≡ T
    D. ¬(p ⇒ q) ≡ p ∧ ¬q
    E. p ∨ q ≡ ¬p ∨ ¬q

    1. A(A), (B) and (D) only
    2. B(A), (C) and (D) only
    3. C(C), (D) and (E) only
    4. D(A), (B) and (C) only

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

    Explanation

    By De Morgan’s laws ¬(p ∧ q) ≡ ¬p ∨ ¬q and ¬(p ∨ q) ≡ ¬p ∧ ¬q, so (A) and (B) hold. For (D), p ⇒ q is ¬p ∨ q, so ¬(p ⇒ q) ≡ ¬(¬p ∨ q) ≡ p ∧ ¬q — also true. (C) fails because p ∧ ¬p is a contradiction, not a tautology, and (E) fails on the row p = q = T. Only (A), (B) and (D) — option A.

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48. Question 48

    What is the output of the code given below :

    #include<iostream.h>
    using namespace std;
    class Base {
    public:
    Base(){
    cout<<"Base Const";
    }
    virtual ~Base() {
    cout<<"Base dest";
    }
    };
    class Derived : public Base {
    public:
    Derived() {
    cout<<"Derived Const";
    }
    ~Derived() {
    cout<<"Derived dest";
    }
    };
    int main() {
    Base *b = new Derived();
    delete b;
    return 0;
    }
    

    1. ABase Const Derived Const Derived dest Base dest
    2. BBase Const Derived Const Base dest Derived dest
    3. CDerived Const Base Const Base dest Derived dest
    4. DBase Const Derived Const Base dest

    Answer: (A) Base Const Derived Const Derived dest Base dest

    Explanation

    Constructing a Derived object runs the Base constructor first and then the Derived one, printing Base Const Derived Const. Because ~Base is declared virtual, deleting through the Base* dispatches to ~Derived first and then ~Base, printing Derived dest Base dest. The full output is option A.

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49. Question 49

    If L₁ and L₂ are context free languages, which one of the following is true about L₁ ∩ L₂?

    1. AL1 L2 is context free.
    2. BL1 L2 is Regular
    3. CL1 L2 is Recursive Enumerable
    4. DL1 L2 is Context Sensitive

    Answer: (C) L1 L2 is Recursive Enumerable

    Explanation

    Context-free languages are not closed under intersection, so L₁ ∩ L₂ need not be context-free, and it certainly need not be regular. Both languages are recursive and therefore recursively enumerable, and that class is closed under intersection, so the intersection is guaranteed to be recursively enumerable. Option C is the only statement that always holds.

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50. Question 50

    Consider the following C code:

    #include<stdio.h>
    int temp =0;
    int fun(int x, int y)
    {
    int z; temp++;
    if(y==3) return (x*x*x);
    else {
    z=fun(x,y/3);
    return (z*z*z);
    }
    }
    int main() {
    fun(4,81);
    printf("%d",temp);
    }
    what will be the output of the following code :
    

    1. A3
    2. B4
    3. C5
    4. D6

    Answer: (B) 4

    Explanation

    fun increments temp on entry and then recurses with y/3 until y reaches 3. The calls use y = 81, 27, 9 and 3, so temp is incremented four times including the base case, leaving temp = 4. Option B is correct.

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51. Question 51

    Which of the following tasks could be attained
    using syntax trees in compiler design ?
    A. Type Checking
    B. Code Generation
    C. Code Optimization
    D. Error Handling

    1. A(A), (B), (C) Only
    2. B(B), (C), (D) Only
    3. C(A), (C), (D) Only
    4. D(A), (B), (D) Only

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

    Explanation

    A syntax tree is the structured representation the compiler walks for type checking (A), code generation (B) and code optimisation (C). It can reveal that an error exists, but recovering from errors is the job of the error-handling machinery rather than the tree itself. Only (A), (B) and (C) — option A.

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52. Question 52

    Match List-I with List-II.

    List-I	List-II
    (A) HTML	(I) Allow for dynamic and interactive web pages.
    (B) DHTML	(II) Defines the structure of web pages.
    (C) XML	(III) Object-oriented programming language for web applications.
    (D) JAVA	(IV) Used for data storage and transport.

    1. A(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
    2. B(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
    3. C(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
    4. D(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

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

    Explanation

    HTML defines page structure, so (A)-(II). DHTML adds dynamic, interactive behaviour, giving (B)-(I). XML carries and stores data, so (C)-(IV). Java is the object-oriented language used for web applications, so (D)-(III). Option C is correct.

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53. Question 53

    Match List-I with List-II.

    List-I (Counter)	List-II (uses/working)
    (A) N-bit Ring Counter	(I) Uses universal clock
    (B) Synchronous Counter	(II) Counts exactly N states
    (C) Asynchronous Counter	(III) Counts 0 to 9
    (D) Decimal Counter	(IV) Main clock is applied to first flip-flop only

    1. A(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
    2. B(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
    3. C(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
    4. D(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

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

    Explanation

    An N-bit ring counter cycles through exactly N states, so (A)-(II). A synchronous counter clocks every flip-flop from one common clock, giving (B)-(I). In an asynchronous (ripple) counter only the first flip-flop sees the main clock, so (C)-(IV). A decimal counter counts 0 to 9, so (D)-(III). Option C is correct.

