UGC NET Computer Science & Applications October 2022 Question PaperPaper 1 & Paper 2 with Answer Key

50 questions with the answer key and explanations.

1. Question 1

   The difference between the number of boys and girls studying in College D is:

   The table gives each college's student share out of 60,000 and girls' share out of 24,000.

   College	Students (% of 60,000)	Girls (% of 24,000)
   A	10%	15%
   B	9%	12%
   C	23%	18%
   D	18%	14%
   E	16%	20%
   F	24%	21%

   1. A4040
   2. B4080
   3. C7440
   4. D3360

   Answer: (B) 4080

   Explanation

   College D has 10,800 students and 3,360 girls. Its boys count is 7,440, so the difference is 7,440 - 3,360 = 4,080.

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2. Question 2

   The number of girls in College F is ____ % more than the number of girls in College A.

   The table gives each college's student share out of 60,000 and girls' share out of 24,000.

   College	Students (% of 60,000)	Girls (% of 24,000)
   A	10%	15%
   B	9%	12%
   C	23%	18%
   D	18%	14%
   E	16%	20%
   F	24%	21%

   1. A25
   2. B30
   3. C40
   4. D50

   Answer: (C) 40

   Explanation

   Girls in F are 21% of 24,000 = 5,040; in A they are 15% of 24,000 = 3,600. The increase is 1,440/3,600 × 100 = 40%.

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3. Question 3

   The ratio of the number of boys in College F to the number of boys in College D is:

   The table gives each college's student share out of 60,000 and girls' share out of 24,000.

   College	Students (% of 60,000)	Girls (% of 24,000)
   A	10%	15%
   B	9%	12%
   C	23%	18%
   D	18%	14%
   E	16%	20%
   F	24%	21%

   1. A31:39
   2. B39:31
   3. C29:37
   4. D37:29

   Answer: (B) 39:31

   Explanation

   F has 14,400 - 5,040 = 9,360 boys and D has 10,800 - 3,360 = 7,440 boys. 9,360:7,440 reduces to 39:31.

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4. Question 4

   The average number of boys studying in Colleges A, B and C is:

   The table gives each college's student share out of 60,000 and girls' share out of 24,000.

   College	Students (% of 60,000)	Girls (% of 24,000)
   A	10%	15%
   B	9%	12%
   C	23%	18%
   D	18%	14%
   E	16%	20%
   F	24%	21%

   1. A4900
   2. B4700
   3. C4400
   4. D4800

   Answer: (D) 4800

   Explanation

   The boys counts are A 6,000-3,600=2,400, B 5,400-2,880=2,520, and C 13,800-4,320=9,480. Their average is (2,400+2,520+9,480)/3 = 4,800.

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5. Question 5

   The number of girl students in College C as a percentage of the number of boys in College E is:

   The table gives each college's student share out of 60,000 and girls' share out of 24,000.

   College	Students (% of 60,000)	Girls (% of 24,000)
   A	10%	15%
   B	9%	12%
   C	23%	18%
   D	18%	14%
   E	16%	20%
   F	24%	21%

   1. A70%
   2. B75%
   3. C80%
   4. D90%

   Answer: (D) 90%

   Explanation

   College C has 18% of 24,000 = 4,320 girls. College E has 9,600 - 4,800 = 4,800 boys, so 4,320/4,800 × 100 = 90%.

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6. Question 6

   Which of the following three numbers in decimal, octal and hexadecimal notations, respectively, is/are equivalent to (11011001)₂?

   A. (217)₁₀
   B. (661)₈
   C. (D9)₁₆

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

   Answer: (D) A and C only

   Explanation

   11011001₂ = 128+64+16+8+1 = 217₁₀, and grouping it as 1101 1001 gives D9₁₆. Its octal form is 331₈, not 661₈.

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7. Question 7

   Human communication involves:

   A. Message transmission
   B. Message reception
   C. Verbal and non-verbal messages
   D. Power-packed messages only
   E. Messages for non-consumption

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

   Answer: (A) A, B and C only

   Explanation

   Communication entails sending and receiving messages, and messages may be verbal or non-verbal. “Power-packed” messages and messages for non-consumption are not defining components.

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8. Question 8

   Assertion (A): Discussion forums are one of the Four Quadrants for MOOCs.

   Reason (R): Discussion forums are an in-built feature of learning management systems (LMS).

   1. ABoth A and R are correct and R is the correct explanation of A
   2. BBoth A and R are correct but R is not the correct explanation of A
   3. CA is correct but R is not correct
   4. DA is not correct but R is correct

   Answer: (B) Both A and R are correct but R is not the correct explanation of A

   Explanation

   Discussion forums are a standard LMS feature, so the reason is true. The MOOC four-quadrant model instead comprises e-tutorials, e-content, discussion forums, and assessment, so an LMS feature does not explain that classification.

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9. Question 9

   Which goal of the 2030 Agenda for Sustainable Development seeks to ensure inclusive and equitable quality education and promote lifelong learning opportunities for all?

   1. AGoal 2
   2. BGoal 4
   3. CGoal 6
   4. DGoal 15

   Answer: (B) Goal 4

   Explanation

   Sustainable Development Goal 4 is the education goal. Goals 2, 6, and 15 address hunger, water and sanitation, and life on land respectively.

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10. Question 10

    Responses to open-ended questions are:

    1. AQuantifiable
    2. BRealistic
    3. CSubjective
    4. DImitative

    Answer: (C) Subjective

    Explanation

    Open-ended responses let participants answer in their own words, so they are qualitative and subjective rather than fixed, directly quantifiable responses.

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11. Question 11

    Communication apprehension is often described as:

    1. ASocial relativity
    2. BPsycho-social barrier
    3. CSocial anxiety
    4. DSocial ambiguity

    Answer: (C) Social anxiety

    Explanation

    Communication apprehension is anxiety or fear associated with actual or anticipated communication. “Social anxiety” captures that condition most directly.

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12. Question 12

    Match List-I with List-II.

    List-I: Learning definitions	List-II: Proponent
    A. Learning is an organization of behaviour	I. Guilford
    B. Learning is the change in behaviour resulting from behaviour	II. Skinner
    C. Learning is selecting the appropriate response and connecting with the stimulus	III. Garret
    D. Learning is a process of progressive behaviour adaptation	IV. Thorndike

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

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

    Explanation

    The correct mapping is A-III, B-I, C-IV, D-II. Garrett describes learning as organisation of behaviour, Guilford as behavioural change, Thorndike as selecting a response to a stimulus, and Skinner as progressive behavioural adaptation.

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13. Question 13

    Find the wrong term in the series: 6, 11, 18, 27, 34, 51, 66, 83.

    1. A34
    2. B51
    3. C83
    4. D11

    Answer: (A) 34

    Explanation

    The intended pattern is successive squares plus 2: 2²+2=6, 3²+2=11, 4²+2=18, 5²+2=27, then 6²+2=38. Therefore 34 is the wrong term.

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14. Question 14

    Two shopkeepers sell machines at the same list price. A gives successive discounts of 20% and 15%; B gives 10% and 25%. Which statement is correct?

    1. AA offers more discount
    2. BB offers more discount
    3. CBoth offer the same discount
    4. DA offers 35% discount and B offers 30% discount

    Answer: (B) B offers more discount

    Explanation

    A's retained price is .80 × .85 = .68, a 32% discount. B's is .90 × .75 = .675, a 32.5% discount, so B offers more.

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15. Question 15

    Which Indian Act makes it illegal to knowingly spread a computer virus?

    1. AData Protection and Security Act, 1997
    2. BInformation Security Act, 1998
    3. CInformation Technology Act, 2000
    4. DComputer Misuse and Cyber Act, 2009

    Answer: (C) Information Technology Act, 2000

    Explanation

    India's Information Technology Act, 2000 addresses computer-related offences, including damage to computer resources. The other titles are not the governing Indian statute.

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16. Question 16

    Statement I: Cache memory is volatile memory and is much slower than RAM.

    Statement II: CDs, DVDs and magnetic tapes are all optical media devices.

    1. ABoth Statement I and Statement II are true
    2. BBoth Statement I and Statement II are false
    3. CStatement I is true but Statement II is false
    4. DStatement I is false but Statement II is true

    Answer: (B) Both Statement I and Statement II are false

    Explanation

    Cache is volatile, but it is faster than main RAM. CDs and DVDs are optical media, whereas magnetic tape is magnetic storage. Both statements are therefore false.

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17. Question 17

    A geothermal field requires a combination of:

    A. A natural underground source of water
    B. A mountain in the vicinity
    C. An impermeable layer
    D. A coal mine in the vicinity
    E. A large mass of hot rock in the vicinity

    1. AA, C and E only
    2. BA, C and D only
    3. CB, C and D only
    4. DB, D and E only

    Answer: (A) A, C and E only

    Explanation

    A usable geothermal reservoir needs water, heat from hot rock, and a cap or impermeable layer that helps retain the heated fluid. A nearby mountain or coal mine is not required.

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18. Question 18

    At what annual compound-interest rate will a sum double in 14 years?

    1. A4%
    2. B5%
    3. C6%
    4. D6.5%

    Answer: (B) 5%

    Explanation

    The exact rate solves (1+r)^14 = 2, so r = 2^(1/14)-1 ≈ 5.08%. Among the offered whole-percentage choices, 5% is the intended nearest value.

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19. Question 19

    The National Education Policy 2020 recommended replacing the UGC with HECI and its four verticals. Which belong to the four verticals?

    A. National Higher Education Regulatory Council
    B. General Education Council
    C. Medical Council of India
    D. National Accreditation Council
    E. Higher Education Grants Council

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

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

    Explanation

    HECI's four verticals are NHERC, NAC, HEGC, and GEC. The Medical Council of India is not one of them.

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20. Question 20

    Which is a tool of grounded theory?

    1. AConstant comparison
    2. BAuditing
    3. CDeconstruction of narratives
    4. DUse of uncritical language

    Answer: (A) Constant comparison

    Explanation

    Grounded theory develops concepts inductively by continually comparing incidents, codes, and categories. This is the constant-comparative method.

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21. Question 21

    Statement I: Natural erosion is the gradual removal of topsoil by natural processes.

    Statement II: Accelerated erosion is caused by human activities and occurs at the same rate as soil formation.

    1. ABoth Statement I and Statement II are true
    2. BBoth Statement I and Statement II are false
    3. CStatement I is true but Statement II is false
    4. DStatement I is false but Statement II is true

    Answer: (C) Statement I is true but Statement II is false

    Explanation

    Natural erosion proceeds through natural agents such as water and wind. Accelerated erosion is human-induced and removes soil faster than it can form, so Statement II is false.

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22. Question 22

    Statement I: Non-probability samples use available respondents without a specific selection procedure.

    Statement II: Non-probability samples accurately reflect population characteristics.

    1. ABoth Statement I and Statement II are true
    2. BBoth Statement I and Statement II are false
    3. CStatement I is true but Statement II is false
    4. DStatement I is false but Statement II is true

    Answer: (C) Statement I is true but Statement II is false

    Explanation

    Convenience and other non-probability samples do not use random selection. Because selection probabilities are unknown, they cannot be assumed to represent the population accurately.

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23. Question 23

    If “All women are honest” is true, which propositions can be inferred?

    A. “No woman is honest” is false
    B. “Some women are honest” is true
    C. “No woman is honest” is undetermined
    D. “Some women are not honest” is false

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

    Answer: (A) A, B and D only

    Explanation

    Under the traditional categorical-logic assumption that the class of women is nonempty, universal inclusion makes “no woman is honest” false, ensures some honest women, and rules out some women not being honest. Without existential import, the inference needs qualification.

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24. Question 24

    In Nyāya syllogism, all three terms stand synthesized at which step of the inferential process?

    1. AExample (Udāharaṇa)
    2. BConclusion (Nigamana)
    3. CApplication (Upanaya)
    4. DReason (Hetu)

    Answer: (C) Application (Upanaya)

    Explanation

    Upanaya applies the universal relation illustrated in the example to the minor term at hand. It is the step in which the relevant terms are brought together before the conclusion.

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25. Question 25

    Which are sources of data in historical research?

    A. Personal observation
    B. Focused group discussion
    C. Oral testimony
    D. Relics
    E. Actuaries

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

    Answer: (D) C, D and E only

    Explanation

    Oral testimony and relics are recognised historical sources. Actuaries' historical records and statistical tables can also supply historical data, whereas personal observation and a focus-group discussion collect contemporary data. Therefore C, D and E is the correct combination.

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26. Question 26

    Which logical informal fallacy is committed in the following argument? “Mr. X used abusive language toward the child who threw a stone at his car. Since child abuse is a crime, he should be reported to the authorities.”

    1. AAppeal to emotion
    2. BHasty generalisation
    3. CEquivocation
    4. DAppeal to force

    Answer: (C) Equivocation

    Explanation

    Equivocation trades on a word or phrase changing meaning. “Abusive language” is not the legal offence of child abuse.

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27. Question 27

    Statement I: The contrapositive replaces the subject by the complement of the predicate and the predicate by the complement of the subject.

    Statement II: All contrapositions are valid.

    1. ABoth statements are true
    2. BBoth statements are false
    3. CStatement I is true but Statement II is false
    4. DStatement I is false but Statement II is true

    Answer: (C) Statement I is true but Statement II is false

    Explanation

    Statement I gives the standard contraposition transformation. In traditional categorical logic, contraposition is valid for A and O propositions, not universally for every form.

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28. Question 28

    Statement I: NEP 2020 changes 10+2 to 5+4+4+3 for ages 3-18.

    Statement II: Before age 5, every child will move to a Preparatory Class or Balvatika with an ECCE-qualified teacher.

