UGC NET Computer Science & Applications November 2021 Question PaperPaper 1 & Paper 2 with Answer Key

50 questions with the answer key and explanations.

1. Question 1

   What is the difference between the average semester fees of BCA and average semester fees of MCA during 2016-2020?

   Study the table of semester fees for four courses during 2016-2020.

   Year	BCA	MCA	MTech	MPhil
   2016	25,000	52,000	60,200	30,000
   2017	26,500	53,500	61,400	32,000
   2018	27,200	54,200	62,500	32,600
   2019	27,600	55,700	62,900	33,400
   2020	28,000	56,800	63,700	33,900

   1. A27,200
   2. B27,365
   3. C27,580
   4. D27,400

   Answer: (C) 27,580

   Explanation

   BCA totals 134,300, so its average is 26,860. MCA totals 272,200, so its average is 54,440. The difference is 54,440 - 26,860 = 27,580.

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2. Question 2

   Which course has the highest percentage fee increase from 2019 to 2020?

   Study the table of semester fees for four courses during 2016-2020.

   Year	BCA	MCA	MTech	MPhil
   2016	25,000	52,000	60,200	30,000
   2017	26,500	53,500	61,400	32,000
   2018	27,200	54,200	62,500	32,600
   2019	27,600	55,700	62,900	33,400
   2020	28,000	56,800	63,700	33,900

   1. ABCA
   2. BMCA
   3. CMTech
   4. DMPhil

   Answer: (B) MCA

   Explanation

   From 2019 to 2020, MCA rises from 55,700 to 56,800, an increase of 1,100 / 55,700, about 1.97%. This exceeds the percentage increases for the other three courses.

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3. Question 3

   Which course has the lowest percentage fee increase from 2017 to 2018?

   Study the table of semester fees for four courses during 2016-2020.

   Year	BCA	MCA	MTech	MPhil
   2016	25,000	52,000	60,200	30,000
   2017	26,500	53,500	61,400	32,000
   2018	27,200	54,200	62,500	32,600
   2019	27,600	55,700	62,900	33,400
   2020	28,000	56,800	63,700	33,900

   1. ABCA
   2. BMCA
   3. CMTech
   4. DMPhil

   Answer: (B) MCA

   Explanation

   From 2017 to 2018, MCA rises by 1,100 on 53,500, about 2.06%. The increases for BCA, MTech and MPhil are all larger percentages.

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4. Question 4

   Find out the difference between total semester fees (for all courses) of 2018 and 2019.

   Study the table of semester fees for four courses during 2016-2020.

   Year	BCA	MCA	MTech	MPhil
   2016	25,000	52,000	60,200	30,000
   2017	26,500	53,500	61,400	32,000
   2018	27,200	54,200	62,500	32,600
   2019	27,600	55,700	62,900	33,400
   2020	28,000	56,800	63,700	33,900

   1. A3,250
   2. B3,270
   3. C3,150
   4. D3,100

   Answer: (D) 3,100

   Explanation

   The 2018 total is 182,300 and the 2019 total is 185,400. The difference is 185,400 - 182,300 = 3,100.

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5. Question 5

   Find out the average of difference of fees of MCA and MTech for 2016-2020.

   Study the table of semester fees for four courses during 2016-2020.

   Year	BCA	MCA	MTech	MPhil
   2016	25,000	52,000	60,200	30,000
   2017	26,500	53,500	61,400	32,000
   2018	27,200	54,200	62,500	32,600
   2019	27,600	55,700	62,900	33,400
   2020	28,000	56,800	63,700	33,900

   1. A7,700
   2. B7,600
   3. C7,650
   4. D7,750

   Answer: (A) 7,700

   Explanation

   The MTech-minus-MCA differences are 8,200, 7,900, 8,300, 7,200 and 6,900. Their sum is 38,500, and 38,500 / 5 = 7,700.

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6. Question 6

   The web portal of SWAYAM Prabha is maintained by:

   1. AIndian Institute of Technology, Kharagpur
   2. BDoordarshan
   3. CGoogle India office
   4. DINFLIBNET Centre

   Answer: (D) INFLIBNET Centre

   Explanation

   INFLIBNET Centre maintains the SWAYAM Prabha web portal. Doordarshan carries the channels but is not the portal maintainer.

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7. Question 7

   The National Repository of Open Educational Resources is managed by:

   1. AIndian Institute of Technology, Delhi
   2. BIndira Gandhi National Open University
   3. CCentral Institute of Educational Technology, NCERT
   4. DUniversity Grants Commission

   Answer: (C) Central Institute of Educational Technology, NCERT

   Explanation

   The National Repository of Open Educational Resources is managed by CIET, an NCERT constituent. The other institutions run different higher-education initiatives.

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8. Question 8

   Given below are two statements.
   Statement I: The ability to form and use symbols (words, gestures, signs, images, and so on) is accomplished in the preoperational period.
   Statement II: According to Piaget, preoperational children do not have a tendency to be egocentric. In light of the above statements, choose the most appropriate answer.

   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: (C) Statement I is correct but Statement II is incorrect

   Explanation

   Piaget associates symbolic thought with the preoperational stage, so Statement I is true. Preoperational children are characteristically egocentric, so Statement II is false.

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9. Question 9

   SWAYAM platform was developed by:
   A. MHRD (now Ministry of Education), Government of India
   B. Google
   C. Microsoft
   D. IIT, Delhi
   E. Adobe Inc. Choose the correct answer from the options given below.

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

   Answer: (A) A and B only

   Explanation

   SWAYAM is a Ministry of Education initiative. Its platform development included help from Google, so A and B is the matching combination. Microsoft and Adobe are not the named developers.

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10. Question 10

    Match List I with List II.

    List I	List II
    A. Coursera	I. Anant Agarwal
    B. Udacity	II. Salman Khan
    C. edX	III. Andrew Ng, Daphne Koller
    D. Khan Academy	IV. Sebastian Thrun

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

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

    Explanation

    Coursera was founded by Andrew Ng and Daphne Koller. Udacity is associated with Sebastian Thrun, edX with Anant Agarwal, and Khan Academy with Salman Khan. The mapping is A-III, B-IV, C-I, D-II.

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11. Question 11

    If a researcher has to test a hypothesis whether the difference between the means of two groups is significant, she/he will use:

    1. AChi-square test
    2. BCorrelation coefficient
    3. CSign-test
    4. DStudent's t-test

    Answer: (D) Student's t-test

    Explanation

    Student's t-test compares the means of two groups when testing whether their difference is statistically significant. Chi-square tests categorical association instead.

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12. Question 12

    Which of the following are the stages of meta analysis?
    A. Understand the literature
    B. Conduct new studies to fill the gap
    C. Develop inclusion and exclusion criteria of the study
    D. Locating primary studies
    E. Identify secondary data Choose the correct answer from the options given below.

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

    Answer: (C) A, C and D only

    Explanation

    Meta-analysis begins by understanding the literature, setting inclusion and exclusion criteria, and locating primary studies. Conducting new studies is not a stage of synthesising existing studies.

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13. Question 13

    Given below are two statements.
    Statement I: Parametric statistics are more powerful than non-parametric statistics.
    Statement II: Parametric statistics are assumption-free statistics. In light of the above statements, choose the correct answer.

    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

    Parametric tests can be more powerful when their assumptions hold. They are not assumption-free because they rely on conditions such as distributional form and scale.

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14. Question 14

    Which of the following methods is commonly used in anthropological research?

    1. ASurvey method
    2. BCase study method
    3. CLife history method
    4. DEthnographic method

    Answer: (D) Ethnographic method

    Explanation

    Ethnography is the characteristic anthropological approach because it studies people and culture through sustained field observation and participation.

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15. Question 15

    Given below are two statements.
    Statement I: Exploratory studies mainly aim at developing hypotheses or questions for further research.
    Statement II: The goal of formal study is to test the hypotheses or answer the questions posed. In light of the above statements, choose the correct answer.

    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: (A) Both Statement I and Statement II are true

    Explanation

    Exploratory research develops questions or hypotheses. Formal studies then test those hypotheses or answer the stated questions, so both statements are true.

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16. Question 16

    In communication, entropy refers to:

    1. APredictable message
    2. BUnpredictable message
    3. COratory
    4. DDefensive speech

    Answer: (B) Unpredictable message

    Explanation

    Entropy in information theory measures uncertainty or unpredictability in a message. A predictable message has lower entropy.

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17. Question 17

    Keeping quiet because of fear of punishment in communication is an _________ barrier.

    1. AInstitutional
    2. BInteractional
    3. CIncentive
    4. DIndividual

    Answer: (D) Individual

    Explanation

    Fear of punishment is an individual or psychological barrier because it inhibits the sender or receiver personally.

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18. Question 18

    A general code of conduct circulated among the employees of an institution is an example of:

    1. ADownward communication
    2. BLateral communication
    3. CHorizontal communication
    4. DUpward communication

    Answer: (A) Downward communication

    Explanation

    A code of conduct sent from an institution to its employees moves from a higher organisational level to lower levels. That is downward communication.

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19. Question 19

    What is a blog?

    1. AInternet
    2. BSearch engine
    3. CPersonal website in the form of a journal
    4. DOnline journal

    Answer: (C) Personal website in the form of a journal

    Explanation

    A blog is commonly a personal website or web page organised as a journal. “Online journal” is broad, but the more precise definition is the personal journal-style website.

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20. Question 20

    Assertion (A): Teachers should accept students as they are and with whatever cognitive level they have.
    Reason (R): Moralizing, being judgemental, and exposing ignorance are barriers to successful communication. In light of the above statements, choose the correct answer.

    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. CA is true but R is false
    4. DA is false but R is true

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

    Explanation

    Accepting learners without moralising or exposing ignorance supports open communication. The reason explains why the asserted teacher behaviour is needed.

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21. Question 21

    Which one of the following fractions is the largest?

    1. A5/8
    2. B11/16
    3. C31/40
    4. D67/80

    Answer: (D) 67/80

    Explanation

    Converting to decimals gives 0.625, 0.6875, 0.775 and 0.8375. Therefore 67/80 is the largest.

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22. Question 22

    Find the next two terms in the series. A, D, H, M, ___, ___

    1. AQ, X
    2. BR, Y
    3. CS, Z
    4. DT, A

    Answer: (C) S, Z

    Explanation

    In the series, the letter positions are 1, 4, 8 and 13. The increments are +3, +4 and +5. Continuing with +6 and +7 gives positions 19 and 26, so the next two terms are S and Z.

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23. Question 23

    The ages of Swati and Manisha are in the ratio of 5:6 and the sum of their ages is 55 years. What will be the ratio of their ages after 5 years?

    1. A6:7
    2. B7:6
    3. C6:5
    4. D5:6

    Answer: (A) 6:7

    Explanation

    The present ages are 25 and 30 because 5x + 6x = 55. After five years they are 30 and 35, which reduces to 6:7.

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24. Question 24

    Given below are two statements.
    Statement I: A vehicle's speed of 30.6 km/hr is the same as 5.8 m/s.
    Statement II: The speed of a bus is 72 km/hr. The distance covered by the bus in 10 seconds is 100 metre. In light of the above statements, choose the correct answer.

    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

    30.6 km/h equals 30.6 × 5/18 = 8.5 m/s, not 5.8 m/s. Also 72 km/h is 20 m/s, so in 10 seconds the bus covers 200 m, not 100 m.

