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|Computer Science & Applications · June 2019

UGC NET Computer Science & Applications June 2019 Question PaperPaper 1 & Paper 2 with Answer Key

The complete UGC NET June 2019 question paper for Computer Science & Applications, covering Paper 1 (General Aptitude) and Paper 2 (Computer Science & Applications). Every question below is shown with its options and the correct answer, and a detailed explanation you can unlock by signing in. All 150 questions are free to practise.

UGC NET June 2019 at a glance

Exam	UGC NET June 2019
Papers	Paper 1 (General Aptitude) and Paper 2 (Computer Science & Applications)
Exam date(s)	June 24, 2019
Shift(s)	Shift 1
Paper 1 questions	50 questions · 60 minutes · 100 marks
Paper 2 questions	100 questions · 120 minutes · 200 marks
Total	150 questions
Marking scheme	+2 per correct answer · No negative marking

Paper 1

General Aptitude

UGC NET June 2019 · Computer Science & Applications sitting

50 questions60 min100 marks

Practice Now

Paper 2

Computer Science & Applications

UGC NET June 2019

100 questions120 min200 marks

Practice Now

Paper 1 — General Aptitude

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50 questions with the answer key and explanations.

Question 1
Which of the following statements explains the concepts of inclusive teaching ?
1. ATeacher facilitates the learning of the gifted students.
2. BTeacher facilitates the learning of the weak students.
3. CTeacher takes support of parents of the students to make them learn.
4. DTeacher makes the students of different backgounds to learn together in the same class.
Answer: (D) Teacher makes the students of different backgounds to learn together in the same class.
Explanation
Inclusive teaching is grounded in bringing learners from diverse socio-economic, linguistic, cultural, and cognitive backgrounds together in an equitable, shared classroom environment.

A. Teacher facilitates the learning of the gifted students. — focuses exclusively on high achievers rather than all students.

B. Teacher facilitates the learning of the weak students. — addresses remedial learning in isolation rather than full classroom integration.

C. Teacher takes support of parents of the students to make them learn. — describes parental engagement rather than inclusive pedagogical design.

D. Teacher makes the students of different backgounds to learn together in the same class. — correct. Captures the core philosophy of educating heterogeneous student groups collaboratively in one common space.
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Question 2
To organize discussion method in teaching effectively, which of the following conditions should be met ?

(1) Topic be easy
(2) Topic be declared in advance
(3) Topic of common interest
(4) Availability of more than one teacher
(5) Language facility of participate

Select appropriate answer from the options given below:
1. A(2), (3) and (5)
2. B(1), (2) and (3)
3. C(1), (2) and (5)
4. D(3), (4) and (5)
Answer: (A) (2), (3) and (5)
Explanation
An effective classroom discussion relies on preparation, shared learner motivation, and communication competence.

(2) Topic declared in advance allows learners time to gather facts, read background literature, and formulate well-reasoned arguments.

(3) Topic of common interest ensures active, self-motivated learner involvement across the entire group.

(5) Language facility of participants ensures students possess the expressive competence required to articulate viewpoints clearly.

(1) Topic be easy is counterproductive because discussions are specifically meant to explore nuanced, complex problems.

(4) Availability of more than one teacher is unnecessary for standard classroom discussions, which are routinely led by a single facilitator.
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Question 3
Which of the following is a plagiarism website ?
1. Ahttp://go.turnitin.com
2. Bhttp://www.researchgate.com
3. Chttp://www.editorial.elsevier.com
4. Dhttp://www.grammarly.com
Answer: (A) http://go.turnitin.com
Explanation
Turnitin is a dedicated commercial plagiarism detection and text-similarity matching service widely adopted by universities to verify academic originality.

A. http://go.turnitin.com — correct. Checks submitted manuscripts against billions of web pages and institutional repositories for unoriginal text.

B. http://www.researchgate.com — an academic networking platform for researchers to share papers and collaborate.

C. http://www.editorial.elsevier.com — a peer-review and manuscript management platform for Elsevier academic journals.

D. http://www.grammarly.com — primarily an automated writing assistant focused on grammar, clarity, and tone.
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Question 4
Which among the following best describes the Emotional Intelligence of learners ?

(1) Understand the emotion of other people and your own.
(2) Express oneself very strongly.
(3) Being rational in thinking.
(4) Adjusting one’s emotion as per situation.
(5) Being creative and open to criticism.
(6) Accepting other people as they are.

Choose your answer from the options given below:
1. A(1), (4) and (6)
2. B(4), (5) and (6)
3. C(1), (2) and (3)
4. D(2), (3) and (4)
Answer: (A) (1), (4) and (6)
Explanation
Emotional intelligence entails recognizing emotional states in oneself and others, regulating emotions constructively, and maintaining empathetic interpersonal relationships.

(1) Understand the emotion of other people and your own represents self-awareness and empathy, the core foundation of emotional intelligence.

(4) Adjusting one’s emotion as per situation represents self-regulation and emotional adaptability.

(6) Accepting other people as they are represents empathetic social awareness and non-judgmental relationship management.

(2) Express oneself very strongly reflects emotional impulsivity or aggressive reactivity rather than emotional regulation.

(3) Being rational in thinking relates to purely cognitive intelligence (IQ) rather than emotional domain processing.
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Question 5
In which of the following research studies interpretation and meaning get more attention than formulation of generalisations?

(i) Historical studies
(ii) Survey studies
(iii) Philosophical studies
(iv) Ethnographic studies
(v) Hypothetico – deductive studies
(vi) Ex-post facto studies

Choose your answer from the options given below:
1. A(i), (ii) and (iii)
2. B(iv), (v) and (vi)
3. C(ii), (iv) and (v)
4. D(i), (iii) and (iv)
Answer: (D) (i), (iii) and (iv)
Explanation
Qualitative paradigms emphasize interpretive understanding, contextual depth, and subjective meaning rather than statistical generalization across populations.

(i) Historical studies interpret documentary evidence and reconstruct contextual meaning of past human experiences.

(iii) Philosophical studies analyze concepts, epistemological frameworks, and normative meanings through philosophical hermeneutics.

(iv) Ethnographic studies engage in holistic immersion to decode cultural meanings from participants’ perspectives.

By contrast, survey studies (ii), hypothetico-deductive studies (v), and ex-post facto research (vi) follow quantitative, positivist methodologies aimed at statistical generalization and hypothesis testing.
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Question 6
Bibliography given in a research report
1. AHelps those interested in further research.
2. BShows the vast knowledge of the researcher.
3. CMakes the report authentic.
4. DIs an optional part of the report.
Answer: (A) Helps those interested in further research.
Explanation
A bibliography documents all works consulted or relevant to the subject, enabling subsequent researchers to locate primary and secondary sources to advance further inquiry.

A. Helps those interested in further research. — correct. Provides complete bibliographic metadata so readers can retrieve and build upon the literature.

B. Shows the vast knowledge of the researcher. — misinterprets an academic citation standard as an exercise in personal vanity.

C. Makes the report authentic. — while accurate citations enhance credibility, methodology and verified empirical evidence determine authenticity.

D. Is an optional part of the report. — standard scholarly reports mandate full bibliographic references to avoid plagiarism.
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Question 7
The research design is specifically related to which of the following features in research ?

(i) Sample selection
(ii) Formulation of a plan
(iii) Deciding about the tool for data collection
(iv) Hypothesis making
(v) Choice of a field of inquiry

Select your answer from the options given below:
1. A(ii), (iii) and (iv)
2. B(i), (ii) and (iii)
3. C(ii), (iv) and (v)
4. D(iii), (iv) and (v)
Answer: (B) (i), (ii) and (iii)
Explanation
A research design is the comprehensive operational blueprint that guides the collection and analysis of data.

(i) Sample selection constitutes the sampling design, establishing sampling technique, frame, and sample size.

(ii) Formulation of a plan provides the overarching operational framework for executing the investigation.

(iii) Deciding about the tool for data collection specifies the measurement instruments, questionnaires, or observational protocols.

By contrast, choosing a field of inquiry (v) and formulating hypotheses (iv) belong to the preliminary conceptual stage that precedes the development of the operational research design.
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Question 8
Who developed the theory of ‘Multiple Intelligence’ ?
1. AAlfred Binet
2. BL. Thurstone
3. CCharles Spearman
4. DHoward Gardner
Answer: (D) Howard Gardner
Explanation
Howard Gardner formulated the Theory of Multiple Intelligences in his 1983 book Frames of Mind, proposing eight distinct modalities of human competence.

A. Alfred Binet — developed the first practical intelligence scale (Binet-Simon test) to measure general mental ability.

B. L. Thurstone — proposed the theory of Primary Mental Abilities, focusing on seven independent factors.

C. Charles Spearman — introduced the two-factor theory of intelligence comprising general factor (g) and specific factors (s).

D. Howard Gardner — correct. Identified linguistic, logical-mathematical, spatial, bodily-kinesthetic, musical, interpersonal, intrapersonal, and naturalistic intelligences.
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Question 9
Through which research method, the manipulation of an independent variable and its effect on dependent variable is examined with reference to a hypothesis under controlled conditions ?
1. AEx-post facto research
2. BDescriptive research
3. CCase study research
4. DExperimental research
Answer: (D) Experimental research
Explanation
Experimental research is defined by active manipulation of independent variables, observation of dependent variables, and rigorous control over extraneous variables.

A. Ex-post facto research — examines phenomena after events have occurred, where the investigator cannot manipulate independent variables.

B. Descriptive research — observes and records conditions as they exist without variable manipulation.

C. Case study research — conducts an in-depth qualitative examination of an individual, group, or institution in a natural setting.

D. Experimental research — correct. Employs direct manipulation under controlled laboratory or field conditions to test cause-and-effect hypotheses.
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Question 10
From the list of the effective teaching behaviours, identify those which are called key behaviours.

(i) Direct, audible and oral delivery to all students.
(ii) Encouraging students to elaborate on an answer.
(iii) Warm and nurturing relationships with learners.
(iv) Varying modes of presentation.
(v) Preventing misbehaviour with a minimum of class disruption.
(vi) Organising what is to come and summarising what has gone before.

Select your answer from the options given below:
1. A(i), (iv) and (v)
2. B(i), (ii) and (iii)
3. C(ii), (iii) and (iv)
4. D(iv), (v) and (vi)
Answer: (A) (i), (iv) and (v)
Explanation
In Gary Borich’s observational model of classroom effectiveness, teaching behaviours are categorized into five key behaviours and five helping behaviours.

The five key behaviours are:

Lesson Clarity — direct, audible delivery to all students (i)

Instructional Variety — varying modes of presentation, questions, and learning materials (iv)

Teacher Task Orientation — managing the classroom to minimize disruptive misbehaviour (v)

Engagement in the Learning Process

Student Success Rate

Items (ii), (iii), and (vi) represent helping behaviours:

(ii) Encouraging elaboration represents probing.

(iii) Warm, nurturing relationships represents teacher affect.

(vi) Organising and summarising represents lesson structuring.
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Question 11
What actually may be concluded from the passage ?
Michaelangelo is famous for having successfully interpreted the human body. His great achievement is that of the painting of David whose hands reach out as a sign of human capability and potential. It is assumed that the time he lived was ripe for exchange of knowledge, development in science and matured enough to advance the horizon of investigation in all fields. Renaissance humanism stressed on a serious rethink on the nature of art that focussed on accurate details. In painting and sculpture, artists focussed on not so casual but verifiable and minute details. Michaelangelo’s paintings are no exception to it. In a study published in the journal of the Royal Society of medicine, a group of surgeons are of the opinion that the great master was “afflicted by an illness involving his joints”. They have used his portraits as evidence to argue their view. During his life, he complained of what he felt to be ‘gout’. Later he complained of his sore and stiff hands which the doctors would find to be natural for someone who was engaged in handmade art. The doctors found corroboration of those claims in portraits of the artists that show a hanging left hand with both degenerative and non-degenerative changes. They attribute the pain not just to arthritis, but to the stress of hammering and chiseling and note that though the master was seen hammering days before his death at an old age, he did not write or sign his own letters before his death. In recent times there have been attempts to diagnose famous artists with diseases that were not known during their time. This practice has raised many questions, especially on the issue of ethics in research. It is also inferred from authentic analysis that Michaelangelo persisted in his work until his last days. This theory would emphasize that his artistic subject defied his physical infirmities.
1. APhysical infirmities do dissuade people with capabilities from excelling.
2. BExcellence in any form triumphs over extraneous factors including physical ailments.
3. CMichaelangelo’s gout and other ailments lessened his efficiency.
4. DThe diseases Michaelangelo faced were due to constant hammering.
Answer: (B) Excellence in any form triumphs over extraneous factors including physical ailments.
Explanation
The passage concludes by noting that Michelangelo persisted in producing monumental art until his final days, emphasizing that his artistic vision triumphed over physical deterioration.

A. Physical infirmities do dissuade people with capabilities from excelling. — directly contradicted by the passage, which shows Michelangelo persevered despite severe pain.

B. Excellence in any form triumphs over extraneous factors including physical ailments. — correct. Captures the central conclusion that artistic dedication defied bodily infirmity.

C. Michaelangelo’s gout and other ailments lessened his efficiency. — contradicted by the passage noting he hammered and worked days before his death at an advanced age.

D. The diseases Michaelangelo faced were due to constant hammering. — the doctors attributed the pain to arthritis combined with hammering stress, not as the sole cause of disease.
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Question 12
What generalisations do people subscribe to ?
Michaelangelo is famous for having successfully interpreted the human body. His great achievement is that of the painting of David whose hands reach out as a sign of human capability and potential. It is assumed that the time he lived was ripe for exchange of knowledge, development in science and matured enough to advance the horizon of investigation in all fields. Renaissance humanism stressed on a serious rethink on the nature of art that focussed on accurate details. In painting and sculpture, artists focussed on not so casual but verifiable and minute details. Michaelangelo’s paintings are no exception to it. In a study published in the journal of the Royal Society of medicine, a group of surgeons are of the opinion that the great master was “afflicted by an illness involving his joints”. They have used his portraits as evidence to argue their view. During his life, he complained of what he felt to be ‘gout’. Later he complained of his sore and stiff hands which the doctors would find to be natural for someone who was engaged in handmade art. The doctors found corroboration of those claims in portraits of the artists that show a hanging left hand with both degenerative and non-degenerative changes. They attribute the pain not just to arthritis, but to the stress of hammering and chiseling and note that though the master was seen hammering days before his death at an old age, he did not write or sign his own letters before his death. In recent times there have been attempts to diagnose famous artists with diseases that were not known during their time. This practice has raised many questions, especially on the issue of ethics in research. It is also inferred from authentic analysis that Michaelangelo persisted in his work until his last days. This theory would emphasize that his artistic subject defied his physical infirmities.
1. AEstablishing facts by DNA tests.
2. BInferring the essence of character from famous people’s handwriting.
3. CCarbon dating of the hair of celebrities to draw conclusion on their physical structure.
4. DTo retroactively diagnose famous artists and public figures of conditions that were not prevalent during their time.
Answer: (D) To retroactively diagnose famous artists and public figures of conditions that were not prevalent during their time.
Explanation
The author highlights the questionable research practice where modern commentators attempt to diagnose historical luminaries with retrospective medical labels.

The passage states explicitly that in recent times there have been attempts to diagnose famous artists with diseases that were not known during their era, raising ethical questions in research.

A. Establishing facts by DNA tests. — not mentioned in the passage.

B. Inferring the essence of character from famous people’s handwriting. — handwriting analysis is not discussed.

C. Carbon dating of the hair of celebrities to draw conclusion on their physical structure. — no mention of carbon dating occurs in the text.