    1. ABoth statements are correct
    2. BBoth statements are incorrect
    3. CStatement I is correct but Statement II is incorrect
    4. DStatement I is incorrect but Statement II is correct

    Answer: (D) Statement I is incorrect but Statement II is correct

    Explanation

    NEP 2020 specifies a 5+3+3+4 design, making Statement I false. It provides for Balavatika before Class 1 with an ECCE-qualified teacher, making Statement II true.

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29. Question 29

    Chlorine in drinking water is used for:

    A. Disinfection
    B. Removal of hardness
    C. Odour treatment
    D. Turbidity control
    E. Removal of iron and manganese

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

    Answer: (A) A, C and E only

    Explanation

    Chlorination disinfects water and can aid odour control and oxidation of iron or manganese. It does not remove hardness or directly control turbidity.

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30. Question 30

    Match the protocol or summit in List-I with its theme in List-II.

    List-I: Protocol/Summit	List-II: Theme
    A. Paris Agreement	I. Emissions trading
    B. Kyoto Protocol	II. Ozone depletion
    C. Rio Declaration	III. INDCs
    D. Montreal Protocol	IV. Environment and development

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

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

    Explanation

    The correct mapping is A-III, B-I, C-IV, D-II: the Paris Agreement uses INDCs, Kyoto established emissions trading, Rio concerns environment and development, and Montreal targets ozone depletion.

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31. Question 31

    Match the communication context in List-I with its related factor in List-II.

    List-I: Communication context	List-II: Related factor
    A. Physical	I. Group norms
    B. Cultural	II. Sequential positioning
    C. Social and Psychological	III. Tangible environment
    D. Temporal	IV. Value system

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

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

    Explanation

    The correct mapping is A-III, B-IV, C-I, D-II: physical context is tangible environment, cultural context reflects a value system, social and psychological context includes group norms, and temporal context is sequential positioning.

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32. Question 32

    The grapevine communication is often driven by:

    1. AExternal professional agencies
    2. BCompeting organisations
    3. CTop management of an organisation
    4. DSocial networks of employees

    Answer: (D) Social networks of employees

    Explanation

    Grapevine communication is informal communication transmitted through employees' social relationships rather than the formal hierarchy.

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33. Question 33

    Which formula entered in D1 will give the shown values when copied to D2 and D3?

    	A	B	C	D
    1		5	10	50
    2			15	50
    3			20	50
    4

    1. AB1 * $C$1
    2. B$B$1 * C1
    3. CB1 * C1
    4. D$B$1 * $C$1

    Answer: (D) $B$1 * $C$1

    Explanation

    The value must remain 5 × 10 = 50 in every copied row. Absolute references $B$1 and $C$1 keep both factors fixed.

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34. Question 34

    Which statements about CERT-In are correct?

    A. It is the national nodal agency for responding to computer-security incidents.
    B. It has operated since January 2014.
    C. Forecast and alert of cyber-security incidents is one of its functions.

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

    Answer: (C) A and C only

    Explanation

    CERT-In is India's national incident-response agency and issues forecasts and alerts. It began operations in 2004, not January 2014.

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35. Question 35

    Which are actions of an ethnographic researcher?

    A. Gets immersed in the social setting
    B. Avoids collecting documents about the group
    C. Observes group members' behaviour
    D. Listens to their conversation
    E. Does not interview participants who are not amenable to observation

    1. AA, B and C only
    2. BA, C and D only
    3. CB, C and D only
    4. DC, D and E only

    Answer: (B) A, C and D only

    Explanation

    Ethnography relies on immersion, observation, and listening in the social setting. Researchers may collect documents and interview participants, so B and E do not describe the method.

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36. Question 36

    Statement I: In symbolic communication, power operates through images.

    Statement II: This advantage of exercising power does not exist in other types of communication.

    1. ABoth statements are true
    2. BBoth statements are false
    3. CStatement I is true but Statement II is false
    4. DStatement I is false but Statement II is true

    Answer: (C) Statement I is true but Statement II is false

    Explanation

    Symbols and images can exercise power by shaping meanings and associations. Power can also operate through language, institutions, and other communication forms, so Statement II is false.

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37. Question 37

    Which is a group-centred teaching-learning method?

    1. AProviding lecture notes
    2. BTeam-teaching
    3. CDemonstration method
    4. DBrainstorming

    Answer: (D) Brainstorming

    Explanation

    Brainstorming has learners generate and build on ideas as a group. Lecture notes and demonstration are chiefly teacher-led, while team teaching refers to teachers collaborating.

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38. Question 38

    Which educational institution was the first to start college classes for women?

    1. ACentral Hindu Girls School
    2. BVasanta College
    3. CMiranda House
    4. DBethune School for Girls

    Answer: (D) Bethune School for Girls

    Explanation

    Bethune School, later Bethune College in Calcutta, pioneered women's college education in India. The other institutions were founded later.

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39. Question 39

    Statement I: Volatile organic chemicals are among the most commonly found groundwater contaminants.

    Statement II: Their concentration in groundwater is much less than in surface waters.

    1. ABoth statements are true
    2. BBoth statements are false
    3. CStatement I is true but Statement II is false
    4. DStatement I is false but Statement II is true

    Answer: (C) Statement I is true but Statement II is false

    Explanation

    VOCs are frequent groundwater contaminants, especially near industrial and fuel sources. They can persist in groundwater, so the claimed lower concentration is not generally true.

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40. Question 40

    Which level of Bloom's taxonomy is achieved by rote learning?

    1. ARemembering
    2. BApplying
    3. CAnalysing
    4. DUnderstanding

    Answer: (A) Remembering

    Explanation

    Rote learning focuses on recalling facts or procedures. That corresponds to the remembering level, not understanding, application, or analysis.

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41. Question 41

    Indira Gandhi National Open University was established in:

    1. A1975
    2. B1980
    3. C1985
    4. D2000

    Answer: (C) 1985

    Explanation

    IGNOU was established by an Act of Parliament in 1985 to advance open and distance learning.

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42. Question 42

    Find the average of the squares of the consecutive odd numbers from 1 to 21.

    1. A162
    2. B159
    3. C161
    4. D160

    Answer: (C) 161

    Explanation

    The odd numbers are the first 11 odds. Their squared sum is 11(2×11-1)(2×11+1)/3 = 1,771, so the average is 1,771/11 = 161.

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43. Question 43

    Two numbers are in the ratio 7:8. If their difference is 30, find the numbers.

    1. A215, 245
    2. B205, 235
    3. C220, 250
    4. D210, 240

    Answer: (D) 210, 240

    Explanation

    The one ratio-unit difference equals 30. Therefore the numbers are 7×30=210 and 8×30=240.

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44. Question 44

    Statement I: Aristotelian syllogism treats deduction and induction as inseparably related.

    Statement II: The Nyāya school treats deduction and induction as two aspects of the same process.

    1. ABoth statements are true
    2. BBoth statements are false
    3. CStatement I is true but Statement II is false
    4. DStatement I is false but Statement II is true

    Answer: (D) Statement I is false but Statement II is true

    Explanation

    Aristotle distinguishes deduction from induction rather than treating them as inseparable. Nyāya inference connects induction-like universal concomitance with deductive application, so Statement II is true.

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45. Question 45

    To develop students holistically, a teacher should emphasise:

    1. ASubject knowledge only
    2. BRemedial classes for slow learners
    3. CExplaining the concept well
    4. DOutcome-based education along with value education

    Answer: (D) Outcome-based education along with value education

    Explanation

    Holistic development includes knowledge, skills, attitudes, and values. Outcome-based education combined with value education addresses this broader aim.

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46. Question 46

    One characteristic of languages is:

    Languages can acquire prestige and power through religion, scholarship, conquest and imperialism. At different points in history, a language spoken by a relatively small minority may become the medium of scholarship, administration, record-keeping or religious ceremony. Latin in medieval Europe is an example. The language of conquerors may similarly become the language of commerce, administration and law. Imperial languages can enter higher education, science and technology, while older or local languages are held back or excluded from these spheres.

    Languages are never static. They change in response to developments in knowledge, technology, social relations, politics and economics. Words change their meanings and acquire new applications; new words enter use and older expressions may decline. Sometimes these changes are gradual and sometimes rapid. Language is therefore not simply a fixed corpus of learning but an instrument continually adapted to human purposes.

    1. AConstant change in meanings of words
    2. BSubjection to different alien powers
    3. CAbility to survive against heavy odds
    4. DMonopoly by elites

    Answer: (A) Constant change in meanings of words

    Explanation

    The passage says meanings change, new words enter use, and older expressions decline. This continual semantic change is its stated characteristic.

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47. Question 47

    Historically, minority languages were the basis of:

    Languages can acquire prestige and power through religion, scholarship, conquest and imperialism. At different points in history, a language spoken by a relatively small minority may become the medium of scholarship, administration, record-keeping or religious ceremony. Latin in medieval Europe is an example. The language of conquerors may similarly become the language of commerce, administration and law. Imperial languages can enter higher education, science and technology, while older or local languages are held back or excluded from these spheres.

    Languages are never static. They change in response to developments in knowledge, technology, social relations, politics and economics. Words change their meanings and acquire new applications; new words enter use and older expressions may decline. Sometimes these changes are gradual and sometimes rapid. Language is therefore not simply a fixed corpus of learning but an instrument continually adapted to human purposes.

    1. AMedieval culture
    2. BCultural independence
    3. CScholarship
    4. DReligious harmony

    Answer: (C) Scholarship

    Explanation

    The passage explains that a language spoken by a relatively small minority could become the medium of scholarship. That directly supports scholarship.

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48. Question 48

    The language of conquerors was used in:

    Languages can acquire prestige and power through religion, scholarship, conquest and imperialism. At different points in history, a language spoken by a relatively small minority may become the medium of scholarship, administration, record-keeping or religious ceremony. Latin in medieval Europe is an example. The language of conquerors may similarly become the language of commerce, administration and law. Imperial languages can enter higher education, science and technology, while older or local languages are held back or excluded from these spheres.

    Languages are never static. They change in response to developments in knowledge, technology, social relations, politics and economics. Words change their meanings and acquire new applications; new words enter use and older expressions may decline. Sometimes these changes are gradual and sometimes rapid. Language is therefore not simply a fixed corpus of learning but an instrument continually adapted to human purposes.

    1. ARuling over landed elite
    2. BAdministration of law
    3. CColonising other countries
    4. DCreating social inclusiveness

    Answer: (B) Administration of law

    Explanation

    The passage explicitly names commerce, administration, and law as domains in which conquerors' languages may be used. Administration of law is the matching choice.

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49. Question 49

    Imperialism was responsible for:

    Languages can acquire prestige and power through religion, scholarship, conquest and imperialism. At different points in history, a language spoken by a relatively small minority may become the medium of scholarship, administration, record-keeping or religious ceremony. Latin in medieval Europe is an example. The language of conquerors may similarly become the language of commerce, administration and law. Imperial languages can enter higher education, science and technology, while older or local languages are held back or excluded from these spheres.

    Languages are never static. They change in response to developments in knowledge, technology, social relations, politics and economics. Words change their meanings and acquire new applications; new words enter use and older expressions may decline. Sometimes these changes are gradual and sometimes rapid. Language is therefore not simply a fixed corpus of learning but an instrument continually adapted to human purposes.

    1. ACodification of law
    2. BRacial equity
    3. CPrimacy of local languages
    4. DExclusion of old languages

    Answer: (D) Exclusion of old languages

    Explanation

    The passage says imperial languages can enter dominant domains while older or local languages are held back or excluded. Thus exclusion of old languages follows.

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50. Question 50

    The passage analyses language as a:

    Languages can acquire prestige and power through religion, scholarship, conquest and imperialism. At different points in history, a language spoken by a relatively small minority may become the medium of scholarship, administration, record-keeping or religious ceremony. Latin in medieval Europe is an example. The language of conquerors may similarly become the language of commerce, administration and law. Imperial languages can enter higher education, science and technology, while older or local languages are held back or excluded from these spheres.

    Languages are never static. They change in response to developments in knowledge, technology, social relations, politics and economics. Words change their meanings and acquire new applications; new words enter use and older expressions may decline. Sometimes these changes are gradual and sometimes rapid. Language is therefore not simply a fixed corpus of learning but an instrument continually adapted to human purposes.

    1. ACorpus of learning
    2. BSymbol of State power
    3. CTool to meet human purposes
    4. DFormal link between old and new generations

    Answer: (C) Tool to meet human purposes

    Explanation

    The passage concludes that language is not merely a fixed corpus of learning. It is continually adapted as an instrument for human purposes.

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100 questions with the answer key and explanations.

1. Question 1

   In a database, a rule is defined as (P1 and P2) or P3: R1 (0.8) and R2 (0.3), where P1, P2, P3 are premises and R1, R2 are conclusions of rules with certainty factors (CF) 0.8 and 0.3 respectively. If any running program has produced P1, P2, P3 with CF as 0.5, 0.8, 0.2 respectively, find the CF of results on the basis of premises.

   1. ACF (R1 = 0.8), CF (R2 = 0.3)
   2. BCF (R1 = 0.40), CF (R2 = 0.15)
   3. CCF (R1 = 0.15), CF (R2 = 0.35)
   4. DCF (R1 = 0.8), CF (R2 = 0.35)

   Answer: (B) CF (R1 = 0.40), CF (R2 = 0.15)

   Explanation

   Certainty factor rules combine premises before scaling by the rule's own CF.

   For an AND premise, CF(P1 and P2) = min(CF(P1), CF(P2)) = min(0.5, 0.8) = 0.5
   For an OR premise, the combined premise strength is max(CF(P1 and P2), CF(P3)) = max(0.5, 0.2) = 0.5

   Each conclusion then scales this combined premise CF by its own rule CF.
   CF(R1) = 0.5 × 0.8 = 0.40
   CF(R2) = 0.5 × 0.3 = 0.15

   This gives CF(R1 = 0.40) and CF(R2 = 0.15).

