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|Computer Science & Applications · December 2019

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

The complete UGC NET December 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 December 2019 at a glance

Exam	UGC NET December 2019
Papers	Paper 1 (General Aptitude) and Paper 2 (Computer Science & Applications)
Exam date(s)	December 4, 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 December 2019 · Computer Science & Applications sitting

50 questions60 min100 marks

Practice Now

Paper 2

Computer Science & Applications

UGC NET December 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 learning outcomes are intended in teaching organized at understanding level?
1. ALonger recall and retention of facts
2. BSeeking of relationships and patterns among facts
3. CCreative construction and critical interpretation of ideas
4. DMastery of facts and information
Answer: (B) Seeking of relationships and patterns among facts
Explanation
Teaching organized at the understanding level focuses on comprehension and seeing the relationships and patterns connecting different facts and principles.

A. Longer recall and retention of facts — characteristic of the memory level of teaching (Herbartian model), which emphasizes rote memorization.

B. Seeking of relationships and patterns among facts — correct. Morrison's understanding-level model emphasizes conceptual grasp, generalization, and identifying relationships among facts.

C. Creative construction and critical interpretation of ideas — belongs to the reflective level of teaching (Hunt's model), which centers on problem-solving and independent critical inquiry.

D. Mastery of facts and information — associated with lower cognitive levels where factual recall dominates.
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Question 2
Which of the following statements differentiate teaching from learning?

(a) Teaching is a social act while learning is a personal act
(b) Teaching implies learning
(c) Teaching is like selling while learning is like buying
(d) Teaching can occur without learning taking place
(e) In teaching, influence is directed towards learning and learner, while in learning it is usually towards oneself

Choose the correct answer from the following options:
1. A(a), (c) and (e)
2. B(a), (b) and (c)
3. C(b), (c) and (d)
4. D(c), (d) and (e)
Answer: (A) (a), (c) and (e)
Explanation
Statements (a), (c), and (e) accurately capture the essential distinctions between teaching as an outward social process and learning as an internal individual process.

(a) Teaching is a social act while learning is a personal act — valid differentiator. Teaching requires interpersonal interaction between instructor and students, whereas learning takes place internally within the cognitive structure of the individual.

(b) Teaching implies learning — invalid differentiator. Teaching does not automatically guarantee that learning occurs — equating or binding them fails to distinguish the two.

(c) Teaching is like selling while learning is like buying — valid differentiator. John Dewey's classic analogy highlights teaching as the offering or facilitation of knowledge and learning as the active acquisition by the learner.

(d) Teaching can occur without learning taking place — although true in practice, this point is not paired with (a) and (e) in the standard options.

(e) In teaching, influence is directed towards learning and learner, while in learning it is usually towards oneself — valid differentiator. Teaching is an outward-directed activity aimed at modifying another person's understanding, whereas learning is self-directed cognitive growth.

Combining the valid differentiators yields statements (a), (c), and (e).
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Question 3
Which of the following are classroom related factors that influence effectiveness of teaching?

(a) Prior task related behaviour of students
(b) Adherence to linear pattern of communication by the teacher
(c) Socio-economic status of the family to which learners belong
(d) Inappropriate use of technological resources by the teacher
(e) Cultural background of students

Choose your answer from the following options:
1. A(a), (b) and (c)
2. B(b), (c) and (d)
3. C(a), (b) and (d)
4. D(c), (d) and (e)
Answer: (C) (a), (b) and (d)
Explanation
Classroom-related factors are immediate pedagogical, communicative, and behavioral variables operating directly inside the learning space during instructional delivery.

(a) Prior task related behaviour of students — includes entry-level readiness, attention span, and task orientation inside the classroom environment.

(b) Adherence to linear pattern of communication by the teacher — represents a rigid one-way communication style within the classroom that stifles interactive feedback.

(c) Socio-economic status of the family to which learners belong — an external socioeconomic home variable rather than an active classroom instructional dynamic.

(d) Inappropriate use of technological resources by the teacher — directly hinders instructional delivery and student engagement during classroom activities.

(e) Cultural background of students — an overarching demographic and contextual attribute rather than an immediate classroom process variable.

Selecting the direct classroom variables yields statements (a), (b), and (d).
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Question 4
In which of the following types of learning materials, the presentations are highly structured and individualised?
1. ATextbooks prescribed by the universities
2. BJournals and the articles recommended for readings
3. CWritings of great thinkers selected for reflective readings
4. DProgrammed instructional and modular learning material
Answer: (D) Programmed instructional and modular learning material
Explanation
Programmed instruction and modular materials are deliberately designed to present small, logically sequenced units with immediate feedback and self-paced individual progression.

A. Textbooks prescribed by the universities — provide broad, standardized reference chapters intended for cohort study rather than individualized self-paced pacing.

B. Journals and the articles recommended for readings — research-oriented literature organized around academic arguments rather than systematic, step-by-step instructional progressions.

C. Writings of great thinkers selected for reflective readings — intended for open-ended, reflective, and philosophical analysis without rigid pre-programmed branching.

D. Programmed instructional and modular learning material — correct. Grounded in Skinnerian behavioral principles and modular design, offering highly structured frames and personalized pacing.
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Question 5
Which of the following types of assessment is conducted periodically with an eye on standards?
1. AFormative assessment
2. BSummative assessment
3. CPortfolio assessment
4. DPerformance assessment
Answer: (B) Summative assessment
Explanation
Summative assessment is administered at predetermined intervals to measure cumulative achievement against established institutional standards and benchmarks.

A. Formative assessment — ongoing and diagnostic, conducted during instruction to provide continuous feedback rather than periodic evaluation against terminal standards.

B. Summative assessment — correct. Conducted periodically at the end of instructional units or semesters to certify student attainment and evaluate performance against defined curricular standards.

C. Portfolio assessment — a continuous collection of student work demonstrating growth and reflection over time.

D. Performance assessment — measures hands-on execution of specific practical tasks or competencies rather than periodic standard-referenced auditing.
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Question 6
A university teacher plans to improve the study habits of students in his/her class. Which type of research paradigm will be helpful in this regard?
1. AEvaluative research paradigm
2. BFundamental research paradigm
3. CApplied research paradigm
4. DAction research paradigm
Answer: (D) Action research paradigm
Explanation
Action research is conducted by practitioners to solve immediate local problems and enhance professional practice within their own specific institutional setting.