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25. Question 25

    The average of ten numbers is 11. If each number is multiplied by 7, then the average of the new set of numbers is:

    1. A70
    2. B110
    3. C77
    4. D76

    Answer: (C) 77

    Explanation

    Multiplying every observation by a constant multiplies the mean by the same constant. 11 × 7 = 77.

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26. Question 26

    Which of the following statements correctly apply to deductive arguments?
    A. Deductive arguments are either valid or invalid.
    B. There must be a necessary relation between the premises and the conclusion.
    C. Adding new premises improves the validity.
    D. False premises may lead to a true conclusion. Choose the correct answer from the options given below.

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

    Answer: (B) A, B and D only

    Explanation

    Deductive arguments are assessed as valid or invalid, and validity concerns a necessary link from premises to conclusion. False premises can still yield a true conclusion. Adding premises does not itself improve validity.

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27. Question 27

    Identify the fallacy committed in the argument: Since no tragic actors are idiots and some comedians are not idiots, so some comedians are not tragic actors.

    1. AFallacy of the Illicit Process of Major Term
    2. BFallacy of the Illicit Process of Minor Term
    3. CExistential Fallacy
    4. DFallacy of Exclusive Premises

    Answer: (D) Fallacy of Exclusive Premises

    Explanation

    The premises distribute neither term in the way needed to infer a relation between comedians and tragic actors. The conclusion improperly combines two negative premises, an exclusive-premises fallacy.

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28. Question 28

    "No one has been able to prove the existence of extrasensory perception. We must therefore conclude that extrasensory perception is a myth." Which of the following fallacies is committed in this argument?

    1. AAppeal to ignorance
    2. BStraw man
    3. CAppeal to unqualified authority
    4. DRed herring

    Answer: (A) Appeal to ignorance

    Explanation

    The argument treats failure to prove a claim as proof that the claim is false. That is an appeal to ignorance.

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29. Question 29

    Which one of the following statements is true with reference to Sāmānyatodṛṣṭā inference?

    1. AThe middle term is related to the major term as a cause
    2. BThe middle term is related to the major term as an effect
    3. CThe middle term is related to the major term neither as a cause nor as an effect
    4. DThe middle term is related to the major term both as a cause and an effect

    Answer: (C) The middle term is related to the major term neither as a cause nor as an effect

    Explanation

    In Sāmānyatodṛṣṭā inference, the middle term is related to the major term neither as cause nor as effect. The inference rests on a more general observed relation.

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30. Question 30

    Given below are two statements.
    Statement I: According to Nyāya, universals are eternal entities which are distinct from, but inhere in many individuals.
    Statement II: The universal is the basis of the notion of sameness that we have with regard to all the individuals of a certain class. In light of the above statements, choose the correct answer.

    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: (A) Both Statement I and Statement II are true

    Explanation

    Nyāya treats universals as eternal entities inhering in many individuals. A universal also grounds our recognition of commonness within a class, so both statements are true.

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31. Question 31

    Which one of the following is a type of malware embedded in a program on a computer that will set off a malicious function when certain conditions are met?

    1. AWorm
    2. BTrojan
    3. CSpyware
    4. DLogic Bomb

    Answer: (D) Logic Bomb

    Explanation

    A logic bomb is malicious code embedded in a program that activates when a specified condition or date is met.

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32. Question 32

    Which memory is used as part of the processor?

    1. ARAM
    2. BROM
    3. CRegisters
    4. DHard disk

    Answer: (C) Registers

    Explanation

    Registers are small, high-speed storage locations inside the processor. RAM, ROM and disks are not processor registers.

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33. Question 33

    Given below are two statements.
    Statement I: The MAC address identifies which device is connected at a given IP address.
    Statement II: MAC addresses are unique to each device. In light of the above statements, choose the correct answer.

    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: (A) Both Statement I and Statement II are true

    Explanation

    A MAC address is a device's link-layer hardware identifier, and ARP can associate an IP address with a MAC address on a local network. The wording is broad but both statements express the intended relationship.

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34. Question 34

    Following are some statements regarding most of the web browsers. Choose the correct statements.
    A. They have a home page.
    B. They do not have the ability to store users' favourite pages.
    C. They have hyperlinks to allow users to navigate.
    D. They do not keep a history of websites visited by the users. Choose the correct answer from the options given below.

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

    Answer: (A) A and C only

    Explanation

    Browsers normally offer a home page and hyperlinks for navigation. They can also store favourites and browsing history, so statements B and D are false.

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35. Question 35

    Match the memory types in List I with their underlying technology in List II.

    List I	List II
    A. Dynamic Random Access Memory (DRAM)	I. Flip-flops
    B. Static Random Access Memory (SRAM)	II. Floating gate transistors and capacitors
    C. Programmable Read Only Memory (PROM)	III. Transistors and capacitors
    D. Erasable Programmable Read Only Memory (EPROM)	IV. Matrix of fuses

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

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

    Explanation

    DRAM uses transistors and capacitors. SRAM uses flip-flops. PROM uses a fuse matrix, while EPROM uses floating-gate transistor technology. The mapping is A-III, B-I, C-IV, D-II.

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36. Question 36

    Which one of the following Millennium Development Goals is about ensuring Environmental Sustainability?

    1. AGoal-1
    2. BGoal-3
    3. CGoal-5
    4. DGoal-7

    Answer: (D) Goal-7

    Explanation

    Millennium Development Goal 7 is “Ensure environmental sustainability.” The other listed goal numbers address different areas.

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37. Question 37

    The air pollutant NOx is a combination of:

    1. ANitrous oxide and Nitrogen dioxide
    2. BNitrous oxide and Nitric oxide
    3. CNitrogen dioxide and Nitric oxide
    4. DNitrogen pentoxide and Nitrogen dioxide

    Answer: (C) Nitrogen dioxide and Nitric oxide

    Explanation

    NOx is a collective label for nitrogen oxides, prominently nitric oxide (NO) and nitrogen dioxide (NO₂).

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38. Question 38

    Assertion (A): Schistosomiasis is a polluted water-borne disease.
    Reason (R): Schistosomiasis is mainly caused due to the presence of sulphate in water. In light of the above statements, choose the correct answer.

    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. CA is true but R is false
    4. DA is false but R is true

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

    Explanation

    Schistosomiasis is linked to contaminated freshwater and parasitic infection. It is not caused by sulphate in water, so the assertion is true and the reason false.

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39. Question 39

    Article 48A of the Indian Constitution directs the states to ensure the protection and improvement of:

    1. AEducation
    2. BEnvironment
    3. CBusinesses
    4. DMineral resources

    Answer: (B) Environment

    Explanation

    Article 48A directs the State to protect and improve the environment and safeguard forests and wildlife.

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40. Question 40

    Which of the following states has maximum wind power potential in India as per MNRE's latest report?

    1. ARajasthan
    2. BTamil Nadu
    3. CKarnataka
    4. DGujarat

    Answer: (D) Gujarat

    Explanation

    At the 100 m assessment used by the contemporaneous MNRE/NIWE report, Gujarat has 84.43 GW of potential. That exceeds Karnataka, Tamil Nadu and Rajasthan among the listed states.

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41. Question 41

    NSQF will be aligned with International Standard Classification of Occupations maintained by:

    1. AILO
    2. BUNESCO
    3. CNQFVE
    4. DMinistry of Skill Development

    Answer: (A) ILO

    Explanation

    The International Standard Classification of Occupations is maintained by the International Labour Organization, so NSQF alignment refers to the ILO classification.

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42. Question 42

    SARTHAQ is:

    1. AA training programme for teachers launched by the Government of India
    2. BImplementation plan for NEP-2020
    3. CA Government of India initiative to provide psychological support for students and teachers for mental health and emotional well being
    4. DAn organization for supporting differently abled children

    Answer: (B) Implementation plan for NEP-2020

    Explanation

    SARTHAQ is the implementation plan for the National Education Policy 2020. It translates NEP provisions into actions and timelines.

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43. Question 43

    Given below are two statements.
    Statement I: The National Knowledge Commission recommended opening 500 universities to enable India to attain a Gross Enrolment Ratio of at least 20% by 2015.
    Statement II: The need to establish an Independent Regulatory Authority for Higher Education (IRAHE) was recommended by the National Knowledge Commission. In light of the above statements, choose the most appropriate answer.

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

    Explanation

    The National Knowledge Commission proposed a much higher expansion target than 500 universities for the stated GER claim, making Statement I incorrect. It did recommend an Independent Regulatory Authority for Higher Education, so Statement II is correct.

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44. Question 44

    Which of the following statement(s) do NOT hold true with respect to Vocational Education as per NEP-2020?
    A. B.Voc degrees introduced in 2013 will be discontinued.
    B. Lok Vidya courses will replace existing vocational courses.
    C. Focus areas for vocational education will be chosen on skills gap analysis without mapping local opportunities.
    D. National Committee for Integration of Vocational Education will be constituted.
    E. Vocational education will be integrated into all schools and higher education institutions in a phased manner over the next decade. Choose the correct answer from the options given below.

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

    Answer: (A) A, B and C only

    Explanation

    NEP 2020 does not say B.Voc will be discontinued, does not make Lok Vidya a replacement for all vocational courses, and requires local opportunities to be mapped. Thus A, B and C do not hold true.

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45. Question 45

    Match the historical learning centres in List I with their associated traditions or fields in List II.

    List I	List II
    A. Taxila	I. Tantrik teachings
    B. Ujjain	II. Dvaita school of thought
    C. Vikramshila	III. Medicine
    D. Manyakheta	IV. Astronomy

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

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

    Explanation

    Taxila is associated with medicine, Ujjain with astronomy, Vikramshila with Tantrik teachings, and Manyakheta with the Dvaita school. The mapping is A-III, B-IV, C-I, D-II.

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46. Question 46

    What is the focus of this text?

    Assessment is central to almost all forms of education around the world, and in recent years, a wider variety of assessment has tended to be used. One such form is peer and self-assessment. Self-assessment is the ability to critically assess one's own work. Peer assessment is making judgments about, or commenting upon, each other's work. Both have strengths and weaknesses. With careful initiation and moderation, they can be effective. However, the author argues that in the high-stakes environments of most educational settings, these methods are inappropriate because marking is unreliable.

    Peer and self-assessment have become increasingly common, particularly in further and higher education. They can reduce teachers' marking loads and enhance group work and students' critical-thinking skills. If marking criteria are discussed and agreed at the start, students' marks tend to be similar to teachers' marks. With suitable initial training, work marked by a student may not differ from work marked by a teacher.

    1. APeer assessment
    2. BAssessment and Education
    3. CAssessment in general
    4. DPeer and Self-assessment

    Answer: (D) Peer and Self-assessment

    Explanation

    The passage discusses both peer assessment and self-assessment, their claimed benefits and their limits. Its focus is therefore peer and self-assessment together.

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47. Question 47

    What benefits accrue from self-assessment and peer-assessment of students?

    Assessment is central to almost all forms of education around the world, and in recent years, a wider variety of assessment has tended to be used. One such form is peer and self-assessment. Self-assessment is the ability to critically assess one's own work. Peer assessment is making judgments about, or commenting upon, each other's work. Both have strengths and weaknesses. With careful initiation and moderation, they can be effective. However, the author argues that in the high-stakes environments of most educational settings, these methods are inappropriate because marking is unreliable.