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2. Question 2

   There are three boxes. First box has 2 white, 3 black and 4 red balls. Second box has 3 white, 2 black and 2 red balls. Third box has 4 white, 1 black and 3 red balls. A box is chosen at random and 2 balls are drawn out of which 1 is white, and 1 is red. What is the probability that the balls came from first box?

   1. A0.237
   2. B0.723
   3. C0.18
   4. D0.452

   Answer: (A) 0.237

   Explanation

   This is a Bayes' theorem problem. Each box is chosen with prior probability 1/3, and the likelihood of drawing exactly one white and one red ball is C(white,1) × C(red,1) / C(total,2) for each box.

   Box 1 (9 balls): (2 × 4) / C(9,2) = 8/36 = 0.2222
   Box 2 (7 balls): (3 × 2) / C(7,2) = 6/21 = 0.2857
   Box 3 (8 balls): (4 × 3) / C(8,2) = 12/28 = 0.4286

   P(box1 | data) = P(data|box1) / [P(data|box1) + P(data|box2) + P(data|box3)]
   = 0.2222 / (0.2222 + 0.2857 + 0.4286) = 0.2222 / 0.9365 ≈ 0.237

   Box 3 has the highest individual likelihood, which is why option D (0.452, a box-3-weighted figure) looks tempting, but Bayes' rule normalizes by the sum over all three boxes, giving box 1 a share of about 0.237.

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3. Question 3

   Consider a memory system having address spaced at a distance of m, T = Bank cycle time and n number of banks, then the average data access time per word access in synchronous organization is

   1. At = { m·T/n for m << n ; T for m >> n }
   2. Bt = { T/n for m << n ; T for m >> n }
   3. Ct = { m·T for m << n ; T for m >> n }
   4. Dt = { m·T for m << n ; m/T for m >> n }

   Answer: (B) t = { T/n for m << n ; T for m >> n }

   Explanation

   In a low-order interleaved memory with n banks and bank cycle time T, the number of words requested (m) relative to n decides how much bank overlap is available.

   When m << n, there are far fewer requests than banks, so successive accesses pipeline almost fully across independent banks, giving an average access time per word of T/n.
   When m >> n, requests exceed the available banks, so each bank must be reused and the access time per word approaches the full bank cycle time T, since overlap can no longer hide the cycle time.

   This gives t = T/n for m << n and t = T for m >> n.

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4. Question 4

   For multiprocessor system, interconnection network – cross bar switch is an example of

   1. ANon blocking network
   2. BBlocking network
   3. CThat varies from connection to connection
   4. DRecurrent network

   Answer: (A) Non blocking network

   Explanation

   A crossbar switch provides a dedicated path between every input and every output through a grid of switching points. Because any free input can always be connected to any free output without contention from other active connections, a crossbar is a non-blocking network by definition.

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5. Question 5

   The representation of 4 bit code 1101 into 7 bit, even parity Hamming code is

   1. A(1010101)
   2. B(1111001)
   3. C(1011101)
   4. D(1110000)

   Answer: (A) (1010101)

   Explanation

   Place data bits 1, 1, 0, 1 at positions 3, 5, 6, 7 and parity bits at positions 1, 2, 4 of a 7-bit codeword.

   p1 covers positions 1, 3, 5, 7 (values 1, 1, 1) so p1 must be 1 to keep the group even
   p2 covers positions 2, 3, 6, 7 (values 1, 0, 1) so p2 must be 0 to keep the group even
   p4 covers positions 4, 5, 6, 7 (values 1, 0, 1) so p4 must be 0 to keep the group even

   Reading positions 1 through 7 gives 1, 0, 1, 0, 1, 0, 1, which is 1010101.

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6. Question 6

   The number of gate inputs, required to realize expression A̅BC + A̅BCD + EF̅ + A̅D is

   1. A12
   2. B13
   3. C14
   4. D15

   Answer: (D) 15

   Explanation

   Count the literal inputs feeding each AND gate for the four product terms exactly as written, then add one input per term for the final OR gate.

   A̅BC needs 3 literal inputs
   A̅BCD needs 4 literal inputs
   EF̅ needs 2 literal inputs
   A̅D needs 2 literal inputs

   Total AND-gate inputs = 3 + 4 + 2 + 2 = 11
   The OR gate combining the 4 product terms needs 4 inputs

   Total gate inputs = 11 + 4 = 15.

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7. Question 7

   Consider a logic gate circuit, with 8 input lines (D0, D1, ….. D7) and 3 output lines (A0, A1, A2) specified by following operations
   A2 = D4 + D5 + D6 + D7
   A1 = D2 + D3 + D6 + D7
   A0 = D1 + D3 + D5 + D0
   Where + indicates logical OR operation. This circuit is

   1. A8 × 8 multiplexer
   2. BDecimal to BCD converter
   3. COctal to Binary encoder
   4. DPriority encoder

   Answer: (C) Octal to Binary encoder

   Explanation

   Each output bit is simply the OR of the input lines whose index has that bit set, which is exactly the standard 8-to-3 (octal to binary) encoder construction: A2 is 1 for D4 through D7 (bit 2 set), A1 is 1 whenever bit 1 of the active line is set, and A0 is 1 whenever bit 0 of the active line is set.

   A priority encoder would need extra logic to resolve which input wins when more than one line is active simultaneously, and none of that priority-resolution logic appears here, so the equations describe a plain octal-to-binary encoder rather than a priority encoder or a multiplexer.

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8. Question 8

   The total storage capacity of a floppy disk having 80 tracks and storing 128 bytes/sector is 163,840 bytes. How many sectors does this disk have?

   1. A27
   2. B2048
   3. C4K
   4. D16

   Answer: (D) 16

   Explanation

   Total capacity equals tracks × sectors per track × bytes per sector.

   Sectors per track = 163,840 / (80 × 128) = 163,840 / 10,240 = 16

   The question's own numbers only resolve consistently for sectors per track, since 80 tracks × 16 sectors per track × 128 bytes = 163,840 bytes matches the given capacity exactly.

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9. Question 9

   In a cache memory, if address has 9 bits in Tag field and 12 bits in index field, the size of main memory and cache memory would be respectively

   1. A2 K, 4 K
   2. B1024 K, 2 K
   3. C4 K, 2048 K
   4. D2048 K, 4 K

   Answer: (D) 2048 K, 4 K

   Explanation

   With no separate block-offset field, the index field directly addresses cache lines and the tag plus index together address main memory words.

   Main memory size = 2^(tag bits + index bits) = 2^(9+12) = 2^21 words = 2048 K
   Cache memory size = 2^(index bits) = 2^12 words = 4 K

   So main memory is 2048 K and cache memory is 4 K.

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10. Question 10

    Consider the primal problem:
    Maximize z = 5x₁ + 12x₂ + 4x₃
    Subject to x₁ + 2x₂ + x₃ = 10
    2x₁ − x₂ + 3x₃ = 8
    x₁, x₂, x₃ ≥ 0
    its dual problem is
    Minimize w = 10y₁ + 8y₂
    Subject to y₁ + 2y₂ ≥ 5
    2y₁ − y₂ ≥ 12
    y₁ + 3y₂ ≥ 4
    Which of the following is correct?

    1. Ay₁ ≥ 0, y₂ unrestricted
    2. By₁ ≥ 0, y₂ ≥ 0
    3. Cy₁ is unrestricted, y₂ ≥ 0
    4. Dy₁ is unrestricted, y₂ restricted

    Answer: (C) y₁ is unrestricted, y₂ ≥ 0

    Explanation

    By the standard primal-dual sign rule, a dual variable's restriction is set by its matching primal constraint's type. Both primal constraints here are equalities, so by that rule both y1 and y2 should be unrestricted in sign, since an equality constraint in the primal corresponds to a free (unrestricted) dual variable rather than a sign-restricted one.

    None of the four listed options states that both y1 and y2 are unrestricted, so this item cannot be answered with full confidence against the printed choices. Option C is kept as the closest listed choice pending a human check against the official source, since it is the only option that lets y1 (the variable tied to the first equality constraint) stay unrestricted.

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11. Question 11

    The logic expression (P̅ ∧ Q) ∨ (P ∧ Q̅) ∨ (P ∧ Q) is equivalent to

    1. AP̅ ∨ Q
    2. BP ∨ Q̅
    3. CP ∨ Q
    4. DP̅ ∨ Q̅

    Answer: (C) P ∨ Q

    Explanation

    Group the last two terms first.

    (P ∧ Q̅) ∨ (P ∧ Q) = P ∧ (Q̅ ∨ Q) = P

    The expression reduces to (P̅ ∧ Q) ∨ P, and by the absorption law P ∨ (P̅ ∧ Q) = P ∨ Q.

    So the full expression is equivalent to P ∨ Q.

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12. Question 12

    The reduced grammar equivalent to the grammar, whose production rules are given below, is
    S → AB | CA
    B → BC | AB
    A → a
    C → aB | b

    1. AS → CA, A → a, C → b
    2. BS → CA | B, B → BC | B, A → a, C → aB | b
    3. CS → CA | B, B → BC, A → a, C → aB | b
    4. DS → AB | AC, B → BC | BA, A → a, C → aB | b

    Answer: (A) S → CA, A → a, C → b

    Explanation

    First remove non-generating symbols, those that can never derive a terminal string.

    Every production for B is B → BC or B → AB, and both alternatives still contain B itself, so B can never bottom out in a terminal string. B is non-generating and must be removed, along with every production that mentions it.

    Removing B eliminates S → AB and C → aB, leaving S → CA and C → b as the only survivors alongside A → a.

    Checking reachability from S confirms every remaining symbol (A, C) and the terminals they produce are reachable, so no further symbols are removed.

    The reduced grammar is S → CA, A → a, C → b.

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13. Question 13

    Consider the production rules of grammer G:
    S → AbB
    A → aAb | λ
    B → bB | λ
    Which of the following language L is generated by grammer G?

    1. AL = {aⁿbᵐ : n ≥ 0, m > n}
    2. BL = {aⁿbᵐ : n ≥ 0, m ≥ 0}
    3. CL = {aⁿbᵐ : n ≥ m}
    4. DL = {aⁿbᵐ : n ≥ m, m > 0}

    Answer: (A) L = {aⁿbᵐ : n ≥ 0, m > n}

    Explanation

    A → aAb | λ generates exactly n matched a's followed by n b's, that is aⁿbⁿ, for any n ≥ 0.

    B → bB | λ independently generates any number of extra trailing b's, say k ≥ 0.

    Since S → A b B, the full derivation is (aⁿbⁿ) followed by one fixed b from the rule itself, followed by bᵏ from B, giving aⁿ b^(n+1+k).

    The total power of b is n + 1 + k, which is always at least n + 1, so it is always strictly greater than n regardless of k.

    This matches L = {aⁿbᵐ : n ≥ 0, m > n}.

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14. Question 14

    Consider the language L = {aⁿbᵐ : n ≥ 4, m ≤ 3}. Which of the following regular expression represents language L?

    1. Aaaaa*(λ + b + bb + bbb)
    2. Baaaaa*(b + bb + bbb)
    3. Caaaaa*(λ + b + bb + bbb)
    4. Daaaa*(b + bb + bbb)

    Answer: (C) aaaaa*(λ + b + bb + bbb)

    Explanation

    The a-part must represent n ≥ 4 a's, and the b-part must represent 0 to 3 b's.

    aaaaa* is four fixed a's followed by a-star on the fifth, so it matches a⁴, a⁵, a⁶, ... which is exactly n ≥ 4. aaaa* only fixes three a's before the star, matching n ≥ 3, which is one too few.

    (λ + b + bb + bbb) covers m = 0, 1, 2, and 3. Dropping the λ term, as in (b + bb + bbb), would exclude the valid case m = 0.

    Only aaaaa*(λ + b + bb + bbb) gets both the a-count and the full 0–3 range of b's correct.

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15. Question 15

    Consider L = {ab, aa, baa}. Which of the following string is NOT in L*?

    1. Abaaaaabaaaaa
    2. Babaabaaabaa
    3. Caaaabaaaa
    4. Dbaaaaabaa

    Answer: (A) baaaaabaaaaa

    Explanation

    Try to tile each string using only the blocks ab, aa, and baa.

    For baaaaabaaaaa, since only baa starts with b, the first block is forced to be baa, leaving aaabaaaaa. The next block must be aa (positions 4–5), leaving abaaaaa, whose only fit is ab (positions 6–7), leaving a plain run of 5 a's. A run of a's can only be tiled by aa blocks, and 5 is odd, so this string cannot be fully tiled and is not in L*.

    By contrast, baaaaabaa splits cleanly as baa + aa + ab + aa, abaabaaabaa splits as ab + aa + baa + ab + aa, and aaaabaaaa splits as aa + aa + baa + aa, so B, C, and D are all in L*.

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16. Question 16

    Consider the following NPDA = ({q₀, q₁, qf}, {a,b}, {1,z}, δ, q₀, z, {qf})
    δ(q₀, λ, z) = {(q₁, z)}
    δ(q₀, a, z) = {(q₁, 11z)}
    δ(q₁, a, 1) = {(q₁, 111)}
    δ(q₁, b, 1) = {(q₁, λ)}
    δ(q₁, λ, z) = {(qf, z)}
    Which of the following Language L is accepted by NPDA?