A. Evaluative research paradigm — assesses the overall impact, efficacy, or worth of an established program or policy.

B. Fundamental research paradigm — aimed at formulating new theories and expanding pure knowledge without immediate practical application.

C. Applied research paradigm — tests and applies theoretical principles to broader real-world problems across diverse general settings.

D. Action research paradigm — correct. Follows a cyclical process (plan, act, observe, reflect) enabling a classroom teacher to intervene directly and improve students' study habits.
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Question 7
A researcher intends to make use of ICT in his/her research. What considerations should weigh most in such a decision?

(a) Appropriateness of the tool used
(b) Cost involved in procuring it
(c) Availability of tools in the department where research is to be undertaken
(d) Willingness of his/her supervisor to offer help
(e) The company from which the ICT equipment has been procured

Choose your answer from the following options:
1. A(a), (b) and (c)
2. B(c), (d) and (e)
3. C(a). (c) and (d)
4. D(b), (c) and (e)
Answer: (A) (a), (b) and (c)
Explanation
Adopting ICT in research requires balancing methodological suitability, budgetary constraints, and institutional infrastructure availability.

(a) Appropriateness of the tool used — essential methodological consideration ensuring the hardware or software matches the research design and data requirements.

(b) Cost involved in procuring it — vital practical factor governing financial viability and resource allocation within the project budget.

(c) Availability of tools in the department where research is to be undertaken — critical operational determinant of accessibility and day-to-day feasibility.

(d) Willingness of his/her supervisor to offer help — personal relational factor rather than an objective technical or infrastructural requirement.

(e) The company from which the ICT equipment has been procured — commercial brand identity that has no bearing on research validity.

Combining the objective infrastructural and methodological criteria yields statements (a), (b), and (c).
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Question 8
Which of the following sampling procedures will be appropriate for conducting researches with empirico-inductive research paradigm?
1. ASimple random sampling procedure
2. BSystematic sampling procedure
3. CStratified sampling procedure
4. DAny of the non-probability sampling procedures
Answer: (D) Any of the non-probability sampling procedures
Explanation
An empirico-inductive paradigm is qualitative and exploratory, aiming to generate concepts and grounded theory from rich observed particulars rather than testing deductive hypotheses.

A. Simple random sampling procedure — a probability sampling technique used in hypothetico-deductive quantitative studies to ensure statistical generalizability.

B. Systematic sampling procedure — a probability method requiring a fixed sampling frame, characteristic of deductive survey research.

C. Stratified sampling procedure — a probability design structuring subgroups for quantitative comparison and population representation.

D. Any of the non-probability sampling procedures — correct. Purposive, quota, snowball, and convenience sampling provide the contextual depth and theoretical relevance required in inductive, qualitative research.
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Question 9
In which of the following formats 'Chapter Scheme of the document has to be formally approved by a research degree committee in the university?
1. AThesis/dissertation
2. BSeminar papers
3. CResearch articles
4. DResearch monographs
Answer: (A) Thesis/dissertation
Explanation
The Research Degree Committee (RDC) or Board of Studies in a university formally evaluates and approves the synopsis and proposed chapter scheme for a doctoral thesis or master's dissertation.

A. Thesis/dissertation — correct. Formal statutory degree requirements demand institutional RDC review and approval of the research design, methodology, and chapter outline.

B. Seminar papers — evaluated by conference organizers and peer reviewers without institutional university RDC statutory approval.

C. Research articles — submitted directly to academic journals and governed by editorial boards and blind peer review.

D. Research monographs — comprehensive scholarly treatises reviewed and accepted by academic publishers rather than university degree committees.
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Question 10
A research scholar while submitting his/her thesis, did not acknowledge in the preface the help and support of the respondents of questionnaires used. This will be called an instance of
1. ATechnical lapse
2. BWilful negligence
3. CUnethical act
4. DLack of respect
Answer: (C) Unethical act
Explanation
Research ethics requires recognizing all human contributors who provided data and cooperation, making the deliberate omission of respondents an unethical act.

A. Technical lapse — refers to minor formatting, typographical, or procedural errors rather than neglecting research participants.

B. Wilful negligence — indicates careless oversight, but in academic governance the failure to credit human participants is recognized as an ethical breach.

C. Unethical act — correct. Academic codes of conduct mandate due acknowledgment of participant contributions and transparency regarding primary data sources.

D. Lack of respect — a social or etiquette evaluation rather than the formal research-methodology classification applied in academic ethics.
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Question 11
When there is an animated discussion between a teacher and his or her students in the classroom, it can be classified as:
1. AHorizontal communication
2. BMechanical communication
3. CLinear communication
4. DCategorical communication
Answer: (A) Horizontal communication
Explanation
An animated, participatory classroom discussion establishes an open two-way exchange where teacher and learners interact on a conversational level, functioning as horizontal communication.

A. Horizontal communication — correct. Characterized by reciprocal dialogue, democratic interaction, and shared participation where participants engage as active conversational equals.

B. Mechanical communication — refers to automated, machine-mediated message transmission lacking interpersonal adaptability.

C. Linear communication — unidirectional, top-down transmission from teacher to passive students without feedback.

D. Categorical communication — dogmatic or rigid statement transmission based on fixed classifications.
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Question 12
Information overload in a classroom environment by a teacher will lead to
1. AHigh level participation
2. BSemantic precision
3. CEffective impression
4. DDelayed feedback
Answer: (D) Delayed feedback
Explanation
Excessive information exceeding students' cognitive processing capacity forces learners to spend extra time decoding content, which results in delayed feedback.

A. High level participation — cognitive exhaustion suppresses active student engagement and questions.

B. Semantic precision — dense message bombardment increases noise, ambiguity, and misunderstanding.

C. Effective impression — cognitive overload causes fatigue and disorientation rather than positive impact.

D. Delayed feedback — correct. Students require additional time to assimilate and filter overflowing inputs before responding.
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Question 13
A verbal communication technique used in teaching is
1. ASlow expression of words
2. BVarying the speed of voice and tone
3. CPresentation without pause
4. DResorting to semantic jugglery
Answer: (B) Varying the speed of voice and tone
Explanation
Vocal modulation through varying speech pace and tone maintains student attention, signals emphasis, and enhances instructional clarity.

A. Slow expression of words — an unnaturally slow, flat delivery induces boredom and disinterest.

B. Varying the speed of voice and tone — correct. Dynamic paralinguistic variations in pitch, pace, and volume retain learner interest and highlight crucial concepts.