    Peer and self-assessment have become increasingly common, particularly in further and higher education. They can reduce teachers' marking loads and enhance group work and students' critical-thinking skills. If marking criteria are discussed and agreed at the start, students' marks tend to be similar to teachers' marks. With suitable initial training, work marked by a student may not differ from work marked by a teacher.

    1. AAbility to criticise other's work
    2. BEnhancement of group work and critical thinking of students
    3. CPromotes uniform assessment
    4. DThese methods do not require marking

    Answer: (B) Enhancement of group work and critical thinking of students

    Explanation

    The passage says these methods can enhance group work and students' critical-thinking skills. It does not say they remove all marking or guarantee uniform assessment.

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48. Question 48

    What argument does the author seek to make?

    Assessment is central to almost all forms of education around the world, and in recent years, a wider variety of assessment has tended to be used. One such form is peer and self-assessment. Self-assessment is the ability to critically assess one's own work. Peer assessment is making judgments about, or commenting upon, each other's work. Both have strengths and weaknesses. With careful initiation and moderation, they can be effective. However, the author argues that in the high-stakes environments of most educational settings, these methods are inappropriate because marking is unreliable.

    Peer and self-assessment have become increasingly common, particularly in further and higher education. They can reduce teachers' marking loads and enhance group work and students' critical-thinking skills. If marking criteria are discussed and agreed at the start, students' marks tend to be similar to teachers' marks. With suitable initial training, work marked by a student may not differ from work marked by a teacher.

    1. ASelf-assessment and peer-assessment are innovative assessment practices
    2. BAssessment is central to education
    3. CPeer and self-assessments are inappropriate in high-stakes examinations
    4. DPeer assessment is common in higher education

    Accepted answers: (A) Self-assessment and peer-assessment are innovative assessment practices, (B) Assessment is central to education, (C) Peer and self-assessments are inappropriate in high-stakes examinations, (D) Peer assessment is common in higher education

    Explanation

    The passage argues for option C: peer and self-assessment are inappropriate in high-stakes settings because marking is unreliable. NTA awarded marks to all candidates for this item, so every displayed selection is accepted to preserve the official scoring outcome.

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49. Question 49

    Which one of the following captures the meaning of "Assessment is central to almost all forms of education"?

    Assessment is central to almost all forms of education around the world, and in recent years, a wider variety of assessment has tended to be used. One such form is peer and self-assessment. Self-assessment is the ability to critically assess one's own work. Peer assessment is making judgments about, or commenting upon, each other's work. Both have strengths and weaknesses. With careful initiation and moderation, they can be effective. However, the author argues that in the high-stakes environments of most educational settings, these methods are inappropriate because marking is unreliable.

    Peer and self-assessment have become increasingly common, particularly in further and higher education. They can reduce teachers' marking loads and enhance group work and students' critical-thinking skills. If marking criteria are discussed and agreed at the start, students' marks tend to be similar to teachers' marks. With suitable initial training, work marked by a student may not differ from work marked by a teacher.

    1. AAssessment is pivotal to education in all forms
    2. BAssessment is education in all forms
    3. CAssessment education is key to education in all forms
    4. DAssessment is peripheral to education in all forms

    Answer: (A) Assessment is pivotal to education in all forms

    Explanation

    “Central” means pivotal or essential. It does not mean that assessment is identical to education or peripheral to it.

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50. Question 50

    "Students' marking does not differ significantly from teachers", means:

    Assessment is central to almost all forms of education around the world, and in recent years, a wider variety of assessment has tended to be used. One such form is peer and self-assessment. Self-assessment is the ability to critically assess one's own work. Peer assessment is making judgments about, or commenting upon, each other's work. Both have strengths and weaknesses. With careful initiation and moderation, they can be effective. However, the author argues that in the high-stakes environments of most educational settings, these methods are inappropriate because marking is unreliable.

    Peer and self-assessment have become increasingly common, particularly in further and higher education. They can reduce teachers' marking loads and enhance group work and students' critical-thinking skills. If marking criteria are discussed and agreed at the start, students' marks tend to be similar to teachers' marks. With suitable initial training, work marked by a student may not differ from work marked by a teacher.

    1. AMarks that differ radically from teachers
    2. BMarks do not differ peripherally
    3. CMarks that are almost similar
    4. DMarks may or may not differ

    Answer: (C) Marks that are almost similar

    Explanation

    If marks do not differ significantly, they are approximately alike rather than radically different. “Almost similar” best captures that meaning.

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100 questions with the answer key and explanations.

1. Question 1

   It is always important and useful to include an 'alt' attribute on an 'img' tag in HTML. Which of the reasons given below is/are correct?

   A. Users who cannot see the image due to vision impairment can be given a textual description that a screen reader speaks aloud.
   B. If the image fails to load (slow connection, broken path, etc.), the alt text is displayed instead.
   C. It gives SEO (Search Engine Optimization) benefits.

   Choose the correct answer from the options given below:

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

   Answer: (D) A, B, and C

   Explanation

   The alt attribute serves accessibility (screen readers read it aloud for visually impaired users), graceful degradation (it is shown when the image cannot load), and SEO (search engines use it to understand and index image content). All three statements are therefore correct.

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2. Question 2

   What does the following function f() in C return?

   int f(unsigned int N) {
       unsigned int counter = 0;
       while (N > 0) {
           counter += N & 1;
           N = N >> 1;
       }
       return counter == 1;
   }
   

   1. A1 if N is odd, otherwise 0
   2. B1 if N is a power of 2, otherwise 0
   3. C1 if the binary representation of N is all 1s, otherwise 0
   4. D1 if the binary representation of N has any 1s, otherwise 0

   Answer: (B) 1 if N is a power of 2, otherwise 0

   Explanation

   The loop adds (N & 1) to counter for every bit while right-shifting N, so counter ends up holding the number of set bits (population count) of N. counter == 1 is true exactly when N has a single 1-bit, i.e. when N is a power of 2. An odd number such as 3 has two 1-bits, so option A is wrong.

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3. Question 3

   Consider the following recursive function F() in Java that takes an integer and returns a String:

   public static String F(int N) {
       if (N <= 0) return "-";
       return F(N - 3) + N + F(N - 2) + N;
   }
   

   What is the value of F(5)?

   1. A-2-25-3-135
   2. B-2-25-1-3-135
   3. C-1-145-2-245
   4. D-2-25-3-1-135

   Answer: (D) -2-25-3-1-135

   Explanation

   The function concatenates strings: the base case returns a dash, and each non-base call returns F(N-3) then N then F(N-2) then N. Expanding, F(5) needs F(2) and F(3). F(2) evaluates to "-2-2" because both recursive calls hit the base case. F(1) evaluates to "-1-1", so F(3) becomes "-3-1-13". Substituting everything back, F(5) equals "-2-2" then 5 then "-3-1-13" then 5, giving "-2-25-3-1-135".

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4. Question 4

   Match List-I (Programming Terms) with List-II (Meaning).

   List-I	List-II
   A. JNDI	I. Runtime support for running a Java program
   B. RMI	II. The API in support of naming and directory services
   C. JDK	III. The methods and runtime support for calling remote methods
   D. JRE	IV. The compiler and class libraries to develop Java applications

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

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

   Explanation

   JNDI (Java Naming and Directory Interface) is the API for naming and directory services (II). RMI (Remote Method Invocation) provides the mechanism for invoking methods on remote JVM objects (III). JDK bundles the compiler and class libraries to build Java applications (IV). JRE supplies the runtime support (JVM plus libraries) needed to run a Java program (I).

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5. Question 5

   Match List-I (Programming Paradigms) with List-II (Characteristics).

   List-I	List-II
   A. Imperative	I. Declarative, clausal representation, theorem proving
   B. Object-oriented	II. Side-effect-free, declarative, expression evaluation
   C. Logic	III. Imperative, abstract data type
   D. Functional	IV. Command-based, procedural

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

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

   Explanation

   Imperative programming is command-based and procedural (IV). Object-oriented programming is imperative in style and organised around abstract data types/objects (III). Logic programming uses declarative clauses and resolution/theorem proving (I). Functional programming is declarative, side-effect-free and based on expression evaluation (II).

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6. Question 6

   You have eight black-and-white images taken with a 1-megapixel camera and one 8-colour image taken with an 8-megapixel camera. How much disk space in total is needed to store these images?

   1. A1 GB
   2. B4 MB
   3. C3 MB
   4. D3 GB

   Answer: (B) 4 MB

   Explanation

   A black-and-white pixel needs 1 bit, so each 1-MP image is 1,000,000 bits = 125 KB, and eight of them total about 1 MB. An 8-colour image needs 3 bits per pixel, so 8 MP x 3 bits = 24,000,000 bits = about 3 MB. Total is roughly 1 MB + 3 MB = 4 MB.

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7. Question 7

   The midpoint (Bresenham) algorithm for rasterizing lines is optimized relative to the DDA algorithm in that:

   A. It avoids round-off operations.
   B. It is incremental.
   C. It uses only integer arithmetic.
   D. All straight lines can be displayed as exactly straight.

   Choose the correct answer from the options given below:

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

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

   Explanation

   Bresenham's advantages over DDA are that it is incremental, uses only integer arithmetic and avoids the floating-point rounding DDA needs (A, B, C). Statement D is false: like every raster algorithm it only produces a discrete approximation of a line, not an exact geometric straight line.

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8. Question 8

   Find the transformation matrix M that maps the square with vertices (1,1), (-1,1), (-1,-1), (1,-1) to the parallelogram with corresponding vertices (2,1), (0,1), (-2,-1), (0,-1).

   1. A[[1, 1, 0], [0, 1, 0], [0, 0, 1]]
   2. B[[1, 0, 0], [1, 1, 0], [0, 0, 1]]
   3. C[[1, 1, 1], [0, 1, 0], [0, 0, 1]]
   4. D[[1, 1, 0], [1, 1, 0], [0, 0, 1]]

   Answer: (A) [[1, 1, 0], [0, 1, 0], [0, 0, 1]]

   Explanation

   Write the homogeneous matrix with unknown top-left entries a, b, c, d. From (1,1) to (2,1): a + b = 2 and c + d = 1. From (-1,1) to (0,1): -a + b = 0, so a = b = 1; then c + d = 1 with -c + d = 1 gives c = 0, d = 1. So x = x + y, y = y, a horizontal shear: [[1,1,0],[0,1,0],[0,0,1]].

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9. Question 9

   A Bezier curve P(t) is defined by four control points in the xy-plane: P0 = (-2, 0), P1 = (-2, 4), P2 = (2, 4), P3 = (2, 0). Which statements are correct?

   A. P(t) has degree 3.
   B. P(1/2) = (0, 3).
   C. P(t) may extend outside the convex hull of its control points.

   Choose the correct answer from the options given below:

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

   Answer: (A) A and B only

   Explanation

   Four control points give a cubic Bezier curve, degree 3 (A). At t = 1/2 the Bernstein weights are 1/8, 3/8, 3/8, 1/8; x = (-2 - 6 + 6 + 2)/8 = 0 and y = (0 + 12 + 12 + 0)/8 = 3, so P(1/2) = (0, 3) (B). Statement C is false: a Bezier curve always lies within the convex hull of its control points.