    1. AL = {a²ⁿbⁿ : n ≥ 0}
    2. BL = {aⁿb²ⁿ : n ≥ 0}
    3. CL = {a²ⁿbⁿ : n > 0}
    4. DL = {aⁿb²ⁿ : n > 0}

    Answer: (B) L = {aⁿb²ⁿ : n ≥ 0}

    Explanation

    Trace the stack contents. Reading the first a replaces z with 11z, pushing 2 ones. Each further a replaces one 1 on top with 111, a net gain of 2 ones per a. So after reading n a's, the stack holds exactly 2n ones above z.

    Each b then pops exactly one 1 from the stack. Acceptance happens only when a λ-move sees z on top, which requires all 2n ones to have been popped first, so the number of b's read must equal 2n.

    This gives the language {aⁿb²ⁿ}. The empty string is also accepted directly through δ(q₀, λ, z) → δ(q₁, λ, z) → qf without reading any symbol, so n = 0 is included.

    The language is L = {aⁿb²ⁿ : n ≥ 0}.

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17. Question 17

    Hidden surface removal problem with minimal 3D pipeline can be solved with

    1. APainter's algorithm
    2. BWindow Clipping algorithm
    3. CBrute force rasterization algorithm
    4. DFlood fill algorithm

    Answer: (A) Painter's algorithm

    Explanation

    Painter's algorithm removes hidden surfaces with minimal extra pipeline machinery by simply depth-sorting polygons from farthest to nearest and drawing them in that order, so nearer surfaces naturally overwrite farther ones on the frame buffer.

    Window clipping only restricts geometry to the viewport and does not resolve visibility, brute-force rasterization scan-converts polygons without any depth logic of its own, and flood fill is a region-filling technique unrelated to depth or visibility at all.

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18. Question 18

    Using 'RSA' algorithm, if p = 13, q = 5 and e = 7, the value of d and cipher value of '6' with (e, n) key are

    1. A7, 4
    2. B7, 1
    3. C7, 46
    4. D55, 1

    Answer: (C) 7, 46

    Explanation

    n = p × q = 65, and φ(n) = (p−1)(q−1) = 12 × 4 = 48.

    Find d such that e·d ≡ 1 (mod φ(n)): 7 × 7 = 49 ≡ 1 (mod 48), so d = 7.

    Encrypt message 6 as C = 6^e mod n = 6^7 mod 65.
    Step through powers mod 65: 6² = 36, 6³ = 21, 6⁴ = 61, 6⁵ = 41, 6⁶ = 51, 6⁷ = 46

    So d = 7 and the cipher value of 6 is 46.

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19. Question 19

    The condition num != 65 cannot be replaced by

    1. Anum > 65 || num < 65
    2. B!(num == 65)
    3. Cnum − 65
    4. D!(num = 65)

    Answer: (D) !(num = 65)

    Explanation

    • A. num > 65 || num < 65 — equivalent, num is not 65 exactly when it is either greater or smaller than 65.
    • B. !(num == 65) — equivalent, this is the direct logical negation of equality.
    • C. num − 65 — equivalent as a truth value, since this expression is nonzero (treated as true in C) precisely when num is not 65, and zero (false) only when num equals 65.
    • D. !(num = 65) — not equivalent. The single = is an assignment, not a comparison, so this always assigns 65 to num first. The expression (num = 65) then evaluates to 65, which is truthy, so !(num = 65) is always false and it also destructively overwrites num's original value, unlike num != 65.

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20. Question 20

    Pointers cannot be used to

    1. Afind the address of a variable in memory
    2. Breference value directly
    3. Csimulate call by reference
    4. Dmanipulate dynamic data structure

    Answer: (B) reference value directly

    Explanation

    A pointer's entire purpose is indirect access, it stores an address and reaches a value only by dereferencing through that address. Referencing a value directly is what an ordinary variable does, not what a pointer does, so pointers cannot be used to reference a value directly.

    Pointers do find a variable's address (via the address-of operator), do simulate call by reference (by passing an address for the callee to dereference), and do manipulate dynamic data structures such as linked lists and trees.

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21. Question 21

    Which mechanism in XML allows organizations to specify globally unique names as element tags in documents?

    1. Aroot
    2. Bheader
    3. Cschema
    4. Dnamespace

    Answer: (D) namespace

    Explanation

    An XML namespace binds a URI to a prefix, and that URI guarantees the element and attribute names using that prefix are globally unique even when documents from different organizations reuse the same tag names, avoiding naming collisions when documents are combined.

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22. Question 22

    If an operating system does not allow a child process to exist when the parent process has been terminated, this phenomenon is called as -

    1. AThreading
    2. BCascading termination
    3. CZombie termination
    4. DProcess killing

    Answer: (B) Cascading termination

    Explanation

    When an operating system's policy forces every child process to be terminated as soon as its parent terminates, so that termination propagates down the process tree, this is called cascading termination. It is a deliberate OS design choice, not the same as a zombie process, which is a terminated process whose exit status has not yet been collected by its parent.

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23. Question 23

    What is called Journalling in Linux operating system?

    1. AProcess scheduling
    2. BFile saving as transaction
    3. CA type of thread
    4. DAn editor

    Answer: (B) File saving as transaction

    Explanation

    A journaling file system logs pending file system changes as transactions in a journal before committing them to the main structures. If the system crashes mid-write, the journal lets it replay or discard the incomplete transaction on recovery, keeping the file system consistent. This is why journaling is described as saving file changes as transactions.

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24. Question 24

    This transformation is called
    [x̅ y̅ z̅ w̅]ᵀ = [[a₁ b₁ c₁ d₁],[a₂ b₂ c₂ d₂],[a₃ b₃ c₃ d₃],[e f g h]] [x y z 1]ᵀ

    1. AScaling
    2. BShear
    3. CHomography
    4. DSteganography

    Answer: (C) Homography

    Explanation

    In a standard affine transform such as scaling or shear, the bottom row of the 4×4 homogeneous matrix is fixed at [0 0 0 1], leaving the homogeneous coordinate w unchanged. Here the bottom row is the general [e f g h], which lets w vary with x, y, and z, producing perspective-style division effects on the transformed coordinates.

    This general projective (perspective-capable) form of a homogeneous coordinate transform is called a homography, distinguishing it from simpler affine operations like pure scaling or shear.

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25. Question 25

    RAD software process model stands for

    1. ARapid Application Development
    2. BRelative Application Development
    3. CRapid Application Design
    4. DRecent Application Development

    Answer: (A) Rapid Application Development

    Explanation

    RAD stands for Rapid Application Development, a process model that emphasizes a very short development cycle using component-based construction, typically completing a working system in 60 to 90 days.

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26. Question 26

    If every requirement can be checked by a cost – effective process, then SRS is called

    1. AVerifiable
    2. BTracable
    3. CModifiable
    4. DComplete

    Answer: (A) Verifiable

    Explanation

    An SRS is verifiable when every stated requirement can be checked by a finite, cost-effective process, such as a test, inspection, or demonstration, to confirm the software actually satisfies it. Traceability is about linking requirements to their origin and to downstream artifacts, and completeness is about covering all needed functions, neither of which is about checkability at reasonable cost.

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27. Question 27

    Fault base testing technique is

    1. AUnit testing
    2. BBeta testing
    3. CStress testing
    4. DMutation testing

    Answer: (D) Mutation testing

    Explanation

    Mutation testing deliberately seeds known faults (mutations) into a program and checks whether the existing test suite detects them, making it the classic fault-based testing technique. Unit, beta, and stress testing are organized around testing scope, release stage, and load conditions rather than around deliberately injected faults.

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28. Question 28

    Alpha and Beta testing are forms of

    1. AWhite – Box Testing
    2. BBlack – Box Testing
    3. CAcceptance Testing
    4. DSystem Testing

    Answer: (C) Acceptance Testing

    Explanation

    Alpha testing happens at the developer's site with actual customers, and beta testing happens at customer sites with real end users, both aiming to gather user feedback before final release. This customer-facing validation makes them forms of acceptance testing rather than a testing style defined by code visibility (white/black box) or by scope (system testing).

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29. Question 29

    The process to gather the software requirements from client, analyze and document is known as:-

    1. ASoftware Engineering Process
    2. BUser Engineering Process
    3. CRequirement Elicitation Process
    4. DRequirement Engineering Process

    Answer: (D) Requirement Engineering Process

    Explanation

    Elicitation is only the gathering step, but this question describes three activities together, gathering, analyzing, and documenting, which together define the broader Requirement Engineering Process. Requirement Engineering encompasses elicitation, analysis, specification, and validation, so it is the correct name for the combined activity described here rather than elicitation alone.

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30. Question 30

    Size and complexity are a part of

    1. APeople Metrics
    2. BProject Metrics
    3. CProcess Metrics
    4. DProduct Metrics

    Answer: (D) Product Metrics

    Explanation

    Product metrics characterize attributes of the software product itself, such as its size, complexity, design features, performance, and quality. Size and complexity describe the delivered artifact, not the people, project schedule, or the process used to build it, so they belong under product metrics.

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31. Question 31

    Which Metrics are derived by normalizing quality and/or productivity measures by considering the size of the software that has been produced?

    1. AFunction – Oriented Metrics
    2. BFunction – Point Metrics
    3. CLine of Code Metrics
    4. DSize Oriented Metrics

    Answer: (D) Size Oriented Metrics

    Explanation

    Size-oriented metrics normalize quality and productivity measures, such as errors found or cost incurred, against a size measure of the delivered software, most commonly lines of code (KLOC). Function-point metrics normalize instead against a size measure derived from information domain values rather than software size directly, which is why the size-normalized family described here is called size-oriented metrics.

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32. Question 32

    The model in which the requirements are implemented by its category is

    1. AEvolutionary Development Model
    2. BWaterfall Model
    3. CPrototyping Model
    4. DIterative Enhancement Model

    Answer: (D) Iterative Enhancement Model

    Explanation

    The Iterative Enhancement Model groups requirements into categories and implements the software in successive increments, each iteration adding one category's worth of functionality on top of what was already built. Evolutionary development instead grows a working system through repeated prototypes shaped by user feedback rather than pre-grouped requirement categories, and the waterfall and prototyping models build the whole requirement set in one linear pass or one exploratory prototype respectively.

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33. Question 33

    Which of the following is an indirect measure of product?

    1. AQuality
    2. BComplexity
    3. CReliability
    4. DAll of these

    Answer: (D) All of these

    Explanation

    Direct measures capture something countable straight from the product, such as lines of code, execution speed, or memory size. Quality, complexity, and reliability cannot be counted directly, they are each derived from combinations of direct measures, which is exactly what makes all three indirect measures of the product.

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34. Question 34

    Modules X and Y operate on the same input and output, then the cohesion is

    1. ALogical cohesion
    2. BSequential cohesion
    3. CProcedural cohesion
    4. DCommunicational cohesion

    Answer: (D) Communicational cohesion

    Explanation

    Communicational cohesion is defined precisely as elements of a module operating on the same input data and/or producing the same output data, which is exactly the relationship described between X and Y. Sequential cohesion instead requires the output of one element to feed directly into the next, and procedural cohesion requires elements to follow a specific execution order without necessarily sharing data.

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35. Question 35

    Which mode is a block cipher implementation as a self synchronizing stream cipher?

    1. ACipher Block Chaining Mode
    2. BCipher Feedback Mode
    3. CElectronic Codebook Mode
    4. DOutput Feedback Mode

    Answer: (B) Cipher Feedback Mode

    Explanation

    In Cipher Feedback (CFB) mode, previously transmitted ciphertext is fed back into the encryption shift register to generate the next keystream, so the decryption side automatically resynchronizes once enough correct ciphertext bits have shifted through, which is why CFB is called self-synchronizing. Output Feedback (OFB) mode instead feeds back the cipher's own output rather than the ciphertext, so a lost bit is never recovered and OFB is a synchronous, not self-synchronizing, stream cipher.

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36. Question 36

    Which one is a connectionless transport – layer protocol that belongs to the Internet protocol family?

    1. ATransmission Control Protocol (TCP)
    2. BUser Datagram Protocol (UDP)
    3. CRouting Protocol (RP)
    4. DDatagram Control Protocol (DCP)

    Answer: (B) User Datagram Protocol (UDP)

    Explanation

    UDP sends independent datagrams without establishing a connection or tracking session state, making it the connectionless transport-layer protocol in the TCP/IP suite. TCP is connection-oriented, and the other two listed names are not real Internet protocol family transport-layer protocols.

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37. Question 37

    Consider an error free 64 kbps satellite channel used to send 512 byte data frames in one direction with very short acknowledgements coming back the other way. What is the maximum throughput for window size of 15?

    1. A32 kbps
    2. B48 kbps
    3. C64 kbps
    4. D70 kbps

    Answer: (C) 64 kbps

    Explanation

    This is a sliding-window pipelining problem using the standard textbook values for a geostationary satellite link, a one-way propagation delay of 270 ms.

    Frame transmission time Tf = (512 × 8 bits) / 64,000 bps = 64 ms
    Round-trip time = 2 × 270 ms = 540 ms
    A window of size w keeps the channel fully busy once w × Tf ≥ Tf + RTT, that is once w ≥ (64 + 540) / 64 ≈ 9.4

    Since the given window of 15 already exceeds this threshold, the sender never has to stall waiting for acknowledgements, so the link runs at its full rated capacity.

    Maximum throughput = 64 kbps.

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38. Question 38

    A classless address is given as 167.199.170.82/27. The number of addresses in the network is

    1. A64 addresses
    2. B32 addresses
    3. C28 addresses
    4. D30 addresses

    Answer: (B) 32 addresses

    Explanation

    A /27 prefix leaves 32 − 27 = 5 host bits.

    Number of addresses = 2^5 = 32

    This counts every address in the block, including the network and broadcast addresses, so the total number of addresses in the network is 32.

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39. Question 39

    Which layer divides each message into packets at the source and re-assembles them at the destination?