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54. Question 54

    Arrange the following steps of file handling in C
    in the correct order :
    A. Close the file
    B. Read from or write to the file
    C. Open the file
    D. Check for error

    1. A(A), (C), (D), (B)
    2. B(D), (B), (C), (A)
    3. C(C), (D), (B), (A)
    4. D(B), (D), (A), (C)

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

    Explanation

    File handling in C runs: open the file (C), check that the open succeeded (D), read from or write to it (B), then close it (A). Checking for error immediately after opening is what makes the subsequent I/O safe. That order is option C.

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55. Question 55

    Match List-I with List-II.

    List-I (Algorithms)	List-II (Characteristics)
    (A) First Come First Served (FCFS)	(I) Each process is assigned a priority.
    (B) Shortest Job First (SJF)	(II) Ensures fair allocation of CPU time by assigning time slice.
    (C) Round Robin (RR)	(III) Processes are executed in the order they arrive.
    (D) Priority Scheduling	(IV) Select the process for execution with smallest next Burst time.

    1. A(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
    2. B(A)-(II), (B)-(IV), (C)-(III), (D)-(I)
    3. C(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
    4. D(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

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

    Explanation

    FCFS runs processes in arrival order, so (A)-(III). SJF picks the process with the smallest next CPU burst, giving (B)-(IV). Round Robin shares the CPU fairly through time slices, so (C)-(II). Priority scheduling assigns each process a priority, so (D)-(I). Option C is correct.

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56. Question 56

    Arrange the following steps in a proper sequence
    for the typical process of a DNS query :
    A. Query authoritative DNS Server
    B. Query local DNS Server
    C. Local DNS Server Check Cache
    D. Query root DNS Server
    E. Query TLD DNS Server

    1. A(A), (C), (E), (B), (D)
    2. B(E), (A), (C), (B), (D)
    3. C(A), (B), (C), (D), (E)
    4. D(B), (C), (D), (E), (A)

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

    Explanation

    Resolution starts at the local recursive resolver (B), which first checks its cache (C). On a miss it queries a root server (D), which refers it to the TLD server (E), which in turn refers it to the authoritative server (A) holding the record. That order is option D.

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57. Question 57

    Match List-I with List-II.

    List-I (operating system concepts)	List-II (characteristics)
    (A) Paging	(I) Evicts least recently used process
    (B) LRU (Least Recently Used)	(II) Extends physical memory
    (C) C-SCAN	(III) Logical to physical mapping
    (D) Virtual Memory	(IV) Circular disk access

    1. A(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
    2. B(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
    3. C(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
    4. D(A)-(I), (B)-(IV), (C)-(III), (D)-(II)

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

    Explanation

    Paging maps logical addresses onto physical frames, so (A)-(III). LRU evicts the least recently used entry, giving (B)-(I). C-SCAN sweeps the disk in one circular direction, so (C)-(IV). Virtual memory makes the address space appear larger than physical memory, so (D)-(II). Option B is correct.

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58. Question 58

    A class B network on the Internet has a subnet
    mask of 255.255.248.0, what is the maximum
    number of hosts per subnet?

    1. A1022
    2. B1023
    3. C2046
    4. D2047

    Answer: (C) 2046

    Explanation

    The mask 255.255.248.0 is 11111111.11111111.11111000.00000000, leaving 11 zero bits for the host field. That gives 2¹¹ = 2048 addresses per subnet, less the network and broadcast addresses, so 2046 usable hosts — option C.

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59. Question 59

    Select the Sorting Algorithms that are stable.
    A. Quick Sort
    B. Bubble Sort
    C. Insertion Sort
    D. Merge Sort
    E. Shell Sort

    1. A(A), (B), (C) and (E) Only
    2. B(A), (D) and (E) Only
    3. C(B) and (C) Only
    4. D(B), (C) and (D) Only

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

    Explanation

    A sort is stable when records with equal keys keep their input order. Bubble sort, insertion sort and merge sort all preserve it in their standard forms. Quick sort and shell sort move elements across long distances and do not. Only (B), (C) and (D) — option D.

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60. Question 60

    ___________is a Self Balancing binary search
    tree, where the path from the root to the furthest
    leaf is no more than twice as long as the path
    from the root to nearest leaf.

    1. AExpression tree
    2. BGame tree
    3. CRed-Black tree
    4. DThreaded tree

    Answer: (C) Red-Black tree

    Explanation

    A red-black tree is the self-balancing binary search tree whose colouring rules guarantee that the longest root-to-leaf path is at most twice the shortest, keeping search, insertion and deletion at O(log n). Expression, game and threaded trees carry no such balance guarantee. Option C is correct.

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61. Question 61

    Let L (x , y) be the statement "x loves y" where
    the domain for both x and y consists of all people
    in the world. Use quantifiers to express "Joy is
    loved by everyone".

    1. A∀x L(x,Joy)
    2. B∀y L(Joy,y)
    3. C∃y∀x L(x,y)
    4. D∃x¬ L(Joy,x)

    Answer: (A) ∀x L(x,Joy)

    Explanation

    “Joy is loved by everyone” says that for every person x, x loves Joy — that is ∀x L(x, Joy). Option B reverses the roles and says Joy loves everyone; option C only asserts that somebody is universally loved; option D is an existential negation. Option A is correct.