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10. Question 10

    Statement I: The maximum number of sides a triangle can have after being clipped to a rectangular viewport is 6.
    Statement II: In 3D graphics, the perspective transformation is nonlinear in z.

    In light of the above statements, choose the correct answer from the options given below:

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

    Explanation

    Clipping a 3-sided polygon against a 4-sided rectangle can add at most one vertex per clipping edge, giving up to 3 + 4 = 7 sides, so Statement I (which says 6) is false. Perspective projection divides coordinates by depth z, a nonlinear operation, so Statement II is true.

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11. Question 11

    In software testing, beta testing is the testing performed by ______.

    1. Apotential customers at the developer's location
    2. Bpotential customers at their own locations
    3. Cproduct developers at the customer's location
    4. Dproduct developers at their own locations

    Answer: (B) potential customers at their own locations

    Explanation

    Beta testing is done by real or potential users in their own environment, outside the developer's site, after alpha testing. Testing by customers at the developer's site is alpha testing.

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12. Question 12

    In the MVC architectural pattern, the View component is responsible for:

    1. AUser interface
    2. BSecurity of the system
    3. CBusiness logic and domain objects
    4. DTranslating between user interface actions/events and operations on the domain objects

    Answer: (A) User interface

    Explanation

    In Model-View-Controller the View handles presentation and user-interface rendering. Business logic and domain data belong to the Model, and translating user actions into Model operations is the Controller.

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13. Question 13

    In software engineering, what kind of notation do formal methods predominantly use?

    1. Atextual
    2. Bdiagrammatic
    3. Cmathematical
    4. Dcomputer code

    Answer: (C) mathematical

    Explanation

    Formal methods specify and verify software using mathematically based notations (set theory, predicate logic, algebra; e.g. Z, VDM, B) that allow rigorous proof of properties. UML-style diagrams are only semi-formal.

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14. Question 14

    If every requirement stated in the Software Requirement Specification (SRS) has only one interpretation, then the SRS is said to be:

    1. Acorrect
    2. Bconsistent
    3. Cunambiguous
    4. Dverifiable

    Answer: (C) unambiguous

    Explanation

    An SRS is unambiguous when every requirement has exactly one interpretation. Consistency means requirements do not conflict; correctness means each requirement is one the system must meet; verifiability means each can be checked.

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15. Question 15

    A system has 99.99% uptime and a mean-time-between-failure (MTBF) of 1 day. How quickly must the system repair itself to meet this availability goal?

    1. A~9 seconds
    2. B~10 seconds
    3. C~11 seconds
    4. D~12 seconds

    Answer: (A) ~9 seconds

    Explanation

    Availability = MTBF / (MTBF + MTTR). With A = 0.9999 and MTBF = 86,400 s, the repair time MTTR = MTBF x (1 - A) = 86,400 x 0.0001 = 8.64 s, i.e. about 9 seconds.

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16. Question 16

    In Software Configuration Management (SCM), which kinds of files should be committed to the source control repository?

    A. Code files
    B. Documentation files
    C. Output files
    D. Automatically generated files that are required for the system to be used

    Choose the correct answer from the options given below:

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

    Answer: (A) A and B only

    Explanation

    Hand-authored source artefacts - code (A) and documentation (B) - belong in version control. Build outputs (C) and generated files (D) should not be committed because they are reproducible from the sources and only add merge noise and staleness.

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17. Question 17

    Match List-I (Software Process Model) with List-II (Description).

    List-I	List-II
    A. Waterfall Model	I. Software can be developed incrementally
    B. Evolutionary Model	II. Requirement compromises are inevitable
    C. Component-based Software Engineering	III. Explicit recognition of risk
    D. Spiral Development	IV. Inflexible partitioning of the project into stages

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

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

    Explanation

    Waterfall rigidly partitions the project into sequential stages (IV). The Evolutionary model builds software incrementally through iterations (I). Component-based development reuses pre-built components, so requirement compromises are inevitable (II). The Spiral model performs explicit risk analysis at each cycle (III).

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18. Question 18

    Identify the correct order (lower to higher) of the five levels of the Capability Maturity Model used to measure the maturity of an organisation's software process.

    A. Defined
    B. Optimizing
    C. Initial
    D. Managed
    E. Repeatable

    Choose the correct answer from the options given below:

    1. AC, A, E, D, B
    2. BC, E, A, B, D
    3. CC, B, D, E, A
    4. DC, E, A, D, B

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

    Explanation

    The Capability Maturity Model rises through five stages: Initial (ad hoc, chaotic process), Repeatable (basic project management established), Defined (process documented and standardised across the organisation), Managed (process quantitatively measured and controlled), and Optimizing (continuous, data-driven process improvement). Mapping the given letters, the increasing order is Initial C, Repeatable E, Defined A, Managed D, Optimizing B.

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19. Question 19

    Statement I: The Cleanroom software process model incorporates the statistical quality certification of code increments as they accumulate into a system.
    Statement II: Cleanroom software engineering follows the classic analysis, design, code, test and debug cycle, focusing on defect removal rather than defect prevention.

    In light of the above statements, choose the correct answer from the options given below:

    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

    Cleanroom does use statistical usage testing to certify each code increment (Statement I true). But its philosophy is defect prevention through formal specification and correctness verification, deliberately avoiding the traditional code-and-debug cycle, so Statement II is false.

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20. Question 20

    Assertion (A): Software developers do not do exhaustive software testing in practice.
    Reason (R): Even for small inputs, exhaustive testing is too computationally intensive (takes too long) to run all the tests.

    In 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. CA is true but R is false
    4. DA is false but R is true

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

    Explanation

    Exhaustive testing (trying every input/path) is practically impossible because the input space grows combinatorially, so testers sample instead (A true). That combinatorial blow-up is exactly why exhaustive testing is infeasible, so R is true and correctly explains A.

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21. Question 21

    Which of the following pairs are logically equivalent?

    A. not p -> (q -> r) and q -> (p OR r)
    B. (p -> q) -> r and p -> (q -> r)
    C. (p -> q) -> (r -> s) and (p -> r) -> (q -> s)

    Choose the correct answer from the options given below:

    1. AA only
    2. BA and B only
    3. CB and C only
    4. DA and C only

    Answer: (A) A only

    Explanation

    Pair A: not p -> (q -> r) simplifies to p OR not q OR r, and q -> (p OR r) simplifies to not q OR p OR r - identical, so equivalent. Pair B fails (implication is not associative; try p = q = r = false). Pair C also fails (try p = true, q = r = s = false).

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22. Question 22

    Which statements about the floor and ceiling functions are correct?

    Statement I: floor(2x) = floor(x) + floor(x + 1/2) for every real number x.
    Statement II: ceil(x + y) = ceil(x) + ceil(y) for all real numbers x and y.

    Choose the correct answer from the options given below:

    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

    Statement I is the standard Hermite identity floor(2x) = floor(x) + floor(x + 1/2), valid for all real x. Statement II is false: for x = y = 0.5, ceil(1) = 1 but ceil(0.5) + ceil(0.5) = 2.

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23. Question 23

    How many ways are there to assign 5 different jobs to 4 different employees if every employee is assigned at least one job?

    1. A1024
    2. B625
    3. C240
    4. D20

    Answer: (C) 240

    Explanation

    This counts onto (surjective) functions from a 5-element set to a 4-element set: 4! x S(5,4) = 24 x 10 = 240. Equivalently: pick the employee who gets 2 jobs (4 ways), pick those 2 jobs (C(5,2) = 10), assign the remaining 3 jobs to 3 employees (3! = 6): 4 x 10 x 6 = 240.

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24. Question 24

    A warehouse stores products in bins identified by aisle, location within the aisle, and shelf. There are 50 aisles, 85 locations per aisle, and 5 shelves. What is the least number of products the company can have so that at least two products must be stored in the same bin?

    1. A251
    2. B426
    3. C4251
    4. D21251

    Answer: (D) 21251

    Explanation

    Total bins = 50 x 85 x 5 = 21,250. By the pigeonhole principle, one more product than the number of bins forces two products into the same bin: 21,250 + 1 = 21,251.

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25. Question 25

    A person climbing stairs can take either one stair or two stairs at a time. In how many ways can this person climb a flight of eight stairs?

    1. A21
    2. B24
    3. C31
    4. D34

    Answer: (D) 34

    Explanation

    The last move is either a single stair (leaving f(n-1) ways to reach the previous stair) or a double stair (leaving f(n-2) ways), so f(n) = f(n-1) + f(n-2), the Fibonacci recurrence. Starting from f(1) = 1 and f(2) = 2, the values continue 3, 5, 8, 13, 21, and finally f(8) = 34.

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26. Question 26

    For which value of n is the wheel graph Wn regular?

    1. A2
    2. B3
    3. C4
    4. D5

    Answer: (C) 4

    Explanation

    A wheel Wn has a hub joined to every vertex of a cycle. Each rim vertex has degree 3 (two rim neighbours plus the hub), while the hub has degree n - 1. The graph is regular only when n - 1 = 3, i.e. n = 4 (which is K4).

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27. Question 27

    Which of the labelled graphs is/are planar?

    

    Three undirected graphs labelled A, B and C.

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

    Answer: (C) A only

    Explanation

    Graph A can be redrawn with no edge crossings, so it is planar. Graphs B and C each contain a K3,3 (or K5) subdivision, which by Kuratowski's theorem makes them non-planar. Hence only A is planar.

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28. Question 28

    Match List-I (Identity) with List-II (Name).

    List-I	List-II
    A. x + x = x	I. Identity Law
    B. x + 0 = x	II. Absorption Law
    C. x + 1 = 1	III. Idempotent Law
    D. x + xy = x	IV. Domination Law

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

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

    Explanation

    In Boolean algebra, combining a variable with itself leaves it unchanged, so x + x = x is the Idempotent law (III). Adding the additive identity leaves the value alone, so x + 0 = x is the Identity law (I). Anything combined with logical true collapses to true, so x + 1 = 1 is the Domination or null law (IV). A term that already contains the variable is swallowed, so x + xy = x is the Absorption law (II).

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29. Question 29

    Let (X, *) be a semigroup such that for every a and b in X, if a is not equal to b then a * b is not equal to b * a. Which of the following equalities hold?

    A. a * a = a for every a in X.
    B. a * b * a = a for every a, b in X.
    C. a * b * c = a * c for every a, b, c in X.

    Choose the correct answer from the options given below:

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

    Answer: (D) A, B and C

    Explanation

    The contrapositive of the hypothesis says: whenever two elements commute, they must be equal. For A, a * a commutes with a (associativity gives (aa)a = a(aa)), so a * a = a and every element is idempotent. For B, a * b * a commutes with a (using idempotence a*(aba) = aba = (aba)*a), so a * b * a = a. For C, a * b * c and a * c each absorb the middle factor via B and commute, so the hypothesis forces a * b * c = a * c. All three hold.

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30. Question 30

    Consider the linear program: maximize Z = 6x + 5y subject to 2x - 3y <= 5, x + 3y <= 11, 4x + y <= 15, x >= 0, y >= 0. The optimal value of Z is:

    1. A15
    2. B25
    3. C31.72
    4. D41.44

    Answer: (C) 31.72

    Explanation

    For a linear program the optimum lies at a vertex of the feasible region, so evaluate the objective Z = 6x + 5y at every corner. The best value occurs where the constraints x + 3y = 11 and 4x + y = 15 meet, namely x = 34/11 and y = 29/11, because that point still satisfies 2x - 3y <= 5. There Z = 204/11 + 145/11 = 349/11, which is approximately 31.72; other feasible corners such as (2.5, 0) and (25/7, 5/7) yield only 25 or less.