    1. ANetwork layer
    2. BTransport layer
    3. CData link layer
    4. DPhysical layer

    Answer: (A) Network layer

    Explanation

    The network layer is responsible for routing individual packets end to end, and in the classic layer-function description it is the layer that divides an outgoing message into packets at the source and reassembles the incoming packets back into the original message at the destination. The data link layer instead frames and error-checks data over one physical hop, and the physical layer only converts bits into electrical or optical signals.

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40. Question 40

    A 4-stage pipeline has the stage delay as 150,120,160 and 140 ns respectively. Registers that are used between the stages have delay of 5 ns. Assuming constant locking rate, the total time required to process 1000 data items on this pipeline is

    1. A160.5 ms
    2. B165.5 ms
    3. C120.5 ms
    4. D590.5 ms

    Answer: (B) 165.5 ms

    Explanation

    The pipeline clock period is set by the slowest stage delay plus the register delay.

    Clock period = max(150, 120, 160, 140) + 5 = 165 ns

    For k = 4 pipeline stages processing n = 1000 items, the number of clock cycles needed is k + (n − 1) = 4 + 999 = 1003.

    Total time = 1003 × 165 ns = 165,495 ns = 165.495 μs

    The answer choices are labeled ms but the intended unit is μs. Read against them, this value rounds to 165.5, matching option B.

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41. Question 41

    Which of the following is correct for the destination address 4A : 30 : 10 : 21 : 10 : 1A?

    1. Aunicast address
    2. Bmulticast address
    3. Cbroadcast address
    4. Dunicast and broadcast address

    Answer: (A) unicast address

    Explanation

    The least significant bit of a MAC address's first byte is the individual/group (I/G) bit, a 0 means unicast and a 1 means multicast or broadcast.

    4A in binary is 0100 1010

    Its last bit is 0, so this address is a unicast address, not a multicast or broadcast address.

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42. Question 42

    Assume that f(n) and g(n) are asymptotically positive. Which of the following is correct?

    1. Af(n)=O(g(n)) and g(n)=O(h(n)) ⇒ f(n)=ω(h(n))
    2. Bf(n)=Ω(g(n)) and g(n)=Ω(h(n)) ⇒ f(n)=O(h(n))
    3. Cf(n)=o(g(n)) and g(n)=o(h(n)) ⇒ f(n)=o(h(n))
    4. Df(n)=ω(g(n)) and g(n)=ω(h(n)) ⇒ f(n)=Ω(h(n))

    Answer: (C) f(n)=o(g(n)) and g(n)=o(h(n)) ⇒ f(n)=o(h(n))

    Explanation

    Little-o is transitive in the exact sense stated. If f(n) = o(g(n)) and g(n) = o(h(n)), then f(n) grows strictly slower than g(n), which in turn grows strictly slower than h(n), so f(n) must also grow strictly slower than h(n), giving f(n) = o(h(n)).

    • A. O∘O ⇒ ω — wrong direction. Composing two upper-bound (O) relations should yield another upper bound O(h(n)), not the strict lower-bound claim ω(h(n)).
    • B. Ω∘Ω ⇒ O — wrong direction. Composing two lower-bound (Ω) relations should yield another lower bound Ω(h(n)), not the opposite upper-bound claim O(h(n)).
    • C. o∘o ⇒ o — correct. This matches the standard transitive closure of the strict little-o relation.
    • D. ω∘ω ⇒ Ω — imprecise. Composing two strict lower-bound (ω) relations should yield the matching strict bound ω(h(n)), not the weaker Ω(h(n)) claim, so this does not state the standard identity being tested.

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43. Question 43

    The solution of the recurrence relation T(n) = 3T(n/4) + n lg n is

    1. Aθ(n² lg n)
    2. Bθ(n lg n)
    3. Cθ(n lg n)²
    4. Dθ(n lg lg n)

    Answer: (B) θ(n lg n)

    Explanation

    Apply the master theorem with a = 3, b = 4, and f(n) = n lg n.

    n^(log_b a) = n^(log₄3) ≈ n^0.792

    Since f(n) = n lg n grows polynomially faster than n^0.792 (the exponent 1 already exceeds 0.792, so the extra lg n factor only helps), this falls under Master Theorem Case 3, and the regularity condition a·f(n/b) ≤ c·f(n) holds for a suitable c < 1.

    So T(n) = θ(f(n)) = θ(n lg n).

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44. Question 44

    The number of nodes of height h in any n-element heap is atmost:

    1. An / 2^(h+1)
    2. Bn / 2^(h−1)
    3. Cn / 2^h
    4. D(n−1) / 2^(h−1)

    Answer: (A) n / 2^(h+1)

    Explanation

    In a binary heap stored as a nearly-complete binary tree, the standard bound states that the number of nodes at height h is at most ⌈n / 2^(h+1)⌉, since each level roughly halves the count of nodes above it and height-h nodes sit h levels above the leaves. This matches option A, n / 2^(h+1).

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45. Question 45

    Consider a B-tree of height h, minimum degree t ≥ 2 that contains any n-key, where n ≥ 1. Which of the following is correct?

    1. Ah ≥ log ((n+1)/2) ₜ
    2. Bh ≤ log ((n+1)/2) ₜ
    3. Ch ≥ log ((n−1)/2) ₜ
    4. Dh ≤ log ((n−1)/2) ₜ

    Answer: (B) h ≤ log ((n+1)/2) ₜ

    Explanation

    For a B-tree of minimum degree t holding n keys, the standard height bound (derived from the minimum number of keys a tree of height h must hold once every non-root node is at its minimum fill) gives

    h ≤ log_t((n+1)/2)

    This is the upper bound on height guaranteed by the B-tree's minimum branching factor, matching option B.

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46. Question 46

    Which of the following algorithm design approach is used in Quick sort algorithm?

    1. ADynamic programming
    2. BBack Tracking
    3. CDivide and conquer
    4. DGreedy approach

    Answer: (C) Divide and conquer

    Explanation

    Quicksort partitions the array around a pivot and then recursively sorts the two resulting sub-arrays independently, which is exactly the divide, conquer, and combine pattern of a divide and conquer algorithm.

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47. Question 47

    Consider the hash table of size 11 that uses open addressing with linear probing. Let h(k)=k mod 11 be the hash function. A sequence of records with keys 43, 36, 92, 87, 11, 47, 11, 13, 14 is inserted into an initially empty hash table, the bins of which are indexed from 0 to 10. What is the index of the bin into which the last record is inserted?

    1. A8
    2. B7
    3. C10
    4. D4

    Answer: (B) 7

    Explanation

    Insert each key with h(k) = k mod 11 and linear probing on collision.

    43 → 10 (empty)
    36 → 3 (empty)
    92 → 4 (empty)
    87 → 10 occupied → 0 (empty)
    11 → 0 occupied → 1 (empty)
    47 → 3 occupied → 4 occupied → 5 (empty)
    11 → 0 occupied → 1 occupied → 2 (empty)
    13 → 2 occupied → 3 occupied → 4 occupied → 5 occupied → 6 (empty)
    14 → 3 occupied → 4 occupied → 5 occupied → 6 occupied → 7 (empty)

    The last record, 14, lands in bin 7.

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48. Question 48

    Consider the traversal of a tree
    Preorder → ABCEIFJDGHKL
    Inorder → EICFJBGDKHLA
    Which of the following is correct post order traversal?

    1. AEIFJCKGLHDBA
    2. BFCGKLHDBUAE
    3. CFCGKLHDBAEIJ
    4. DIEJFCGKLHDBA

    Answer: (D) IEJFCGKLHDBA

    Explanation

    Rebuild the tree recursively from preorder and inorder. A is the root since it is the first preorder symbol, and since A is the very last symbol in the inorder sequence, A has no right subtree, everything else falls under its left child B.

    Splitting B's inorder region (EICFJBGDKHL) around B gives a left part EICFJ and a right part GDKHL, matching the next preorder symbols C,E,I,F,J for the left and D,G,H,K,L for the right.

    Working through each smaller region the same way places C with left child E (E has right child I) and right child F (F has right child J), and places D with left child G and right child H (H has left child K and right child L).

    Walking this reconstructed tree in left, right, root order gives

    I, E, J, F, C, G, K, L, H, D, B, A

    which is IEJFCGKLHDBA.

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49. Question 49

    How many rotations are required during the construction of an AVL tree if the following elements are to be added in the given sequence?
    35, 50, 40, 25, 30, 60, 78, 20, 28

    1. A2 left rotations, 2 right rotations
    2. B2 left rotations, 3 right rotations
    3. C3 left rotations, 2 right rotations
    4. D3 left rotations, 1 right rotation

    Answer: (C) 3 left rotations, 2 right rotations

    Explanation

    Insert each key and rebalance as soon as any node's balance factor exceeds 1.

    Inserting 40 after 35, 50 creates a Right-Left imbalance at 35, fixed with a right rotation at 50 followed by a left rotation at 35 (40 becomes the new subtree root).

    Inserting 30 after 25 creates a Left-Right imbalance at 35 (now under 40), fixed with a left rotation at 25 followed by a right rotation at 35 (30 becomes the new subtree root).

    Inserting 78 after 60 creates a Right-Right imbalance at 50, fixed with a single left rotation at 50.

    Inserting 20 and 28 afterward keeps every node balanced with no further rotations.

    Counting individual rotations across the two double rotations and one single rotation gives 3 left rotations and 2 right rotations in total.

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50. Question 50

    Match List I with List II regarding types of interrupts.

    List-I	List-II
    A. Stack overflow	I. Software interrupt
    B. Timer	II. Internal interrupt
    C. Invalid opcode	III. External interrupt
    D. Superior call	IV. Machine check interrupt

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

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

    Explanation

    Interrupts split into four classic categories.

    Stack overflow is a program-execution error detected by the processor itself, an internal interrupt.
    Timer expiry is signaled by external hardware, an external interrupt.
    Invalid opcode is a fault the machine's hardware raises against a bad instruction, grouped here with machine-check style interrupts.
    A supervisor call is a deliberately executed trap instruction used to request OS services, the definition of a software interrupt.

    Matching each item gives A-II, B-III, C-IV, D-I.

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51. Question 51

    Let R (ABCDEFGH) be a relation schema and F be the set of dependencies F = {A → B, ABCD → E, EF → G, EF → H and ACDF → EG}. The minimal cover of a set of functional dependencies is

    1. AA → B, ACD → E, EF → G, and EF → H
    2. BA → B, ACD → E, EF → G, EF → H and ACDF → G
    3. CA → B, ACD → E, EF → G, EF → H and ACDF → E
    4. DA → B, ABCD → E, EF → H and EF → G

    Answer: (A) A → B, ACD → E, EF → G, and EF → H

    Explanation

    Build the minimal cover step by step.

    First split ACDF → EG into ACDF → E and ACDF → G, since every right-hand side must be a single attribute.

    Next remove extraneous left-hand-side attributes. In ABCD → E, computing the closure of {A,C,D} already yields B (through A → B) and then E (through the reduced ABCD → E), so B is extraneous and this becomes ACD → E.

    Now check ACDF → E: since ACD → E already holds, ACDF → E follows automatically by augmentation, so it is redundant and can be dropped. The same closure argument, extended with EF → G, shows ACDF → G is also implied once ACD → E and EF → G are present, so it drops too.

    What remains, A → B, ACD → E, EF → G, EF → H, has no further redundant or extraneous parts, since removing any one of these four breaks the ability to derive E, G, or H.

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52. Question 52

    A trigger is

    1. AA statement that enables to start DBMS.
    2. BA statement that is executed by the user when debugging an application program.
    3. CA condition the system tests for the validity of the database user.
    4. DA statement that is executed automatically by the system as a side effect of modification to the database.

    Answer: (D) A statement that is executed automatically by the system as a side effect of modification to the database.

    Explanation

    A trigger is a stored procedure that the database management system fires automatically whenever a specified modification, an insert, update, or delete, occurs on a table, without any explicit user invocation. It has nothing to do with starting the DBMS, manual debugging, or user-login validation.

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53. Question 53

    For the following page reference string 4, 3, 2, 1, 4, 3, 5, 4, 3, 2, 1, 5, the number of page faults that occur in Least Recently Used (LRU) page replacement algorithm with frame size 3 is

    1. A6
    2. B8
    3. C10
    4. D12

    Answer: (C) 10

    Explanation

    Simulate LRU with 3 frames, evicting the least recently used page on every miss.

    4→fault {4}
    3→fault {4,3}
    2→fault {4,3,2}
    1→fault, evict 4 (LRU) {3,2,1}
    4→fault, evict 3 {2,1,4}
    3→fault, evict 2 {1,4,3}
    5→fault, evict 1 {4,3,5}
    4→hit (recency updates to 3,5,4)
    3→hit (recency updates to 5,4,3)
    2→fault, evict 5 {4,3,2}
    1→fault, evict 4 {3,2,1}
    5→fault, evict 3 {2,1,5}

    Counting the faults across all 12 references gives 10 faults and 2 hits.

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54. Question 54

    A magnetic tape drive has transport speed of 200 inches per second and a recording density of 1600 bytes per inch. The time required to write 600000 bytes of data grouped in 100 characters record with a blocking factor 10 is

    1. A2.0625 sec
    2. B2.6251 sec
    3. C2.0062 sec
    4. D2.6150 sec

    Answer: (B) 2.6251 sec

    Explanation

    With a blocking factor of 10 and a 100-byte logical record, each physical block holds 1000 bytes, so the 600000 bytes span 600000 / 1000 = 600 blocks.