C. Presentation without pause — continuous rapid delivery prevents cognitive processing and reflection.

D. Resorting to semantic jugglery — confusing wordplay and pretentious jargon obscure instructional meaning.
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Question 14
Which of the following modes of communication can be employed in a classroom for effective teaching?

(a) Top-down
(b) Iconic
(c) Associational
(d) Dissociational
(e) Symbolic
(f) Abstract

Choose the most appropriate option from the following:
1. A(a), (b) and (f)
2. B(c), (e) and (f)
3. C(b), (c) and (e)
4. D(a), (c) and (d)
Answer: (C) (b), (c) and (e)
Explanation
Effective classroom communication employs visual representations, conceptual linkages, and symbolic language systems to facilitate learning.

(a) Top-down — hierarchical, one-way dissemination that reduces active learner engagement.

(b) Iconic — employs visual images, diagrams, models, and illustrations to make concepts concrete.

(c) Associational — links new concepts to prior knowledge and everyday experiences to enhance retention.

(d) Dissociational — disrupts connections between ideas, impeding comprehension.

(e) Symbolic — uses verbal codes, written words, and mathematical symbols for structured academic discourse.

(f) Abstract — unsupported abstraction without concrete grounding confuses introductory learners.

Combining the constructive representational modes yields statements (b), (c), and (e).
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Question 15
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R:

Assertion (A): Use of slang in formal teaching makes communication lively and interesting.
Reason (R): Academic decency demands the avoidance of slang in the classroom environment.

In the light of the above statements, choose the correct option:
1. ABoth (A) and (R) are true and (R) is the correct explanation of (A)
2. BBoth (A) and (R) are true and (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
Formal pedagogical standards require professional decorum and precise academic language, making slang inappropriate in formal classroom instruction.

Assertion (A) is false — informal slang degrades instructional clarity, invites ambiguity, and undermines the rigor expected in formal educational environments.

Reason (R) is true — academic norms and classroom decorum require standard linguistic conventions to ensure respectful, professional, and universally understandable communication.

Therefore, (A) is false, but (R) is true.
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Question 16
In a certain coding language, AEIOU is written as TNHDZ. Using the same coding language, 'BFJPV' will be written as
1. AUOIEA
2. BCGKQW
3. CAEIOU
4. DWQKGC
Answer: (A) UOIEA
Explanation
The coding rule reverses the letters of the word and shifts each letter backward by one position in the alphabet.

Analysis of AEIOU:
Reversing the word yields U-O-I-E-A.
Subtracting 1 from each letter position:
U (21) - 1 = T (20)
O (15) - 1 = N (14)
I (9) - 1 = H (8)
E (5) - 1 = D (4)
A (1) - 1 = Z (26)
This generates TNHDZ.

Applying the same rule to BFJPV:
Reversing the word yields V-P-J-F-B.
Subtracting 1 from each letter position:
V (22) - 1 = U (21)
P (16) - 1 = O (15)
J (10) - 1 = I (9)
F (6) - 1 = E (5)
B (2) - 1 = A (1)
This generates UOIEA.

A. UOIEA — correct. Exactly matches the reverse and shift-minus-one transformation.

B. CGKQW — represents shifting forward by +1 without word reversal.

C. AEIOU — simply repeats the vowel string from the premise.

D. WQKGC — reflects forward shifts combined with incorrect reverse ordering.
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Question 17
If 3% of (a + b) = 7% of (ab) and 5% of (a − b) = 4% of (ab), then what percentage of b is a?
1. A47/23 %
2. B23/47 %
3. C4700/23 %
4. D2300/47 %
Answer: (C) 4700/23 %
Explanation
Converting the two percentage equations into reciprocal form yields the ratio of a to b as 47/23, which equals 4700/23 percent.

First equation:
3(a + b) = 7(ab)
Dividing both sides by ab:
3(1/b + 1/a) = 7
1/a + 1/b = 7/3

Second equation:
5(a - b) = 4(ab)
Dividing both sides by ab:
5(1/b - 1/a) = 4
-1/a + 1/b = 4/5

Let x = 1/a and y = 1/b.
Adding the equations:
2y = 7/3 + 4/5 = 47/15
y = 47/30
Subtracting the equations:
2x = 7/3 - 4/5 = 23/15
x = 23/30

Finding the percentage of b that is a:
a / b = y / x = (47/30) / (23/30) = 47/23
Percentage = (a / b) × 100 = (47/23) × 100% = 4700/23 %

A. 47/23 % — an unmultiplied ratio that omits the factor of 100.

B. 23/47 % — inverted ratio without the percentage scaling.

C. 4700/23 % — correct. Evaluates the ratio (a / b) × 100%.

D. 2300/47 % — inverts the ratio, computing (b / a) × 100% instead of (a / b) × 100%.
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Question 18
The sum of a number and its inverse is -4. The sum of their cubes is :
1. A-52
2. B52
3. C64
4. D-64
Answer: (A) -52
Explanation
Using the algebraic cubic expansion formula for a number and its reciprocal:

Identity:
(x + 1/x)^3 = x^3 + 1/x^3 + 3(x + 1/x)

Given that x + 1/x = -4:
(-4)^3 = x^3 + 1/x^3 + 3(-4)
-64 = x^3 + 1/x^3 - 12

Solving for the sum of cubes:
x^3 + 1/x^3 = -64 + 12 = -52

A. -52 — correct. Derived from -64 - (-12) = -52.

B. 52 — ignores the negative sign on the base value of -4.

C. 64 — incorrectly cubes 4 directly without accounting for the negative sign and identity expansion.

D. -64 — directly cubes the sum without subtracting the 3(x + 1/x) cross term.
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Question 19
A certain principal invested at compound interest payable yearly amounts to Rs. 10816.00 in 3 years and Rs. 11248.64 in 4 years. What is the rate of interest?
1. A3%
2. B4%
3. C4.5%
4. D5.5%
Answer: (B) 4%
Explanation
Under annual compounding, the difference in amounts between year 3 and year 4 equals one year of simple interest on the year 3 amount.