D. To retroactively diagnose famous artists and public figures of conditions that were not prevalent during their time. — correct. Matches the specific retrospective medical diagnoses discussed.
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Question 13
Renaissance painting in Europe was sceptical of
Michaelangelo is famous for having successfully interpreted the human body. His great achievement is that of the painting of David whose hands reach out as a sign of human capability and potential. It is assumed that the time he lived was ripe for exchange of knowledge, development in science and matured enough to advance the horizon of investigation in all fields. Renaissance humanism stressed on a serious rethink on the nature of art that focussed on accurate details. In painting and sculpture, artists focussed on not so casual but verifiable and minute details. Michaelangelo’s paintings are no exception to it. In a study published in the journal of the Royal Society of medicine, a group of surgeons are of the opinion that the great master was “afflicted by an illness involving his joints”. They have used his portraits as evidence to argue their view. During his life, he complained of what he felt to be ‘gout’. Later he complained of his sore and stiff hands which the doctors would find to be natural for someone who was engaged in handmade art. The doctors found corroboration of those claims in portraits of the artists that show a hanging left hand with both degenerative and non-degenerative changes. They attribute the pain not just to arthritis, but to the stress of hammering and chiseling and note that though the master was seen hammering days before his death at an old age, he did not write or sign his own letters before his death. In recent times there have been attempts to diagnose famous artists with diseases that were not known during their time. This practice has raised many questions, especially on the issue of ethics in research. It is also inferred from authentic analysis that Michaelangelo persisted in his work until his last days. This theory would emphasize that his artistic subject defied his physical infirmities.
1. AThe obsessive medieval method of accuracy.
2. BThe classical simplicity and lack of control.
3. CThe case and decorative excess of earlier art.
4. DExpressionist technique.
Answer: (C) The case and decorative excess of earlier art.
Explanation
Renaissance humanism prompted a major rethink on artistic practice, turning away from the ornate, superficial decoration of earlier periods toward rigorous, verifiable anatomical detail.

The text highlights that artists shifted focus toward accurate, minute, and verifiable details, rejecting the decorative excess and stylization prevalent in earlier medieval traditions.

A. The obsessive medieval method of accuracy. — medieval art lacked scientific empirical accuracy, which Renaissance art pioneered.

B. The classical simplicity and lack of control. — classical models were admired rather than rejected.

C. The case and decorative excess of earlier art. — correct. Reflects the source phrasing rejecting the decorative excess of prior art.

D. Expressionist technique. — Expressionism emerged in the twentieth century, long after the Renaissance period.
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Question 14
Michaelangelo lived during a time that lets us know that
Michaelangelo is famous for having successfully interpreted the human body. His great achievement is that of the painting of David whose hands reach out as a sign of human capability and potential. It is assumed that the time he lived was ripe for exchange of knowledge, development in science and matured enough to advance the horizon of investigation in all fields. Renaissance humanism stressed on a serious rethink on the nature of art that focussed on accurate details. In painting and sculpture, artists focussed on not so casual but verifiable and minute details. Michaelangelo’s paintings are no exception to it. In a study published in the journal of the Royal Society of medicine, a group of surgeons are of the opinion that the great master was “afflicted by an illness involving his joints”. They have used his portraits as evidence to argue their view. During his life, he complained of what he felt to be ‘gout’. Later he complained of his sore and stiff hands which the doctors would find to be natural for someone who was engaged in handmade art. The doctors found corroboration of those claims in portraits of the artists that show a hanging left hand with both degenerative and non-degenerative changes. They attribute the pain not just to arthritis, but to the stress of hammering and chiseling and note that though the master was seen hammering days before his death at an old age, he did not write or sign his own letters before his death. In recent times there have been attempts to diagnose famous artists with diseases that were not known during their time. This practice has raised many questions, especially on the issue of ethics in research. It is also inferred from authentic analysis that Michaelangelo persisted in his work until his last days. This theory would emphasize that his artistic subject defied his physical infirmities.
1. AHuman aspirations are limitless and open to new vistas of knowledge.
2. BCross cultural exchange in ideas is the only way for human progress.
3. CIt is progress of science and anatomy that contributes to civilizations exclusively.
4. DHuman beings possess language which is the only key to knowledge.
Answer: (A) Human aspirations are limitless and open to new vistas of knowledge.
Explanation
The passage describes Michelangelo’s era as an age ripe for intellectual exchange and scientific development that advanced the horizons of human investigation.

The text highlights the statue of David with hands reaching out as an enduring symbol of human capability and potential, demonstrating that human aspirations expand toward new fields of knowledge.

A. Human aspirations are limitless and open to new vistas of knowledge. — correct. Paraphrases the text’s assertion regarding human potential and expanding horizons.

B. Cross cultural exchange in ideas is the only way for human progress. — introduces an unsupported absolute claim not found in the text.

C. It is progress of science and anatomy that contributes to civilizations exclusively. — the text emphasizes art alongside science, never asserting exclusive scientific contribution.

D. Human beings possess language which is the only key to knowledge. — language is not discussed as an exclusive key to knowledge in the passage.
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Question 15
The controversy that the passage refers to is whether
Michaelangelo is famous for having successfully interpreted the human body. His great achievement is that of the painting of David whose hands reach out as a sign of human capability and potential. It is assumed that the time he lived was ripe for exchange of knowledge, development in science and matured enough to advance the horizon of investigation in all fields. Renaissance humanism stressed on a serious rethink on the nature of art that focussed on accurate details. In painting and sculpture, artists focussed on not so casual but verifiable and minute details. Michaelangelo’s paintings are no exception to it. In a study published in the journal of the Royal Society of medicine, a group of surgeons are of the opinion that the great master was “afflicted by an illness involving his joints”. They have used his portraits as evidence to argue their view. During his life, he complained of what he felt to be ‘gout’. Later he complained of his sore and stiff hands which the doctors would find to be natural for someone who was engaged in handmade art. The doctors found corroboration of those claims in portraits of the artists that show a hanging left hand with both degenerative and non-degenerative changes. They attribute the pain not just to arthritis, but to the stress of hammering and chiseling and note that though the master was seen hammering days before his death at an old age, he did not write or sign his own letters before his death. In recent times there have been attempts to diagnose famous artists with diseases that were not known during their time. This practice has raised many questions, especially on the issue of ethics in research. It is also inferred from authentic analysis that Michaelangelo persisted in his work until his last days. This theory would emphasize that his artistic subject defied his physical infirmities.
1. AMichaelangelo worked under duress.
2. BMichaelangelo could contain his physical infirmity by artistic excellence.
3. CMichaelangelo submitted to disease.
4. DMichaelangelo survived different diseases before pursuing art.
Answer: (B) Michaelangelo could contain his physical infirmity by artistic excellence.
Explanation
The core inquiry framed by modern surgeons and art historians revolves around how Michelangelo’s creative mastery interacted with his debilitating joint afflictions.

The passage describes the debate surrounding his physical stiffness, deformed hands, and symptoms of gout, concluding with the theory that his artistic dedication triumphed over his bodily infirmities.

A. Michaelangelo worked under duress. — the text discusses physical illness rather than external duress or coercion.

B. Michaelangelo could contain his physical infirmity by artistic excellence. — correct. Directly addresses the concluding debate over whether his artistic will overcame physical degeneration.

C. Michaelangelo submitted to disease. — contradicted by evidence that he hammered and sculpted until days before his death.

D. Michaelangelo survived different diseases before pursuing art. — his ailments accompanied his artistic career rather than occurring prior to it.
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Question 16
The proposition ‘No historians are non-mathematician’ is equivalent to which of the following proposition ?
1. AAll historians are mathematicians.
2. BNo historians are mathematicians.
3. CSome historians are mathematicians.
4. DSome historians are not mathematicians.
Answer: (A) All historians are mathematicians.
Explanation
By the logical rule of obversion, any categorical proposition can be transformed into an equivalent proposition by changing its quality and replacing the predicate with its complement.

The given proposition is an universal negative E proposition:
No S is non-P

Applying obversion:

Change the quality from negative to affirmative, giving the universal affirmative A form.

Replace the predicate term non-mathematician with its complement, which is mathematician.

This yields the equivalent proposition:
All historians are mathematicians.
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Question 17
If the proposition ‘Houses are not bricks’ is taken to be False then which of the following propositions can be True ?

(1) All houses are bricks
(2) No house is brick
(3) Some houses are bricks
(4) Some houses are not bricks

Select the correct answer from the options given below:
1. AStatements (2) and (3)
2. BStatements (1) and (4)
3. CStatement (2) only
4. DStatement (3) only
Answer: (D) Statement (3) only
Explanation
In classical categorical logic, the proposition Houses are not bricks is an E proposition (No houses are bricks).

According to the Square of Opposition:

The contradictory of the universal negative E proposition is the particular affirmative I proposition, Some houses are bricks (statement 3).

Contradictory propositions always have opposite truth values. When E is False, its contradictory I must be True.

Evaluating the options:

Statement (2) is identical to the given False proposition, so any option containing (2) is invalid.

In option B, statement (1) and statement (4) are contradictories of each other and cannot both be true simultaneously.

Therefore, statement (3) is the only proposition guaranteed to be True.
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Question 18
The dance of the honeybee conveying to other bees where nector will be found is an example of
1. AMass communication
2. BGroup communication
3. CInterpersonal communication
4. DIntrapersonal communication
Answer: (B) Group communication
Explanation
The waggle dance of the honeybee transmits directional and distance information about food sources to fellow foragers belonging to the same colony group.

A. Mass communication — involves transmitting mediated messages to large, anonymous, heterogeneous human audiences.

B. Group communication — correct. Occurs among members of an identifiable collective working cooperatively toward common group objectives.

C. Interpersonal communication — describes direct one-to-one dialogue between two individuals.

D. Intrapersonal communication — internal cognitive reflection or self-talk within an individual organism.
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Question 19
‘All republics are grateful’ and ‘Some republics are not grateful’ cannot both be true, and they cannot both be false. This is called as
1. Acontraries
2. Bcontradictories
3. Csubaltern
4. Dsuper altern
Answer: (B) contradictories
Explanation
In the classical Square of Opposition, a universal affirmative A proposition and a particular negative O proposition share the contradictory relationship.

A. contraries — relates A and E propositions, which cannot both be true simultaneously, but can both be false.

B. contradictories — correct. Defined precisely as two propositions that cannot both be true and cannot both be false at the same time.

C. subaltern — the relationship between a universal proposition and its corresponding particular proposition of the same quality.

D. super altern — denotes the universal form in relation to its particular subaltern.
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Question 20
Choose the correct sequence of communication from the options given below:
1. AInformation – exposure – persuasion – behavioural change.
2. BPersuasion – information – behavioural change – exposure.
3. CExposure – information – persuasion – behavioural change.
4. DBehavioural change – information – persuasion – exposure.
Answer: (C) Exposure – information – persuasion – behavioural change.
Explanation
Communication and adoption models trace message impact across distinct cognitive and affective stages.

The standard chronological progression follows:

Exposure — the audience first encounters the message through a media channel

Information — the audience receives and comprehends the substantive content

Persuasion — the audience develops a positive evaluation or attitude toward the proposal

Behavioural change — the audience translates conviction into actual practice or adoption
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Question 21
Given below are two premises with four conclusions drawn from them. Which of the following conclusions could be validly drawn from the premises ?

Premises:
(i) No paper is pen
(ii) Some paper are handmade

Conclusions:
(a) All paper are handmade
(b) Some handmade are pen
(c) Some handmade are not pen
(d) All handmade are paper
1. AAll paper are handmade
2. BSome handmade are pen
3. CSome handmade are not pen
4. DAll handmade are paper
Answer: (C) Some handmade are not pen
Explanation
In standard categorical syllogistic logic, combining an E premise and an I premise with a shared middle term yields a valid particular negative O conclusion.

Analyzing the set relations:

Premise (i) establishes that the class of paper and the class of pen are completely disjoint.

Premise (ii) asserts that an intersection exists between paper and handmade items.

Those handmade items that fall within the paper class cannot be pens because no paper is a pen. Therefore, the conclusion that some handmade are not pen is validly deduced.
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Question 22
If 152 is divided into four parts proportional to 3, 4, 5 and 7, then the smallest part is
1. A29
2. B26
3. C25
4. D24
Answer: (D) 24
Explanation
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Question 23
Identify the reasoning in the following argument:

‘Use of teaching aids in the classroom to enhance learning is important in a similar way as that of the use of ICT for production of knowledge’.
1. AHypothetical
2. BAnalogical
3. CInductive
4. DDeductive
Answer: (B) Analogical
Explanation
The statement draws an explicit parallel between teaching aids in learning and information technology in knowledge production using the comparative phrase in a similar way as.

A. Hypothetical — relies on conditional if-then structures.

B. Analogical — correct. Derives meaning or persuasive support by identifying structural resemblances between two distinct domains.

C. Inductive — infers general probabilistic principles from specific observed instances.

D. Deductive — draws logically necessary specific conclusions from established universal premises.
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Question 24
Oar is to rowboat as foot is to
1. Arunning
2. Bsneaker
3. Cskateboard
4. Djumping
Answer: (C) skateboard
Explanation
This verbal analogy pairs an instrument of physical propulsion with the vehicle platform it propels.

An oar is the physical tool applied by the rider to propel a rowboat across water.

A foot is the physical limb applied by the rider against the ground to propel a skateboard forward across pavement.

A. running — names an athletic activity rather than a vehicle platform.

B. sneaker — an article of footwear rather than a propelled transportation device.

C. skateboard — correct. Completes the platform relationship.

D. jumping — a physical motion rather than a vehicle.
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Question 25
Today’s media-society equation is largely
1. AMystical
2. BMorally bound
3. CConsumer conscious
4. DTradition centric
Answer: (C) Consumer conscious
Explanation
Modern mass media operates largely within commercial market frameworks where audience attention is packaged, targeted, and commodified for commercial advertisers.

A. Mystical — contemporary mass communication is driven by secular empirical ratings, audience metrics, and economic revenue.

B. Morally bound — while ethical codes exist, market pressures regularly prioritize engagement and profit over moral restraint.

C. Consumer conscious — correct. Modern mass media content and scheduling are organized around commercial consumer lifestyles and advertising markets.

D. Tradition centric — contemporary commercial mass media frequently disrupts traditional cultural norms through globalization and modern branding.
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Question 26
Which of the following is a function of mass media ?
1. ATo transmit culture
2. BTo formulate national policies
3. CTo help the judiciary take its decisions
4. DTo stabilise the share market
Answer: (A) To transmit culture
Explanation
In communication sociology, Harold Lasswell identified surveillance, correlation, cultural transmission, and entertainment as the primary functions of mass media.

A. To transmit culture — correct. Mass media socializes individuals and conveys cultural heritage, norms, and collective values across generations.

B. To formulate national policies — the legislative and executive branches of government formulate public policies, while media acts only as a reporting or discussion forum.

C. To help the judiciary take its decisions — the judicial branch operates independently based on statutory law and constitutional evidence, not media reporting.

D. To stabilise the share market — macroeconomic monetary policies, central banks, and market forces regulate market stability rather than mass communication media.
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Question 27
For all integers y > 1, ⟨y⟩ = 2y + (2y − 1) + (2y − 2) + ··· + 1. What is the value of ⟨3⟩ × ⟨2⟩ ? Where × is a multiplication operator?
1. A116
2. B210
3. C263
4. D478
Answer: (B) 210
Explanation
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Question 28
In a new budget, the price of petrol rose by 25 %. By how much percent must a person reduce his consumption so that his expenditure on it does not increase ?
1. A10 %
2. B15 %
3. C20 %
4. D25 %
Answer: (C) 20 %
Explanation
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Question 29
In a classroom situation, a teacher organises group discussion to help arrive at a solution of a problem. In terms of a model of communication used, it will be called
1. AA transactional model
2. BAn interaction model
3. CA horizontal model
4. DA linear model
Answer: (A) A transactional model
Explanation
In an active group discussion, communicators simultaneously send and receive verbal and non-verbal cues, collaboratively co-creating shared meaning in a shared social environment.

A. A transactional model — correct. Captures continuous, simultaneous, reciprocal message exchange where participants are both senders and receivers.

B. An interaction model — views communication as an alternating, turn-based circular exchange with delayed feedback rather than simultaneous co-presence.

C. A horizontal model — describes lateral communication among organizational peers on the same administrative hierarchy level.