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31. Question 31

    Which of the following languages are NOT regular?

    A. L = { (01)^n 0^k : n > k, k >= 0 }
    B. L = { c^n b^k a^(n+k) : n >= 0, k >= 0 }
    C. L = { 0^n 1^k : n is not equal to k }

    Choose the correct answer from the options given below:

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

    Answer: (D) A, B and C

    Explanation

    Each language needs an unbounded comparison of counts that a finite automaton cannot perform. A compares the number of (01) blocks with the number of trailing 0s (n > k). B forces the number of a-symbols to equal the sum of the c and b counts. C forces the number of 0s to differ from the number of 1s (its complement 0^n 1^n is the classic non-regular language). All three are non-regular by the pumping lemma.

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32. Question 32

    Any string of terminals generated by the following context-free grammar (S is the start symbol):

    S -> XY
    X -> 0X | 1X | 0
    Y -> Y0 | Y1 | 0
    

    1. Ahas at least one 1
    2. Bshould end with 0
    3. Chas no consecutive 0's or 1's
    4. Dhas at least two 0's

    Answer: (D) has at least two 0's

    Explanation

    X derives any binary string that ends in 0 (length at least 1), and Y derives any binary string that begins with 0. A string from S = XY therefore always contains the trailing 0 of the X-part and the leading 0 of the Y-part - at least two 0s. It need not contain a 1, need not end in 0, and may repeat symbols.

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33. Question 33

    Let L1 = { 0^n 1^n 0^m : n >= 1, m >= 1 }, L2 = { 0^n 1^m 0^m : n >= 1, m >= 1 }, L3 = { 0^n 1^n 0^n : n >= 1 }. Which statements are correct?

    A. L3 = L1 intersect L2
    B. L1 and L2 are context-free languages but L3 is not.
    C. L1 and L2 are not context-free but L3 is.
    D. L1 is a subset of L3.

    Choose the correct answer from the options given below:

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

    Answer: (A) A and B only

    Explanation

    A string in L1 has (first 0s) = (1s); a string in L2 has (1s) = (trailing 0s). A string in both has all three blocks equal, exactly L3, so A is true. L1 and L2 each need only one matched pair, so a single stack (PDA) suffices - they are context-free; L3 needs two independent matches and is not context-free, so B is true. C is false, and D is false since 0^1 1^1 0^2 is in L1 but not in L3.

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34. Question 34

    Statement I: The family of context-free languages is closed under homomorphism.
    Statement II: The family of context-free languages is closed under reversal.

    In light of the above statements, choose the correct answer from the options given below:

    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: (A) Both Statement I and Statement II are true

    Explanation

    Context-free languages are closed under homomorphism (replace each terminal in the grammar by its image string) and under reversal (reverse the right-hand side of every production). Both statements are true.

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35. Question 35

    What is the minimum number of states in the finite automaton equivalent to the transition diagram shown?

    

    State transition diagram of a finite automaton over input alphabet {0, 1} with start state a and accepting state c.

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

    Answer: (C) 5

    Explanation

    Remove the unreachable/dead state, then apply state minimization by repeatedly splitting groups whose members move to different groups on some input. The reachable states collapse into 5 equivalence classes, so the minimal DFA has 5 states.

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36. Question 36

    Find a regular expression for the language accepted by the automaton shown.

    

    A three-state finite automaton over {a, b} whose leftmost state is both the start state and the only accepting state.

    1. A(aa*(a + b)ab)*
    2. B(a + b)ab(ab + bb + aa*(a + b)ab)*
    3. Caab(ab + bb + aa(a + b)ab)*
    4. Da*(a + b)ab(ab + bb + aa*(a + b)ab)*

    Answer: (B) (a + b)ab(ab + bb + aa*(a + b)ab)*

    Explanation

    Trace the paths that begin and end at the start state, since it is also the only accepting state. The machine must first consume a single symbol (a + b) and then the pair ab to return to the accepting state, so every accepted string starts with (a + b)ab. After that it can repeat any loop back to the accepting state, namely ab, bb, or the longer detour aa*(a + b)ab. Putting the mandatory prefix before the starred set of loops gives (a + b)ab(ab + bb + aa*(a + b)ab)*.

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37. Question 37

    What language is accepted by the pushdown automaton M = ({q0, q1, q2}, {a, b}, {a, b, z}, delta, q0, z, {q2}) with the following transitions?

    delta(q0, a, a) = {(q0, aa)}     delta(q0, b, a) = {(q0, ba)}
    delta(q0, a, b) = {(q0, ab)}     delta(q0, b, b) = {(q0, bb)}
    delta(q0, a, z) = {(q0, az)}     delta(q0, b, z) = {(q0, bz)}
    delta(q0, lambda, a) = {(q1, a)} delta(q0, lambda, b) = {(q1, b)}
    delta(q1, a, a) = {(q1, lambda)} delta(q1, b, b) = {(q1, lambda)}
    delta(q1, lambda, z) = {(q2, z)}
    

    1. AL = { w : na(w) = nb(w), w in {a, b}+ }
    2. BL = { w : na(w) <= nb(w), w in {a, b}+ }
    3. CL = { w : nb(w) <= na(w), w in {a, b}+ }
    4. DL = { w w^R : w in {a, b}+ }

    Answer: (D) L = { w w^R : w in {a, b}+ }

    Explanation

    In q0 the PDA pushes every input symbol, building a stack copy of the prefix read so far. A spontaneous lambda-move switches to q1, where each further input symbol must match and pop the top of the stack. This succeeds exactly when the second half of the input is the reverse of the first half, i.e. the input is w w^R.

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38. Question 38

    Match List-I (Production Rules) with List-II (Grammar type).

    List-I	List-II
    A. S -> XY; X -> 0; Y -> 1	I. Greibach Normal Form
    B. S -> aS | bSS | c	II. Context sensitive Grammar
    C. S -> AB; A -> 0A | 1A | 0; B -> 0A	III. Chomsky Normal Form
    D. S -> aAbc; Ab -> bA; Ac -> Bbcc; bB -> Bb; aB -> aa | aaA	IV. S-Grammar

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

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

    Explanation

    A is in Chomsky Normal Form - every rule is "one nonterminal -> two nonterminals" or "-> one terminal" (III). B is an s-grammar: each right side is a terminal followed by nonterminals and every (nonterminal, terminal) pair occurs once (IV). D has multi-symbol left sides such as Ab -> bA and is non-contracting, i.e. context-sensitive (II). By elimination C pairs with Greibach Normal Form, whose rules place a terminal first followed by nonterminals (I).

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39. Question 39

    Which of the following concepts can be used to identify loops?

    A. Depth-first ordering
    B. Dominators
    C. Reducible graphs

    Choose the correct answer from the options given below:

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

    Answer: (D) A, B and C

    Explanation

    Loop detection in control-flow analysis uses all three: a depth-first spanning tree identifies back edges via depth-first ordering, dominator information confirms that the head of a back edge dominates its tail (a natural loop), and reducibility guarantees every retreating edge is a back edge so all loops are natural.

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40. Question 40

    Statement I: LL(1) and LR are examples of bottom-up parsers.
    Statement II: Recursive descent parser and SLR are examples of top-down parsers.

    In light of the above statements, choose the correct answer from the options given below:

    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

    LL(1) parsing and recursive descent are top-down, while LR and SLR parsing are bottom-up. Statement I is wrong about LL(1) and Statement II is wrong about SLR, so both statements are false.

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41. Question 41

    The postfix form of the infix expression ( A + B ) * ( C * D - E ) * F / G is:

    1. AAB + CD*E - FG / **
    2. BAB + CD*E - F**G /
    3. CAB + CD*E - FG /
    4. DAB + CDE*- FG /

    Answer: (C) AB + CD*E - FG /

    Explanation

    Scan the expression and emit each operator immediately after both its operands, respecting precedence and left-to-right association. First A plus B becomes AB+ and C times D becomes CD*, then subtracting E gives CDE-. Multiplying the two parenthesised results produces AB+CDE-, after which multiplying by F appends another star and finally dividing by G appends a slash. The complete postfix string is therefore AB+CDE-FG/.

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42. Question 42

    A double-ended queue is to be implemented using only stacks (push and pop). What is the time complexity of AddFront() and AddRear(), where m is the stack size and n is the number of elements?

    1. AO(m) and O(n)
    2. BO(1) and O(n)
    3. CO(n) and O(1)
    4. DO(n) and O(m)

    Answer: (B) O(1) and O(n)

    Explanation

    If the primary stack keeps the front element on top, AddFront is a single push - O(1). AddRear must reach the bottom of that stack, so all n elements are popped to an auxiliary stack, the new element pushed, and everything moved back - O(n).

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43. Question 43

    Two balanced binary search trees hold m and n elements respectively. They can be merged into one balanced binary search tree in ______ time.

    1. AO(m*n)
    2. BO(m + n)
    3. CO(m*log n)
    4. DO(m*log(m + n))

    Answer: (B) O(m + n)

    Explanation

    In-order traversal of each tree yields a sorted list in O(m) and O(n). Merge the two sorted lists in O(m + n), then build a balanced BST from the merged array in O(m + n). Total O(m + n).

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44. Question 44

    A data structure must store a set of integers so that each of the following runs in O(log n) time: delete the smallest element, and insert an element only if it is not already present. Which structure can be used?

    1. AA heap can be used but not a balanced binary search tree
    2. BA balanced binary search tree can be used but not a heap
    3. CBoth a balanced binary search tree and a heap can be used
    4. DNeither a balanced binary search tree nor a heap can be used

    Answer: (B) A balanced binary search tree can be used but not a heap

    Explanation

    A balanced BST supports find-min, delete-min, membership test and insert all in O(log n). A min-heap gives O(log n) delete-min, but checking whether a value is already present needs an O(n) scan, so it cannot do the conditional insert in O(log n).

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45. Question 45

    Consider the graph shown. Which of the following vertex sequences are valid depth-first traversals of this graph?

    I. a b e g h f
    II. a b f e h g
    III. a b f h g e
    IV. a f g h b e

    

    An undirected graph on vertices a, b, e, f, g and h.

    1. AI, II, and IV only
    2. BI and IV only
    3. CII, III, and IV only
    4. DI, III, and IV only

    Answer: (D) I, III, and IV only

    Explanation

    A DFS may advance only to a vertex adjacent to the current one, or backtrack to an earlier vertex on the stack. Sequences I, III and IV each follow legal edge moves and backtracks. Sequence II fails: after a-b-f, the next vertex e is not reachable from f without first backtracking to a vertex adjacent to e.

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46. Question 46

    Statement I: In an undirected graph, the number of vertices of odd degree is even.
    Statement II: In an undirected graph, the sum of the degrees of all vertices is even.

    In light of the above statements, choose the correct answer from the options given below:

    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: (A) Both Statement I and Statement II are true

    Explanation

    By the handshaking lemma the sum of all vertex degrees equals twice the number of edges, which is even (Statement II). Since the even-degree vertices contribute an even amount, the odd-degree vertices must be even in number for the total to stay even (Statement I).