    Pure data transfer time per block = block size / (density x speed) = 1000 / (1600 x 200) = 3.125 ms

    Matching option B also requires an inter-block gap (IBG) traversal time per block. Using an IBG of 0.25 inch gives gap time = 0.25 / 200 = 1.25 ms

    Total time per block = 3.125 + 1.25 = 4.375 ms
    Total write time = 600 x 4.375 ms = 2625 ms, approximately 2.625 sec, which matches option B (2.6251 sec)

    The 0.25-inch gap is the value needed to reach the listed answer. It is not given directly in the extracted question text, so verify it against the original source PDF before relying on this figure.

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55. Question 55

    Consider two lists A and B of three strings on {0,1}.
    X: (1,111), (10111,10), (10,0)
    Y: (10,101), (011,11), (101,011)
    Which of the following is true?

    1. AOnly PCP in X has solution.
    2. BOnly PCP in Y has solution.
    3. CPCP in both X and Y has solution.
    4. DPCP neither in X nor in Y has solution.

    Answer: (A) Only PCP in X has solution.

    Explanation

    For X, using dominoes in the order 2, 1, 1, 3 gives a matching solution.

    Top: 10111 + 1 + 1 + 10 = 101111110
    Bottom: 10 + 111 + 111 + 0 = 101111110

    Both strings equal 101111110, so X has a solution.

    For Y, the only tile whose top and bottom share a starting prefix is (10,101), so every solution attempt must start there, leaving the bottom one character ahead. From that point, the only tile whose top can continue matching is (101,011), and repeating it always regenerates the exact same one-character gap rather than closing it, so no sequence of Y's tiles can ever produce equal strings.

    Only the PCP instance in X has a solution.

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56. Question 56

    Consider the properties of recursively enumerable sets:
    A. Finiteness
    B. Context Freedom
    C. Emptiness
    Which of the following is true?

    1. AOnly (A) and (B) are not decidable
    2. BOnly (B) and (C) are not decidable
    3. COnly (C) and (A) are not decidable
    4. DAll (A), (B) and (C) are not decidable

    Answer: (D) All (A), (B) and (C) are not decidable

    Explanation

    By Rice's theorem, every non-trivial semantic property of the language recognized by a Turing machine is undecidable. Whether the recognized language is finite, whether it is context-free, and whether it is empty are all non-trivial properties of the language itself, not properties that can be read off the machine's description mechanically, so all three, finiteness, context-freedom, and emptiness, are undecidable.

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57. Question 57

    Match List I with List II.

    List-I	List-II
    A. Activation record	I. Linking Loader
    B. Location counter	II. Garbage Collection
    C. Reference count	III. Subroutine Call
    D. Address relocation	IV. Assembler

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

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

    Explanation

    An activation record is the runtime frame created for each subroutine call, matching Subroutine Call.
    The location counter is the running address pointer an assembler maintains while assembling code, matching Assembler.
    A reference count is the technique reference-counting garbage collectors use to know when an object can be reclaimed, matching Garbage Collection.
    Address relocation is performed by a linking loader when it places a program at its final load address, matching Linking Loader.

    Printed answer key mapping: A-III, B-IV, C-I, D-II.

    A-III (Activation record with Subroutine Call) and B-IV (Location counter with Assembler) are unambiguous. The printed key pairs C with Linking Loader and D with Garbage Collection, the reverse of the standard textbook definitions, where Reference count belongs with Garbage Collection and Address relocation belongs with Linking Loader. This item is kept for a human check against the source PDF and official key.

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58. Question 58

    Consider the following related to Fourth Generation Technique (4GT):
    A. It controls efforts.
    B. It controls resources.
    C. It controls cost of development.
    Choose the correct answer from the options given below:

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

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

    Explanation

    4GT tools dramatically shorten development time, but without discipline that speed can hide poor engineering practice. The standard caution about 4GT is that effort, resources, and cost of development must all still be actively controlled and measured, exactly as in any other process model, rather than left unmanaged just because the tools are fast. All three, effort, resources, and cost, are the things 4GT still needs to control.

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59. Question 59

    Consider the grammer S → SbS | a.
    Consider the following statements:
    The string abababa has
    A. two parse trees
    B. two left most derivations
    C. two right most derivations
    Which of the following is correct?

    1. AAll (A), (B) and (C) are true
    2. BOnly (B) is true
    3. COnly (C) is true
    4. DOnly (A) is true

    Answer: (A) All (A), (B) and (C) are true

    Explanation

    The grammar S → SbS | a is ambiguous because the string abababa can be parenthesized as ((a b a) b a) b a) or a b (a b (a b a)), among other groupings, giving genuinely distinct parse trees.

    Since a parse tree corresponds to exactly one leftmost derivation and exactly one rightmost derivation, having two distinct parse trees for abababa necessarily produces two distinct leftmost derivations and two distinct rightmost derivations as well. All three statements hold together.

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60. Question 60

    In a game playing search tree, upto which depth α – β pruning can be applied?
    A. Root (0) level
    B. 6 level
    C. 8 level
    D. Depends on utility value in a breadth first order
    Choose the correct answer from the options given below:

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

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

    Explanation

    Alpha-beta pruning is not tied to any fixed depth such as 6 or 8, it can cut off a branch at any level of the tree, including right at the root, the moment a value is found that makes a branch provably irrelevant. Whether and where pruning actually happens depends entirely on the utility values encountered as the search explores nodes, not on reaching some predetermined depth. This makes (A), pruning can occur at the root, and (D), it depends on the utility values, correct, while singling out a fixed depth like 6 or 8 is not.

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61. Question 61

    Consider α, β, γ as logical variables. Identify which of the following represents correct logical equivalence:
    A. (α ∧ (β ∨ γ)) ≡ ((α ∧ β) ∨ (α ∧ γ))
    B. (α ∨ β̅) ≡ ¬α ∨ β
    C. (α ⇒ β) ≡ (¬β ⇒ ¬α)
    D. ¬(α ∨ β) ≡ (¬α ⇒ ¬β)
    Choose the correct answer from the options given below:

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

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

    Explanation

    • (A) — true. This is the standard distributive law of ∧ over ∨.
    • (B) — false. Testing α = true, β = false gives α ∨ β̅ = true ∨ true = true, but ¬α ∨ β = false ∨ false = false, so the two sides disagree.
    • (C) — true. This is the contrapositive law, always logically equivalent to the original implication.
    • (D) — false. Testing α = true, β = false gives ¬(α ∨ β) = ¬true = false, but ¬α ⇒ ¬β = false ⇒ true = true, so the two sides disagree.

    Only (A) and (C) hold as genuine logical equivalences.

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62. Question 62

    Let ({a,b},) be a semigroup, where aa=b.
    A. ab = ba
    B. b*b = b
    Choose the most appropriate answer from the options given below:

    1. A(A) only true
    2. B(B) only true
    3. CBoth (A) and (B) true
    4. DNeither (A) nor (B) true

    Answer: (C) Both (A) and (B) true

    Explanation

    Associativity applied to aaa forces (aa)a = a(aa), that is ba = ab, so statement (A) holds regardless of what a*b actually equals.

    Call this common value x = ab = ba. Now consider aaaa, which is well defined because * is associative, and can be grouped two ways.
    Grouping as (aa)(aa) gives bb.
    Grouping as a(a*(aa)) gives a(ab) = ax.

    Since the set is only {a,b}, x is either a or b. If x = a, then ax = aa = b. If x = b, then ax = ab = x = b. Either way, a*x = b.

    So bb = ax = b, which proves statement (B) as well. Both (A) and (B) are forced to be true by associativity alone, with no further assumption needed.

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63. Question 63

    Consider the following graph.
    For the graph, the following sequences of depth first search (DFS) are given
    A. abcghf
    B. abfchg
    C. abfhgc
    D. afghbc
    Which of the following is correct?

    

    Undirected graph with vertices a, b, c, f, g, h for the DFS traversal question

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

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

    Explanation

    The graph's edges are a-c, a-b, a-f, c-b, c-g, b-f, b-g, b-h, f-g, f-h, g-h. A DFS order is valid whenever, at each step, the next vertex is an unvisited neighbor of the current path's active vertex, and a vertex with an unvisited neighbor left cannot be skipped before that neighbor is visited.

    • A. abcghf — valid. a→b (edge a-b), b→c (edge b-c), c→g (edge c-g), g→h (edge g-h), h→f (edge f-h).
    • B. abfchg — invalid. After a→b→f, vertex f still has unvisited neighbors g and h, so DFS must visit one of those next, but c is not adjacent to f at all.
    • C. abfhgc — valid. a→b, b→f (edge b-f), f→h (edge f-h), h→g (edge g-h), g→c (edge c-g).
    • D. afghbc — valid. a→f (edge a-f), f→g (edge f-g), g→h (edge g-h), h→b (edge b-h), b→c (edge b-c).

    Only A, C, and D describe valid depth-first traversals of this graph.

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64. Question 64

    Let ε=0.0005. and Let Re be the relation {(x,y) ∈ R² : |x−y| < ε}. Re could be interpreted as the relation approximately equal. Re is
    A. Reflexive
    B. Symmetric
    C. transitive
    Choose the correct answer from the options given below:

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

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

    Explanation

    Reflexive holds because |x − x| = 0 < ε for every x.
    Symmetric holds because |x − y| = |y − x|, so the condition is unaffected by swapping x and y.
    Transitive fails. Take x = 0, y = 0.0004, and z = 0.0008. Then |x − y| = 0.0004 < 0.0005 and |y − z| = 0.0004 < 0.0005, so both pairs satisfy the relation, but |x − z| = 0.0008, which is not less than 0.0005, so (x,z) does not satisfy the relation.

    Re is reflexive and symmetric but not transitive, which is exactly why a tolerance-style approximately-equal relation is not a true equivalence relation.

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65. Question 65

    In reference to Big data, consider the following database:
    A. Memcached
    B. Couch DB
    C. Infinite graph
    Choose the most appropriate answer from the options given below:

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

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

    Explanation

    Big data NoSQL systems are commonly grouped by data model, and all three named systems are standard examples across those categories. Memcached is a key-value store, CouchDB is a document-oriented database, and InfiniteGraph is a graph database, so all three belong on a list of Big Data database technologies.

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66. Question 66

    Match List I with List II regarding normal forms.

    List-I	List-II
    A. BCNF	I. Removes multivalued dependency
    B. 3NF	II. Not always dependency preserving
    C. 2NF	III. Removes transitive dependency
    D. 4NF	IV. Removes partial functional dependency

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

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

    Explanation

    BCNF decomposition is well known for not always preserving every original functional dependency.
    3NF eliminates transitive dependencies on the key.
    2NF eliminates partial functional dependencies on part of a composite key.
    4NF eliminates multivalued dependencies.

    Matching gives A-II, B-III, C-IV, D-I.

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67. Question 67

    Match List I with List II regarding software design concepts.

    List-I	List-II
    A. Localization	I. Encapsulation
    B. Packaging or binding of a collection of items	II. Abstraction
    C. Mechanism that enables designer to focus on essential details of a program component	III. Characteristic of software that indicates the manner in which information is concentrated in program
    D. Information hiding	IV. Suppressing the operational details of a program component

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

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

    Explanation

    Localization describes the characteristic of software indicating how information is concentrated within it.
    Packaging or binding a collection of items together is the definition of encapsulation.
    A mechanism that lets a designer focus on essential details while ignoring lower-level ones is abstraction.
    Information hiding is precisely the suppression of a component's operational details from the rest of the system.

    Matching gives A-III, B-I, C-II, D-IV.

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68. Question 68

    Match List I with List II regarding the Chomsky hierarchy.

    List-I	List-II
    A. Type 0	I. Finite automata
    B. Type 1	II. Turing machine
    C. Type 2	III. Linear bound automata
    D. Type 3	IV. Pushdown automata

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

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

    Explanation

    In the Chomsky hierarchy, Type 0 (unrestricted grammars) corresponds to the Turing machine, Type 1 (context-sensitive) corresponds to the linear bounded automaton, Type 2 (context-free) corresponds to the pushdown automaton, and Type 3 (regular) corresponds to the finite automaton.

    Matching gives A-II, B-III, C-IV, D-I.

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69. Question 69

    Match List I with List II regarding knowledge representation.

    List-I	List-II
    A. Ontological Engineering	I. Organizing subclass relations
    B. Taxonomy Hierarchy	II. Organizing knowledge into category and sub category
    C. Inheritance	III. Attaches a number with each possibility
    D. Probability mode	IV. Representing concepts, events, time, physical concepts of different domains

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

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

    Explanation

    Ontological engineering deals with representing very general concepts, events, time, and physical objects across many domains, not just a narrow task.
    A probability model attaches a numeric likelihood to each possibility.

    Printed answer key mapping: A-IV, B-I, C-II, D-III.

    A-IV (Ontological Engineering with representing concepts across domains) and D-III (Probability model with attaching a number to each possibility) are unambiguous. Taxonomy hierarchy and inheritance are closely related terms, organizing knowledge into categories and organizing subclass relations, that some sources pair differently. Option D is kept as the best-supported match because it correctly places ontological engineering and the probability model, but this item is worth a human check against the source PDF for the exact intended pairing of taxonomy hierarchy and inheritance.

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70. Question 70

    Match List I with List II regarding operating system concepts.

    List-I	List-II
    A. Least frequently used	I. Memory is distributed among processors
    B. Critical Section	II. Page replacement policy in cache memory
    C. Loosely coupled multiprocessor system	III. Program section that once begin must complete execution before another processor access the same shared resource
    D. Distributed operating system organization	IV. O/S routines are distributed among available processors

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

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

    Explanation

    Least frequently used is a page replacement policy applied in cache memory.
    A critical section is the program section that, once begun, must run to completion before another processor can access the same shared resource.
    A loosely coupled multiprocessor system is one where memory is distributed among the processors rather than shared.
    A distributed operating system organization is one where OS routines themselves are distributed among the available processors.