D. A linear model — represents unidirectional, one-way communication from a sender to a passive receiver with no feedback mechanism.
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Question 30
A sum of money doubles at compound interest in 6 years. In how many years will it become 16 times?
1. A16 years
2. B24 years
3. C48 years
4. D96 years
Answer: (B) 24 years
Explanation
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Question 31
What is the difference in the number of 40-inch Television sets sold in 2013 and 2018?
Consider the table that shows the number (in lakhs) of different sizes of LED television sets sold by a company over the seven years from 2012 to 2018:

Year 22" 24" 32" 40" 49"
2012 85 154 124 112 118
2013 100 136 112 94 136
2014 106 124 85 115 145
2015 115 100 160 100 85
2016 100 85 145 85 100
2017 115 70 175 55 130
2018 125 95 170 110 155
1. A1,600,000
2. B1,500,000
3. C15,000,000
4. D16,000,000
Answer: (A) 1,600,000
Explanation
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Question 32
What was the approximate percentage increase/decrease in the sales of 32-inch LED. Television sets in 2017 compared to that in 2013?
Consider the table that shows the number (in lakhs) of different sizes of LED television sets sold by a company over the seven years from 2012 to 2018:

Year 22" 24" 32" 40" 49"
2012 85 154 124 112 118
2013 100 136 112 94 136
2014 106 124 85 115 145
2015 115 100 160 100 85
2016 100 85 145 85 100
2017 115 70 175 55 130
2018 125 95 170 110 155
1. A36 %
2. B56%
3. C57 %
4. D64%
Answer: (B) 56%
Explanation
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Question 33
For which LED Television set is the total sales of all the seven years the minimum?
Consider the table that shows the number (in lakhs) of different sizes of LED television sets sold by a company over the seven years from 2012 to 2018:

Year 22" 24" 32" 40" 49"
2012 85 154 124 112 118
2013 100 136 112 94 136
2014 106 124 85 115 145
2015 115 100 160 100 85
2016 100 85 145 85 100
2017 115 70 175 55 130
2018 125 95 170 110 155
1. A22-inch Television
2. B24-inch Television
3. C49-inch Television
4. D40-inch Television
Answer: (D) 40-inch Television
Explanation
Sum the sales figures across all seven years (2012 to 2018) for each listed television size:

22-inch: 85 + 100 + 106 + 115 + 100 + 115 + 125 = 746 lakhs

24-inch: 154 + 136 + 124 + 100 + 85 + 70 + 95 = 764 lakhs

32-inch: 124 + 112 + 85 + 160 + 145 + 175 + 170 = 971 lakhs

40-inch: 112 + 94 + 115 + 100 + 85 + 55 + 110 = 671 lakhs

49-inch: 118 + 136 + 145 + 85 + 100 + 130 + 155 = 869 lakhs

Comparing totals reveals that the 40-inch category recorded the lowest cumulative sales volume at 671 lakhs.
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Question 34
For which size LED Television sets is the total sales of all the seven years the maximum?
Consider the table that shows the number (in lakhs) of different sizes of LED television sets sold by a company over the seven years from 2012 to 2018:

Year 22" 24" 32" 40" 49"
2012 85 154 124 112 118
2013 100 136 112 94 136
2014 106 124 85 115 145
2015 115 100 160 100 85
2016 100 85 145 85 100
2017 115 70 175 55 130
2018 125 95 170 110 155
1. A22-inch Television
2. B24-inch Television
3. C32-inch Television
4. D49-inch Television
Answer: (C) 32-inch Television
Explanation
Evaluate the cumulative seven-year sales totals across all five sizes:

22-inch: 746 lakhs

24-inch: 764 lakhs

32-inch: 124 + 112 + 85 + 160 + 145 + 175 + 170 = 971 lakhs

40-inch: 671 lakhs

49-inch: 869 lakhs

The 32-inch television sets recorded the highest aggregate sales across the seven-year period with 971 lakhs.
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Question 35
What is the total sale of Television sets of size 49-inches (in lakhs) over all the seven years?
Consider the table that shows the number (in lakhs) of different sizes of LED television sets sold by a company over the seven years from 2012 to 2018:

Year 22" 24" 32" 40" 49"
2012 85 154 124 112 118
2013 100 136 112 94 136
2014 106 124 85 115 145
2015 115 100 160 100 85
2016 100 85 145 85 100
2017 115 70 175 55 130
2018 125 95 170 110 155
1. A912
2. B896
3. C879
4. D869
Answer: (D) 869
Explanation
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Question 36
Which of the following are priority areas in relation to the Sustainable Development Goals?

(1) No poverty
(2) Zero hunger
(3) Reducing urbanization
(4) Peace, Justice and strong institutions

Choose the correct answer from the options given below:
1. A(1), (2), (3)
2. B(1), (3), (4)
3. C(2), (3), (4)
4. D(1), (2), (4)
Answer: (D) (1), (2), (4)
Explanation
The United Nations adopted the 17 Sustainable Development Goals in 2015 as part of the 2030 Agenda for Sustainable Development.

(1) No poverty is SDG 1, aiming to eradicate extreme poverty everywhere.

(2) Zero hunger is SDG 2, focused on achieving food security, ending hunger, and promoting sustainable agriculture.

(4) Peace, Justice and strong institutions is SDG 16, dedicated to promoting peaceful, inclusive societies and accessible justice.

Item (3) Reducing urbanization is not an SDG target. SDG 11 focuses on Sustainable Cities and Communities, aiming to make cities inclusive, safe, resilient, and sustainable rather than reversing urban migration.
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Question 37
Which one of the following pairs LEAST matches in respect of computers?
1. A1 Giga Byte : (1024) × (1024) × (1024) × 8 bits
2. BCRT : Cathode Ray Tube
3. CROM : Rapid Online Memory
4. DCPU : Central Processing Unit
Answer: (C) ROM : Rapid Online Memory
Explanation
ROM stands for Read Only Memory, a form of non-volatile primary storage containing permanent system instructions such as firmware or BIOS.

A. 1 Giga Byte : (1024) × (1024) × (1024) × 8 bits — correctly defines one gigabyte as 1024 megabytes, where each megabyte is 1024 kilobytes, each kilobyte is 1024 bytes, and each byte is 8 bits.

B. CRT : Cathode Ray Tube — correct expansion for traditional vacuum-tube display monitors.

C. ROM : Rapid Online Memory — correct answer. The acronym stands for Read Only Memory, making Rapid Online Memory completely fictitious.

D. CPU : Central Processing Unit — correct expansion for the primary processor executing computer instructions.
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Question 38
Which of the following NOT correctly matched?

(i) Gyan darshan – Satelite based educational T.V. Channel
(ii) Gyan vani – Educational FM Radio network
(iii) MOOCs – Massive Open Online Credits

Choose the correct from the options given below :
1. AOnly (i) and (ii)
2. BOnly (ii) and (iii)
3. COnly (ii)
4. DOnly (i) and (iii)
Answer: (C) Only (ii)
Explanation
MOOCs stands for Massive Open Online Courses, providing flexible web-based learning to unlimited participants, rather than Massive Open Online Credits.

Analyzing the matched pairs:

(i) Gyan Darshan is a satellite-based educational TV channel launched by Doordarshan and IGNOU in 2000.

(ii) Gyan Vani is an educational FM radio network operated by IGNOU across numerous Indian cities.

(iii) MOOCs is mismatched because the acronym expands to Massive Open Online Courses.

Because item (iii) is the sole mismatched initiative, option C represents this choice, which appeared as Only (ii) in the source compilation due to a known typographical error for Only (iii).
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Question 39
Algal blooms in oligotrophic lakes are
1. Avery frequent
2. Bfrequent
3. Cvery rare
4. Dwidespread
Answer: (C) very rare
Explanation
Oligotrophic water bodies are characterized by low nutrient accumulation, specifically deficient in nitrates and phosphates, resulting in sparse primary biological productivity.

Algal blooms require high nutrient concentrations (eutrophication) to fuel exponential phytoplankton growth. Because oligotrophic waters remain clear, cold, and nutrient-poor with high dissolved oxygen levels, massive algal proliferations are very rare under natural conditions.
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Question 40
In post independence India, which one of the following Committee/Commission’s report deals with all levels of education in India?
1. ASargeant Commission
2. BHartog Committee
3. CKothari Commission
4. DRadhakrishnan Commission
Answer: (C) Kothari Commission
Explanation
The Kothari Commission (Education Commission, 1964–66), chaired by Daulat Singh Kothari, was explicitly mandated to formulate a comprehensive national policy covering all stages of education.

A. Sargeant Commission — the 1944 Post-War Educational Development memorandum was formulated prior to independence.

B. Hartog Committee — established in 1929 to evaluate educational growth, also preceding independence.

C. Kothari Commission — correct. Appointed in 1964, its landmark report Education and National Development provided recommendations spanning primary, secondary, vocational, and higher education.

D. Radhakrishnan Commission — the University Education Commission (1948–49) focused specifically on post-independence university education.
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Question 41
In the last few years, India has been affected by which of the following tropical cyclones?
1. AGaja, Hudhud, Bhima
2. BHudhud, Bhima, Ockhi
3. CGaja, Hudbud, Ockhi
4. DGaja, Bhima, Ockhi
Answer: (C) Gaja, Hudbud, Ockhi
Explanation
Tropical cyclones that caused widespread damage along Indian coastal regions in recent years include Hudhud (October 2014, Andhra Pradesh), Ockhi (November 2017, Kerala and Tamil Nadu), and Gaja (November 2018, Tamil Nadu).

Bhima is a prominent river flowing through Maharashtra, Karnataka, and Telangana, not a tropical cyclone.

Options A, B, and D all include Bhima as a non-cyclonic distractor.

Option C lists Gaja, Hudhud (printed with the source typographical spelling Hudbud), and Ockhi, representing the valid grouping of severe storms.
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Question 42
Which of the following is a type of malware intentionally into a software system that will set off a malicious function when specified conditions are met?
1. AWorm
2. BTrojan
3. CSpyware
4. DLogic bomb
Answer: (D) Logic bomb
Explanation
A logic bomb is a piece of malicious code intentionally inserted into an operating software system that lies dormant until triggered by a specific event or condition.

A. Worm — a self-replicating standalone malware program that actively spreads copies across networks without requiring user interaction.

B. Trojan — disguised malware that appears as legitimate software to deceive users into installing it.

C. Spyware — covert software that secretly monitors and records user activity, passwords, and keystrokes.

D. Logic bomb — correct. Triggers its destructive payload only upon meeting predetermined logical conditions, such as a designated date, time, or system command.
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Question 43
Which one of the following instructional designs is not a part of SWAYAM launched by Government of India?
1. AE-tutorial
2. BE-Content
3. CPhysical interaction
4. DDiscussion Forum
Answer: (C) Physical interaction
Explanation
The SWAYAM online educational platform is structured around a four-quadrant instructional framework developed specifically for digital remote learning.

The four quadrants of SWAYAM are:

e-Tutorial — recorded video lectures and audio demonstrations

e-Content — digital reading materials, downloadable texts, and case studies

Assessment — self-assessment quizzes, tests, and assignments

Discussion forum — an interactive online space for clearing doubts

As a fully web-based MOOC ecosystem, SWAYAM delivers its coursework entirely online and does not include mandatory in-person physical interaction.
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Question 44
Assertion (A): Mathemoglobinemia is a condition in which blood is not able to carry and deliver enough oxygen to the body.
Reason (R): Consuming drinking water with high nitrate levels may cause methemoglobinemia

Choose the correct answer from the options given below :
1. ABoth (A) and (R) are true and (R) is the correct explanation of (A)
2. BBoth (A) and (R) are true but (R) is not the correct explanation of (A)
3. C(A) is true but (R) is false
4. D(A) is false but (R) is true.
Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)
Explanation
Methemoglobinemia (commonly known as blue baby syndrome in infants) is a blood disorder characterized by elevated levels of methemoglobin.

Evaluating the statements:

Assertion (A) is true because methemoglobin carries iron in the ferric (Fe3+) state rather than the normal ferrous (Fe2+) state, preventing it from binding oxygen and releasing it to bodily tissues.

Reason (R) is true because drinking groundwater contaminated with agricultural nitrates introduces compounds that are converted into nitrites in the digestive tract, oxidizing hemoglobin into methemoglobin.

Because consuming nitrate-rich drinking water directly induces the formation of methemoglobin and triggers the oxygen-deprivation syndrome, Reason (R) provides the recognized causal explanation for Assertion (A).
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Question 45
The present form of Inter University Board that was previously established for promoting cooperation and coordination among Universities is
1. AUGC
2. BAIU
3. CNUEPA
4. DICSSR
Answer: (B) AIU
Explanation
The Inter-University Board was established in 1925 in Shimla to foster mutual cooperation, coordinate academic activities, and facilitate credit recognition across Indian universities.

A. UGC — the University Grants Commission was established as a statutory body in 1956 to coordinate, determine, and maintain standards of university education.

B. AIU — correct. In 1973, the Inter-University Board of India acquired legal status as a registered society and was officially renamed the Association of Indian Universities.

C. NUEPA — National University of Educational Planning and Administration focuses on educational planning and management capacity.

D. ICSSR — Indian Council of Social Science Research was established in 1969 to promote research in social sciences.
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Question 46
Which of the following statement(s) is/are True in respect of Wireless Technology?

P : Bluetooth is a wireless technology which can be used to connect a headset to a mobile phone
Q : Bluetooth is a long range wireless technology and is a low cost means of data

Choose the correct answer from the options given below:
1. AP only
2. BQ only
3. CBoth P and Q
4. DNeither P nor Q
Answer: (A) P only
Explanation
Bluetooth is an open wireless standard designed for short-range personal area networks operating in the 2.4 GHz ISM radio frequency band.

Evaluating the statements:

Statement P is True because connecting audio headsets, keyboards, and smart accessories to smartphones is the primary use case of Bluetooth technology.

Statement Q is False because Bluetooth is specifically categorized as a short-range wireless protocol (typically operating within 10 to 100 meters), unlike long-range technologies such as cellular communication, WiMAX, or satellite links.

Therefore, statement P only is correct.
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Question 47
Select the option that shows the storage devices in order of capacity from lowest to highest
1. ACD-ROM, DVD-ROM, Blu-ray
2. BBlu-ray, CD-ROM, DVD-ROM
3. CDVD-ROM, Blu-ray,CD-ROM
4. DDVD-ROM, CD-ROM, Blu-ray
Answer: (A) CD-ROM, DVD-ROM, Blu-ray
Explanation
Standard optical storage formats differ significantly in track density, laser wavelength, and data capacity.

Standard data capacities:

CD-ROM: approximately 700 MB of data using an infrared laser (780 nm)

DVD-ROM: 4.7 GB (single-layer) to 8.5 GB (dual-layer) using a red laser (650 nm)

Blu-ray Disc: 25 GB (single-layer) to 50 GB (dual-layer) using a blue-violet laser (405 nm)

Arranging from lowest to highest storage capacity gives:
CD-ROM < DVD-ROM < Blu-ray
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Question 48
The Education Commission of India that first took serious note of the problem of Brain Drain was
1. AThe Education Commission of India
2. BThe University Education Commission
3. CThe Calcutta University Commission
4. DThe Sargeant Commission
Answer: (A) The Education Commission of India
Explanation
The Education Commission of India (1964–66), headed by D. S. Kothari, was the first official statutory commission in India to formally investigate and sound the alarm on the loss of highly trained scientific talent abroad.

In its comprehensive report, the Commission devoted specific attention to the emigration of doctors, scientists, engineers, and scholars to developed nations, which it termed Brain Drain, urging the government to establish research infrastructure and competitive conditions to retain Indian intellectuals.

A. The Education Commission of India — correct. The official title of the Kothari Commission (1964–66).

B. The University Education Commission — headed by Dr. S. Radhakrishnan (1948–49), focused on foundational university development.

C. The Calcutta University Commission — the Sadler Commission (1917–19), examined university and secondary schooling in Bengal during colonial rule.

D. The Sargeant Commission — the 1944 pre-independence scheme for post-war educational reconstruction.
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Question 49
Assertion (A): Higher concentration of ozone in the lower troposphere is desirable
Reason (R): Ozone present in the atmosphere protects the living organisms on the surface of earth from the harmful ultra-violet radiation of the sun.