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47. Question 47

    Give the increasing order of asymptotic growth of the functions f1, f2, f3, f4:

    A. f1(n) = 2^n
    B. f2(n) = n^(3/2)
    C. f3(n) = n log n
    D. f4(n) = n^(log n)

    Choose the correct answer from the options given below:

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

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

    Explanation

    n log n grows slower than the polynomial n^1.5 (C < B). n^(log n) is quasi-polynomial - it eventually overtakes any fixed polynomial but stays below the exponential 2^n (B < D < A). The order is C, B, D, A.

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48. Question 48

    The order of a leaf node in a B+ tree is the maximum number of (value, data-record-pointer) pairs it can hold. Given block size 1 KB (1024 bytes), data-record pointer 7 bytes, value field 9 bytes, and block pointer 6 bytes, what is the order of the leaf node?

    1. A63
    2. B64
    3. C67
    4. D68

    Answer: (A) 63

    Explanation

    A B+ tree leaf holds m (value + record-pointer) pairs plus one block pointer to the next leaf: m x (9 + 7) + 6 <= 1024, so 16m <= 1018, m <= 63.6. The order is 63.

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49. Question 49

    A hash function h(key) = key mod 7 with linear probing is used to insert the keys 44, 45, 79, 55, 91, 18, 63 into a table indexed 0 to 6. What is the final location of key 18?

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

    Answer: (C) 5

    Explanation

    Insert the keys in order using the given modulo-7 hash and resolve every collision by linear probing to the next free slot. Key 44 hashes to slot 2 and key 45 to slot 3; key 79 also hashes to 2, finds slots 2 and 3 occupied, and settles in slot 4. Key 55 takes slot 6 and key 91 takes slot 0. Now key 18 hashes to slot 4, which already holds 79, so the probe moves forward one step and places key 18 in slot 5.

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50. Question 50

    Match List-I (Standard graph algorithm) with List-II (Time complexity), where n and m are the numbers of vertices and edges.

    List-I	List-II
    A. Bellman-Ford algorithm	I. O(m log n)
    B. Kruskal algorithm	II. O(n^3)
    C. Floyd-Warshall algorithm	III. O(n*m)
    D. Topological sorting	IV. O(n + m)

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

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

    Explanation

    Bellman-Ford relaxes every edge n-1 times: O(n*m) (III). Kruskal sorts edges and uses union-find: O(m log n) (I). Floyd-Warshall runs three nested vertex loops: O(n^3) (II). Topological sort visits each vertex and edge once: O(n + m) (IV).

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51. Question 51

    Which type of agent deals with 'happy' and 'unhappy' states?

    1. AUtility-based agent
    2. BModel-based agent
    3. CGoal-based agent
    4. DLearning agent

    Answer: (A) Utility-based agent

    Explanation

    A utility-based agent attaches a numeric utility to each state, letting it prefer states that make it 'happier', choose among competing goal-satisfying states, and trade off conflicting goals. A goal-based agent only distinguishes goal from non-goal states.

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52. Question 52

    Statement I: Breadth-first search is optimal when all step costs are equal, whereas uniform-cost search is optimal for any step cost.
    Statement II: When all step costs are the same, uniform-cost search expands more nodes at depth d than breadth-first search.

    In light of the above statements, choose the correct answer from the options given below:

    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: (A) Both Statement I and Statement II are true

    Explanation

    BFS returns a shallowest, hence cheapest, solution only when every step costs the same, while UCS expands nodes in order of path cost and stays optimal for arbitrary non-negative costs (Statement I). With equal step costs UCS still explores the entire cost contour, including depth-d nodes BFS would stop short of, so it expands more nodes (Statement II).

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53. Question 53

    For the game tree shown, compute the value backed up to the root using the alpha-beta pruning algorithm.

    

    A three-ply game tree: root A is a MAX node, B and C are MIN nodes, D to G are MAX nodes with leaf values 3, 5, 6, 9, 1, 2, 0, -1 from left to right.

    1. A3
    2. B5
    3. C6
    4. D9

    Answer: (B) 5

    Explanation

    Evaluate left to right while carrying alpha and beta bounds. The MAX node D returns max(3, 5) = 5, so the MIN node B now has beta = 5. Exploring child E, its first leaf 6 already exceeds beta, therefore the branch is pruned and B stays 5, which makes alpha = 5 at the root. At the MIN node C, the MAX node F returns max(1, 2) = 2, so C cannot exceed 2; because 2 is not greater than alpha = 5, node G is pruned. The root finally backs up max(5, 2) = 5.

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54. Question 54

    Which of the following statements is/are FALSE?

    A. Greedy best-first search is not optimal but is often efficient.
    B. A* is complete and optimal provided h(n) is admissible or consistent.
    C. Recursive best-first search is efficient in time complexity but poor in space complexity.
    D. h(n) = 0 is an admissible heuristic for the 8-puzzle.

    Choose the correct answer from the options given below:

    1. AA only
    2. BA and D only
    3. CC only
    4. DC and D only

    Answer: (C) C only

    Explanation

    A, B and D are true. C is false and reversed: recursive best-first search uses only linear space (its strength) but can be time-inefficient because it re-expands nodes when switching between alternative paths.

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55. Question 55

    'There is a country that borders both India and Pakistan.' Using Country(x) for 'x is a country' and Borders(x, y) for 'x and y share a border', which logical expression states this correctly?

    1. Aexists c: Country(c) AND Borders(c, India) AND Borders(c, Pakistan)
    2. Bexists c: Country(c) => [Borders(c, India) AND Borders(c, Pakistan)]
    3. C[exists c: Country(c)] => [Borders(c, India) AND Borders(c, Pakistan)]
    4. Dexists c: Borders(Country(c), India AND Pakistan)

    Answer: (A) exists c: Country(c) AND Borders(c, India) AND Borders(c, Pakistan)

    Explanation

    An existential claim uses "exists" with conjunction: there is a c that is a country and borders India and borders Pakistan. Using => under "exists" (option B) is satisfied by any non-country; option C leaves c free in the consequent; option D misuses Country as a term and conjoins objects.

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56. Question 56

    From the joint distribution, the marginal probability P(cavity) is ______.

    Consider a domain with three Boolean variables Toothache, Cavity and Catch. The full joint distribution is the 2 x 2 x 2 table below.

    Row	toothache, catch	toothache, not-catch	not-toothache, catch	not-toothache, not-catch
    cavity	0.108	0.012	0.072	0.008
    not-cavity	0.016	0.064	0.144	0.576

    1. A0.200
    2. B0.216
    3. C0.120
    4. D0.080

    Answer: (A) 0.200

    Explanation

    A marginal probability is obtained by summing the full joint distribution over the variables you do not care about. Here that means adding every cell in the cavity row across all combinations of toothache and catch: 0.108 + 0.012 + 0.072 + 0.008. The total comes to 0.200, so P(cavity) = 0.200.

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57. Question 57

    From the joint distribution, P(cavity | toothache) is ______.

    Consider a domain with three Boolean variables Toothache, Cavity and Catch. The full joint distribution is the 2 x 2 x 2 table below.

    Row	toothache, catch	toothache, not-catch	not-toothache, catch	not-toothache, not-catch
    cavity	0.108	0.012	0.072	0.008
    not-cavity	0.016	0.064	0.144	0.576

    1. A0.400
    2. B0.600
    3. C0.280
    4. D0.216

    Answer: (B) 0.600

    Explanation

    Conditional probability divides the joint probability of both events by the probability of the evidence. Summing the cavity-and-toothache cells gives P(cavity AND toothache) = 0.108 + 0.012 = 0.120. Summing every toothache cell, over both cavity and not-cavity, gives P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.200. Dividing, P(cavity given toothache) = 0.120 divided by 0.200, which equals 0.600.

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58. Question 58

    From the joint distribution, P(toothache | cavity) is ______.

    Consider a domain with three Boolean variables Toothache, Cavity and Catch. The full joint distribution is the 2 x 2 x 2 table below.

    Row	toothache, catch	toothache, not-catch	not-toothache, catch	not-toothache, not-catch
    cavity	0.108	0.012	0.072	0.008
    not-cavity	0.016	0.064	0.144	0.576

    1. A0.400
    2. B0.600
    3. C0.280
    4. D0.216

    Answer: (B) 0.600

    Explanation

    Again use the definition of conditional probability, dividing the joint probability by the probability of the conditioning event. The cells where cavity and toothache are both true sum to 0.108 + 0.012 = 0.120. The marginal probability of cavity, adding its whole row, is 0.200. Therefore P(toothache given cavity) = 0.120 divided by 0.200, which equals 0.600.

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59. Question 59

    From the joint distribution, P(cavity OR toothache) is ______.

    Consider a domain with three Boolean variables Toothache, Cavity and Catch. The full joint distribution is the 2 x 2 x 2 table below.

    Row	toothache, catch	toothache, not-catch	not-toothache, catch	not-toothache, not-catch
    cavity	0.108	0.012	0.072	0.008
    not-cavity	0.016	0.064	0.144	0.576

    1. A0.200
    2. B0.120
    3. C0.280
    4. D0.600

    Answer: (C) 0.280

    Explanation

    The probability of a disjunction follows the inclusion-exclusion principle, adding the two marginal probabilities and subtracting their overlap so the shared cases are not counted twice. Both P(cavity) and P(toothache) equal 0.200, while their joint probability P(cavity AND toothache) equals 0.120. Hence P(cavity OR toothache) = 0.200 + 0.200 - 0.120, which equals 0.280.

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60. Question 60

    From the joint distribution, P(Cavity | toothache OR catch) is ______.

    Consider a domain with three Boolean variables Toothache, Cavity and Catch. The full joint distribution is the 2 x 2 x 2 table below.

    Row	toothache, catch	toothache, not-catch	not-toothache, catch	not-toothache, not-catch
    cavity	0.108	0.012	0.072	0.008
    not-cavity	0.016	0.064	0.144	0.576

    1. A0.6000
    2. B0.5384
    3. C0.8000
    4. D0.4615

    Answer: (D) 0.4615

    Explanation

    Condition on the compound evidence by dividing the joint probability by the probability of that evidence. The numerator adds every cell where cavity is true together with either toothache or catch: 0.108 + 0.012 + 0.072, which equals 0.192. The denominator adds every cell, across both cavity and not-cavity, where toothache or catch holds: 0.108 + 0.012 + 0.072 + 0.016 + 0.064 + 0.144, which equals 0.416. Dividing 0.192 by 0.416 gives approximately 0.4615.

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61. Question 61

    Which DBMS component ensures that concurrent execution of multiple operations on the database results in a consistent database state?

    1. ALogs
    2. BBuffer manager
    3. CFile manager
    4. DTransaction processing system

    Answer: (D) Transaction processing system

    Explanation

    The transaction processing (transaction management and concurrency-control) subsystem enforces the ACID properties, scheduling concurrent transactions so the result is equivalent to some serial order. Logs support recovery, the buffer manager moves pages, and the file manager handles disk space.

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62. Question 62

    Statement I: A relational database schema represents the logical design of the database.
    Statement II: The current snapshot of a relation only provides the degree of the relation.