    Matching gives A-II, B-III, C-I, D-IV.

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71. Question 71

    Match List I with List II regarding systems concepts.

    List-I	List-II
    A. Firmware	I. Number of logical records into physical blocks
    B. Batch file	II. ASCII format
    C. Packing	III. Resource allocation
    D. Banker's Algorithm	IV. ROM

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

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

    Explanation

    Firmware refers to software permanently held in ROM.
    A batch file is a plain ASCII-format text file of commands.
    Packing is the grouping of a number of logical records into physical blocks.
    Banker's algorithm is a resource allocation technique used to avoid deadlock.

    Matching gives A-IV, B-II, C-I, D-III.

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72. Question 72

    Match List I with List II regarding cryptographic key sizes.

    List-I	List-II
    A. DES	I. Key size - 256
    B. AES	II. Key size - 1024
    C. 3 DES	III. Key size - 56
    D. RSA	IV. Key size - 168

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

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

    Explanation

    DES uses a 56-bit key.
    AES, as referenced here, uses a 256-bit key.
    Triple DES (3DES) has an effective key size of 168 bits, three 56-bit DES keys applied in sequence.
    RSA is commonly cited in this context with a 1024-bit key.

    Matching gives A-III, B-I, C-IV, D-II.

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73. Question 73

    Match List I with List II regarding socket programming calls.

    List-I	List-II
    A. BIND	I. Block the caller until a connection attempt arrives
    B. LISTEN	II. Give a local address to a socket
    C. ACCEPT	III. Show willingness to accept connections
    D. SOCKET	IV. Create a new point

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

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

    Explanation

    BIND assigns a local address to a socket.
    LISTEN marks a socket as willing to accept incoming connections.
    ACCEPT blocks the caller until a connection attempt actually arrives.
    SOCKET creates a brand new communication endpoint.

    Matching gives A-II, B-III, C-I, D-IV.

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74. Question 74

    Match List I with List II regarding OSI layer functions.

    List-I	List-II
    A. Physical layer	I. Routing of the signals divide the outgoing message into packets, to act as network controller for routing data
    B. Data link layer	II. Make and break connections, define voltages and data rates, convert data bits into electrical signal
    C. Network layer	III. Synchronization, error detection and correction. To assemble outgoing message into frames
    D. Presentation layer	IV. It works as a translating layer

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

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

    Explanation

    The physical layer makes and breaks connections, defines voltages and data rates, and converts bits into electrical signals.
    The data link layer handles synchronization and error detection and correction, assembling the outgoing message into frames.
    The network layer routes signals and divides the outgoing message into packets.
    The presentation layer works as a translating layer between applications and the network format.

    Matching gives A-II, B-III, C-I, D-IV.

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75. Question 75

    Match each algorithm with its running-time complexity.

    Algorithm	Complexity
    A. Breadth First Search	II. O(V+E)
    B. Rabin-Karp Algorithm	III. θ((n−m+1)m)
    C. Depth-First Search	I. θ(V+E)
    D. Heap sort (worst case)	V. O(n lg n)
    E. Quick sort (worst case)	IV. O(n²)

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

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

    Explanation

    Breadth-first search and depth-first search both run in O(V+E) (equivalently θ(V+E)) time on an adjacency-list graph.
    Rabin-Karp string matching has worst-case complexity θ((n−m+1)m).
    Heap sort runs in O(n lg n) in every case, including the worst case, since it always maintains a balanced heap.
    Quick sort's worst case is O(n²), which happens when the pivot repeatedly gives the most unbalanced possible split.

    The key distinguishing check here is D and E. Heap sort's worst case is never quadratic, and quick sort's worst case is never O(n lg n), so D must be O(n lg n) and E must be O(n²), which only option C gets right.

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76. Question 76

    Match List I with List II regarding classic operating system algorithms.

    List-I	List-II
    A. Stack algorithm	I. Deadlock
    B. Elevator algorithm	II. Disk scheduling
    C. Priority scheduling algorithm	III. Page replacement
    D. Havender's algorithm	IV. CPU scheduling

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

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

    Explanation

    Stack algorithms, page replacement policies that satisfy the stack property such as LRU and OPT, belong to page replacement.
    The elevator algorithm is another name for the SCAN disk-scheduling algorithm.
    A priority scheduling algorithm assigns the CPU based on priority, a CPU scheduling technique.
    Havender's algorithm imposes an ordering on resource requests to prevent deadlock.

    Matching gives A-III, B-II, C-IV, D-I.

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77. Question 77

    Consider the following statements of approximation algorithm:
    Statement I: Vertex-cover is a polynomial time 2-approximation algorithm.
    Statement II: TSP-tour is a polynomial time 3-approximation algorithm for travelling salesman problem with the triangle inequality.
    Which of the following is correct?

    1. AStatement I true and Statement II false
    2. BStatement I and Statement II true
    3. CStatement I false and Statement II true
    4. DStatement I and Statement II false

    Answer: (A) Statement I true and Statement II false

    Explanation

    Statement I is true. The standard vertex-cover approximation algorithm, which greedily picks both endpoints of an uncovered edge, runs in polynomial time and always returns a cover at most twice the size of the optimal one, a 2-approximation.

    Statement II is false. For metric TSP, the well-known polynomial-time approximation guarantee is a 2-approximation using a minimum spanning tree, and Christofides' algorithm improves this to a 3/2-approximation. Neither of the standard results gives a 3-approximation, so Statement II's claimed ratio is incorrect.

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78. Question 78

    Consider the following statements:
    Statement I: Conservative 2 PL is a deadlock-free protocol.
    Statement II: Thomas's write rule enforces conflict serializability.
    Statement III: Timestamp ordering protocol ensures serializability based on the order of transaction timestamps.
    Which of the following is correct?

    1. AStatement I, Statement II true and Statement III false
    2. BStatement I, Statement III true and Statement II false
    3. CStatement I, Statement II false and Statement III true
    4. DStatement I, Statement II and Statement III true

    Answer: (B) Statement I, Statement III true and Statement II false

    Explanation

    Statement I is true. Conservative (static) two-phase locking requires a transaction to acquire all its locks before starting execution, which removes the wait-and-hold pattern that causes deadlock.

    Statement II is false. Thomas's write rule deliberately relaxes conflict serializability by allowing certain outdated writes to be ignored (the thin thomas write rule) rather than treated as a conflict, so it guarantees view serializability, not conflict serializability.

    Statement III is true. Basic timestamp ordering enforces an execution order equivalent to the order in which transactions received their timestamps, which is exactly its serializability guarantee.

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79. Question 79

    Consider the following statements:
    Statement I: Composite attributes cannot be divided into smaller subparts.
    Statement II: Complex attribute is formed by nesting composite attributes and multi-valued attributes in an arbitrary way.
    Statement III: A derived attribute is an attribute whose values are computed from other attribute.
    Which of the following is correct?

    1. AStatement I, Statement II and Statement III are true
    2. BStatement I true and Statement II, Statement III false
    3. CStatement I, Statement II true and Statement III false
    4. DStatement I false and Statement II, Statement III true

    Answer: (D) Statement I false and Statement II, Statement III true

    Explanation

    Statement I is false. It describes a simple (atomic) attribute, not a composite one. A composite attribute is defined by the fact that it can be divided into smaller subparts, such as splitting Name into First Name and Last Name.

    Statement II is true. Nesting composite and multivalued attributes together in any combination is exactly how a complex attribute is built.

    Statement III is true. A derived attribute, such as Age computed from Date of Birth, has its value calculated from another stored attribute rather than stored directly.

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80. Question 80

    A top down approach to programming calls for :
    Statement I: Working from the general to the specific.
    Statement II: Postpone the minor decisions.
    Statement III: A systematic approach.
    Statement IV: Intermediate coding of the problem.
    Which of the following is true?

    1. AStatement I only
    2. BStatement I and Statement II only
    3. CStatement I, Statement II and Statement III only
    4. DStatement I, Statement II and Statement IV only

    Answer: (C) Statement I, Statement II and Statement III only

    Explanation

    Top-down design proceeds by stepwise refinement, and three of the four listed statements describe that process correctly.

    • Statement I. Working from the general to the specific. — true. This is the core idea of stepwise refinement: start from the overall structure and progressively add detail.
    • Statement II. Postpone the minor decisions. — true. Implementation-level details are deferred until the higher-level structure is settled.
    • Statement III. A systematic approach. — true. Refinement proceeds through an orderly sequence of steps rather than ad hoc changes.
    • Statement IV. Intermediate coding of the problem. — false. Top-down design defers coding until the structure is fully refined, and there is no recognized stage called intermediate coding.

    Statements I, II, and III together describe top-down design, so option C is correct.

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81. Question 81

    Consider the following statements:
    Statement I: LALR parser is more powerful than canonical LR Parser.
    Statement II: SLR parser is more powerful than LALR
    Which of the following is correct?

    1. AStatement I true and Statement II false
    2. BStatement I false and Statement II true
    3. CBoth Statement I and Statement II false
    4. DBoth Statement I and Statement II true

    Answer: (C) Both Statement I and Statement II false

    Explanation

    The standard power hierarchy among these LR-family parsers is LR(0) ⊆ SLR ⊆ LALR ⊆ canonical LR(1).

    Statement I is false. Canonical LR is the most powerful of the four, strictly more powerful than LALR, not the other way around.
    Statement II is false. LALR is strictly more powerful than SLR, not less.

    Both statements invert the actual hierarchy, so both are false.

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82. Question 82

    Consider the following statements about Context Free Language (CFL):
    Statement I: CFL is closed under homomorphism.
    Statement II: CFL is closed under complement.
    Which of the following is correct?

    1. AStatement I is true and Statement II is false
    2. BStatement II is true and Statement I is false
    3. CBoth Statement I and Statement II are true
    4. DNeither Statement I nor Statement II is true

    Answer: (A) Statement I is true and Statement II is false

    Explanation

    Context-free languages are closed under homomorphism, applying a homomorphism to every string of a CFL always yields another CFL, so Statement I is true.

    Context-free languages are not closed under complement in general, there exist context-free languages whose complement is not context-free, so Statement II is false.

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83. Question 83

    Consider the following in Boolean Algebra:
    X : a ∨ (b ∧ (a ∨ c)) = (a ∨ b) ∧ (a ∨ c)
    Y : a ∧ (b ∨ (a ∧ c)) = (a ∧ b) ∨ (a ∧ c)
    a ∨ (b ∧ c) = (a ∨ b) ∧ c is satisfied if

    1. AX is true
    2. BY is true
    3. CBoth X and Y are true
    4. DIt does not depend on X and Y

    Answer: (D) It does not depend on X and Y

    Explanation

    X and Y are each separately true identities. Expanding X gives a ∨ (b ∧ (a ∨ c)) = (a ∨ b) ∧ (a ∨ (a ∨ c)) = (a ∨ b) ∧ (a ∨ c), and expanding Y gives a ∧ (b ∨ (a ∧ c)) = (a ∧ b) ∨ (a ∧ (a ∧ c)) = (a ∧ b) ∨ (a ∧ c), so both hold unconditionally for all a, b, c.

    But the equation actually being asked about, a ∨ (b ∧ c) = (a ∨ b) ∧ c, is a different equation from X or Y, and it is not a valid Boolean identity at all. Taking a = 1, b = 0, c = 0 gives a ∨ (b ∧ c) = 1 ∨ 0 = 1 on the left, but (a ∨ b) ∧ c = 1 ∧ 0 = 0 on the right, so the two sides disagree.

    Since this equation is simply false for some values of a, b, and c regardless of what X or Y state, whether it holds cannot depend on X or Y being true.

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84. Question 84

    A good software requirement specification does NOT have the characteristic

    1. ACompleteness
    2. BConsistency
    3. CClarity
    4. DReliability

    Answer: (D) Reliability

    Explanation

    The standard characteristics of a good SRS are correctness, completeness, consistency, clarity (unambiguity), verifiability, modifiability, and traceability. Reliability describes a quality of the running software system itself, not a property used to judge whether a requirements specification document is well written, so it is not one of the recognized SRS characteristics.

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85. Question 85

    Assertion (A): p̅
    Reason (R): (r → q̅, r ∨ s, s → q̅, p → q)
    In the light of the above statements, choose the correct answer from the options given below:

    1. ABoth (A) and (R) are true and (R) is the correct explanation of (A)
    2. BBoth (A) and (R) are true but (R) is NOT the correct explanation of (A)
    3. C(A) is true but (R) is false
    4. D(A) is false but (R) is true

    Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)

    Explanation

    Reason (R) gives four premises: r → ¬q, r ∨ s, s → ¬q, and p → q.

    From r ∨ s together with r → ¬q and s → ¬q, both possible cases lead to the same conclusion ¬q, a constructive dilemma, so ¬q follows validly from the first three premises.

    Taking the contrapositive of p → q gives ¬q → ¬p. Combining this with the derived ¬q by modus ponens yields ¬p, which is exactly Assertion (A), p̅.

    So (R) is not just true, it is a valid and complete derivation of (A), making (R) the correct explanation of (A).

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86. Question 86

    Of the following, which is NOT a logical error?

    1. AUsing the '=', instead of '==' to determine if two values are equal
    2. BDivide by zero
    3. CFailing to initialize counter and total variables before the body of loop
    4. DUsing commas instead of two required semicolon in a for loop header

    Answer: (B) Divide by zero

    Explanation

    Confusing = with ==, forgetting to initialize a counter or accumulator, and mixing up separators in a for-loop header are all classic logic-level mistakes, code that compiles and runs but produces the wrong result because of a flawed program design.

    Divide by zero is different in kind. It is a runtime error, an operation that fails or crashes during execution rather than one that silently produces an incorrect result, which is what sets logic errors apart.