Interest earned in the 4th year:
11248.64 - 10816.00 = Rs. 432.64

Rate of interest formula:
Rate = (Interest / Principal at start of year) × 100
Rate = (432.64 / 10816.00) × 100
Rate = 432.64 / 108.16 = 4%

A. 3% — yields 324.48 in interest, understating the actual growth.

B. 4% — correct. Exactly matches 432.64 / 108.16 = 4%.

C. 4.5% — would produce 486.72 in interest.

D. 5.5% — excessive growth rate resulting in 594.88 in interest.
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Question 20
What is the exact equivalent discount of three successive discounts of 10%, 15% and 20% by sale of an article?
1. A35.8%
2. B38.5%
3. C35.5%
4. D38.8%
Answer: (D) 38.8%
Explanation
The effective discount is calculated by determining the net multiplying factor of the remaining price after each successive discount.

Net price multiplier:
(1 - 0.10) × (1 - 0.15) × (1 - 0.20)
0.90 × 0.85 × 0.80 = 0.612

Equivalent single discount:
(1 - 0.612) × 100% = 38.8%

A. 35.8% — transposition error swapping digits 8 and 5.

B. 38.5% — arithmetic miscalculation when evaluating 0.90 × 0.85.

C. 35.5% — an erroneous estimate that approximates successive compounding.

D. 38.8% — correct. Exactly matches 100% - 61.2% = 38.8%.
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Question 21
Which one of the following hetwabhasa (fallacy) is involved in the argument, "Sound is element because it is caused"?
1. AViruddha or contradictory middle
2. BSatpratipaksa or inferentially contradicted middle
3. CSadhyasama or the unproved middle
4. DBadhita or non-inferentially contradicted middle
Answer: (A) Viruddha or contradictory middle
Explanation
The fallacy of Viruddha occurs when the reason (hetu) proves the exact opposite of the conclusion (sadhya) it intends to establish.

A. Viruddha or contradictory middle — correct. In Nyaya epistemology, being caused or produced is invariably concomitant with non-eternality and impermanence, directly contradicting the sadhya.

B. Satpratipaksa or inferentially contradicted middle — occurs when another equally valid inference exists that establishes the opposite conclusion.

C. Sadhyasama or the unproved middle — occurs when the middle term is itself as unproven as the conclusion to be established.

D. Badhita or non-inferentially contradicted middle — occurs when the conclusion is disproved by stronger non-inferential evidence such as direct sensory perception.
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Question 22
Consider the following argument:

Statements: Some chairs are curtains.
All curtains are bedsheets.

Conclusion: Some chairs are bedsheets.

What is the Mood of the above proposition?
1. AIAI
2. BIIA
3. CIAA
4. DAII
Answer: (A) IAI
Explanation
The mood of a categorical syllogism is determined by the standard letter types (A, E, I, O) of its constituent premises and conclusion in order.

Premise 1: "Some chairs are curtains" is a particular affirmative statement, which is type I.

Premise 2: "All curtains are bedsheets" is a universal affirmative statement, which is type A.

Conclusion: "Some chairs are bedsheets" is a particular affirmative statement, which is type I.

Reading the three propositions in sequence gives the mood IAI.

A. IAI — correct. Identifies the ordered sequence of particular affirmative (I), universal affirmative (A), and particular affirmative (I).

B. IIA — incorrectly classifies Premise 2 as particular and the conclusion as universal.

C. IAA — misidentifies the particular conclusion as a universal affirmative.

D. AII — reverses the order of the two premises.
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Question 23
Which of the following is converse of 'Some S is P'?
1. ASome S is not P
2. BSome P is not S
3. CSome P is S
4. DNo P is S
Answer: (C) Some P is S
Explanation
Conversion is an immediate inference that transposes the subject and predicate terms while preserving the quality and truth value.

Original proposition: "Some S is P" is a particular affirmative (I) proposition.

For an I proposition, simple conversion is valid by interchanging subject and predicate: "Some P is S".

A. Some S is not P — represents sub-contrary or obverse relationships rather than conversion.

B. Some P is not S — introduces a negative quality, changing the proposition type to O.

C. Some P is S — correct. Preserves the particular affirmative type (I) while transposing subject and predicate.

D. No P is S — a universal negative (E) proposition, contradicting the original particular affirmative.
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Question 24
Which one of the following fallacious hetu (middle term) is not uniformly concomitant with the major term?
1. AAsatpratipaksa
2. BAuyatireki
3. CAnyonya-Asiddha
4. DSuyabhicara
Answer: (D) Suyabhicara
Explanation
In Nyaya logic, the fallacy of Savyabhicara (printed as Suyabhicara in the source) is the irregular or inconstant middle that fails to maintain uniform concomitance (vyapti) with the major term.

A. Asatpratipaksa — refers to a reason that lacks a counter-balancing inference rather than inconstant concomitance.

B. Auyatireki — an idiosyncratic variant of Anvaya-Vyatireki relating to positive and negative agreement methods.

C. Anyonya-Asiddha — an unproved reason (Asiddha) where the locus is mutual or non-existent.

D. Suyabhicara — correct. Designates the erratic middle term that leads to more than one conclusion because it is present both where the sadhya exists and where it is absent.
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Question 25
According to classical Indian school of logic, what is the correct sequence of steps involved in Anumana (influence)?
1. AUpanaya, Pratijna, Hetu, Udaharana, Nigmana
2. BPratijna, Hetu, Upanaya, Udaharana, Nigmana
3. CPratijna, Upanaya, Hetu, Udaharana, Nigmana
4. DPratijna, Hetu. Udaharana, Upanaya, Nigmana
Answer: (D) Pratijna, Hetu. Udaharana, Upanaya, Nigmana
Explanation
The Nyaya syllogism consists of five sequential members known as pancha-avayava:

Pratijna (the initial proposition to be proved)

Hetu (the statement of reason)

Udaharana (the universal statement with a concrete example)

Upanaya (the application of the universal rule to the specific case)

Nigamana (the final conclusion derived from the preceding steps)

A. Upanaya, Pratijna, Hetu, Udaharana, Nigmana — starts incorrectly with application rather than the proposition.

B. Pratijna, Hetu, Upanaya, Udaharana, Nigmana — swaps the positions of Upanaya and Udaharana.

C. Pratijna, Upanaya, Hetu, Udaharana, Nigmana — misplaces Upanaya before the reason and example.

D. Pratijna, Hetu. Udaharana, Upanaya, Nigmana — correct. Follows the established classical Nyaya sequence.
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Question 26
What is the overall percentage of marks obtained by Amar?
The following table gives the percentage of marks obtained by 6 students in 5 different subjects in an examination. The numbers in the parenthesis give the maximum marks in each subject. Answer the given questions based on the table:

Subject (Max. Marks) / Name of student Maths (150) Chemistry (130) Physics (120) Biology (100) English (100)
Ankit 90 50 80 60 70
Amar 100 60 70 40 80
Sanya 90 60 85 80 60
Rahul 80 75 65 90 90
Puneet 80 80 70 85 50
Pooja 70 75 50 80 40
1. A70%
2. B72%
3. C60%
4. D58.33%
Answer: (B) 72%
Explanation
Amar's overall percentage is calculated by determining his actual marks in each subject, summing them, and dividing by the total maximum marks across all subjects.