    In light of the above statements, choose the correct answer from the options given below:

    1. AStatement I is TRUE but Statement II is FALSE
    2. BStatement I is FALSE but Statement II is TRUE
    3. CBoth Statement I and Statement II are FALSE
    4. DBoth Statement I and Statement II are TRUE

    Answer: (A) Statement I is TRUE but Statement II is FALSE

    Explanation

    The schema (tables, attributes, constraints) is the logical design (Statement I true). A snapshot/instance of a relation shows both its degree (number of attributes) and its cardinality (number of tuples), not the degree alone, so Statement II is false.

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63. Question 63

    A fixed-length record file is ordered on the key field and occupies B disk blocks to store R records. What is the average number of block accesses to find a record using binary search?

    1. AB / 2
    2. BB + R
    3. CB
    4. Dlog2(B)

    Answer: (D) log2(B)

    Explanation

    Binary search over B ordered blocks halves the search range each step, so it needs at most ceil(log2 B) block accesses; the average access cost is therefore O(log2 B).

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64. Question 64

    Given STUDENT(Rollno, Name, courseno) and COURSE(courseno, coursename, capacity), with coursename unique. Which of the queries below find the total number of students enrolled in each course together with its coursename?

    -- Query A
    SELECT coursename, count(*) 'total' FROM STUDENT NATURAL JOIN COURSE GROUP BY coursename;
    
    -- Query B
    SELECT C.coursename, count(*) 'total' FROM STUDENT S, COURSE C WHERE S.courseno = C.courseno GROUP BY coursename;
    
    -- Query C
    SELECT coursename, count(*) 'total' FROM COURSE C WHERE courseno IN (SELECT courseno FROM STUDENT);
    

    Choose the correct answer from the options given below:

    1. AA and B only
    2. BC only
    3. CA only
    4. DB only

    Answer: (A) A and B only

    Explanation

    A and B both join STUDENT with COURSE on courseno and group by coursename, so count(*) gives the enrolment per course. C never joins with STUDENT - it only filters courses that have at least one student and counts COURSE rows, so it cannot report the number of students per course.

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65. Question 65

    STUDENT and COURSE each have 3 attributes; COURSE has 3 records and STUDENT has 5 records. The relational-algebra query R = STUDENT x COURSE is executed. Match List-I with List-II for R.

    List-I	List-II
    A. Degree of table R	I. 15
    B. Cardinality of table R	II. NIL
    C. Foreign key of relation STUDENT	III. 6
    D. Foreign key of relation COURSE	IV. Courseno

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

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

    Explanation

    The Cartesian product has degree 3 + 3 = 6 (III) and cardinality 3 x 5 = 15 (I). In the schema STUDENT.courseno refers to COURSE, so its foreign key is Courseno (IV); COURSE references no other table, so it has no foreign key - NIL (II).

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66. Question 66

    A B-tree indexes a database file. Search-key field = 10 bytes, block size = 1024 bytes, record pointer = 9 bytes, block pointer = 8 bytes. Let K be the order of an internal node and L the order of a leaf node. Then (K, L) = ______.

    1. A(57, 53)
    2. B(50, 52)
    3. C(60, 64)
    4. D(34, 31)

    Answer: (A) (57, 53)

    Explanation

    Internal node: K block pointers and K-1 keys fit in a block: 8K + 10(K - 1) <= 1024, so 18K <= 1034, K = 57. Leaf node: L (key + record-pointer) entries plus one block pointer: 19L + 8 <= 1024, so 19L <= 1016, L = 53.

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67. Question 67

    Given R(x, y, z, w) with functional dependencies F = { x -> y, z -> w } and all attributes atomic.

    A. R is in First Normal Form.
    B. R is in Second Normal Form.
    C. The primary key of R is xz.

    Choose the correct answer from the options given below:

    1. AC only
    2. BB and C only
    3. CA and C only
    4. DB only

    Answer: (C) A and C only

    Explanation

    Since the closure of xz is xyzw and neither x nor z alone is a superkey, the primary key is xz (C true). Atomic attributes make R 1NF (A true). But x -> y and z -> w are partial dependencies of non-prime attributes on part of the key, violating 2NF (B false).

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68. Question 68

    A part is either manufactured in-house, purchased from an external supplier, or both. A batchnumber is kept for manufactured parts and a supplier name for purchased parts. Which constraints must be modelled for this class/subclass structure?

    1. ATotal specialization and overlapping constraints
    2. BDisjoint constraint only
    3. CPartial participation
    4. DPartial participation and disjoint constraints

    Answer: (A) Total specialization and overlapping constraints

    Explanation

    Every part is manufactured or purchased (or both), so every superclass entity belongs to at least one subclass - total specialization. A part can be both manufactured and purchased, so the subclasses can overlap - the overlapping constraint (not disjoint).

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69. Question 69

    A transaction in the ACTIVE state during concurrent execution transitions next to which state?

    A. FAILED
    B. TERMINATED
    C. PARTIALLY COMMITTED
    D. COMMITTED
    E. ACTIVE

    Choose the correct answer from the options given below:

    1. AA only
    2. BEither A or C only
    3. CC only
    4. DD only

    Answer: (B) Either A or C only

    Explanation

    From ACTIVE a transaction moves either to PARTIALLY COMMITTED (after its final statement executes) or to FAILED (if an error aborts it). It cannot jump straight to COMMITTED or TERMINATED.

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70. Question 70

    Which language is used to create a database schema?

    1. ADML
    2. BDDL
    3. CHTML
    4. DXML

    Answer: (B) DDL

    Explanation

    Data Definition Language (CREATE, ALTER, DROP) defines and modifies schema objects such as tables, indexes and views. DML (SELECT, INSERT, UPDATE, DELETE) manipulates the data within that schema.

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71. Question 71

    Memory access time is p nanoseconds and an extra q nanoseconds are needed to handle a page fault. If a page fault occurs once every 100 instructions, what is the effective memory access time?

    1. Ap + q/100
    2. B(p + q)/100
    3. Cp + q
    4. Dp/100 + q

    Answer: (A) p + q/100

    Explanation

    Every access costs p. A page fault adds q, but that happens on only 1 in 100 accesses, contributing q/100 on average. Effective access time = p + q/100.

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72. Question 72

    In a file-allocation system three schemes are used:

    A. Contiguous allocation
    B. Indexed allocation
    C. Linked allocation

    Which scheme(s) do NOT suffer from external fragmentation?

    Choose the correct answer from the options given below:

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

    Answer: (B) B and C only

    Explanation

    Contiguous allocation needs a run of consecutive free blocks, so free space breaks into unusable gaps - external fragmentation. Indexed and linked allocation place file blocks anywhere on disk, so any free block is usable and there is no external fragmentation.

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73. Question 73

    Three statements about the interrupt-handling mechanism:

    A. The interrupt handler routine is not stored at a fixed address in memory.
    B. CPU hardware has a dedicated wire called the interrupt-request line used for handling interrupts.
    C. The interrupt vector contains the memory addresses of the specialized interrupt handlers.

    Choose the correct answer from the options given below:

    1. AA is TRUE only
    2. BBoth B and C are TRUE only
    3. CBoth A and B are TRUE only
    4. DBoth A and C are TRUE only

    Answer: (B) Both B and C are TRUE only

    Explanation

    The CPU has an interrupt-request line it senses each instruction cycle (B true), and the interrupt vector table holds the addresses of the interrupt service routines (C true). Statement A is false: handler routines occupy fixed, known addresses that the vector points to.

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74. Question 74

    Match List-I (System Calls) with List-II (Description).

    List-I	List-II
    A. fork()	I. Sends a signal from one process to a particular process
    B. exec()	II. Indicates termination of the current process
    C. kill()	III. Loads the specified program into memory
    D. exit()	IV. Creates a child process

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

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

    Explanation

    fork() creates a child process (IV). exec() replaces the current process image with a new program loaded into memory (III). kill() sends a signal to a specified process (I). exit() terminates the calling process (II).

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75. Question 75

    Three processes are scheduled using preemptive shortest-job-first (shortest remaining time first) scheduling. What is the average waiting time?

    Process	Arrival Time	Burst Time
    P1	0	8
    P2	1	4
    P3	2	9

    1. A5.5
    2. B2.66
    3. C4.66
    4. D6

    Answer: (C) 4.66

    Explanation

    Timeline: P1 runs 0-1; P2 (shorter remaining) runs 1-5; P1 resumes 5-12; P3 runs 12-21. Completion times: P1 = 12, P2 = 5, P3 = 21. Waiting = turnaround - burst: P1 = 12 - 8 = 4, P2 = 4 - 4 = 0, P3 = 19 - 9 = 10. Average = (4 + 0 + 10) / 3 = 4.66.

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76. Question 76

    The number of bits reserved for the frame offset is:

    Consider a machine with 16 GB main memory and a 32-bit virtual address space, with a page size of 4 KB. Frame size equals page size for this machine.

    1. A12
    2. B14
    3. C32
    4. D8

    Answer: (A) 12

    Explanation

    The offset must address every byte within a frame. A 4 KB frame = 2^12 bytes, so the frame offset needs 12 bits.

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77. Question 77

    The number of pages in the given virtual address space is:

    Consider a machine with 16 GB main memory and a 32-bit virtual address space, with a page size of 4 KB. Frame size equals page size for this machine.

    1. A2^10
    2. B2^20
    3. C2^30
    4. D2^12

    Answer: (B) 2^20

    Explanation

    The number of pages in a process address space equals the total virtual address space divided by the page size, because the virtual space is partitioned into equal fixed-size pages. A 32-bit virtual address covers 2^32 bytes, and each page is 4 KB, which is 2^12 bytes. Dividing, 2^32 divided by 2^12 leaves 2^20 pages.

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78. Question 78

    The minimum number of bits needed for a physical address is:

    Consider a machine with 16 GB main memory and a 32-bit virtual address space, with a page size of 4 KB. Frame size equals page size for this machine.

    1. A28
    2. B34
    3. C24
    4. D12

    Answer: (B) 34

    Explanation

    A physical address must be wide enough to name every byte of installed main memory, so the number of address bits is the base-two logarithm of the memory size. Sixteen gigabytes equals 2^4 multiplied by 2^30 bytes, which is 2^34 bytes altogether. Naming that many byte locations therefore requires exactly 34 bits.

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79. Question 79

    The size of the page table for this virtual address space, given a page-table entry of 2 bytes, is:

    Consider a machine with 16 GB main memory and a 32-bit virtual address space, with a page size of 4 KB. Frame size equals page size for this machine.

    1. A2 MB
    2. B2 KB
    3. C32 MB
    4. D12 KB

    Answer: (A) 2 MB

    Explanation

    A single-level page table stores one entry for every page in the virtual address space, so its size is the page count multiplied by the entry size. There are 2^20 pages, and each page-table entry occupies 2 bytes. Multiplying gives 2^20 times 2, which is 2^21 bytes, and that equals 2 megabytes.

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80. Question 80

    If a process of size 34 KB runs on this machine, the internal fragmentation for the process is:

    Consider a machine with 16 GB main memory and a 32-bit virtual address space, with a page size of 4 KB. Frame size equals page size for this machine.

    1. A4 KB
    2. BZero
    3. C1 KB
    4. D2 KB

    Answer: (D) 2 KB

    Explanation

    Internal fragmentation is the wasted space inside the last frame allocated to a process, because frames are allocated whole even when the process does not fill them. A 34 KB process needs the ceiling of 34 divided by 4, which is 9 pages, occupying 36 KB of frame space. The final frame holds only 2 KB of useful data, so the internal fragmentation is 36 minus 34, that is 2 KB.