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62. Question 62

    What will be the output of the following C code?

    #include<stdio.h>
    void main()
    {
    int arr[5] = {10,20,30,40,50};
    int *p = (int*) (&arr +1);
    printf("%d%d", *(arr+1), *(p-1));
    }
    

    1. A10 50
    2. B20 50
    3. C30 40
    4. D20 40

    Answer: (B) 20 50

    Explanation

    (arr+1) is arr[1] = 20. Because &arr has type int()[5], &arr + 1 steps a whole array forward, so p points just past arr[4]; p − 1 therefore points at arr[4] = 50. The program prints 20 50, so option B is correct.

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63. Question 63

    Arrange the following stages of a Turing Machine
    (TM) operation in the correct order as they occur
    during computation.
    A. Writing a symbol on the tape
    B. Moving the tape head left to right
    C. Reading a symbol from the tape
    D. Transitioning to a new state based on the
    current state and symbol read
    E. Halting and accepting or rejecting the input

    1. A(C), (A), (B), (D), (E)
    2. B(C), (B), (A), (D), (E)
    3. C(C), (D), (A), (B), (E)
    4. D(C), (D), (B), (A), (E)

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

    Explanation

    One Turing machine step reads the symbol under the head (C), combines it with the current state to choose a transition (D), writes a symbol (A) and moves the head (B). The machine halts and accepts or rejects when no further transition applies (E). That order is option C.

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64. Question 64

    Match List-I with List-II.

    List-I (software design principles)	List-II (Definition)
    (A) Cohesion	(I) Degree to which one module relies on another module.
    (B) Coupling	(II) Dividing a software system into distinct modules.
    (C) Abstraction	(III) Degree to which elements of a module belong together.
    (D) Modularity	(IV) Simplifying complex reality by modeling classes appropriate to the problem.

    1. A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
    2. B(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
    3. C(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
    4. D(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

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

    Explanation

    Cohesion measures how well the elements inside a single module belong together, so (A)-(III). Coupling measures how much one module depends on another, giving (B)-(I). Abstraction simplifies reality by modelling only the classes that matter, so (C)-(IV). Modularity is the division of the system into distinct modules, so (D)-(II). Option C is correct.

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65. Question 65

    Which of the following languages can be recognized by a Non-Deterministic Finite Automaton (NFA) but cannot be recognized by a Deterministic Finite Automaton (DFA)?

    A. L₁ = {w ∈ {0,1}* | the length of w is even}
    B. L₂ = {w ∈ {0,1}* | the length of w is odd}
    C. L₃ = {w ∈ {0,1}* | all 0s precede all 1s in w}
    D. L₄ = {w ∈ {0,1}* | w contains an equal number of 0s and 1s}
    E. L₅ = {w ∈ {0,1}* | all 1s precede all 0s in w}

    1. A(A) and (B) only
    2. B(B) and (C) only
    3. C(C) and (D) only
    4. DNone of the above

    Answer: (D) None of the above

    Explanation

    NFAs and DFAs recognize exactly the same regular languages. L₁, L₂, L₃, and L₅ are regular and recognized by both, while L₄ is non-regular and recognized by neither. Therefore option D is correct.

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66. Question 66

    Given a project with an estimated effort of 1500
    person-hours and a team of 5 people, how many
    days will it take to complete the project, if each
    person works 8 hours a day?

    1. A30
    2. B37.5
    3. C40
    4. D45.5

    Answer: (B) 37.5

    Explanation

    Five people working eight hours a day supply 40 person-hours per day. Dividing 1500 person-hours by 40 gives 37.5 days, so option B is correct.

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67. Question 67

    Arrange the following phases of database design
    in the correct order :
    A. Physical Design
    B. Conceptual Design
    C. Logical Design
    D. Requirement Analysis

    1. A(B), (D), (A), (C)
    2. B(C), (A), (B), (D)
    3. C(D), (B), (C), (A)
    4. D(A), (D), (C), (B)

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

    Explanation

    Database design proceeds: requirement analysis (D), conceptual design (B) producing an ER model, logical design (C) mapping that model to relations, then physical design (A) choosing storage structures and indexes. That order is option C.

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68. Question 68

    Consider the three way hand shaking process
    followed during TCP connection establishment
    between two hosts A and B . Let S and R be two
    random 32 -bit starting sequence numbers
    chosen by A and B, respectively. Suppose A
    sends a TCP segment having SYN bit =1, SEQ
    number = S and ACK bit =0 and B accepts the
    connection request.
    Which one of the following choices represents
    the information present in the TCP segment
    header that is sent by B to A ?

    1. ASYN bit =1, SEQ number =S+1, ACK bit =0, ACK number =R, FiN bit =0
    2. BSYN bit =0, SEQ number =R, ACK bit =1, ACK number =S+1, FiN bit =0
    3. CSYN bit =1, SEQ number =R, ACK bit =1, ACK number =S+1, FiN bit =0
    4. DSYN bit =1, SEQ number = R, ACK bit =1, ACK number =S, FiN bit =0

    Answer: (C) SYN bit =1, SEQ number =R, ACK bit =1, ACK number =S+1, FiN bit =0

    Explanation

    B’s reply is a SYN-ACK. It carries B’s own initial sequence number R, sets SYN = 1 because B is opening its own direction of the connection, sets ACK = 1, and acknowledges A’s stream with ACK number S + 1. FIN stays 0. Option C matches all five fields.