Calculation of Amar's marks:

Maths (150 max): 100% of 150 = 150 marks

Chemistry (130 max): 60% of 130 = 78 marks

Physics (120 max): 70% of 120 = 84 marks

Biology (100 max): 40% of 100 = 40 marks

English (100 max): 80% of 100 = 80 marks

Total marks obtained by Amar:
150 + 78 + 84 + 40 + 80 = 432 marks

Total maximum marks:
150 + 130 + 120 + 100 + 100 = 600 marks

Overall percentage:
(432 / 600) × 100% = 72%

A. 70% — simple arithmetic average of percentage scores (350 / 5 = 70%), which incorrectly ignores the unequal subject maximum marks.

B. 72% — correct. Weighted percentage accounting for differing subject maximum marks (432 / 600 × 100% = 72%).

C. 60% — underestimates Amar's performance by overlooking his full score in Mathematics.

D. 58.33% — an arithmetic error during aggregation.
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Question 27
What was the aggregate of marks obtained by Sanya in all the five subjects?
The following table gives the percentage of marks obtained by 6 students in 5 different subjects in an examination. The numbers in the parenthesis give the maximum marks in each subject. Answer the given questions based on the table:

Subject (Max. Marks) / Name of student Maths (150) Chemistry (130) Physics (120) Biology (100) English (100)
Ankit 90 50 80 60 70
Amar 100 60 70 40 80
Sanya 90 60 85 80 60
Rahul 80 75 65 90 90
Puneet 80 80 70 85 50
Pooja 70 75 50 80 40
1. A375
2. B395
3. C455
4. D475
Answer: (C) 455
Explanation
Sanya's aggregate marks are calculated by converting each subject percentage into absolute marks based on that subject's maximum marks.

Calculation of Sanya's marks:

Maths: 90% of 150 = 135 marks

Chemistry: 60% of 130 = 78 marks

Physics: 85% of 120 = 102 marks

Biology: 80% of 100 = 80 marks

English: 60% of 100 = 60 marks

Aggregate marks obtained by Sanya:
135 + 78 + 102 + 80 + 60 = 455 marks

A. 375 — represents the sum of raw percentage points (90 + 60 + 85 + 80 + 60 = 375) without converting to subject-weighted marks.

B. 395 — computation error omitting 60 marks from English or undercounting Physics.

C. 455 — correct. Exactly matches the sum of subject scores 135 + 78 + 102 + 80 + 60 = 455.

D. 475 — calculation error overstating Chemistry or Biology marks.
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Question 28
What is the average marks obtained by all the six students in Chemistry?
The following table gives the percentage of marks obtained by 6 students in 5 different subjects in an examination. The numbers in the parenthesis give the maximum marks in each subject. Answer the given questions based on the table:

Subject (Max. Marks) / Name of student Maths (150) Chemistry (130) Physics (120) Biology (100) English (100)
Ankit 90 50 80 60 70
Amar 100 60 70 40 80
Sanya 90 60 85 80 60
Rahul 80 75 65 90 90
Puneet 80 80 70 85 50
Pooja 70 75 50 80 40
1. A86.66
2. B76.66
3. C66.66
4. D60.00
Answer: (A) 86.66
Explanation
The average marks in Chemistry (maximum marks 130) is calculated by finding the total marks obtained by all six students and dividing by 6.

Chemistry marks per student:

Ankit: 50% of 130 = 65.0 marks

Amar: 60% of 130 = 78.0 marks

Sanya: 60% of 130 = 78.0 marks

Rahul: 75% of 130 = 97.5 marks

Puneet: 80% of 130 = 104.0 marks

Pooja: 75% of 130 = 97.5 marks

Total Chemistry marks obtained:
65.0 + 78.0 + 78.0 + 97.5 + 104.0 + 97.5 = 520.0 marks

Average marks:
520.0 / 6 = 86.666... ≈ 86.66 marks

Alternatively, sum of percentage points = 50 + 60 + 60 + 75 + 80 + 75 = 400%
Average percentage = 400% / 6 = 66.666%
Average marks = 66.666% of 130 = (400 × 130) / 600 = 520 / 6 = 86.66 marks

A. 86.66 — correct. Exactly matches 520 / 6 = 86.66 marks.

B. 76.66 — calculation mistake subtracting 10 marks from the true average.

C. 66.66 — tempting distractor representing the average percentage rather than the average marks.

D. 60.00 — arbitrary median score among the lower performers.
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Question 29
How many students obtained 60% or higher marks in all the subjects?
The following table gives the percentage of marks obtained by 6 students in 5 different subjects in an examination. The numbers in the parenthesis give the maximum marks in each subject. Answer the given questions based on the table:

Subject (Max. Marks) / Name of student Maths (150) Chemistry (130) Physics (120) Biology (100) English (100)
Ankit 90 50 80 60 70
Amar 100 60 70 40 80
Sanya 90 60 85 80 60
Rahul 80 75 65 90 90
Puneet 80 80 70 85 50
Pooja 70 75 50 80 40
1. AThree
2. BTwo
3. COne
4. DZero
Answer: (B) Two
Explanation
Evaluating each student across all five subject percentage columns against the threshold of 60% or higher:

Student-by-student verification:

Ankit: Maths 90%, Chemistry 50%, Physics 80%, Biology 60%, English 70% — disqualified due to Chemistry (50% < 60%).

Amar: Maths 100%, Chemistry 60%, Physics 70%, Biology 40%, English 80% — disqualified due to Biology (40% < 60%).

Sanya: Maths 90%, Chemistry 60%, Physics 85%, Biology 80%, English 60% — qualified, every subject is 60% or higher.

Rahul: Maths 80%, Chemistry 75%, Physics 65%, Biology 90%, English 90% — qualified, every subject is 60% or higher.