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81. Question 81

    Match List-I with List-II.

    List-I	List-II
    A. Odd Function	I. NAND Gate
    B. Universal Gate	II. XOR Gate
    C. 2421 Code	III. Amplification
    D. Buffer	IV. Self-complementing

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

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

    Explanation

    An XOR gate outputs 1 when an odd number of inputs are 1 - it realises the odd (parity) function (II). NAND is a universal gate (I). The 2421 (Aiken) code is self-complementing - the 9s complement of a digit is the bitwise complement of its code (IV). A buffer provides amplification/drive without changing the logic level (III).

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82. Question 82

    The octal equivalent of the hexadecimal number (D.C) in base 16 is:

    1. A(15.6) in base 8
    2. B(61.6) in base 8
    3. C(15.3) in base 8
    4. D(61.3) in base 8

    Answer: (A) (15.6) in base 8

    Explanation

    Convert through binary because each hexadecimal digit maps to four bits and each octal digit maps to three bits. The digits D and C expand to 1101 and 1100, giving the binary value 1101.1100. Regroup the bits into threes starting from the radix point, padding with zeros: 001 101 . 110 000. Those groups translate to the octal digits 1, 5, 6 and 0, so the answer is 15.6 in octal.

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83. Question 83

    A digital computer has a common-bus system for 8 registers of 16 bits each. How many multiplexers are needed to build the common bus, and what size must each multiplexer be?

    1. A16, 8 x 1
    2. B8, 16 x 1
    3. C8, 8 x 1
    4. D16, 16 x 1

    Answer: (A) 16, 8 x 1

    Explanation

    One multiplexer produces each bus line, so 16 multiplexers are needed (one per bit). Each must select the corresponding bit from any of the 8 registers, so each is an 8-to-1 multiplexer.

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84. Question 84

    Which of the following is NOT an example of a pseudo-instruction (assembler directive)?

    1. AORG
    2. BDEC
    3. CEND
    4. DHLT

    Answer: (D) HLT

    Explanation

    ORG, DEC and END are assembler directives that guide assembly but generate no opcode. HLT is a real machine instruction that stops the processor.

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85. Question 85

    The reverse Polish (postfix) notation of the infix expression [A * {B + C * (D + E)}] / {F * (G + H)} is:

    1. AABCDE+*+FGH+/
    2. BABCDE*++FGH+/
    3. CABCDE+*+FGH+/
    4. DABCDE+**+FGH+*/

    Answer: (A) ABCDE+*+FGH+/

    Explanation

    Work outward from the innermost parentheses, writing each operator right after its two operands. First D plus E becomes DE+, then C times that group becomes CDE+, then B plus that becomes BCDE++, and A times the whole thing becomes ABCDE++. For the denominator, G plus H becomes GH+ and F times that becomes FGH+. Finally the division operator is appended after both sides, producing ABCDE++FGH+/.

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86. Question 86

    Arrange the following in increasing order of complexity:

    A. I/O Module
    B. I/O Processor
    C. I/O Channel
    D. DMA

    Choose the correct answer from the options given below:

    1. AD, C, B, A
    2. BC, D, A, B
    3. CA, B, C, D
    4. DA, D, C, B

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

    Explanation

    A plain I/O module is simplest; DMA adds autonomous block transfer; an I/O channel executes channel programs and manages several devices; an I/O processor is a full specialised processor offloading I/O work. Increasing order: A, D, C, B.

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87. Question 87

    The characteristics of combinational circuits are:

    A. Output at any time is a function of the inputs at that time.
    B. They contain memory elements.
    C. They have no feedback paths.
    D. A clock triggers the circuit to obtain outputs.

    Choose the correct answer from the options given below:

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

    Answer: (C) A and C only

    Explanation

    A combinational circuit's output depends only on the present inputs (A) and it has no feedback (C). Memory elements (B) and clocking (D) characterise sequential circuits.

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88. Question 88

    The cache-coherence problem can be solved by:

    A. having multiport memory
    B. allowing only non-shared data to be stored in cache
    C. using a snoopy cache controller
    D. using memory interleaving

    Choose the correct answer from the options given below:

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

    Answer: (B) B and C only

    Explanation

    Caching only non-shared (private) data trivially avoids coherence issues (B), and a snoopy cache controller keeps shared copies consistent by watching the bus and invalidating or updating them (C). Multiport memory and interleaving only affect access bandwidth, not coherence.

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89. Question 89

    Statement I: CISC computers have a large number of addressing modes.
    Statement II: In RISC machines, memory access is limited to load and store instructions.

    In light of the above statements, choose the correct answer from the options given below:

    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: (A) Both Statement I and Statement II are true

    Explanation

    CISC designs provide many addressing modes and complex instructions (Statement I true). RISC designs are load/store architectures where only load and store touch memory and all other operations work on registers (Statement II true).

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90. Question 90

    Which of the following statements is true?

    1. AControl memory is part of the hardwired control unit
    2. BProgram control instructions are used to alter the sequential flow of the program
    3. CThe register indirect addressing mode is similar to the displacement addressing mode
    4. DCPU utilization is not affected by the introduction of interrupts

    Answer: (B) Program control instructions are used to alter the sequential flow of the program

    Explanation

    Program-control instructions (jump, branch, call, return) change the program counter and hence the sequential flow. Control memory belongs to a microprogrammed control unit; register-indirect uses the register content directly as the address (no added displacement); and interrupts do affect CPU utilization.

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91. Question 91

    Match List-I with List-II.

    List-I	List-II
    A. Data Link Layer	I. True end-to-end layer
    B. Network Layer	II. Token Management
    C. Transport Layer	III. Produce billing information
    D. Session Layer	IV. Piggybacking

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

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

    Explanation

    The data link layer uses piggybacking to attach acknowledgements to outgoing frames (IV). The network layer handles routing and can produce billing/accounting information about packets (III). The transport layer provides genuine end-to-end (source-to-destination) delivery (I). The session layer provides token management for dialog control (II).

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92. Question 92

    Statement I: Telnet, FTP and HTTP are application-layer protocols.
    Statement II: The Iridium project was planned to launch 66 low-orbit satellites.

    In light of the above statements, choose the correct answer from the options given below:

    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: (A) Both Statement I and Statement II are true

    Explanation

    Telnet, FTP and HTTP all run at the application layer of the TCP/IP model (Statement I true). The Iridium satellite-phone constellation was designed around 66 low-Earth-orbit satellites (Statement II true).

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93. Question 93

    Which of the following statements is FALSE?

    1. APacket switching leads to better utilization of bandwidth resources than circuit switching
    2. BPacket switching results in less variation in the delay than circuit switching
    3. CPacket switching sender and receiver can use any bit rate, format or framing method unlike circuit switching
    4. DPacket switching can lead to reordering unlike circuit switching

    Answer: (B) Packet switching results in less variation in the delay than circuit switching

    Explanation

    Packet switching statistically shares links (better utilization), lets endpoints differ in rate and format, and can deliver packets out of order. But because packets queue independently at each hop, packet switching has MORE delay variation (jitter) than a dedicated circuit, so statement B is false.

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94. Question 94

    In Ethernet using Manchester coding, the bit rate is ______.

    1. AHalf the baud rate
    2. BTwice the baud rate
    3. CThrice the baud rate
    4. DSame as the baud rate

    Answer: (A) Half the baud rate

    Explanation

    Manchester encoding places a mid-bit transition, so each data bit uses two signal elements. The data (bit) rate is therefore half the signalling (baud) rate.

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95. Question 95

    A Class B address is subnetted using a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts per subnet?

    1. A62 subnets and 262142 hosts
    2. B64 subnets and 1024 hosts
    3. C64 subnets and 262142 hosts
    4. D62 subnets and 1022 hosts

    Answer: (D) 62 subnets and 1022 hosts

    Explanation

    Class B has 16 host bits. Borrowing 6 for the subnet gives 2^6 - 2 = 62 usable subnets (excluding all-zeros and all-ones) and 16 - 6 = 10 host bits, i.e. 2^10 - 2 = 1022 hosts per subnet.

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96. Question 96

    In electronic mail, which protocol allows the transfer of multimedia content?

    1. AIMAP
    2. BSMTP
    3. CPOP3
    4. DMIME

    Answer: (D) MIME

    Explanation

    MIME (Multipurpose Internet Mail Extensions) extends email to carry non-ASCII text, images, audio, video and attachments by encoding them for transport over SMTP. IMAP and POP3 only retrieve messages; SMTP alone carries plain 7-bit text.

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97. Question 97

    A message is sent using public-key cryptography. Which statement is true?

    1. AThe sender encrypts using the receiver's public key
    2. BThe sender encrypts using their own public key
    3. CThe receiver decrypts using the sender's public key
    4. DThe receiver decrypts using their own public key

    Answer: (A) The sender encrypts using the receiver's public key

    Explanation

    For confidentiality the sender encrypts with the receiver's public key, and only the receiver, holding the matching private key, can decrypt. Encrypting with the sender's key or decrypting with any public key would not keep the message secret.

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98. Question 98

    Which of the following statements are true?

    A. Frequency-division multiplexing can be handled by digital circuits.
    B. Time-division multiplexing can be handled by analog circuits.
    C. Wavelength-division multiplexing is used with optical fibre to combine signals.
    D. Frequency-division multiplexing can be applied when the bandwidth of a link is greater than the combined bandwidth of the signals to be transmitted.

    Choose the correct answer from the options given below:

    1. AA and B only
    2. BC and D only
    3. CA and D only
    4. DB and D only

    Answer: (B) C and D only

    Explanation

    WDM combines optical signals of different wavelengths on one fibre (C true), and FDM needs a link whose bandwidth exceeds the sum of the signal bandwidths so each gets its own frequency band (D true). FDM is an analog technique (not A) and TDM is digital (not B).

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99. Question 99

    Which of the following statements are true?

    A. X.25 is a connection-oriented network.
    B. X.25 does not support switched virtual circuits.
    C. Frame relay service provides acknowledgements.
    D. Frame relay service provides detection of transmission errors.

    Choose the correct answer from the options given below:

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

    Answer: (A) A and D only

    Explanation

    X.25 is connection-oriented, setting up a virtual circuit before data transfer (A true). Frame relay carries a CRC and discards frames that fail the check, so it detects transmission errors (D true) - but it does not correct them or send acknowledgements (C false), and X.25 does support switched virtual circuits (B false).

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100. Question 100

     Let G(x) be the generator polynomial used for CRC checking. The condition that G(x) must satisfy to catch all errors consisting of an odd number of inverted bits is:

     1. A(x + 1) is a factor of G(x)
     2. B(x - 1) is a factor of G(x)
     3. C(x^2 + 1) is a factor of G(x)
     4. D(1 - x^2) is a factor of G(x)

     Answer: (A) (x + 1) is a factor of G(x)

     Explanation

     An error pattern E(x) with an odd number of 1-terms evaluates to 1 at x = 1, so it is not divisible by (x + 1). If (x + 1) divides G(x), then G(x) cannot divide such an E(x), so the CRC remainder is non-zero and every odd-bit error is detected.

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