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69. Question 69

    Two computers A and B are configured as follows
    : A has IP address 203.197.2.53 and subnet mask
    255.255.128.0, B has IP address 203.197.75.201
    and subnet mask 255.255.192.0. What one of the
    following statement is true?

    1. AA and B both assume they are on the same network
    2. BB assumes, A is on same network, A assumes B is on different network
    3. CA assumes, B is on same network, but B assumes, A is on a different network
    4. DA and B both assume they are on different network

    Answer: (C) A assumes, B is on same network, but B assumes, A is on a different network

    Explanation

    A applies its own mask to both addresses: 203.197.2.53 and 203.197.75.201 both reduce to 203.197.0.0, so A concludes B is local. B applies its own mask: its address reduces to 203.197.64.0 but A’s reduces to 203.197.0.0, so B concludes A is remote. Option C describes exactly this asymmetry.

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70. Question 70

    In Software configuration management, what is
    the primary purpose of version control ?

    1. ATo control the changes made to software
    2. BTo document user requirements
    3. CTo estimate project cost
    4. DTo design the software architecture

    Answer: (A) To control the changes made to software

    Explanation

    Within software configuration management, version control exists to record and govern changes to software artefacts over time — tracking history, enabling rollback, and supporting branching and merging. Requirements documentation, cost estimation and architectural design are separate activities. Option A is correct.

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71. Question 71

    Arrange the following stages of parsing in the
    correct order as they typically occur in the
    compilation process.
    A. Lexical Analysis
    B. Sematic Analysis
    C. Syntax Analysis
    D. Intermediate Code Generation
    E. Code Optimization

    1. A(A), (B), (C), (D), (E)
    2. B(A), (C), (B), (D), (E)
    3. C(A), (D), (B), (C), (E)
    4. D(A), (C), (D), (B), (E)

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

    Explanation

    Compilation runs lexical analysis (A) to produce tokens, syntax analysis (C) to build the parse tree, semantic analysis (B) to check types and meaning, intermediate code generation (D), then code optimisation (E). That order is option B.

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72. Question 72

    Match List-I with List-II.

    List-I (Queries)	List-II (Probability)
    (A) A bag contains 6 white and 4 red balls. Two balls are drawn at random. What is the chance they will be the same colour?	(I) 3/68
    (B) In a pack of 52 cards, one card is drawn at random. What is the probability that it is either a king or a queen?	(II) 14/68
    (C) A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability of 1 red and 2 white balls.	(III) 2/13
    (D) A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability of 2 blue and 1 red balls.	(IV) 7/15

    1. A(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
    2. B(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
    3. C(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
    4. D(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

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

    Explanation

    There are C(10,2) = 45 equally likely pairs and C(6,2) + C(4,2) = 21 same-colour pairs, so the first probability is 21/45 = 7/15 and (A)-(IV). A pack holds 4 kings and 4 queens, so 8/52 = 2/13 and (B)-(III). From 18 balls C(18,3) = 816; one red with two white gives C(6,1)C(4,2) = 36, so 36/816 = 3/68 and (C)-(I). Two blue with one red gives C(8,2)C(6,1) = 168, so 168/816 = 14/68 and (D)-(II). Option D is correct.

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73. Question 73

    Match List-I with List-II.

    List-I	List-II
    (A) Natural language processing	(I) A method of training algorithm by rewarding desired behaviour and/or punishing undesired one.
    (B) Reinforcement learning	(II) System designed to emulate the decision making abilities of a human expert.
    (C) Support vector machine	(III) A branch of AI focused on understanding and generating human language.
    (D) Expert system	(IV) A machine learning technique that finds the hyperplane that best separates different classes in a feature space.

    1. A(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
    2. B(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
    3. C(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
    4. D(A)-(II), (B)-(IV), (C)-(III), (D)-(I)

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

    Explanation

    Natural language processing is the AI branch for understanding and generating human language, so (A)-(III). Reinforcement learning trains by rewarding desired and punishing undesired behaviour, giving (B)-(I). A support vector machine finds the separating hyperplane, so (C)-(IV). An expert system emulates a human expert’s decisions, so (D)-(II). Option C is correct.

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74. Question 74

    Match List-I with List-II.

    List-I	List-II
    (A) Dijkstra’s Algorithm	(I) Find the shortest path between all pairs of vertices in a graph with positive or negative edge weights.
    (B) Floyd-Warshall Algorithm	(II) Finds the shortest path in a weighted graph with non-negative edge weights.
    (C) Bellman-Ford Algorithm	(III) Finds single-source shortest paths with possible negative weights.
    (D) Prim’s Algorithm	(IV) Finds the Minimum Spanning Tree (MST).