Puneet: Maths 80%, Chemistry 80%, Physics 70%, Biology 85%, English 50% — disqualified due to English (50% < 60%).

Pooja: Maths 70%, Chemistry 75%, Physics 50%, Biology 80%, English 40% — disqualified due to Physics (50%) and English (40%).

Exactly two students (Sanya and Rahul) scored 60% and above in all five subjects.

A. Three — mistakenly counts Puneet or Ankit despite their failing subjects.

B. Two — correct. Sanya and Rahul are the only two students meeting the criteria in all subjects.

C. One — overlooks either Sanya or Rahul.

D. Zero — assumes no student qualified across all five disciplines.
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Question 30
In which subject the performance of the students is worst in terms of percentage of marks?
The following table gives the percentage of marks obtained by 6 students in 5 different subjects in an examination. The numbers in the parenthesis give the maximum marks in each subject. Answer the given questions based on the table:

Subject (Max. Marks) / Name of student Maths (150) Chemistry (130) Physics (120) Biology (100) English (100)
Ankit 90 50 80 60 70
Amar 100 60 70 40 80
Sanya 90 60 85 80 60
Rahul 80 75 65 90 90
Puneet 80 80 70 85 50
Pooja 70 75 50 80 40
1. AChemistry
2. BPhysics
3. CBiology
4. DEnglish
Answer: (D) English
Explanation
Comparing the aggregate percentage points obtained across all six students for each of the four listed subjects:

Sum of percentages per subject:

Chemistry: 50 + 60 + 60 + 75 + 80 + 75 = 400 percentage points (average 66.67%)

Physics: 80 + 70 + 85 + 65 + 70 + 50 = 420 percentage points (average 70.00%)

Biology: 60 + 40 + 80 + 90 + 85 + 80 = 435 percentage points (average 72.50%)

English: 70 + 80 + 60 + 90 + 50 + 40 = 390 percentage points (average 65.00%)

English has the lowest sum and lowest average percentage among the cohort at 65.00%.

A. Chemistry — average is 66.67%, which is higher than English.

B. Physics — average is 70.00%, well above English.

C. Biology — average is 72.50%, representing the highest average among the four options.

D. English — correct. Lowest aggregate percentage score (390 points across 6 students, averaging 65.00%).
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Question 31
Which one of the following tasks is associated with changing appearance of a document in word processing?
1. AEditing
2. BFormatting
3. CWriting
4. DPrinting
Answer: (B) Formatting
Explanation
In word processing, formatting refers to modifying visual presentation and document layout without altering the textual content itself.

A. Editing — involves modifying textual content, such as adding, deleting, rephrasing, or correcting spelling and grammar.

B. Formatting — correct. Controls typography, font styles, font sizes, margins, line spacing, text alignment, and colors to alter document appearance.

C. Writing — the creative composition and drafting of the actual text.

D. Printing — generating a physical hard copy of the digital document on paper.
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Question 32
The term one gigabyte refers to:
1. A1024 petabytes
2. B1024 megabytes
3. C1024 kilobytes
4. D1024 bytes
Answer: (B) 1024 megabytes
Explanation
In binary data measurement units, computer memory and storage capacities scale in powers of 2 (2^10 = 1024).

Hierarchy of binary prefixes:

1 Kilobyte (KB) = 1024 Bytes

1 Megabyte (MB) = 1024 Kilobytes

1 Gigabyte (GB) = 1024 Megabytes

1 Terabyte (TB) = 1024 Gigabytes

1 Petabyte (PB) = 1024 Terabytes

A. 1024 petabytes — equals 1 Exabyte (EB), vastly larger than a gigabyte.

B. 1024 megabytes — correct. Exactly defines 1 Gigabyte (2^30 bytes or 1024 MB).

C. 1024 kilobytes — defines 1 Megabyte (MB).

D. 1024 bytes — defines 1 Kilobyte (KB).
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Question 33
is a wireless technology built in electronic gadgets for transferring data over short distance
1. AWiFi
2. BModem
3. CBluetooth
4. DUSB
Answer: (C) Bluetooth
Explanation
Bluetooth is a short-range wireless communication standard designed to connect and exchange data between electronic devices in a personal area network (PAN).

A. WiFi — wireless local area network (WLAN) standard providing high-speed network connectivity over wider operational areas (typically tens of meters).

B. Modem — hardware device that modulates and demodulates analog and digital communication signals, traditionally wired to external lines.

C. Bluetooth — correct. Designed specifically for short-range ad-hoc wireless data transmission between gadgets (usually within 10 meters) using low-power 2.4 GHz radio waves.

D. USB — Universal Serial Bus, a wired physical connection standard rather than wireless technology.
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Question 34
PCI stands for
1. APartial Component Interconnect
2. BPartial Component Interaction
3. CPeripheral Component Interconnect
4. DPeripheral Component Interaction
Answer: (C) Peripheral Component Interconnect
Explanation
PCI stands for Peripheral Component Interconnect, a local computer bus standard for attaching hardware devices to a computer's motherboard.

A. Partial Component Interconnect — incorrect modifier. PCI devices connect full hardware components rather than partial ones.

B. Partial Component Interaction — misidentifies the bus architecture as an interaction protocol.

C. Peripheral Component Interconnect — correct. Developed by Intel in the 1990s as a high-speed parallel bus connecting peripheral expansion cards (sound, video, network) to the microprocessor.

D. Peripheral Component Interaction — incorrect expansion replacing interconnect with interaction.
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Question 35
The national agency for responding to computer security incidents as and when they occur, is
1. ACAT
2. BCCA
3. CCDAC
4. DICERT
Answer: (D) ICERT
Explanation
The Indian Computer Emergency Response Team (CERT-In / ICERT) is the designated national nodal agency under the Ministry of Electronics and Information Technology (MeitY) for responding to cybersecurity threats and security breaches.

A. CAT — refers to the Central Administrative Tribunal or Centre for Advanced Technology, not cybersecurity incident response.

B. CCA — Controller of Certifying Authorities, which licenses and regulates digital signature certificate authorities under the IT Act.

C. CDAC — Centre for Development of Advanced Computing, an R&D organization specializing in supercomputing and IT hardware/software development.