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87. Question 87

    Assertion (A): A load-and-go assembler avoids the overhead of writing the object program out and reading it back in.
    Reason (R): This can be done with either one-pass or two pass assembler.
    In the light of the above statements, choose the correct answer from the options given below:

    1. ABoth (A) and (R) are true and (R) is the correct explanation of (A)
    2. BBoth (A) and (R) are true but (R) is NOT the correct explanation of (A)
    3. C(A) is true but (R) is false
    4. D(A) is false but (R) is true

    Answer: (C) (A) is true but (R) is false

    Explanation

    Assertion (A) is true, a load-and-go assembler places generated code directly into memory for immediate execution, which is exactly what lets it skip writing an object file to disk and reading it back in.

    Reason (R) is false. Load-and-go is characteristically a one-pass assembler technique, since a one-pass design can resolve addresses and emit executable code in memory as it scans the source once. A conventional two-pass assembler instead builds a complete symbol table on the first pass and typically produces a separate object module rather than loading directly into memory, so claiming this works equally well with either design misstates how two-pass assemblers are normally used.

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88. Question 88

    Which statement is false?

    1. AAll function calls in C pass arguments using call by value.
    2. BCall by reference enables a called function to modify a variable in calling function.
    3. CCall by value is always more efficient than call by reference.
    4. DProgrammers use pointers and indirection operation to simulate call by reference.

    Answer: (C) Call by value is always more efficient than call by reference.

    Explanation

    • A. All function calls in C pass arguments using call by value. — true. Even when a pointer is passed, the pointer's own value (the address) is copied, C has no native pass-by-reference mechanism.
    • B. Call by reference enables a called function to modify a variable in calling function. — true, that is the defining behavior of reference-style parameter passing.
    • C. Call by value is always more efficient than call by reference. — false. Copying a large struct or array by value is more expensive than passing a small pointer or reference to it, so call by value is not always the more efficient choice.
    • D. Programmers use pointers and indirection operation to simulate call by reference. — true, this is exactly how C achieves reference-like behavior without a native reference type.

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89. Question 89

    Given below are two statements:
    Statement I: "Grandparent is a parent of one's parent".
    Statement II: First Order Predicate Logic (FOPL) representation of above statement is
    ∀ g, c grandparent(g, c) ⇔ ∃ p parent(g, p) ∧ parent(p, c)
    In the light of the above statements, choose the most appropriate answer from the options given below:

    1. ABoth Statement I and Statement II are correct
    2. BBoth Statement I and Statement II are incorrect
    3. CStatement I is correct but Statement II is incorrect
    4. DStatement I is incorrect but Statement II is correct

    Answer: (A) Both Statement I and Statement II are correct

    Explanation

    Statement I correctly captures the informal meaning of grandparent, someone who is a parent of one of your parents.

    Statement II formalizes it faithfully. It reads as: for every g and c, g is a grandparent of c exactly when there exists some p such that g is a parent of p, and p is a parent of c. This is precisely the chain, g parents p, and p parents c, that makes g a grandparent of c, matching Statement I's definition.

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90. Question 90

    Given below are two statements:
    Statement I: Consider 20 bit 'Branch' microinstruction code format given below:
    F1(3) | F2(3) | F3(3) | CD(2) | BR(2) | AD(7)
    F1, F2, F3 : Micro-operation fields
    CD : Condition for branching
    BR : Branch field
    AD : Address field
    Statement II: Instruction represented in above format can perform branch in 4 conditions.
    In the light of the above statements, choose the most appropriate answer from the options given below:

    1. ABoth Statement I and Statement II are correct
    2. BBoth Statement I and Statement II are incorrect
    3. CStatement I is correct but Statement II is incorrect
    4. DStatement I is incorrect but Statement II is correct

    Answer: (A) Both Statement I and Statement II are correct

    Explanation

    Statement I is correct. Adding up the field widths, 3 + 3 + 3 + 2 + 2 + 7, gives exactly 20 bits, matching the stated format.

    Statement II is correct. The CD (condition for branching) field is 2 bits wide, and 2 bits can encode 2² = 4 distinct condition codes, so the format supports exactly 4 branch conditions.

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91. Question 91

    Which of the following relational algebra query computes the Sid's of sailors with age over 20 who have not reserved a red boat?

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. Aπ sid (σ age > 20 Sailors) − π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors)
    2. Bπ sid ((σ color ≠ red ∧ age > 20 (Boats ⋈ Sailors ⋈ Reserves)))
    3. Cπ sid (σ age > 20 Sailors) − π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors)
    4. Dπ sid (σ age > 20 Sailors) ∧ π sid ((σ color ≠ red Boats) ⋈ Reserves ⋈ Sailors)

    Accepted answers: (A) π sid (σ age > 20 Sailors) − π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors), (C) π sid (σ age > 20 Sailors) − π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors)

    Explanation

    The correct strategy is set difference: start from every sailor with age over 20, then subtract out every sailor who has reserved at least one red boat, regardless of age.

    π sid (σ age > 20 Sailors) gives the sids of sailors older than 20.
    π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors) gives the sids of sailors who reserved a red boat.

    Subtracting the second set from the first gives exactly the sailors over 20 who have not reserved a red boat.

    Options 1 and 3 are printed as byte-identical text in the source PDF itself, confirmed directly against the page image, and both show this correct expression. The color-comparison approaches in options 2 and 4 are unsound, since a sailor who has reserved both a red boat and a non-red boat would still wrongly satisfy a per-row color ≠ red condition on some other reservation.

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92. Question 92

    Which of the following relational algebra query/queries computes/compute the names of sailors who have reserved a red boat?
    Q1: π sname ((σ color = red Boats) ⋈ Reserves ⋈ Sailors)
    Q2: π sname (π sid ((π bid σ color = red Boats) ⋈ Reserves) ⋈ Sailors)
    Q3: π sname ((σ color = red Reserves) ⋈ Boats ⋈ Sailors)

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. ABoth Q1 and Q2
    2. BBoth Q2 and Q3
    3. COnly Q1
    4. DOnly Q2

    Answer: (A) Both Q1 and Q2

    Explanation

    Q1 filters Boats to red boats, joins with Reserves and Sailors, and projects the name, a direct and correct approach.

    Q2 first finds the bids of red boats, joins those bids with Reserves to get the sids who reserved one, then joins with Sailors for the name, an equivalent, correct two-step version of the same logic.

    Q3 applies the condition color = red directly to Reserves, but Reserves has no color attribute at all, that attribute only exists on Boats, so this expression is malformed and cannot be correct.

    Only Q1 and Q2 correctly compute the required names.

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93. Question 93

    Which of the following relational algebra query/queries computes/compute the name of sailors who have reserved boat 103?
    Q1: π sname ((σ bid = 103 Boats) ⋈ Sailors)
    Q2: π sname (σ bid = 103 (Reserves ⋈ Sailors))
    Q3: π sname ((σ bid = 103 Reserves) ⋈ Sailors)

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. ABoth Q1 and Q3
    2. BBoth Q2 and Q3
    3. COnly Q3
    4. DOnly Q2

    Answer: (B) Both Q2 and Q3

    Explanation

    Boats and Sailors share no join attribute, so Q1's direct join of a filtered Boats with Sailors is not a meaningful relational algebra expression and cannot produce the right names.

    Q2 joins Reserves with Sailors on sid first (bringing bid along from Reserves), then filters the joined result to bid = 103, which correctly isolates sailors who reserved that boat.

    Q3 filters Reserves to bid = 103 first, then joins the smaller filtered result with Sailors on sid, reaching the same correct answer more efficiently.

    Only Q2 and Q3 give the correct names.

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94. Question 94

    Which of the following relational algebra query computes the names of sailor who have reserved all boats?

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. Aρ (Tempsids, (π bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors)
    2. Bρ (Tempsids, (π sid,bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors)
    3. Cρ (Tempsids, (π sid Sailors) / π bid Boats) π sname ((Tempsids) ⋈ Sailors)
    4. Dρ (Tempsids, (π sid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Boats)

    Answer: (B) ρ (Tempsids, (π sid,bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors)

    Explanation

    Relational division needs a dividend that pairs the sid and bid attributes together, so the dividend must come from Reserves and include both columns.

    • A. ρ (Tempsids, (π bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors) — wrong. The dividend keeps only the bid column, so dividing it by Boats bid column cannot produce any sid values.
    • B. ρ (Tempsids, (π sid,bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors) — correct. Dividing the sid-bid pairs from Reserves by every bid in Boats yields exactly the sids that appear with all boats, and joining with Sailors retrieves their names.
    • C. ρ (Tempsids, (π sid Sailors) / π bid Boats) π sname ((Tempsids) ⋈ Sailors) — wrong. The dividend is drawn from Sailors instead of Reserves, so it divides the wrong relation and cannot capture which boats a sailor reserved.
    • D. ρ (Tempsids, (π sid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Boats) — wrong. The dividend keeps only sid with no bid to divide against, and the final join is with Boats instead of Sailors, so sname could not even be retrieved.

    Only option B pairs sid with bid in the dividend and joins the result with Sailors, so it correctly computes sailors who reserved all boats.

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95. Question 95

    Which of the following relational algebra query computes the names of sailors who have reserved a red and a green boat?

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. Aρ (Tempred, π sid ((σ color = 'red' Boats) ⋈ Reserves)), ρ (Tempgreen, π sid ((σ color = 'green' Boats) ⋈ Reserves)), π sname (Tempred ∩ Tempgreen) ⋈ Sailors
    2. Bρ (Tempboats 2, (σ color = 'red' Boats) ∩ (σ color = 'green' Boats)) π sname (Tempboats 2 ⋈ Reserves)
    3. Cπ sname ((σ color = 'red' Boats) ∩ (σ color = 'green' Boats) (Tempboats 2 ⋈ Sailors ⋈ Reserves))
    4. Dρ (Tempboats 2, (σ color = 'red' ∧ color = 'green' Boats)) π sname (Tempboats 2 ⋈ Sailors)

    Answer: (A) ρ (Tempred, π sid ((σ color = 'red' Boats) ⋈ Reserves)), ρ (Tempgreen, π sid ((σ color = 'green' Boats) ⋈ Reserves)), π sname (Tempred ∩ Tempgreen) ⋈ Sailors

    Explanation

    The correct approach separately computes which sailors reserved a red boat and which reserved a green boat, and only then intersects the two sets of sailor ids.

    Tempred holds the sids of sailors who reserved a red boat, and Tempgreen holds the sids of sailors who reserved a green boat. Intersecting them gives exactly the sailors who did both, and joining with Sailors retrieves their names.

    The other options intersect the Boats relation directly on color, but no single boat row can be both red and green at once, so intersecting (σ color = 'red' Boats) with (σ color = 'green' Boats), or selecting color = 'red' AND color = 'green' on Boats, always produces an empty relation. Those approaches can never return a sailor.

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96. Question 96

    Based on the passage, the transmission and propagation delays are respectively

    A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.

    1. AT = 333.33 μ sec, T = 18000 μ sec
    2. BT = 300 μ sec, T = 15360 μ sec
    3. CT = 83.33 μ sec, T = 1800 μ sec
    4. DT = 1800 μ sec, T = 33.33 μ sec

    Answer: (A) T = 333.33 μ sec, T = 18000 μ sec

    Explanation

    Transmission delay = frame size / bandwidth = (64 × 8 bits) / 1,536,000 bps = 512 / 1,536,000 ≈ 333.33 μs
    Propagation delay = distance × propagation speed per km = 3000 km × 6 μs/km = 18,000 μs

    So the transmission delay is 333.33 μs and the propagation delay is 18,000 μs.

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97. Question 97

    Based on the passage, the minimum number of bits required in the sequence number field of the packet is

    A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.

    1. A6 bits
    2. B7 bits
    3. C5 bits
    4. D4 bits

    Answer: (B) 7 bits

    Explanation

    The bandwidth-delay ratio a = propagation delay / transmission delay = 18000 / 333.33 = 54.

    The window size needed for maximum efficiency is 1 + 2a = 1 + 108 = 109.

    A sequence number field needs enough distinct values to cover this window, and 2^6 = 64 is too small while 2^7 = 128 comfortably covers 109, so the minimum sequence number field width is 7 bits.

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98. Question 98

    Based on the passage, the sender window size to get the maximum efficiency is

    A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.

    1. A108
    2. B109
    3. C55
    4. D56

    Answer: (B) 109

    Explanation

    With a = propagation delay / transmission delay = 18000 / 333.33 = 54, the sender window size that keeps the link continuously busy is

    W = 1 + 2a = 1 + 2(54) = 109

    This is the smallest window size at which the sender never has to stall waiting for an acknowledgement.

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99. Question 99

    Based on the passage, if only 6 bits are reserved for sequence number field, then the efficiency of the system is:

    A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.

    1. A0.587
    2. B0.875
    3. C0.578
    4. D0.50

    Answer: (A) 0.587

    Explanation

    A 6-bit sequence number field allows a maximum window of 2^6 = 64, which is smaller than the ideal window of 109 found earlier, so the link cannot be kept continuously busy.

    Efficiency = window / (1 + 2a) = 64 / 109 ≈ 0.587

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100. Question 100

     Based on the passage, the maximum achievable throughput is

     A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.

     1. A0.768
     2. B0.678
     3. C0.901
     4. D0.887

     Answer: (C) 0.901

     Explanation

     Using the 6-bit sequence number scenario, the link's efficiency was 64/109 ≈ 0.587.

     Throughput = efficiency × channel bandwidth = 0.587 × 1.536 mbps ≈ 0.901 mbps

     So the maximum achievable throughput under this constraint is about 0.901 mbps.

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