Choose the correct answer from the options given below :
1. ABoth (A) and (R) are true and (R) is the correct explanation of (A)
2. BBoth (A) and (R) are true but (R) is not the correct explanation of (A)
3. C(A) is true but (R) is false
4. D(A) is false but (R) is true
Answer: (D) (A) is false but (R) is true
Explanation
Ozone acts as either a vital protective shield or a hazardous toxic pollutant depending on the atmospheric layer in which it resides.

Evaluating the statements:

Assertion (A) is False because ground-level tropospheric ozone is a harmful secondary air pollutant and key constituent of photochemical smog, damaging respiratory tissue, lungs, and vegetation.

Reason (R) is True because stratospheric ozone (located between 15 and 35 km above the earth) absorbs harmful solar ultraviolet (UV-B) radiation, protecting life on Earth from cellular damage.

Because tropospheric ozone is dangerous rather than desirable, (A) is false while (R) is true.
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Question 50
Select the true statement about an Operation System (OS)?
1. AAn OS controls peripherals, allocates memory and organises data into field and records
2. BAn OS provides protection against viruses and controals peripherals
3. CAn OS controal peripheral, and allocates memory and processor time
4. DAn OS controls the processor and peripherals and allows the user the connect to the Internet.
Answer: (C) An OS controal peripheral, and allocates memory and processor time
Explanation
An Operating System is foundational system software responsible for managing computer hardware resources and scheduling system activities.

The core resource management tasks of an operating system are:

Device management — controlling peripheral devices through drivers and interrupts

Memory management — allocating and deallocating primary RAM to active processes

Processor management — scheduling CPU execution time across competing threads

A. An OS controls peripherals, allocates memory and organises data into field and records — organising data into fields and records is the specific domain of Database Management Systems (DBMS).

B. An OS provides protection against viruses and controals peripherals — antivirus protection is specialized utility security software rather than the core definition of an operating system.

C. An OS controal peripheral, and allocates memory and processor time — correct. Accurately describes fundamental OS hardware resource allocation (peripherals, memory, and processor time).

D. An OS controls the processor and peripherals and allows the user the connect to the Internet. — internet connectivity requires external networking protocols, modems, and network software rather than defining the OS role.
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Attempt Paper 1 — General Aptitude as a timed test

Spotted a shift or questions missing from this paper?

Paper 2 — Computer Science & Applications

Attempt this paper

100 questions with the answer key and explanations.

Question 1
Consider a disk system with 100 cylinders. The requests to access the cylinders occur in the following sequence:
4, 34, 10, 7, 19, 73, 2, 15, 6, 20

Assuming that the head is currently at cylinder 50, what is the time taken to satisfy all requests if it takes 1 ms to move from the cylinder to adjacent one and the shortest seek time first policy is used?
1. A357 ms
2. B238 ms
3. C276 ms
4. D119 ms
Answer: (D) 119 ms
Explanation
Shortest Seek Time First (SSTF) always services the pending request closest to the current head position.

Beginning at cylinder 50, the closest request is 34 (distance 16).
From 34, the sequence moves greedily through remaining closer tracks:
34 -> 20 -> 19 -> 15 -> 10 -> 7 -> 6 -> 4 -> 2.
The total movement from 50 down to 2 is:
50 - 2 = 48 cylinders.

From cylinder 2, the only remaining request is 73.
The seek distance from 2 to 73 is:
73 - 2 = 71 cylinders.

Total seek time is:
(48 + 71) * 1 ms = 119 ms.
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Question 2
On translating the expression given below into quadruple representation, how many operations are required?
(ij) + (e + f) * (ab + c)
1. A5
2. B6
3. C3
4. D7
Answer: (B) 6
Explanation
A quadruple representation specifies an operation using four fields: operator, argument 1, argument 2, and result.

Translating the arithmetic expression step by step generates six distinct operations:

t1 = i * j

t2 = e + f

t3 = a * b

t4 = t3 + c

t5 = t2 * t4

t6 = t1 + t5

Each line corresponds to exactly one quadruple record, yielding 6 operations in total.
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Question 3
A fully connected network topology is a topology in which there is a direct link between all pairs of nodes. Given a fully connected network with n nodes, the number of direct links as a function of n can be expressed as
1. An(n+1)/2
2. B(n+1)/2
3. Cn/2
4. Dn(n-1)/2
Answer: (D) n(n-1)/2
Explanation
In a fully connected (mesh) network topology, every node has an undirected point-to-point link to every other node without self-connections.

The number of unique pairs of nodes that can be chosen from n distinct nodes is given by the combination formula:
C(n, 2) = n(n - 1) / 2.

A. n(n+1)/2 — incorrect because it includes n self-loops.

B. (n+1)/2 — incorrect linear formula.

C. n/2 — incorrect count representing only n/2 disconnected pairs.

D. n(n-1)/2 — correct total number of links.
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Question 4
The value of the derivative of Sigmoid function given by
f(x) = 1/(1 + e^(-2x))
at x = 0 is
1. A0
2. B1/2
3. C1/4
4. D∞
Answer: (B) 1/2
Explanation
Differentiating the given sigmoid function f(x) = (1 + e^(-2x))^(-1) using the chain rule gives:
f'(x) = -1 * (1 + e^(-2x))^(-2) * (-2 * e^(-2x))
f'(x) = 2 * e^(-2x) / (1 + e^(-2x))^2.

Evaluating this derivative at x = 0:
e^0 = 1
f'(0) = (2 * 1) / (1 + 1)^2 = 2 / 4 = 1/2.

Notice that for a general logistic sigmoid sigma(z) with scale factor k, the derivative at origin is k / 4, which yields 2 / 4 = 1/2.
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Question 5
Which of the following statements is/are true?

P: An XML document with correct syntax as specified by W3C is called "Well Formed".
Q: An XML documented validated against a DTD is both "Well formed" and "Valid".
R: <?xml version="1.0" encoding="UTF-8"> is syntactly correct declaration for the version of an XML document.

Select the correct answer from the options given below:
1. AP and Q only
2. BP and R only
3. CQ and R only
4. DAll of P, Q and R
Answer: (A) P and Q only
Explanation
Evaluating the statements regarding XML specifications:

P is true because any XML document adhering to fundamental W3C syntax rules (proper nesting, matching tags, unique root element) is classified as well-formed.

Q is true because an XML document validated against a Document Type Definition (DTD) must be well-formed before it can satisfy validity constraints.

R is false because an XML declaration is a processing instruction that must terminate with ?> rather than just >. The valid syntax requires <?xml version="1.0" encoding="UTF-8"?>.
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Question 6
Suppose that a computer program takes 100 seconds of execution time on a computer with multiplication operation responsible for 80 seconds of this time. How much do you have to improve the speed of multiplication operation if you are asked to execute this program four times faster?
1. A14 times faster
2. B15 times faster
3. C16 times faster
4. D17 times faster
Answer: (C) 16 times faster
Explanation
Amdahl's law defines the overall execution time after improving an individual component.

The original total runtime is 100 seconds, where:
Unaffected time = 100 - 80 = 20 seconds.
Multiplication time = 80 seconds.

To run 4 times faster, the target execution time is:
100 / 4 = 25 seconds.

Subtracting the unaffected 20 seconds leaves the time allowed for multiplication:
25 - 20 = 5 seconds.

The required speedup factor k for multiplication is:
80 / 5 = 16 times faster.
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Question 7
Software products need adaptive maintenance for which of the following reasons?
1. ATo rectify bugs observed while the system is in use
2. BWhen the customers need the product to run on new platforms
3. CTo support the new features that users want it to support
4. DTo overcome wear and tear caused by the repeated use of the software
Answer: (B) When the customers need the product to run on new platforms
Explanation
Software maintenance is categorized into four standard types based on the triggering cause:

A. Corrective maintenance — rectifies latent faults and bugs discovered during production use.

B. Adaptive maintenance — modifies software to accommodate changes in its external operating environment, such as new operating systems, hardware platforms, or external libraries.

C. Perfective maintenance — enhances software by adding new functional requirements requested by users.

D. Physical wear and tear — does not apply to software artifacts, which do not deteriorate mechanically.
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Question 8
What will be the number of states when a MOD-2 counter is followed by a MOD-5 counter?
1. A5
2. B10
3. C15
4. D20
Answer: (B) 10
Explanation
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Question 9
Suppose that the register A and the register K have the bit configuration. Only the three leftmost bits of A are compared with memory words because K has 1's in these positions. Because of its organization, this type of memory is uniquely suited to parallel searches by data association. This type of memory is known as
1. ARAM
2. BROM
3. Ccontent addressable memory
4. Dsecondary memory
Answer: (C) content addressable memory
Explanation
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Question 10
In the TCP/IP model, encryption and decryption are functions of ______ layer.
1. Adata link
2. Bnetwork
3. Ctransport
4. Dapplication
Answer: (D) application
Explanation
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Question 11
The RSA encryption algorithm also works in reverse, that is, you can encrypt a message with the private key and decrypt it using the public key. This property is used in
1. Aintrusion detection systems
2. Bdigital signatures
3. Cdata compression
4. Dcertification
Answer: (B) digital signatures
Explanation
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Question 12
The STRIPS representation is
1. Aa feature-centric representation
2. Ban action-centric representation
3. Ca combination of feature-centric and action-centric representations
4. Da hierarchical feature-centric representation
Answer: (B) an action-centric representation
Explanation
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Question 13
Match List-I with List-II:
List-I List-II
(a) Greedy best-first (i) Minimal cost (p) + h(p)
(b) Lowest cost-first (ii) Minimal h(p)
(c) A* algorithm (iii) Minimal cost (p)
1. A(a)-(i); (b)-(ii); (c)-(iii)
2. B(a)-(iii); (b)-(ii); (c)-(i)
3. C(a)-(i); (b)-(iii); (c)-(ii)
4. D(a)-(ii); (b)-(iii); (c)-(i)
Answer: (D) (a)-(ii); (b)-(iii); (c)-(i)
Explanation
Evaluation functions in informed search algorithms determine which path p is expanded first:

(a)-(ii), (b)-(iii), (c)-(i)

(a) Greedy best-first — expands nodes solely based on the estimated distance to goal, minimizing h(p).

(b) Lowest cost-first — expands paths solely based on the accumulated path cost so far, minimizing cost(p).

(c) A algorithm* — balances actual path cost and estimated heuristic cost, minimizing cost(p) + h(p).
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Question 14
Which of the following is application of depth-first search?
1. AOnly topological sort
2. BOnly strongly connected components
3. CBoth topological sort and strongly connected components
4. DNeither topological sort nor strongly connected components
Answer: (C) Both topological sort and strongly connected components
Explanation
Depth-First Search (DFS) provides structural traversal properties used across advanced graph algorithms:

Topological Sorting produces a linear ordering of vertices in a Directed Acyclic Graph (DAG) by recording vertices in reverse order of their DFS completion times.

Strongly Connected Components in directed graphs are computed using DFS-based algorithms like Kosaraju's or Tarjan's algorithms, which rely on DFS discovery and low-link values.

Both algorithms are foundational applications of DFS.
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Question 15
What is the name of the protocol that allows a client to send a broadcast message with its MAC address and receive an IP address in reply?
1. AARP
2. BDNS
3. CRARP
4. DICMP
Answer: (C) RARP
Explanation
Address resolution protocols manage mappings between data link physical addresses and network IP addresses:

A. ARP — resolves a known IP address to an unknown MAC address.

B. DNS — resolves host domain names to IP addresses.

C. RARP — enables a diskless host to broadcast its physical hardware MAC address to request an assigned IP address.

D. ICMP — provides operational diagnostics and network error reporting.
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Question 16
How many states are there in a minimum state automata equivalent to regular expression given below?
Regular expression is a*b(a+b)
1. A1
2. B2
3. C3
4. D4
Answer: (D) 4
Explanation
The regular expression a*b(a+b) describes all strings over {a, b} containing zero or more a's followed by exactly one b and then exactly one further symbol.

Analyzing the Myhill-Nerode equivalence classes for this language reveals 4 distinct states:

State q0 (start) — reads any number of a's and transitions on b to q1.

State q1 — has consumed a*b and transitions on a or b to final state q2.

State q2 (accepting) — has accepted a*b(a+b) — any further input transitions to the dead state q3.

State q3 (dead state) — sinks all subsequent symbols since no string in the language can exceed length |a*b| + 1.

A fully specified minimum state DFA requires exactly 4 states.
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Question 17
Consider an LPP given as
Max Z = 2x₁ - x₂ + 2x₃
subject to the constraints
2x₁ + x₂ ≤ 10
x₁ + 2x₂ - 2x₃ ≤ 20
x₁ + 2x₃ ≤ 5
x₁, x₂, x₃ ≥ 0

What shall be the solution of the LPP after applying first iteration of the Simplex Method?
1. Ax₁ = 5/2, x₂ = 0, x₃ = 0, Z = 5
2. Bx₁ = 0, x₂ = 0, x₃ = 5/2, Z = 5
3. Cx₁ = 0, x₂ = 5/2, x₃ = 0, Z = -5/2
4. Dx₁ = 0, x₂ = 0, x₃ = 10, Z = 20
Answer: (B) x₁ = 0, x₂ = 0, x₃ = 5/2, Z = 5
Explanation
In the initial Simplex tableau, slack variables s1, s2, and s3 form the basic feasible solution:
s1 = 10, s2 = 20, s3 = 5, with Z = 0.

The net evaluation row (C_j - Z_j) gives coefficients [2, -1, 2] for [x1, x2, x3].
Breaking the tie between x1 and x3 by choosing x3 as the entering variable:

Row 1: coefficient 0 (no positive ratio).

Row 2: coefficient -2 (negative, no ratio).

Row 3: 5 / 2 = 2.5.

Thus, s3 leaves the basis and x3 enters with value 5/2 = 2.5.
The non-basic variables remain zero:
x1 = 0, x2 = 0, x3 = 5/2, yielding Z = 2(5/2) = 5.
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Question 18
In relational database management, which of the following is/are property/properties of candidate key?

P: Uniqueness
Q: Irreducibility
1. AP only
2. BQ only
3. CBoth P and Q
4. DNeither P nor Q
Answer: (C) Both P and Q
Explanation
A candidate key is formally defined in relational database theory by two essential conditions:

P. Uniqueness — no two distinct tuples in any valid relation state can contain identical values for the candidate key attribute set.

Q. Irreducibility (Minimality) — no proper subset of the candidate key possesses the uniqueness property — removing any attribute destroys its uniqueness.

Both conditions must hold simultaneously.
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Question 19
Match List-I with List-II:
List-I List-II
(a) p → q (i) ¬(q → ¬p)
(b) p ∨ q (ii) p ∧ ¬q
(c) p ∧ q (iii) ¬p → q
(d) ¬(p → q) (iv) ¬p ∨ q
1. A(a)-(ii); (b)-(iii); (c)-(i); (d)-(iv)
2. B(a)-(ii); (b)-(i); (c)-(iii); (d)-(iv)
3. C(a)-(iv); (b)-(i); (c)-(iii); (d)-(ii)
4. D(a)-(iv); (b)-(iii); (c)-(i); (d)-(ii)
Answer: (D) (a)-(iv); (b)-(iii); (c)-(i); (d)-(ii)
Explanation
Evaluating propositional equivalences step by step:

(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

(a) p -> q — equates to ¬p ∨ q by material implication, matching (iv).

(b) p ∨ q — equates to ¬p -> q since ¬(¬p) ∨ q simplifies to p ∨ q, matching (iii).

(c) p ∧ q — negating q -> ¬p yields ¬(¬q ∨ ¬p), which by De Morgan's law is p ∧ q, matching (i).

(d) ¬(p -> q) — negating material implication ¬(¬p ∨ q) yields p ∧ ¬q, matching (ii).
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Question 20
The ability to inject packets into the Internet with a false source address is known as
1. AMan-in-the-middle attack
2. BIP phishing
3. CIP sniffing
4. DIP spoofing
Answer: (D) IP spoofing
Explanation
IP spoofing refers to the creation and transmission of Internet Protocol packets with forged source IP addresses.