D. ICERT — correct. Established under Section 70B of the Information Technology Act to handle incident collection, analysis, alerts, and cybersecurity emergency response.
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Question 36
Identify from the options given below, the co-benefit of Montreal Protocol
1. AImpetus to development of energy efficient systems
2. BReduction in carbon dioxide (equivalent) emissions
3. CConvergence of efforts of international community in addressing air pollution
4. DControl of transboundary movement of hazardous waste
Answer: (B) Reduction in carbon dioxide (equivalent) emissions
Explanation
The primary objective of the Montreal Protocol is phasing out ozone-depleting substances (ODSs), but because substances like CFCs and HCFCs have high global warming potential, their reduction yielded an enormous climate co-benefit in avoided greenhouse gas emissions.

A. Impetus to development of energy efficient systems — energy efficiency co-benefits arose as an indirect later outcome (notably under the Kigali Amendment for refrigeration) rather than the primary treaty co-benefit.

B. Reduction in carbon dioxide (equivalent) emissions — correct. Phasing out potent heat-trapping halogenated gases avoided gigatons of CO2-equivalent emissions, making the Montreal Protocol one of the most effective historical climate mitigation agreements.

C. Convergence of efforts of international community in addressing air pollution — air pollution is addressed under treaties like the LRTAP Convention rather than the Montreal Protocol.

D. Control of transboundary movement of hazardous waste — regulated by the Basel Convention.
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Question 37
Under Green India Mission. how many hectares of degraded forest land is to be brought under afforestation?
1. A02 million hectares
2. B06 million hectares
3. C08 million hectares
4. D20 million hectares
Answer: (B) 06 million hectares
Explanation
The National Mission for a Green India (GIM) targets 10 million hectares of total forest and non-forest land, within which improving the quality of degraded forest ecosystems covers 6 million hectares.

Mission targets breakdown:

Improving quality of moderately dense and open degraded forest cover: 1.5 million hectares

Open forests with signs of degradation: 3.0 million hectares

Degraded grasslands: 0.4 million hectares

Scrub and barren lands: 1.1 million hectares
Total degraded forest and grassland area earmarked for eco-restoration and afforestation: 6 million hectares (06 million hectares).

A. 02 million hectares — far below the designated mission targets.

B. 06 million hectares — correct. Specific target for eco-restoration and quality improvement of degraded forest and associated ecosystems.

C. 08 million hectares — incorrect intermediate figure.

D. 20 million hectares — exceeds the overall 10-million-hectare total mandate of the mission.
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Question 38
The themes of some Sustainable Development Goals are

(a) Climate action
(b) Sustainable cities and communities
(c) Peace. justice and strong institutions
(d) Skill development and decent employment
(e) Green agriculture
(f) Responsible consumption and production

Choose the most appropriate from those given below:
1. A(a). (b), (c). (e) and (f)
2. B(b), (c). (e) and (f)
3. C(b). (c). (d), (e) and (f)
4. D(a), (b), (c) and (f)
Answer: (D) (a), (b), (c) and (f)
Explanation
Evaluating the listed items against the 17 official United Nations Sustainable Development Goals (SDGs):

(a) Climate action — official title of SDG 13.

(b) Sustainable cities and communities — official title of SDG 11.

(c) Peace, justice and strong institutions — official title of SDG 16.

(d) Skill development and decent employment — invalid title. SDG 8 is titled 'Decent Work and Economic Growth'.

(e) Green agriculture — invalid title. SDG 2 is titled 'Zero Hunger'.

(f) Responsible consumption and production — official title of SDG 12.

Selecting the authentic SDG titles gives statements (a), (b), (c), and (f).

A, B, C are incorrect because they include non-existent SDG titles (e) 'Green agriculture' or (d) 'Skill development and decent employment'.

D. (a), (b), (c) and (f) — correct. Identifies the four genuine SDG goals in the list.
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Question 39
The most dominant source of Benzene emissions in ambient air is
1. ACement industry
2. BCar exhausts
3. CCigarettes
4. DPaints and varnish
Accepted answers: (B) Car exhausts, (C) Cigarettes
Explanation
Benzene emissions are prominent in both outdoor ambient air and overall human exposure assessments, making car exhausts and cigarettes both defensible answers depending on the investigative context.

In ambient (outdoor) atmosphere studies, motor vehicle exhaust and fuel evaporation from car exhausts constitute the largest dominant contributor, accounting for over 70% of urban outdoor benzene emissions.

In total personal and population exposure assessments (such as the EPA Total Exposure Assessment Methodology study), cigarette smoke represents the single largest source of direct benzene intake among smokers and passive recipients.

A. Cement industry — emits particulates and carbon dioxide with minimal relative benzene contribution.

B. Car exhausts — correct. Dominant source of benzene in outdoor ambient air due to incomplete gasoline combustion.

C. Cigarettes — correct. Widely recognized in public health curricula as the dominant source of personal benzene exposure.

D. Paints and varnish — release volatile organic compounds such as toluene and xylene, with lower benzene concentrations under modern chemical standards.
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Question 40
The most important pollutants that cause degradation of water quality in rivers and streams are

(a) Bacteria
(b) Nutrients
(c) Metals
(d) Total dissolved solids
(e) Algae

Choose the most appropriate answer from the options given below:
1. A(a), (b) and (c)
2. B(a), (b) and (d)
3. C(a), (b), (d) and (e)
4. D(a), (b), (c), (d) and (e)
Answer: (A) (a), (b) and (c)
Explanation
Bacteria, excess nutrients, and heavy metals represent the primary direct anthropogenic pollutant categories responsible for severe water quality degradation in river systems.

(a) Bacteria (pathogens) — enter water bodies through untreated sewage, municipal wastewater, and livestock runoff, causing widespread biological contamination.

(b) Nutrients (nitrogen and phosphorus) — originate from agricultural fertilizers and domestic detergents, driving excessive weed growth and eutrophication.

(c) Metals — discharge from industrial effluents, mining, and urban runoff, presenting persistent toxic hazards to aquatic biotas and human health.

(d) Total dissolved solids — while an indicator of mineralization, TDS includes naturally occurring harmless mineral salts alongside contaminants.

(e) Algae — living photosynthetic organisms whose proliferation is a symptom or consequence of nutrient pollution rather than an primary originating pollutant.

Selecting the primary pollutant categories yields statements (a), (b), and (c).
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Question 41
Database WOS stands for
1. AWeb of Science
2. BWorld of Science
3. CWeb of Sources
4. DWeb of Service
Answer: (A) Web of Science
Explanation
Web of Science (WOS) is a globally recognized multidisciplinary citation indexing and bibliographic database maintained by Clarivate Analytics.