    1. A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
    2. B(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
    3. C(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
    4. D(A)-(III), (B)-(II), (C)-(IV), (D)-(I)

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

    Explanation

    Dijkstra’s algorithm needs non-negative edge weights, so (A)-(II). Floyd-Warshall solves all-pairs shortest paths, giving (B)-(I). Bellman-Ford handles single-source shortest paths when negative weights are present, so (C)-(III). Prim’s algorithm builds a minimum spanning tree, so (D)-(IV). Option B is correct.

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75. Question 75

    Which of the following statements are TRUE
    about Process Control Block (PCB) ?
    A. The PCB contains information about the
    process state, such as whether it is running,
    waiting or terminated
    B. The PCB includes the program code and data
    segments of the process
    C. The PCB stores the process's memory
    management information, such as page tables
    and segment tables
    D. The PCB is used to track process scheduling
    information and CPU registers for process
    execution

    1. A(A), (B) and (C) Only
    2. B(B), (C) and (D) Only
    3. C(A), (C) and (D) Only
    4. D(A) and (B) Only

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

    Explanation

    The process control block records the process state (A), its memory-management data such as page and segment tables (C), and its scheduling information together with the saved CPU registers (D). The program’s code and data segments live in the process’s address space, not in the PCB, so (B) is false. Only (A), (C) and (D) — option C.

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76. Question 76

    #include<stdio.h>
    void main()
    {
    int arr[ ] = {1, 2, 3, 4, 5};
    int *p = arr;
    printf("%d", *p++);
    printf("%d", *(p+1));
    }
    Find the output of the above code?
    

    1. A1,2
    2. B1,3
    3. C2,3
    4. D1,4

    Answer: (B) 1,3

    Explanation

    *p++ yields the current target arr[0] = 1 and then advances p to arr[1]. The second statement evaluates *(p + 1), which is arr[2] = 3. The output is 1 followed by 3, so option B is correct.

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77. Question 77

    Arrange the following Language Classes in
    ascending order according to their expressive
    power, as defined by Chomsky hierarchy:
    A. Context-free languages
    B. Context-sensitive languages
    C. Regular languages
    D. Unrestricted Grammars

    1. A(C), (A), (B), (D)
    2. B(C), (A), (D), (B)
    3. C(A), (C), (B), (D)
    4. D(A), (D), (B), (C)

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

    Explanation

    By expressive power the Chomsky hierarchy runs regular ⊂ context-free ⊂ context-sensitive ⊂ unrestricted, recognised respectively by finite automata, pushdown automata, linear bounded automata and Turing machines. Ascending order is therefore (C), (A), (B), (D) — option A.

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78. Question 78

    An undirected graph has vertex set V and edge set E. If it has l edges, what is the sum of the degrees of all vertices?

    1. A2l
    2. Bl/2
    3. Cl²
    4. D√l

    Answer: (A) 2l

    Explanation

    Each undirected edge contributes one to the degree of each of its two endpoints. By the handshaking lemma, the total degree is 2l, so option A is correct.

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79. Question 79

    40 software professionals were interviewed for a
    job. 25 knew PYTHON 20 knew JAVA and 7 knew
    neither language. How many knew both
    languages ?

    1. A12
    2. B53
    3. C10
    4. D88

    Answer: (A) 12

    Explanation

    Seven knew neither language, so 33 of the 40 knew at least one. By inclusion-exclusion |P ∪ J| = |P| + |J| − |P ∩ J|, giving 33 = 25 + 20 − |P ∩ J|, so |P ∩ J| = 12. Option A is correct.

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80. Question 80

    A. In IPv4 addressing, a block of address can be
    define as x.y.z.t/ n in which x.y.z.t define one of
    the address and / n define the mask
    B. The first address in the block can be found by
    setting the rightmost 32-n bits to 0s
    C. Address of class C is used for multicast
    communication
    D. There are five classes in IPv4 address
    E. Supernetting combine several networks into
    one large network

    1. A(A), (B) and (C) Only
    2. B(A), (B), (D) and (E) Only
    3. C(A), (C), (D) and (E) Only
    4. D(C), (D) and (E) Only

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

    Explanation

    (A) is true — x.y.z.t/n names a block by one address and a prefix length. (B) is true — zeroing the rightmost 32 − n bits yields the first address of the block. (D) is true — IPv4 defines classes A through E. (E) is true — supernetting aggregates contiguous blocks into one larger route. (C) is false: multicast uses class D, not class C. So (A), (B), (D) and (E) — option B.

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81. Question 81

    Match List-I with List-II.

    List-I (Testing Type)	List-II (Description)
    (A) Unit testing	(I) Testing individual components of the software.
    (B) Integration testing	(II) Testing the interaction between integrated components.
    (C) System testing	(III) Testing to verify the system meets business needs.
    (D) Acceptance testing	(IV) Testing the complete system to ensure it meets requirements.

    1. A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
    2. B(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
    3. C(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
    4. D(A)-(I), (B)-(IV), (C)-(II), (D)-(III)

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

    Explanation

    Unit testing exercises individual components, so (A)-(I). Integration testing checks how integrated components interact, giving (B)-(II). System testing validates the assembled system against its requirements, so (C)-(IV). Acceptance testing confirms the system meets business needs, so (D)-(III). Option B is correct.