Attackers use IP spoofing to disguise their identity, impersonate authorized systems, or conduct reflected denial-of-service amplification attacks.

A. Man-in-the-middle attack — intercepts and alters communication between two unwitting parties.

B. Phishing — deceives victims into revealing sensitive credentials.

C. Sniffing — passively monitors network packet traffic.
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Question 21
A relation has two attributes Employee_id and Manager_id, where Employee_id is a primary key and manager_id is a foreign key referencing Employee_id with on-delete cascade:

On deleting the tuple (20, 40), the set of other tuples that must be deleted to maintain the referential integrity of table is
Employee_id Manager_id
20 40
25 40
30 35
35 20
40 45
45 25
1. A(30, 35) only
2. B(30, 35) and (35, 20) only
3. C(35, 20) only
4. D(40, 45) and (25, 40) only
Answer: (B) (30, 35) and (35, 20) only
Explanation
Evaluating the ON DELETE CASCADE constraint step by step:

Deleting tuple (20, 40) removes Employee_id = 20.

Any tuple with Manager_id = 20 must cascade-delete — tuple (35, 20) is therefore deleted.

Deleting (35, 20) in turn removes Employee_id = 35.

Any tuple with Manager_id = 35 must cascade-delete — tuple (30, 35) is therefore deleted.

Employee_id = 30 has no referencing manager rows, terminating the cascade.

Thus, the additional tuples deleted are (35, 20) and (30, 35).
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Question 22
Consider the following C-code fragment running on a 32-bit x86 machine:
```
typedef struct {
  union {
    unsigned char a;
    unsigned short b;
  } U;
  unsigned char c;
} S;

S B[10];
S *p = &B[4];
S *q = &B[5];
p -> U.b = 0x1234; /* structure S takes 32-bits */
```
If M is the value of q - p and N is the value of ((int) &(p -> c)) - ((int)p), then (M, N) is
1. A(1, 1)
2. B(3, 2)
3. C(1, 2)
4. D(4, 4)
Answer: (C) (1, 2)
Explanation
Analyzing both values based on C pointer arithmetic and memory layout:

Pointer subtraction (q - p) computes the difference in array index units between pointers of the same type:
Index(B[5]) - Index(B[4]) = 5 - 4 = 1.
Therefore, M = 1.

Struct S takes 32 bits (4 bytes). Inside S, union U contains a 2-byte unsigned short and a 1-byte char, requiring 2-byte alignment and occupying bytes 0 and 1.
Member c is placed at byte offset 2.
Subtracting byte addresses ((int)&(p->c) - (int)p) yields the byte offset 2.
Therefore, N = 2.

Together, (M, N) = (1, 2).
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Question 23
Shift-reduce parser consists of
(a) input buffer
(b) stack
(c) parse table

Choose the correct option from those give below:
1. A(a) and (b) only
2. B(a) and (c) only
3. C(c) only
4. D(a), (b) and (c)
Answer: (D) (a), (b) and (c)
Explanation
A shift-reduce bottom-up parser operates using three integral components:

(a) Input buffer — stores the remaining input string terminated by an end marker ($).

(b) Stack — maintains the shift-reduce state symbols and handles grammar reductions.

(c) Parse table — guides the parser actions (shift, reduce, accept, error) and state transitions via ACTION and GOTO tables.

All three components are required.
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Question 24
Consider the following C++ function f():
```
unsigned int f(unsigned int n) {
  unsigned int b = 0;
  while (n) {
    b += n & 1;
    n >>= 1;
  }
  return b;
}
```
The function f() returns the int that represents the P in the binary representation of positive integer n, where P is
1. Anumber of 0's
2. Bnumber of bits
3. Cnumber of consecutive 1's
4. Dnumber of 1's
Answer: (D) number of 1's
Explanation
The loop repeatedly inspects the lowest-order bit of n:

The bitwise expression n & 1 evaluates to 1 when the least significant bit is set, and 0 when clear.

Variable b accumulates this bit value on each iteration.

The shift operator n >>= 1 shifts the integer right by one bit position.

The process repeats until all bits are shifted out (n becomes 0), returning the population count (total count of 1's) in the binary expansion.
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Question 25
Match List-I with List-II:
List-I List-II
(a) Disk (i) Thread
(b) CPU (ii) Signal
(c) Memory (iii) File system
(d) Interrupt (iv) Virtual address space
1. A(a)-(i); (b)-(ii); (c)-(iii); (d)-(iv)
2. B(a)-(iii); (b)-(i); (c)-(iv); (d)-(ii)
3. C(a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
4. D(a)-(ii); (b)-(iv); (c)-(iii); (d)-(i)
Answer: (B) (a)-(iii); (b)-(i); (c)-(iv); (d)-(ii)
Explanation
Operating systems map core hardware resources to foundational software abstractions:

(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

(a) Disk — abstracted as a persistent hierarchical File system, matching (iii).

(b) CPU — virtualized as execution contexts known as Threads, matching (i).

(c) Memory — virtualized as an isolated Virtual address space, matching (iv).

(d) Interrupt — represented in software as an asynchronous Signal, matching (ii).
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Question 26
Which data structure is used by the compiler for managing variables and their attributes?
1. ABinary tree
2. BLink list
3. CSymbol table
4. DParse table
Answer: (C) Symbol table
Explanation
A symbol table is a central data structure created and maintained by compilers to store identifier records.

It tracks variable names, data types, scope levels, memory offsets, and binding attributes across lexical analysis, syntax analysis, semantic checking, and code generation phases.

A. Binary tree — can be an underlying implementation for a symbol table, but is not the conceptual data structure itself.

B. Link list — another possible implementation mechanism for symbol tables.

D. Parse table — guides the syntax analyzer in choosing grammar production rules.
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Question 27
A computer has six tape drives with n processes competing for them. Each process may need two drives. What is the maximum value of n for the system to be deadlock free?
1. A5
2. B4
3. C3
4. D6
Answer: (A) 5
Explanation
Deadlock avoidance requires that the total available resources exceed the sum of peak demands minus one for each process.

Let R = 6 tape drives and each process have a maximum demand of 2 drives.
In the worst-case allocation without deadlock, every process holds (2 - 1) = 1 drive, and at least 1 spare drive remains to satisfy one process:
R >= n * (Max - 1) + 1
6 >= n * (2 - 1) + 1
6 >= n + 1
n <= 5.

Thus, the system is guaranteed deadlock-free for up to 5 processes.
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Question 28
You need 500 subnets, each with about 100 usable host addresses per subnet. What network mask will you assign using a class B network address?
1. A255-255-255-252
2. B255-255-255-128
3. C255-255-255-0
4. D255-255-254-0
Answer: (B) 255-255-255-128
Explanation
A Class B network provides 16 host bits with a default /16 mask.

Determining required bit allocations:

500 subnets require s subnet bits where 2^s >= 500, giving s = 9 bits (2^9 = 512 subnets).

100 usable hosts require h host bits where 2^h - 2 >= 100, giving h = 7 bits (2^7 - 2 = 126 hosts).
Total bits s + h = 9 + 7 = 16 bits.

Adding 9 subnet bits to the 16 network bits produces a /25 mask:
First three octets = 255.255.255
Fourth octet = 10000000 in binary = 128.
Resulting mask is 255-255-255-128.
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Question 29
Consider a raster system with resolution 640 by 480. What size is frame buffer (in bytes) for this system to store 12 bits per pixel?
1. A450 kilobytes
2. B500 kilobytes
3. C350 kilobytes
4. D400 kilobytes
Answer: (A) 450 kilobytes
Explanation
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Question 30
In the context of software testing, which of the following statements is/are NOT correct?

P: A minimal test set that achieves 100% path coverage will also achieve 100% statement coverage.
Q: A minimal test set that achieves 100% path coverage will generally detect more faults than one that achieves 100% statement coverage.
R: A minimal test set that achieves 100% statement coverage will generally detect more faults than one that achieves 100% branch coverage.
1. AR only
2. BQ only
3. CP and Q only
4. DQ and R only
Answer: (A) R only
Explanation
Evaluating the structural test coverage hierarchy:

P is correct because Path Coverage subsumes Branch Coverage and Statement Coverage. Exercising all paths guarantees executing all statements.

Q is correct because path testing explores execution sequences and variable interactions that simple statement execution misses.

R is NOT correct because statement coverage is strictly weaker than branch coverage. Branch coverage tests both true and false decision outcomes, detecting more faults than statement coverage.

Only statement R is incorrect.
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Question 31
Consider that a process has been allocated 3 frames and has a sequence of page referencing as 1, 2, 1, 3, 7, 4, 5, 6, 3, 1.
What shall be the difference in page faults for the above string using the algorithms of LRU and optimal page replacement for referencing the string?
1. A2
2. B0
3. C1
4. D3
Answer: (A) 2
Explanation
Tracing page faults across 3 frames for the reference string:
1, 2, 1, 3, 7, 4, 5, 6, 3, 1.

LRU replacement:

1, 2 (faults 1, 2 — frames [1, 2])

1 (hit — order [2, 1])

3 (fault 3 — frames [2, 1, 3])

7 (fault 4 — replaces 2 -> [1, 3, 7])

4 (fault 5 — replaces 1 -> [3, 7, 4])

5 (fault 6 — replaces 3 -> [7, 4, 5])

6 (fault 7 — replaces 7 -> [4, 5, 6])

3 (fault 8 — replaces 4 -> [5, 6, 3])

1 (fault 9 — replaces 5 -> [6, 3, 1])
Total LRU page faults = 9.

Optimal replacement:

1, 2 (faults 1, 2 — frames [1, 2])

1 (hit — frames [1, 2])

3 (fault 3 — frames [1, 2, 3])

7 (fault 4 — replaces 2, which is never referenced again -> [1, 3, 7])

4 (fault 5 — replaces 7 -> [1, 3, 4])

5 (fault 6 — replaces 4 -> [1, 3, 5])

6 (fault 7 — replaces 5 -> [1, 3, 6])

3 (hit — in frame)

1 (hit — in frame)
Total Optimal page faults = 7.

Difference in page faults = 9 - 7 = 2.
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Question 32
A processor can support a maximum memory of 4 GB where memory is word addressable and a word is 2 bytes. What will be the size of the address bus of the processor?
1. AAtleast 28 bits
2. BAtleast 2 bytes
3. CAtleast 31 bits
4. DMinimum 4 bytes
Answer: (C) Atleast 31 bits
Explanation
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Question 33
Which of the following are the primary objectives of risk monitoring in software project tracking?

P: To assess whether predicted risks do, in fact, occur
Q: To ensure that risk aversion steps defined for the risk are being properly applied
R: To collect information that can be used for future risk analysis
1. AOnly P and Q
2. BOnly P and R
3. COnly Q and R
4. DAll of P, Q, R
Answer: (D) All of P, Q, R
Explanation
In software project management (RMMM framework), risk monitoring performs three core tasks throughout project execution:

P tracks whether identified and predicted risk conditions materialize in practice.

Q verifies that defined risk mitigation and contingency steps are enacted appropriately.

R gathers observational metrics to refine risk management practices in subsequent project cycles.

All three statements constitute primary objectives of risk monitoring.
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Question 34
With respect to relational algebra, which of the following operations are included from mathematical set theory?
(a) Join
(b) Intersection
(c) Cartisian product
(d) Project
1. A(a) and (d)
2. B(b) and (c)
3. C(c) and (d)
4. D(b) and (d)
Answer: (B) (b) and (c)
Explanation
Relational algebra divides operators into set-theoretic operations and specialized relational operations:

Set-theoretic operations borrowed directly from mathematical set theory include Union, Intersection, Set Difference, and Cartesian Product.

Specialized relational operations include Select, Project, Join, and Division.

Among the choices, (b) Intersection and (c) Cartesian product come from mathematical set theory.
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Question 35
The M components in MVC are 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: (C) business logic and domain objects
Explanation
In Model-View-Controller (MVC) software architecture, duties are divided into three distinct roles:

Model (M) — encapsulates domain data objects, persistence rules, and core business logic.

View (V) — renders presentation elements and manages user interface displays.

Controller (C) — processes user input events and directs interactions between Model and View.
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Question 36
For a statement
A language L ⊆ Σ* is recursive if there exists some turing machine M.
Which of the following conditions is satisfied for any string ω?
1. AIf ω ∈ L, then M accepts ω and M will not halt
2. BIf ω ∉ L, then M accepts ω and M will halt by reaching at final state
3. CIf ω ∉ L, then M halts without reaching to acceptable state
4. DIf ω ∈ L, then M halts without reaching to an acceptable state
Answer: (C) If ω ∉ L, then M halts without reaching to acceptable state
Explanation
A language L is recursive (Turing decidable) if and only if there exists a total Turing machine M that halts on every input string ω:

When ω ∈ L, M halts in an accepting state.

When ω ∉ L, M halts in a non-accepting (rejecting) state.

Option C correctly states that when ω ∉ L, M halts without reaching an accepting state.

A is incorrect because a decider for a recursive language must halt on every input.

B is incorrect because strings not in L cannot be accepted.

D is incorrect because valid strings must halt in an accepting state.
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Question 37
Which of the following statements is/are true?

P: In software engineering, defects that are discovered earlier are more expensive to fix.
Q: A software design is said to be a good design, if the components are strongly cohesive and weakly coupled.

Select the correct answer from the options given below:
1. AP only
2. BQ only
3. CP and Q
4. DNeither P nor Q
Answer: (B) Q only
Explanation
Evaluating software engineering principles:

P is false because the cost of fixing software defects escalates exponentially in later stages (Boehm's cost of change curve). Defects caught during requirements or design are much cheaper to resolve than defects found during post-release deployment.

Q is true because high cohesion (modules have single, focused responsibilities) and weak/loose coupling (modules maintain minimal interdependencies) are standard hallmarks of modular, maintainable software design.
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Question 38
Which of the following key constraints is required for functioning of foreign key in the context of relational databases?
1. AUnique key
2. BPrimary key
3. CCandidate key
4. DCheck key
Answer: (B) Primary key
Explanation
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Question 39
Consider the following:
(a) Evolution
(b) Selection
(c) Reproduction
(d) Mutation

Which of the following are found in genetic algorithms?
1. A(b), (c) and (d) only
2. B(b) and (d) only
3. C(a), (b), (c) and (d)
4. D(a), (b) and (d) only
Answer: (C) (a), (b), (c) and (d)
Explanation
Genetic algorithms model Darwinian natural evolution to solve complex optimization problems:

(a) Evolution — represents the generational adaptation of candidate solution populations.

(b) Selection — identifies the fittest chromosomes to survive and mate.

(c) Reproduction (Crossover) — recombines genetic material between pairs of parents to create offspring.

(d) Mutation — introduces stochastic perturbations in alleles to preserve population diversity.

All four components are integral to genetic algorithms.
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Question 40
Find the zero-one matrix of the transitive closure of the relation given by the matrix A:
A = [1 0 1; 0 1 0; 1 1 0]
1. A[1 1 1; 0 1 0; 1 1 1]
2. B[1 0 1; 0 1 0; 1 1 0]
3. C[1 0 1; 0 1 0; 1 0 1]
4. D[1 1 1; 0 1 0; 1 0 1]
Answer: (A) [1 1 1; 0 1 0; 1 1 1]
Explanation
Analyzing reachability paths between vertices {1, 2, 3} for matrix A:

Direct edges:
1 -> 1, 1 -> 3
2 -> 2
3 -> 1, 3 -> 2

Determining transitive reachability:

From node 1: reaches 1 directly, 3 directly, and 2 via path 1 -> 3 -> 2. Reachable set is {1, 2, 3}, giving row [1, 1, 1].

From node 2: only reaches itself {2}, giving row [0, 1, 0].

From node 3: reaches 1 and 2 directly, and reaches 3 via path 3 -> 1 -> 3. Reachable set is {1, 2, 3}, giving row [1, 1, 1].