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82. Question 82

    Consider the following statement regarding
    special purpose registers.
    A. Program Counter (PC) keeps the track of next
    instruction executed
    B. Instruction register holds the address of first
    instruction to be executed
    C. Accumulator holds the output of ALU
    D. Program Counter ( PC ) keeps the track of
    only first instruction of the program

    1. A(A) and (C) Only
    2. B(A) and (D) Only
    3. C(B) and (D) Only
    4. D(A), (B) and (C) Only

    Answer: (A) (A) and (C) Only

    Explanation

    (A) is true — the program counter holds the address of the next instruction to be executed. (C) is true — the accumulator holds the ALU’s result. (B) is false: the instruction register holds the instruction currently being executed, not an address. (D) is false: the PC tracks the next instruction throughout execution, not only the first. Only (A) and (C) — option A.

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83. Question 83

    Arrange the following steps in the correct order
    to solve the Knapsack problem using Dynamic
    Programming.
    A. Define the base case when the capacity is
    zero (0) or no items are left to consider
    B. Compute the maximum value that can be
    obtained using items up to the i-th item and a
    knapsack capacity of 0
    C. Identify subproblems and their dependencies
    based on items weights and values
    D. Initialize a table to store results of
    subproblems
    E. Iterate through each item and each possible
    Capacity to fill the table

    1. A(C), (D), (A), (E), (B)
    2. B(D), (C), (A), (E), (B)
    3. C(A), (C), (D), (E), (B)
    4. D(D), (A), (C), (E), (B)

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

    Explanation

    The dynamic-programming solution starts from the base case of zero capacity or no items left (A), identifies the subproblems and how they depend on item weights and values (C), initialises the table that stores them (D), iterates over every item and capacity to fill it (E), and finally reads the maximum value out of the completed table (B). That order is option C.

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84. Question 84

    A multi processor system with 16 processors is
    used to execute a parallelizable task. If the serial
    portion of the task takes 200 clock cycles and
    the parallel portion take 800 clock cycles. When
    all 16 processor are used how many total clock
    cycles are required to complete the task ?

    1. A250
    2. B300
    3. C400
    4. D450

    Answer: (A) 250

    Explanation

    The 200-cycle serial portion cannot be sped up. The 800-cycle parallel portion divides across 16 processors, taking 800/16 = 50 cycles. The total is 200 + 50 = 250 cycles, so option A is correct.

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85. Question 85

    Arrange the given steps required for a Direct
    Memory Access (DMA) transfer in the correct
    order.
    A. Initiate DMA transfer request
    B. Transfer data directly between peripheral and
    memory
    C. Processor grants DMA control over the
    system bus
    D. DMA controller completes data transfer and
    signals completion

    1. A(C), (A), (B), (D)
    2. B(A),(C),(B),(D)
    3. C(A), (B), (C), (D)
    4. D(C), (B), (A), (D)

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

    Explanation

    A DMA transfer begins with the peripheral raising a transfer request (A). The processor then releases the bus and grants the controller DMA control (C). The controller moves data directly between peripheral and memory (B), and finally signals completion (D). That order is option B.

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86. Question 86

    Match List-I with List-II.

    List-I (Recurrence Relation)	List-II (Complexity)
    (A) T(n) = 2T(n/2) + n	(I) T(n) = θ(n log n) {exact solution}
    (B) T(n) = T(n/2) + 1	(II) O(n²)
    (C) T(n) = 2T(n/2) + 1	(III) T(n) = θ(n) {exact solution}
    (D) T(n) = T(n−1) + n	(IV) O(n)

    1. A(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
    2. B(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
    3. C(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
    4. D(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

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

    Explanation

    By the Master Theorem, T(n) = 2T(n/2) + n is θ(n log n), so (A)-(I). T(n) = T(n−1) + n sums to θ(n²), so (D)-(II). T(n) = 2T(n/2) + 1 is θ(n), and T(n) = T(n/2) + 1 is θ(log n) — which is not listed, so the official key pairs (B) with the remaining O(n)-family entry. The official NTA key is option C. Note that (B) = T(n/2) + 1 is θ(log n), not θ(n): option A pairs the exact-θ(n) entry with (C) and is the mathematically tighter reading. Answer as per the official key.

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87. Question 87

    Translate ∀x ∃y (x < y) into English. Consider the domain to be the real numbers for both variables.

    1. AFor all real numbers x, there exists a real number y such that x is less than y
    2. BFor every real numbers y, there exists a real number x such that x is less than y
    3. CFor some real numbers x, there exists a real number y such that x is less than y
    4. DFor each and every real numbers x and y, x is less than y

    Answer: (A) For all real numbers x, there exists a real number y such that x is less than y

    Explanation

    ∀x binds every real number x, and for each of them ∃y asserts that some real y exists with x < y. Read literally: for all real numbers x there exists a real number y such that x is less than y. Option A states this; option B swaps the quantified variables, option C weakens ∀ to “for some”, and option D collapses both quantifiers into a single universal claim.

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88. Question 88

    A vector processor with 16 lanes can perform an
    operation on 1024 elements with each operation
    taking 5 clock cycles. How many cycles are
    needed to complete the operation?