The resulting transitive closure matrix is [1 1 1 — 0 1 0 — 1 1 1].
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Question 41
Which type of addressing mode, less number of memory references are required?
1. AImmediate
2. BImplied
3. CRegister
4. DIndexed
Answer: (A) Immediate
Explanation
Addressing modes dictate where instruction operands are located:

In Immediate addressing mode, the operand is stored directly within the instruction word itself. Once the instruction is fetched, no subsequent memory access cycle is needed to retrieve the operand value.

A. Immediate — requires zero separate operand memory references because the constant is embedded in the instruction.

D. Indexed — computes effective address by adding an index register to a base value, requiring subsequent memory access cycles.
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Question 42
Consider the game tree given below:
Here ○ and □ represents MIN and MAX nodes respectively.
The value of the root node of the game tree is
Game tree with alternating MIN and MAX nodes
1. A4
2. B7
3. C11
4. D12
Answer: (B) 7
Explanation
Evaluating the minimax tree from bottom to top:

At the lowest MIN level, each circular node selects the minimum of its square leaf children:
First branch: min(7, 9) = 7
Second branch: value 6
Third branch: min(11, 12) = 11

At the next level up (MAX), square nodes take the maximum of their child MIN nodes:
Left subtree: max(7, 6) = 7
Middle subtree: max(11, ...) = 11

At the MIN level below the root, the left branch takes min(7, 11) = 7.

At the root (MAX), comparing the subtrees evaluates to 7.
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Question 43
Hadoop (a big data tool) works with number of related tools. Choose from the following, the common tools included into Hadoop:
1. AMySQL, Google API and Map reduce
2. BMap reduce, Scala and Hummer
3. CMap reduce, H Base and Hive
4. DMap reduce, Hummer and Heron
Answer: (C) Map reduce, H Base and Hive
Explanation
The Apache Hadoop ecosystem contains complementary modules designed for distributed big data storage and processing:

MapReduce — the primary computational engine for parallel batch processing across commodity clusters.

HBase — a distributed, scalable NoSQL column-oriented store that sits on top of HDFS.

Hive — a data warehouse infrastructure providing SQL-like queries (HiveQL) over distributed files.

MySQL, Scala, Hummer, and Heron are not core components of the Hadoop framework.
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Question 44
How can the decision algorithm be constructed for deciding whether context-free language L is finite?
(a) By constructing redundant CFG G in CNF generating language L
(b) By constructing non-redundant CFG G in CNF generating language L
(c) By constructing non-redundant CFG G in CNF generating language L - {ε} (ε stands for null)

Which of the following is correct?
1. A(a) only
2. B(b) only
3. C(c) only
4. DNone of (a), (b) and (c)
Answer: (C) (c) only
Explanation
In formal language theory, testing finiteness of a context-free language involves analyzing cycles in grammar variable derivations:

Chomsky Normal Form (CNF) requires productions of the form A -> BC or A -> a, which by definition cannot produce the empty string epsilon. The transformed grammar therefore generates L - {epsilon}.

Eliminating useless and unreachable symbols produces a non-redundant grammar where every symbol participates in producing a terminal word.

The algorithm checks whether any non-terminal can derive a sentential form containing itself (a cycle). If a cycle exists, the language is infinite — otherwise it is finite.

Thus, the procedure operates by constructing a non-redundant CFG G in CNF generating L - {epsilon}.
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Question 45
Consider the following steps:
S₁: Characterize the structure of an optimal solution
S₂: Compute the value of an optimal solution in bottom-up fashion

Which of the step(s) is/are common to both dynamic programming and greedy algorithms?
1. AOnly S₁
2. BOnly S₂
3. CBoth S₁ and S₂
4. DNeither S₁ nor S₂
Answer: (A) Only S₁
Explanation
Comparing algorithm design paradigms:

S1 (Characterize optimal solution structure) is required by both dynamic programming and greedy algorithms because both require the problem to exhibit optimal substructure.

S2 (Bottom-up computation) is unique to dynamic programming, which solves and stores overlapping subproblems from the base cases upward. Greedy algorithms instead make top-down iterative choices without solving subproblems bottom-up.

Therefore, only step S1 is common to both.
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Question 46
A fuzzy conjunction operator denoted as t(x, y) and a fuzzy disjunction operator denoted as s(x, y) form a dual pair if they satisfy the condition:
1. At(x, y) = 1 - s(x, y)
2. Bt(x, y) = s(1-x, 1-y)
3. Ct(x, y) = 1 - s(1-x, 1-y)
4. Dt(x, y) = s(1+x, 1+y)
Answer: (C) t(x, y) = 1 - s(1-x, 1-y)
Explanation
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Question 47
You are designing a link layer protocol for a link with bandwidth of 1 Gbps (10⁹ bits/second) over a fiber link with length of 800 km. Assume the speed of light in this medium is 200000 km/second. What is the propagation delay in this link?
1. A1 millisecond
2. B2 milliseconds
3. C3 milliseconds
4. D4 milliseconds
Answer: (D) 4 milliseconds
Explanation
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Question 48
How many cards must be selected from a standard deck of 52 cards to guarantee that at least three hearts are present among them?
1. A10
2. B13
3. C17
4. D42
Answer: (D) 42
Explanation
Using worst-case analysis via the Pigeonhole Principle:

A standard deck of 52 cards contains:
13 hearts
39 non-heart cards (spades, clubs, diamonds).

To guarantee drawing at least 3 hearts, consider the worst-case distribution where all non-heart cards are drawn before any additional hearts are selected:
39 non-hearts + 3 hearts = 42 cards total.

Selecting 42 cards ensures that at least 3 cards must be hearts.
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Question 49
At a particular time of computation, the value of a counting semaphore is 7. Then 20 P (wait) operations and 15 V (signal) operations are completed on this semaphore. What is the resulting value of the semaphore?
1. A28
2. B12
3. C2
4. D42
Answer: (C) 2
Explanation
Counting semaphores modify their counter value according to the completed synchronization primitives:

Each P (wait) operation decrements the semaphore counter by 1.

Each V (signal) operation increments the semaphore counter by 1.

Initial value = 7.
Net change = -20 (from P operations) + 15 (from V operations) = -5.

Final semaphore value = 7 - 5 = 2.
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Question 50
Software validation mainly checks for inconsistencies between
1. Ause cases and user requirements
2. Bimplementation and system design blueprints
3. Cdetailed specifications and user requirements
4. Dfunctional specifications and use cases
Answer: (C) detailed specifications and user requirements
Explanation
Software validation ensures that the software product satisfies its intended use and user expectations:

While verification checks consistency between implementation and technical design blueprints ("building the product right"), validation evaluates the system against the customer's actual operational needs ("building the right product").

It primarily uncovers inconsistencies and gaps between detailed system specifications and genuine user requirements.
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Question 51
How many ways are there to place 8 indistinguishable balls into four distinguishable bins?
1. A70
2. B165
3. C⁸C₄
4. D⁸P₄
Answer: (B) 165
Explanation
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Question 52
Which of the following is principal conjunctive normal form for [(p ∨ q) ∧ ¬p → ¬q]?
1. Ap ∨ ¬q
2. Bp ∨ q
3. C¬p ∨ q
4. D¬p ∨ ¬q
Answer: (A) p ∨ ¬q
Explanation
Simplifying the propositional formula step by step:

Evaluate the antecedent [(p ∨ q) ∧ ¬p]:
(p ∧ ¬p) ∨ (q ∧ ¬p) = F ∨ (¬p ∧ q) = ¬p ∧ q.

Rewrite the implication [¬p ∧ q] -> ¬q using the equivalence A -> B ≡ ¬A ∨ B:
¬(¬p ∧ q) ∨ ¬q
Applying De Morgan's law:
(p ∨ ¬q) ∨ ¬q = p ∨ ¬q.

This expression is a single maxterm containing both literals p and q, which is its Principal Conjunctive Normal Form (PCNF).
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Question 53
Consider the poset ({3, 5, 9, 15, 24, 45}, |).
Which of the following is correct for the given poset?
1. AThere exists a greatest element and a least element
2. BThere exists a greatest element but not a least element
3. CThere exists a least element but not a greatest element
4. DThere does not exist a greatest element and a least element
Answer: (D) There does not exist a greatest element and a least element
Explanation
In a partially ordered set (poset) ordered by divisibility a | b:

A least element must divide every element in the set. Here, 3 does not divide 5, and 5 does not divide 3, so no element divides all others. Hence, no least element exists (minimal elements are 3 and 5).

A greatest element must be a multiple of every element in the set. The maximal elements are 24 and 45. Neither divides the other, so no unique upper bound exists in the set.

Therefore, neither a greatest element nor a least element exists.
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Question 54
Which of the following problems is/are decidable problem(s) (recursively enumerable) on turing machine M?
(a) G is a CFG with L(G) = φ
(b) There exist two TMs M₁ and M₂ such that L(M) ⊆ {L(M₁) ∪ L(M₂)} = language of all TMs
(c) M is a TM that accepts ω using at most 2^{|ω|} cells of tape
1. A(a) and (b) only
2. B(a) only
3. C(a), (b) and (c)
4. D(c) only
Answer: (B) (a) only
Explanation
Analyzing decidability across the given problems:

(a) Emptiness of a Context-Free Grammar (L(G) = φ) is decidable. The standard marking algorithm systematically checks whether any sequence of grammar productions can derive terminal strings from the start variable.

(b) Language containment and universality properties for general Turing machines are undecidable by Rice's theorem.

(c) While tape-bounded acceptance is decidable when formulated for linear bounded automata, general unbounded Turing machine decision problems are undecidable.

Only problem (a) is decidable.
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Question 55
How many address lines and data lines are required to provide a memory capacity of 16K × 16?
1. A10, 4
2. B16, 16
3. C14, 16
4. D4, 16
Answer: (C) 14, 16
Explanation
Evaluating memory bus widths from the given capacity specification 16K x 16:

The second term gives the word size (data bus width):
Each word stores 16 bits, requiring 16 data lines.

The first term gives the total number of words:
16K words = 16 * 1024 = 2^4 * 2^10 = 2^14 words.
Addressing 2^14 separate locations requires 14 address lines.

Thus, 14 address lines and 16 data lines are required.
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Question 56
In relational databases, if relation R is in BCNF, then which of the following is true about relation R?
1. AR is in 4NF
2. BR is not in 1NF
3. CR is in 2NF and not in 3NF
4. DR is in 2NF and 3NF
Answer: (D) R is in 2NF and 3NF
Explanation
Relational normal forms follow a strict containment hierarchy:

1NF ⊇ 2NF ⊇ 3NF ⊇ BCNF ⊇ 4NF ⊇ 5NF.

Boyce-Codd Normal Form (BCNF) is a stricter variant of Third Normal Form that requires every determinant in a functional dependency to be a superkey. Any relation in BCNF is guaranteed to satisfy all conditions for 3NF, 2NF, and 1NF.

It does not automatically satisfy 4NF, which requires resolving multivalued dependencies.
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Question 57
Match List-I with List-II:
where L₁: Regular language
L₂: Context-free language
L₃: Recursive language
L₄: Recursively enumerable language
List-I List-II
(a) L̄₃ ∪ L₄ (i) Context-free language
(b) L̄₂ ∪ L₃ (ii) Recursively enumerable language
(c) L₁* ∩ L₂ (iii) Recursive language
1. A(a)-(ii); (b)-(i); (c)-(iii)
2. B(a)-(ii); (b)-(iii); (c)-(i)
3. C(a)-(iii); (b)-(i); (c)-(ii)
4. D(a)-(i); (b)-(ii); (c)-(iii)
Answer: (B) (a)-(ii); (b)-(iii); (c)-(i)
Explanation
Analyzing closure properties across Chomsky hierarchy levels:

(a)-(ii), (b)-(iii), (c)-(i)

(a) L̄₃ ∪ L₄ — recursive languages are closed under complement, so L̄₃ is recursive and hence recursively enumerable (RE). The union of two RE languages is RE, matching (ii).

(b) L̄₂ ∪ L₃ — every context-free language is recursive. The complement of a CFL is recursive, and the union of two recursive languages is recursive, matching (iii).

(c) L₁ ∩ L₂* — regular languages are closed under Kleene star, so L₁* is regular. The intersection of a context-free language with a regular language is context-free, matching (i).
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Question 58
K-mean clustering algorithm has clustered the given 8 observations into 3 clusters after 1st iteration as follows:
C1: {(3, 3), (5, 5), (7, 7)}
C2: {(0, 6), (6, 0), (3, 0)}
C3: {(8, 8), (4, 4)}

What will be the Manhattan distance for observation (4, 4) from cluster centroid C1 in second iteration?
1. A2
2. B√2
3. C0
4. D18
Answer: (A) 2
Explanation
Calculating the centroid of cluster C1 and Manhattan distance:

Cluster C1 contains {(3, 3), (5, 5), (7, 7)}.
The centroid coordinates are the mean of these points:
x_c1 = (3 + 5 + 7) / 3 = 15 / 3 = 5
y_c1 = (3 + 5 + 7) / 3 = 15 / 3 = 5
Centroid C1 = (5, 5).

The Manhattan (L1) distance from point (4, 4) to centroid (5, 5) is:
|x1 - x2| + |y1 - y2| = |4 - 5| + |4 - 5| = 1 + 1 = 2.
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Question 59
Reinforcement learning can be formalized in terms of ______ in which the agent initially only knows the set of possible ______ and the set of possible actions.
1. AMarkov decision processes, objects
2. BHidden states, objects
3. CMarkov decision processes, states
4. Dobjects, states
Answer: (C) Markov decision processes, states
Explanation
Reinforcement learning models agent-environment interactions through the formal framework of Markov Decision Processes (MDPs).

An MDP is formally defined by states, actions, transition dynamics, and rewards. In an unknown environment, the learning agent starts knowing only the set of possible states and the set of executable actions, discovering transition models and reward values through trial-and-error exploration.
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Question 60
There are many sorting algorithms based on comparison. The running time of heapsort algorithm is O(n lg n). Like P, but unlike Q, heapsort sorts in place where (P, Q) is equal to
1. AMerge sort, Quick sort
2. BQuick sort, insertion sort
3. CInsertion sort, Quick sort
4. DInsertion sort, Merge sort
Answer: (D) Insertion sort, Merge sort
Explanation
In-place sorting algorithms sort elements without requiring extra auxiliary storage proportional to the input size (requiring O(1) auxiliary space):

Heapsort is strictly an in-place comparison sort.

Insertion sort operates in-place using O(1) auxiliary memory, matching property P.

Merge sort requires O(n) auxiliary array space for merging, meaning it is not in-place, matching property Q.

Hence, like Insertion sort, but unlike Merge sort, heapsort sorts in place.
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Question 61
Consider the Euler’s phi function given by
φ(n) = n ∏(1 - 1/p)
where p runs over all the primes dividing n. What is the value of φ(45)?
1. A3
2. B12
3. C6
4. D24
Answer: (D) 24
Explanation
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Question 62
Consider the following properties with respect to a flow network G = (V, E), in which a flow is a real-valued function f : V × V → R:
P₁: For all u, v ∈ V, f(u, v) = -f(v, u)
P₂: ∑ᵤ ∈ V f(u, v) = 0 for all u ∈ V

Which one of the following is/are correct?
1. AOnly P₁
2. BOnly P₂
3. CBoth P₁ and P₂
4. DNeither P₁ nor P₂
Answer: (A) Only P₁
Explanation
Evaluating network flow properties:

P1 (Skew symmetry) states that f(u, v) = -f(v, u) for all vertex pairs, which is a standard defining property of network flows.

P2 (Flow conservation) states that total net flow is zero, but this holds only for intermediate nodes V \ {s, t}. Source vertex s produces net positive outflow, and sink vertex t absorbs net positive inflow, violating the claim that conservation holds for all vertices.

Therefore, only property P1 is correct.
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Question 63
For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to
1. Anon-uniform distribution of requests
2. Barm starting or stopping inertia
3. Chigher capacity of tracks on the periphery of the platter
4. Duse of unfair arm scheduling policies
Answer: (B) arm starting or stopping inertia
Explanation
Seek time accounts for the mechanical movement of the disk head arm across cylinder tracks.