    1. A64
    2. B80
    3. C100
    4. D128

    Answer: (A) 64

    Explanation

    With 16 lanes, 1024 elements are covered in 1024/16 = 64 passes through the vector pipeline, so option A is the official key. Note the wording: each pass is stated to take 5 clock cycles, so the number of cycles would be 64 × 5 = 320. The key counts operations rather than cycles.

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89. Question 89

    Given a project that uses the COCOMO model
    with an estimated effort of 2000 person-months
    and a productivity rate of 5 person-month per
    KLOC, what is the estimated size of the project in
    KLOC ?

    1. A200
    2. B400
    3. C100
    4. D50

    Answer: (B) 400

    Explanation

    Dividing effort by the productivity rate gives 2000 person-months ÷ 5 person-months per KLOC = 400 KLOC, so option B is correct.

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90. Question 90

    Which of the following languages are context-free?

    A. {wⁱxʲyᵏzˡ | i + k = j + l; i, j, k, l ≥ 0}
    B. {wⁱxʲyᵏzˡ | i = j and k = l; i, j, k, l ≥ 0}
    C. {wⁱxʲyᵏzˡ | i = j = k and k ≠ l; i, j, k, l ≥ 0}
    D. {wⁱxʲyᵏzˡ | i = j = k + l; i, j, k, l ≥ 0}
    E. {wⁱxʲyᵏzˡ | i = j = l and k ≠ l; i, j, k, l ≥ 0}

    1. A(A) and (B) only
    2. B(B) and (C) only
    3. C(C) and (D) only
    4. D(D) and (E) only

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

    Explanation

    Language A has one balance condition that can be handled by a stack, and language B is an intersection/concatenation of simple equality constraints expressible as a context-free language. The remaining languages impose simultaneous multi-count dependencies that are not context-free, so option A is correct.

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91. Question 91

    What is the role of Back Propagation
    Algorithm?

    Artificial Neural Networks (ANNs) are computational models inspired by the human brain’s neural networks. They consist of interconnected nodes, or neurons, organized into layers: an input layer, one or more hidden layers and an output layer. Each connection between neurons has a weight that adjusts as learning progresses, allowing the network to adapt and improve its performance. ANNs are particularly effective in recognizing patterns, making them valuable for tasks such as image and speech recognition, natural language processing and predictive analytics. Learning in ANNs typically involves training algorithms like back propagation, which minimize the error by adjusting the weights. As a subset of machine learning, ANNs have revolutionized the field of Artificial Intelligence by providing solutions to complex problems that traditional algorithms struggle with.

    1. ATo reduce error
    2. BTo secure network
    3. CTo control speed of data
    4. DTo add different layers

    Answer: (A) To reduce error

    Explanation

    Back-propagation computes the gradient of the loss with respect to every weight by applying the chain rule backwards through the network, so that the weights can then be adjusted to reduce the error. Its purpose is error reduction, not securing the network, controlling data speed or adding layers. Option A is correct.

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92. Question 92

    What is the role of weights in an ANN ?

    Artificial Neural Networks (ANNs) are computational models inspired by the human brain’s neural networks. They consist of interconnected nodes, or neurons, organized into layers: an input layer, one or more hidden layers and an output layer. Each connection between neurons has a weight that adjusts as learning progresses, allowing the network to adapt and improve its performance. ANNs are particularly effective in recognizing patterns, making them valuable for tasks such as image and speech recognition, natural language processing and predictive analytics. Learning in ANNs typically involves training algorithms like back propagation, which minimize the error by adjusting the weights. As a subset of machine learning, ANNs have revolutionized the field of Artificial Intelligence by providing solutions to complex problems that traditional algorithms struggle with.

    1. ATo store data
    2. BTo adjust and improve network performance
    3. CTo control the speed
    4. DTo secure the network

    Answer: (B) To adjust and improve network performance

    Explanation

    The weight on each connection scales the signal passing along it, and learning consists precisely of adjusting these weights so the network’s predictions improve. They are trainable parameters, not a data store, a speed control or a security mechanism. Option B is correct.

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93. Question 93

    Which of the following is/are the
    application area(s) of ANN ?
    A. Natural Language Processing
    B. Image Processing
    C. Pattern Recognition
    D. Speech Recognition
    Choose the correct answer from the
    options given below :

    Artificial Neural Networks (ANNs) are computational models inspired by the human brain’s neural networks. They consist of interconnected nodes, or neurons, organized into layers: an input layer, one or more hidden layers and an output layer. Each connection between neurons has a weight that adjusts as learning progresses, allowing the network to adapt and improve its performance. ANNs are particularly effective in recognizing patterns, making them valuable for tasks such as image and speech recognition, natural language processing and predictive analytics. Learning in ANNs typically involves training algorithms like back propagation, which minimize the error by adjusting the weights. As a subset of machine learning, ANNs have revolutionized the field of Artificial Intelligence by providing solutions to complex problems that traditional algorithms struggle with.

    1. A(A) and (B) Only
    2. B(B) and (C) Only
    3. C(A), (B) and (C) Only
    4. D(A), (B), (C) and (D)

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

    Explanation

    The passage lists pattern recognition together with image recognition, speech recognition, natural language processing and predictive analytics among the tasks artificial neural networks handle well. All four listed areas therefore apply, so option D is correct.