The physical arm assembly must accelerate from rest, cruise across intermediate tracks, decelerate, and settle precisely onto the target track. Because of arm starting and stopping inertia, mechanical latency follows a non-linear trajectory (proportional to the square root of seek distance for short distances) rather than a linear rate.
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Question 64
The fault can be easily diagnosed in the micro-program control unit using diagnostic tools by maintaining the contents of
1. Aflags and counters
2. Bregisters and counters
3. Cflags and registers
4. Dflags, registers and counters
Answer: (D) flags, registers and counters
Explanation
In microprogrammed control units, microinstructions dictate hardware sequencing and micro-operations.

Diagnostic tools locate faults during execution by capturing the comprehensive architectural state:

Condition flags reflect micro-branching criteria.

Internal registers (such as the Control Address Register and Control Data Register) reflect execution pointers.

Step counters track sequencing and loop bounds.

Maintaining all three components enables complete trace diagnosis.
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Question 65
Replacing the expression 4 * 2.14 by 8.56 is known as
1. Aconstant folding
2. Binduction variable
3. Cstrength reduction
4. Dcode reduction
Answer: (A) constant folding
Explanation
Compiler optimizations simplify code constructs during intermediate code generation:

A. Constant folding — computes arithmetic expressions with known constant operands at compile time, substituting 8.56 in place of 4 * 2.14.

B. Induction variable elimination — simplifies variables that increment by fixed steps inside loops.

C. Strength reduction — substitutes computationally expensive operations with cheaper alternatives, such as replacing multiplication by addition.

D. Code reduction — generic term for eliminating redundant instructions.
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Question 66
Suppose that a connected planar graph has six vertices, each of degree four. Into how many regions is the plane divided by a planar representation of this graph?
1. A6
2. B8
3. C12
4. D20
Answer: (B) 8
Explanation
Applying Euler's planar graph formula V - E + R = 2:

The graph has V = 6 vertices, each of degree 4.
By the Handshaking lemma, the sum of vertex degrees equals 2E:
6 * 4 = 2E
24 = 2E
E = 12 edges.

Substituting into Euler's formula:
6 - 12 + R = 2
-6 + R = 2
R = 8 regions.
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Question 67
For which values of m and n does the complete bipartite graph K_{m,n} have a Hamilton circuit?
1. Am ≠ n, m, n ≥ 2
2. Bm ≠ n, m, n ≥ 3
3. Cm = n, m, n ≥ 2
4. Dm = n, m, n ≥ 3
Answer: (C) m = n, m, n ≥ 2
Explanation
In any bipartite graph, every edge alternates between the two disjoint vertex subsets V1 and V2.

A Hamiltonian circuit visits every vertex exactly once before returning to the start vertex. Because the circuit must alternate between partitions at every step, both subsets must have an equal number of vertices (m = n).

Furthermore, to form a simple cycle containing distinct vertices, each partition must contain at least 2 vertices (m, n >= 2).
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Question 68
How many different Boolean functions of degree n are there?
1. A2^(2^n)
2. B(2²)^n
3. C2^(2^n) - 1
4. D2^n
Answer: (A) 2^(2^n)
Explanation
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Question 69
Consider the following statements regarding 2D transforms in computer graphics:
S1: [1 0; 0 -1] is a 2×2 matrix that reflects (mirrors) any 2D point about the X-axis.
S2: A 2×2 matrix which mirrors any 2D point about the X-axis, is a rotation matrix.

What can you say about the statements S1 and S2?
1. ABoth S1 and S2 are true
2. BOnly S1 is true
3. COnly S2 is true
4. DBoth S1 and S2 are false
Answer: (B) Only S1 is true
Explanation
Analyzing geometric transformation matrices:

S1 is true because reflection across the X-axis leaves x unchanged (x' = x) and inverts y (y' = -y), which corresponds precisely to the transformation matrix [1 0 — 0 -1].

S2 is false because a pure 2D rotation matrix has determinant +1 and preserves coordinate orientation. The reflection matrix has determinant (1)(-1) - (0)(0) = -1, representing an orientation-reversing reflection that cannot be produced by any rotation.
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Question 70
Which of the following statements is/are true with regard to various layers in the Internet stack?
P: At the link layer, a packet of transmitted information is called a frame.
Q: At the network layer, a packet of transmitted information is called a segment.
1. AP only
2. BQ only
3. CP and Q
4. DNeither P nor Q
Answer: (A) P only
Explanation
Packet data units are named according to their architectural layer:

P is true because data link layer packets encapsulated with MAC headers and trailers are called frames.

Q is false because packets at the network layer are called datagrams (or IP packets). Segments are protocol data units formed at the transport layer (e.g. TCP segments).

Only statement P is true.
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Question 71
Which of the following is best running time to sort n integers in the range 0 to n² - 1?
1. AO(lg n)
2. BO(n)
3. CO(n lg n)
4. DO(n²)
Answer: (B) O(n)
Explanation
Sorting n numbers in the range 0 to n^2 - 1 can be achieved in linear time using Radix Sort:

Representing numbers in base n:
Each number requires at most d = 2 digits in base n because the maximum value is n^2 - 1.

Using Counting Sort as the stable sub-routine:
Each digit pass takes O(n + n) = O(n) time.
With d = 2 passes, total runtime is 2 * O(n) = O(n).

This achieves linear O(n) execution time.
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Question 72
What percentage (%) of the IPv4, IP address space do all class C addresses consume?
1. A12.5%
2. B25%
3. C37.5%
4. D50%
Answer: (A) 12.5%
Explanation
In classful IPv4 addressing, leading bit prefixes allocate total address space proportions:

Class A starts with 0 (1/2 of space = 50%).

Class B starts with 10 (1/4 of space = 25%).

Class C starts with 110 (1/8 of space = 12.5%).

Class D starts with 1110 (1/16 of space = 6.25%).

Class E starts with 1111 (1/16 of space = 6.25%).

Class C networks account for 12.5% of all IPv4 addresses.
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Question 73
Consider the equation (146)ᵦ + (313)ᵦ₋₂ = (246)₈. Which of the following is the value of b?
1. A8
2. B7
3. C10
4. D16
Answer: (B) 7
Explanation
Converting each positional representation into decimal:

Right-hand side:
(246)8 = 2 * 8^2 + 4 * 8 + 6 = 128 + 32 + 6 = 166.

Left-hand side terms:
(146)b = b^2 + 4b + 6
(313)b-2 = 3(b - 2)^2 + 1(b - 2) + 3 = 3(b^2 - 4b + 4) + b + 1 = 3b^2 - 11b + 13.

Equating sum to 166:
(b^2 + 4b + 6) + (3b^2 - 11b + 13) = 166
4b^2 - 7b + 19 = 166
4b^2 - 7b - 147 = 0.

Testing b = 7:
4(49) - 7(7) - 147 = 196 - 49 - 147 = 0.
Thus, b = 7.
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Question 74

Match List-I with List-II:
List-I (Software Process Models)
(a) Waterfall model
(b) Incremental development
(c) Prototyping
(d) RAD
List-II (Software Systems)
(i) e-business software that starts with only the basic functionalities and then moves on to more advanced features
(ii) An inventory control system for a super-market to be developed within three months
(iii) A virtual reality system for simulating vehicle navigation in a highway
(iv) Automate the manual system for student record maintenance in a school

List-I (Software Process Models)	List-II (Software Systems)
(a) Waterfall model	(iv) Automate the manual system for student record maintenance in a school
(b) Incremental development	(i) e-business software that starts with only the basic functionalities and then moves on to more advanced features
(c) Prototyping	(iii) A virtual reality system for simulating vehicle navigation in a highway
(d) RAD	(ii) An inventory control system for a super-market to be developed within three months

A(a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
B(a)-(i); (b)-(iii); (c)-(iv); (d)-(ii)
C(a)-(iii); (b)-(ii); (c)-(iv); (d)-(i)
D(a)-(iv); (b)-(i); (c)-(iii); (d)-(ii)

Answer: (D) (a)-(iv); (b)-(i); (c)-(iii); (d)-(ii)

Explanation

Matching software development process models to ideal project scenarios:

(a)-(iv), (b)-(i), (c)-(iii), (d)-(ii)

(a) Waterfall model — fits well-defined, stable systems with predictable requirements like automating an existing manual student record system, matching (iv).
(b) Incremental development — suits staged delivery where core functionality ships first and advanced features follow incrementally, matching (i).
(c) Prototyping — clarifies ambiguous user interaction and novel technologies through early exploratory models like virtual reality navigation, matching (iii).
(d) RAD — delivers functional systems under strict, short time constraints (typically within 60 to 90 days), matching (ii).

Question 75
Software reuse is
1. Athe process of analysing software with the objective of recovering its design and specification
2. Bthe process of using existing software artifacts and knowledge to build new software
3. Cconcerned with reimplementing legacy system to make them more maintainable
4. Dthe process of analysing software to create a representation of a higher level of abstraction and breaking software down into its parts to see how it works
Answer: (B) the process of using existing software artifacts and knowledge to build new software
Explanation
Software reuse is the software engineering discipline of building new applications using pre-existing software assets, components, architectures, and design patterns rather than constructing everything from scratch.

A and D describe reverse engineering.

C describes software re-engineering / refactoring.
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Question 76
Which of the following UNIX/Linux pipes will count the number of lines in all the files having .c and .h as their extension in the current working directory?
1. Acat *.ch | wc -l
2. Bcat *.[c-h] | wc -l
3. Ccat *.[ch] | ls -l
4. Dcat *.[ch] | wc -l
Answer: (D) cat *.[ch] | wc -l
Explanation
Evaluating Unix shell wildcards and pipe operators:

The character class glob pattern *.[ch] matches any file ending with either .c or .h in the current working directory.

The cat command concatenates and streams the file contents to standard output.

The pipe operator | redirects that output into wc -l, which counts the total number of lines.

cat *.ch looks for extension .ch, *.[c-h] matches any character from c through h, and piping into ls -l ignores incoming stdin stream data.
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Question 77
Consider the following pseudo-code fragment in which an invariant for the loop is "m*x^k = p^n and k ≥ 0" (here, p and n are integer variables that have been initialized):
```
/* Pre-conditions: p ≥ 1 ∧ n ≥ 0 */
/* Assume that overflow never occurs */
int x = p; int k = n; int m = 1;
while (k <> 0) {
  if (k is odd) then m = m*x;
  x = x*x;
  k = [k/2]; /* floor(k/2) */
}
```
Which of the following must be true at the end of the while loop?
1. Ax = p^n
2. Bm = p^n
3. Cp = x^n
4. Dp = m^n
Answer: (B) m = p^n
Explanation
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Question 78
Which of the following statements is/are true?

P: In a scripting language like JavaScript, types are typically associated with values, not variables.
Q: It is not possible to show images on a web page without the <img> tag of HTML.

Select the correct answer from the options given below:
1. AP only
2. BQ only
3. CBoth P and Q
4. DNeither P nor Q
Answer: (A) P only
Explanation
Evaluating programming language and web presentation principles:

P is true because JavaScript is dynamically typed. Variables hold references to values of any type, and type definitions attach dynamically to the runtime values themselves.

Q is false because images can be displayed without <img> tags through CSS background-image properties, SVG graphics, HTML5 <canvas>, <picture>, or embedded data URIs.

Only statement P is true.
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Question 79
Which of the following features is supported in the relational database model?
1. AComplex data-types
2. BMultivalued attributes
3. CAssociations with multiplicities
4. DGeneralization relationships
Answer: (C) Associations with multiplicities
Explanation
In relational data modeling:

First normal form requires attribute values to be strictly atomic, disallowing multivalued attributes and complex nested objects.
Generalization and inheritance hierarchies belong to object-oriented or extended entity-relationship models.

The relational model directly represents associations between entities with defined multiplicities (one-to-one, one-to-many, and many-to-many) using primary key and foreign key constraints across relations.
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Question 80
Match List-I with List-II:
List-I List-II
(a) Prim's algorithm (i) O(V³ log V)
(b) Dijkstra's algorithm (ii) O(VE²)
(c) Faster all-pairs shortest path (iii) O(E lg V)
(d) Edmonds-Karp algorithm (iv) O(V²)
1. A(a)-(ii); (b)-(iv); (c)-(i); (d)-(iii)
2. B(a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
3. C(a)-(ii); (b)-(i); (c)-(iv); (d)-(iii)
4. D(a)-(iii); (b)-(i); (c)-(iv); (d)-(ii)
Answer: (B) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii)
Explanation
Matching graph algorithms with their standard worst-case time complexities:

(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

(a) Prim's algorithm — runs in O(E lg V) using a binary heap implementation, matching (iii).

(b) Dijkstra's algorithm — runs in O(V²) when implemented with an array or adjacency matrix, matching (iv).

(c) Faster all-pairs shortest path — repeated matrix squaring evaluates all-pairs shortest paths in O(V³ log V), matching (i).

(d) Edmonds-Karp algorithm — implements Ford-Fulkerson via BFS in O(V E²), matching (ii).
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Question 81
How many bit strings of length ten either start with a 1 bit or end with two bits 00?
1. A320
2. B480
3. C640
4. D768
Answer: (C) 640
Explanation
Applying the Principle of Inclusion-Exclusion:

Let A be the set of 10-bit strings starting with 1:
Remaining 9 bits can take any value, so |A| = 2^9 = 512.

Let B be the set of 10-bit strings ending with 00:
Remaining 8 bits can take any value, so |B| = 2^8 = 256.

The intersection A ∩ B contains strings starting with 1 and ending with 00:
Remaining 10 - 3 = 7 bits can take any value, so |A ∩ B| = 2^7 = 128.

Total matching bit strings:
|A ∪ B| = |A| + |B| - |A ∩ B| = 512 + 256 - 128 = 640.
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Question 82
Consider double hashing of the form
h(k, i) = (h₁(k) + i h₂(k)) mod m
where h₁(k) = k mod m
h₂(k) = 1 + (k mod n)
where n = m - 1 and m = 701

For k = 123456, what is the difference between first and second probes in terms of slots?
1. A255
2. B256
3. C257
4. D258
Answer: (C) 257
Explanation
In double hashing, the offset between successive probe locations is given by the secondary hash value h2(k):

Given:
m = 701
n = m - 1 = 700
k = 123456.

The difference between the first probe (i = 0) and second probe (i = 1) equals h2(k):
h2(k) = 1 + (k mod n).

Computing k mod 700:
123456 / 700 = 176 with remainder 256
(176 * 700 = 123200, and 123456 - 123200 = 256).

Thus:
h2(123456) = 1 + 256 = 257 slots.
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Question 83
Consider the following methods:
M₁: Mean of maximum
M₂: Centre of area
M₃: Height method

Which of the following is/are defuzzification method(s)?
1. AOnly M₁
2. BOnly M₁ and M₂
3. COnly M₂ and M₃
4. DM₁, M₂ and M₃
Answer: (D) M₁, M₂ and M₃
Explanation
Defuzzification converts fuzzy membership sets produced by inference engines into crisp numerical values:

M1 (Mean of Maximum, MOM) calculates the average of all domain values that reach peak membership grade.

M2 (Centre of Area, COA / Centroid) calculates the geometric center of gravity across the aggregate membership curve.

M3 (Height method) evaluates crisp outputs by weighting the peak heights and center locations of individual fuzzy subsets.

All three are recognized defuzzification techniques.
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Question 84
Using the phong reflectance model, the strength of the specular highlight is determined by the angle between
1. Athe view vector and the normal vector
2. Bthe light vector and the normal vector
3. Cthe light vector and the reflected vector
4. Dthe reflected vector and the view vector
Answer: (D) the reflected vector and the view vector
Explanation
The Phong illumination model calculates specular reflection using the cosine of the angle between the viewer's line of sight and the ideal reflection path:

I_spec = I_s * k_s * (R · V)^n

Here, R is the unit vector along the direction of ideal mirror reflection of the incident light, and V is the unit vector pointing toward the observer viewpoint.
The highlight intensity depends directly on the angle between the reflected vector and the view vector.
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Question 85
Let Aα₀ denotes the α-cut of a fuzzy set A at α₀. If α₁ < α₂, then
1. AAα₁ ⊇ Aα₂
2. BAα₁ ⊃ Aα₂
3. CAα₁ ⊆ Aα₂
4. DAα₁ ⊂ Aα₂
Answer: (A) Aα₁ ⊇ Aα₂
Explanation
In fuzzy set theory, the alpha-cut A_alpha consists of all elements whose membership degree is at least alpha:
A_alpha = { x ∈ X | μ_A(x) >= alpha }.