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94. Question 94

    Artificial Neural Networks (ANNs) are
    inspired by :

    Artificial Neural Networks (ANNs) are computational models inspired by the human brain’s neural networks. They consist of interconnected nodes, or neurons, organized into layers: an input layer, one or more hidden layers and an output layer. Each connection between neurons has a weight that adjusts as learning progresses, allowing the network to adapt and improve its performance. ANNs are particularly effective in recognizing patterns, making them valuable for tasks such as image and speech recognition, natural language processing and predictive analytics. Learning in ANNs typically involves training algorithms like back propagation, which minimize the error by adjusting the weights. As a subset of machine learning, ANNs have revolutionized the field of Artificial Intelligence by providing solutions to complex problems that traditional algorithms struggle with.

    1. AQuantum mechanics
    2. BHuman brain's neural network
    3. CComputer Hardware architecture
    4. DGenetic algorithm

    Answer: (B) Human brain's neural network

    Explanation

    Artificial neural networks are computational models inspired by the human brain’s neural networks: interconnected nodes, or neurons, arranged in layers with weighted connections that adjust as learning progresses. Option B is correct.

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95. Question 95

    Which of the following layers may be more
    than one in number ?

    Artificial Neural Networks (ANNs) are computational models inspired by the human brain’s neural networks. They consist of interconnected nodes, or neurons, organized into layers: an input layer, one or more hidden layers and an output layer. Each connection between neurons has a weight that adjusts as learning progresses, allowing the network to adapt and improve its performance. ANNs are particularly effective in recognizing patterns, making them valuable for tasks such as image and speech recognition, natural language processing and predictive analytics. Learning in ANNs typically involves training algorithms like back propagation, which minimize the error by adjusting the weights. As a subset of machine learning, ANNs have revolutionized the field of Artificial Intelligence by providing solutions to complex problems that traditional algorithms struggle with.

    1. AInput layer
    2. BHidden layer
    3. COutput layer
    4. DPhysical layer

    Answer: (B) Hidden layer

    Explanation

    A network has exactly one input layer and one output layer, but it may stack one or more hidden layers between them — that depth is what lets it model complex relationships. The physical layer belongs to networking models, not to artificial neural networks. Option B is correct.

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96. Question 96

    Which of the following is NOT a
    requirement for Banker's algorithm to
    grant a resource request ?

    The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.

    1. AThe requested resources must be available.
    2. BThe system must be in a safe state after granting the request
    3. CThe request must not exceed the maximum resources the process can request.
    4. DThe process must be the only one requesting resources.

    Answer: (D) The process must be the only one requesting resources.

    Explanation

    Banker’s algorithm grants a request only when the resources are available, when the request stays within the process’s previously declared maximum, and when the resulting state is safe. It never requires a process to be the only one requesting resources — several processes hold and request concurrently. Option D is therefore the statement that is NOT a requirement.

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97. Question 97

    Which data structure does the Banker's
    Algorithm use to maintain the state of
    available, maximum and allocated
    resources ?

    The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.

    1. APriority Queue
    2. BHash table
    3. CWait-for-Graph
    4. DMatrices and Vectors

    Answer: (D) Matrices and Vectors

    Explanation

    The algorithm tracks Available as a vector, and Maximum, Allocation and Need as matrices indexed by process and by resource type. Priority queues, hash tables and wait-for graphs appear elsewhere in deadlock handling but not in this algorithm. Option D is correct.

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98. Question 98

    What is the significance of the Banker's
    algorithm in terms of resource
    management?

    The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.

    1. AIt enures that all processes can finish their execution without deadlock.
    2. BIt eliminates the need for processes to request resources.
    3. CIt accelerates the execution of critical sections in processes.
    4. DIt minimizes the number of context switches between processes.

    Answer: (A) It enures that all processes can finish their execution without deadlock.

    Explanation

    Its significance is that by admitting only requests which leave the system in a safe state, it guarantees a sequence exists in which every process can run to completion — deadlock is avoided rather than merely detected afterwards. It does not remove the need to request resources, accelerate critical sections, or reduce context switches. Option A is correct.

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99. Question 99

    What is the primary goal of the Banker's
    Algorithm?

    The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.

    1. ATo allocate resources optimally
    2. BTo prevent processes from requesting resources
    3. CTo detect and recover from deadlocks
    4. DTo maximise CPU utilization

    Answer: (A) To allocate resources optimally

    Explanation

    Banker’s algorithm is a deadlock-avoidance method: its goal is to allocate resources in the way that keeps the system in a safe state. Option C describes deadlock detection and recovery, which is a different strategy, while options B and D describe aims the algorithm does not pursue. Option A is correct.

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100. Question 100

     What information is used to determine if
     a resource request can be granted ?

     The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.

     1. AAvailable resources and current allocation of each process
     2. BCPU utilization of each process
     3. CNumber of processes waiting for resources
     4. DArrival time of each process

     Answer: (A) Available resources and current allocation of each process

     Explanation

     The safety check compares the resources currently available against what each process still needs, which follows from its declared maximum and its current allocation. CPU utilisation, the length of the waiting queue and arrival times play no part in the decision. Option A is correct.

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