When alpha1 < alpha2:
Any element whose membership grade satisfies μ_A(x) >= alpha2 automatically satisfies μ_A(x) >= alpha1.
Every element in A_alpha2 is therefore contained in A_alpha1.

This establishes the superset inclusion relation A_alpha1 ⊇ A_alpha2.
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Question 86
What is the output of the following JAVA program?
```
public class Good {
  private int m;
  public Good(int m) { this.m = m; }
  public Boolean equals(Good n) { return n.m == m; }
  public static void main (String args []){
    Good m₁ = new Good(22);
    Good m₂ = new Good(22);
    Object s₁ = new Good(22);
    Object s₂ = new Good(22);
    System.out.println(m₁.equals(m₂));
    System.out.println(s₁.equals(s₂));
    System.out.println(m₁.equals(s₂));
    System.out.println(s₁.equals(m₁));
  }
}
```
1. ATrue, True, False, False
2. BTrue, False, True, False
3. CTrue, True, False, True
4. DTrue, False, False, False
Answer: (D) True, False, False, False
Explanation
Tracing method overloading versus overriding in Java:

Class Good defines public Boolean equals(Good n), which overloads rather than overrides Object.equals(Object).

m₁.equals(m₂) — static types are both Good, so compiler selects Good.equals(Good). Comparing 22 == 22 evaluates to True.

s₁.equals(s₂) — static type of s1 is Object. Compiler invokes Object.equals(Object), which checks reference equality. Since s1 and s2 are separate instances, this evaluates to False.

m₁.equals(s₂) — argument s2 has static type Object. The compiler cannot bind to equals(Good), resolving to inherited Object.equals(Object). Different references yield False.

s₁.equals(m₁) — static type of s1 is Object. Invokes Object.equals(Object). Different references yield False.

Output is True, False, False, False.
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Question 87
Consider the following grammar:
S → XY
X → YaY | a and Y → bbX

Which of the following statements is/are true about the above grammar?
(a) Strings produced by the grammar can have consecutive three a's.
(b) Every string produced by the grammar have alternate a and b.
(c) Every string produced by the grammar have at least two a's.
(d) Every string produced by the grammar have b's in multiple of 2.
1. A(a) only
2. B(b) and (c) only
3. C(d) only
4. D(c) and (d) only
Answer: (D) (c) and (d) only
Explanation
Analyzing derivations for grammar S -> XY where X -> YaY | a and Y -> bbX:

Non-terminal Y only introduces terminal b's in pairs via the rule Y -> bbX. Every occurrence of b in any generated string is produced in blocks of two, guaranteeing that total b's are a multiple of 2. Thus statement (d) is true.

Any derivation of S must terminate non-terminal symbols X and Y. Non-terminal X derives at least one a, while Y derives bbX, whose X must also derive at least one a. Every valid sentence therefore contains at least two a's. Thus statement (c) is true.

Consecutive three a's cannot occur because Y always intervenes between recursive expansions.

Statements (c) and (d) are true.
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Question 88
The parallel bus arbitration technique uses an external priority encoder and a decoder. Suppose, a parallel arbiter has 5 bus arbiters. What will be the size of priority encoder and decoder respectively?
1. A4×2, 2×4
2. B2×4, 4×2
3. C3×8, 8×3
4. D8×3, 3×8
Answer: (D) 8×3, 3×8
Explanation
In centralized parallel bus arbitration circuits:

Five bus request inputs must be encoded into a unique binary priority identifier.
Since 2² = 4 < 5 <= 2³ = 8, at least 3 bits are required to encode 5 lines.
The smallest standard priority encoder accommodating 5 lines is an 8-to-3 (8x3) priority encoder.

The external decoder takes the 3-bit priority code and decodes it back to assert one of the bus grant lines.
A 3-to-8 (3x8) decoder performs this decoding function.

Hence, the sizes are 8x3 for the encoder and 3x8 for the decoder.
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Question 89
Which of the following is an example of unsupervised neural network?
1. ABack-propagation network
2. BHebb network
3. CAssociative memory network
4. DSelf-organizing feature map
Answer: (D) Self-organizing feature map
Explanation
Neural network architectures are classified by learning paradigm:

A. Back-propagation network — relies on supervised learning using error backpropagation computed from target output labels.

B. Hebb network — adjusts weights based on co-activation correlations.

C. Associative memory — associates patterns using stored vector matrices.

D. Self-organizing feature map (Kohonen SOM) — uses competitive unsupervised learning to map multi-dimensional input data onto a low-dimensional grid without supervisory feedback.
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Question 90
Which of the following are NOT shared by the threads of the same process?
(a) Stack
(b) Registers
(c) Address space
(d) Message queue
1. A(a) and (d)
2. B(b) and (c)
3. C(a) and (b)
4. D(a), (b) and (c)
Answer: (C) (a) and (b)
Explanation
Thread-level resource isolation versus sharing in multithreaded systems:

Threads within the same process share common process resources:

Address space (c)

Global variables and static memory

Operating system resources such as open file descriptors and IPC message queues (d)

Each thread maintains its own private execution state:

Stack (a) for local activation frames and function calls

Register set (b) including the program counter and stack pointer

Thus, stack and registers are not shared across threads.
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Question 91
Consider the following two statements with respect to IPv4 in computer networking:
P: The loopback (IP) address is a member of class B network.
Q: The loopback (IP) address is used to send a packet from host to itself.

What can you say about the statements P and Q?
1. AP-True; Q-False
2. BP-False; Q-True
3. CP-True; Q-True
4. DP-False; Q-False
Answer: (B) P-False; Q-True
Explanation
Evaluating IPv4 loopback network properties:

P is false because the entire 127.0.0.0/8 address block is designated for loopback testing. In classful addressing, first-octet values from 0 through 127 fall under Class A, not Class B (Class B ranges from 128 to 191).

Q is true because loopback packets short-circuit within the host network stack, allowing a machine to transmit IP packets directly to its own services without placing frames onto physical network media.

Thus, P is false and Q is true.
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Question 92
Which of the following statements are DML statements?
(a) Update [tablename]
Set [columnname] = VALUE
(b) Delete [tablename]
(c) Select * from [tablename]
1. A(a) and (b)
2. B(a) and (d)
3. C(a), (b) and (c)
4. D(b) and (c)
Answer: (C) (a), (b) and (c)
Explanation
Data Manipulation Language (DML) commands retrieve and modify table data in relational databases:

(a) UPDATE modifies existing column values within table rows.

(b) DELETE removes specific tuples from a table.

(c) SELECT queries and retrieves data sets from relations (classified as DML or Data Query Language sub-category).

All three statements are DML operations.
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Question 93
A Web application and its support environment has not been fully fortified against attack. Web engineers estimate that the likelihood of repelling an attack is only 30 percent. The application does not contain sensitive or controversial information, so the threat probability is 25 percent. What is the integrity of the web application?
1. A0.625
2. B0.725
3. C0.775
4. D0.825
Answer: (D) 0.825
Explanation
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Question 94
Consider the following statements:
S₁: For any integer n > 1, a^φ(n) ≡ 1 (mod n) for all a ∈ Z_n*, where φ(n) is Euler's phi function.
S₂: If p is prime, then a^p ≡ 1 (mod p) for all a ∈ Z_p*.

Which one of the following is/are correct?
1. AOnly S₁
2. BOnly S₂
3. CBoth S₁ and S₂
4. DNeither S₁ nor S₂
Answer: (A) Only S₁
Explanation
Evaluating modular arithmetic theorems:

S1 is Euler's Totient Theorem: for any integer n > 1 and any integer a coprime to n (a ∈ Z_n*), a^phi(n) ≡ 1 (mod n). This statement is correct.

S2 is incorrect. Fermat's Little Theorem states that if p is prime and a is not divisible by p, then a^(p-1) ≡ 1 (mod p), or a^p ≡ a (mod p). It does not assert a^p ≡ 1 (mod p).

Therefore, only statement S1 is correct.
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Question 95
The minimum number of page frames that must be allocated to a running process in a virtual memory environment is determined by
1. Apage size
2. Bphysical size of memory
3. Cthe instruction set architecture
4. Dnumber of processes in memory
Answer: (C) the instruction set architecture
Explanation
In virtual memory management, a process must hold enough physical frames in memory to complete execution of any single machine instruction without generating recurrent page faults.

Because instructions may feature multiple memory references (such as indirect addressing modes referencing instructions, pointers, and operands across separate page boundaries), the minimum number of frames required is determined by the computer's Instruction Set Architecture (ISA).
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Question 96
Consider the complexity class CO - NP as the set of languages L such that L̄ ∈ NP, and the following two statements:
S₁: P ⊆ CO-NP
S₂: If NP ≠ CO-NP, then P ≠ NP

Which of the following is/are correct?
1. AOnly S₁
2. BOnly S₂
3. CBoth S₁ and S₂
4. DNeither S₁ nor S₂
Answer: (C) Both S₁ and S₂
Explanation
Evaluating structural complexity class relations:

S1 is true because class P is closed under complement (P = co-P). Since P ⊆ NP, taking complements gives co-P ⊆ co-NP, which means P ⊆ co-NP.

S2 is true by contrapositive. If P = NP, then taking complements yields co-P = co-NP. Since P = co-P, this would imply NP = co-NP. By contrapositive, if NP ≠ co-NP, then P ≠ NP.

Both statements are correct.
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Question 97
Consider three CPU intensive processes, which require 10, 20 and 30 units of time and arrive at times 0, 2 and 6 respectively. How many context switches are needed if the operating system implements a shortest remaining time first scheduling algorithm? Do not count the context switches at time zero and at the end.
1. A4
2. B2
3. C3
4. D1
Answer: (B) 2
Explanation
Simulating Shortest Remaining Time First (SRTF) scheduling across timeline t:

At t = 0: P1 arrives with burst 10 and begins running.

At t = 2: P2 arrives with burst 20. Remaining time of P1 is 8 < 20, so P1 continues.

At t = 6: P3 arrives with burst 30. Remaining time of P1 is 4 < 30, so P1 continues.

At t = 10: P1 finishes. CPU context switches to P2 (switch 1).

At t = 30: P2 finishes. CPU context switches to P3 (switch 2).

At t = 60: P3 finishes.

Excluding boundaries at start and end, exactly 2 context switches take place.
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Question 98
Which of the following has same expressive power with regard to relational query language?
(a) Relational algebra and domain relational calculus
(b) Relational algebra and tuples relational calculus
(c) Relational algebra and domain relational calculus restricted to safe expression
(d) Relational algebra and tuples relational calculus restricted to safe expression
1. A(a) and (b) only
2. B(c) and (d) only
3. C(a) and (c) only
4. D(b) and (d) only
Answer: (B) (c) and (d) only
Explanation
Codd's theorem establishes equivalence in expressive power across relational query languages:

Unrestricted relational calculi can express unsafe queries that evaluate to infinite relations.
When restricted to safe expressions, Tuple Relational Calculus (TRC) and Domain Relational Calculus (DRC) possess identical expressive power to basic Relational Algebra.

Thus, statements (c) and (d) correctly identify equivalent expressive power.
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Question 99
In the context of 3D computer graphics, which of the following statements is/are true?
P: Orthographic transformations keep parallel lines parallel.
Q: Orthographic transformations are affine transformations.

Select the correct answer from the options given below:
1. ABoth P and Q
2. BNeither P nor Q
3. COnly P
4. DOnly Q
Answer: (A) Both P and Q
Explanation
Evaluating orthographic projection characteristics in 3D computer graphics:

P is true because orthographic projections project along parallel lines onto the viewing plane, preserving parallelism so that parallel lines remain parallel.

Q is true because orthographic projections are linear transformations (followed by translation if needed), satisfying the definition of affine transformations that preserve lines and parallelism.

Both statements are true.
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Question 100
Which of the following terms best describes Git?
1. AIssue Tracking System
2. BIntegrated Development Environment
3. CDistributed Version Control System
4. DWeb-based Repository Hosting Service
Answer: (C) Distributed Version Control System
Explanation
Git is a Distributed Version Control System (DVCS) designed to record and manage changes across file trees over time:

A. Issue tracking systems — track bug reports and tasks (e.g. Jira).

B. Integrated development environments — provide code authoring and debugging tools (e.g. VS Code).

C. Distributed Version Control System — allows developers to maintain complete local clones of repositories and collaborate without mandatory central connection.

D. Hosting services — provide remote cloud repositories (e.g. GitHub).
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Attempt Paper 2 — Computer Science & Applications as a timed test

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How many questions were in the UGC NET Computer Science & Applications June 2019 exam?

Paper 1 had 50 questions and Paper 2 (Computer Science & Applications) had 100 questions, for 150 in total across 3 hours.

Is there negative marking in UGC NET?

No. Each correct answer earns 2 marks and there is no penalty for a wrong answer, so leaving a question blank is never better than attempting it.

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Answers follow the official NTA answer key for this session. If you spot a discrepancy, use the report option inside the practice test and we will review it.

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Every Computer Science & Applications PYQ on one topic

Artificial Intelligence 133 Computer System Architecture 133 Data Communication and Computer Networks 170 Data Structures and Algorithms 143 Database Management Systems 137 Discrete Structures and Optimization 173 Programming Languages and Computer Graphics 141 Software Engineering 146 System Software and Operating System 156 Theory of Computation and Compilers 167

More Computer Science & Applications practice

All 2019 sessions All Computer Science & Applications PYQs Computer Science & Applications topic-wise practice Paper 1 topic-wise practice

Practice Paper 1 from other subjects

Every subject's exam sitting has its own Paper 1 (General Aptitude) set — more real PYQs to solve.

Paper 1 · English Literature sittings Paper 1 · Economics sittings Paper 1 · Management sittings Paper 1 · Commerce sittings

Year	22"	24"	32"	40"	49"
2012	85	154	124	112	118
2013	100	136	112	94	136
2014	106	124	85	115	145
2015	115	100	160	100	85
2016	100	85	145	85	100
2017	115	70	175	55	130
2018	125	95	170	110	155

List-I	List-II
(a) Greedy best-first	(i) Minimal cost (p) + h(p)
(b) Lowest cost-first	(ii) Minimal h(p)
(c) A* algorithm	(iii) Minimal cost (p)

List-I	List-II
(a) p → q	(i) ¬(q → ¬p)
(b) p ∨ q	(ii) p ∧ ¬q
(c) p ∧ q	(iii) ¬p → q
(d) ¬(p → q)	(iv) ¬p ∨ q

List-I	List-II
(a) Disk	(i) Thread
(b) CPU	(ii) Signal
(c) Memory	(iii) File system
(d) Interrupt	(iv) Virtual address space

List-I	List-II
(a) L̄₃ ∪ L₄	(i) Context-free language
(b) L̄₂ ∪ L₃	(ii) Recursively enumerable language
(c) L₁* ∩ L₂	(iii) Recursive language

List-I	List-II
(a) Prim's algorithm	(i) O(V³ log V)
(b) Dijkstra's algorithm	(ii) O(VE²)
(c) Faster all-pairs shortest path	(iii) O(E lg V)
(d) Edmonds-Karp algorithm	(iv) O(V²)

Employee_id	Manager_id
20	40
25	40
30	35
35	20
40	45
45	25

UGC NET Computer Science & Applications June 2019 Question PaperPaper 1 & Paper 2 with Answer Key

UGC NET June 2019 at a glance

General Aptitude

Computer Science & Applications

Paper 1 — General Aptitude

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Question 35

Question 36

Question 37

Question 38

Question 39

Question 40

Question 41

Question 42

Question 43

Question 44

Question 45

Question 46

Question 47

Question 48

Question 49

Question 50