A. Web of Science — correct. Originally produced by Eugene Garfield's Institute for Scientific Information (ISI), providing comprehensive citation tracking and Journal Impact Factor metrics.

B. World of Science — plausible sounding title but not the actual name of the indexing service.

C. Web of Sources — incorrect expansion.

D. Web of Service — confusing database indexing with service-oriented computing architectures.
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Question 42
The major barriers for access to higher education in India are :

(a) more opportunities of employment for less educated
(b) government policies
(c) language of instruction
(d) economic status
(e) competition from foreign universities
(f) gender discrimination in society

Choose the correct answer from the options given below:
1. A(a), (b) and (c)
2. B(b). (c) and (e)
3. C(c). (d) and (f)
4. D(d), (e) and (f)
Answer: (C) (c). (d) and (f)
Explanation
Linguistic medium, household economic constraints, and societal gender inequities are the most pronounced structural barriers hindering equitable access to higher education in India.

(a) More opportunities of employment for less educated — incorrect. Structural youth unemployment is high among less educated individuals.

(b) Government policies — affirmative action, scholarships, and fee waivers aim to expand access rather than act as primary systemic barriers.

(c) Language of instruction — English dominance creates severe hurdles for students transitioning from vernacular-medium regional schools.

(d) Economic status — high tuition, accommodation costs, and opportunity costs severely restrict low-income family enrollment.

(e) Competition from foreign universities — foreign universities enroll a tiny fraction of elite students and do not limit domestic higher education access.

(f) Gender discrimination in society — patriarchal attitudes and early marriage pressures disproportionately suppress female higher education participation.

Selecting the structural barriers yields statements (c), (d), and (f).
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Question 43
According to GATS (General Agreement on Trade and Services), higher education should be a commodity in the
1. Adomestic public sector
2. Bdomestic private sector
3. Cnon-trading sector
4. Dglobal marketplace
Answer: (D) global marketplace
Explanation
The World Trade Organization's General Agreement on Trade in Services (GATS) classifies higher education as a commercially tradable service within the global marketplace.

Under GATS framework, educational services are treated as economic commodities open to international cross-border trade, commercial presence, and global investment across member nations.

A. Domestic public sector — traditional state-funded education model that GATS seeks to open to international private trade.

B. Domestic private sector — focuses solely on national private providers rather than the global trade regimes established under WTO.

C. Non-trading sector — GATS specifically removes higher education from non-trading exemptions by treating it as commercial service.

D. Global marketplace — correct. GATS conceptualizes education as an internationally marketable commodity traded across borders.
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Question 44
Which of the following statements is true in the Indian context?
1. AAutonomous colleges can grant degrees independent of universities
2. BAutonomous colleges can grant only bachelor's degree independent of universities
3. CExcept doctoral degrees, all other degrees and diplomas can be granted by autonomous colleges
4. DWhatever may be the degree or diploma, autonomous colleges can grant them under their own name but under the seal of an affiliated university
Answer: (D) Whatever may be the degree or diploma, autonomous colleges can grant them under their own name but under the seal of an affiliated university
Explanation
Under UGC guidelines on autonomous colleges, autonomy grants academic and administrative freedom over curriculum and examinations, but degree-awarding power remains vested in the affiliating parent university.

A, B, C are incorrect because autonomous colleges in India do not hold university status and cannot award independent academic degrees on their own authority.

D. Whatever may be the degree or diploma, autonomous colleges can grant them under their own name but under the seal of an affiliated university — correct. The diploma or degree bears the college's name, but must be issued under the formal seal, authority, and accreditation of the affiliating university.
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Question 45
Rashtriya Uchhatar Shiksha Abhiyan (RUSA) aims to achieve the following in Higher Education System:

(a) Equity
(b) Access
(c) 50% GER
(d) Excellence

Choose the correct option from those given below:
1. A(a), (b) and (c)
2. B(a), (c) and (d)
3. C(a), (b) and (d)
4. D(b), (c) and (d)
Answer: (C) (a), (b) and (d)
Explanation
Rashtriya Uchhatar Shiksha Abhiyan (RUSA) is built upon three foundational strategic pillars: Access, Equity, and Excellence in state higher education institutions.

(a) Equity — target pillar providing targeted funding to socially disadvantaged groups, women, and unserved educationally backward districts.

(b) Access — target pillar aimed at expanding institutional capacity, infrastructure, and geographical reach.

(c) 50% GER — incorrect target. RUSA 1.0 targeted raising the gross enrolment ratio to 30% by 2020 (50% GER by 2035 was later introduced by NEP 2020).

(d) Excellence — target pillar enhancing academic faculty quality, accreditation, governance, and research infrastructure.

Combining the three foundational pillars yields statements (a), (b), and (d).
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Question 46
Which of the following played a role in the liberal critique?
Read the passage carefully and answer questions that follow :

There is no doubt that the market as a reality and political economy as a theory played an important role in the liberal critique. But liberalism is neither the consequence nor the development of these: rather, the market played. in the liberal critique. the role of a "test", a locus of privileged experience where one can identify the effects of excessive governmentality and even weigh their significance: the analysis of the mechanisms of "dearth" or more generally, of the grain trade in the middle of the eighteenth century. was meant to show the point at which governing was always governing too much. Therefore, an analysis to make visible, in the form of evidence, the formation of the value and circulation of wealth-or, on the contrary, an analysis pre-supposing the intrinsic invisibility of the connection between individual profit-seeking and the growth of collective wealth-economics, in any case, shows a basic incompatibility between the optimal development of the economic process and maximisation of government procedures. It is by this. more than the play of ideas. The French or English economists broke away from mercantilism and commercialism: they freed reflection on economic practice from the hegemony of the "reason of state" and from the saturation of governmental intervention. By using it as a measure of "governing too much". they placed it at the limit of governmental action. Liberalism does not derive from juridical thought any more than it does from an economic analysis. It is not the idea of a political society, but the result of search for a liberal technology of government.
1. ALiberalism as a consequence of market forces
2. BLiberalism as an offshoot of political economy
3. CReality of market
4. DPolitical economy as a practice
Answer: (C) Reality of market
Explanation
The passage states explicitly in its opening sentence that 'the market as a reality and political economy as a theory played an important role in the liberal critique.'

A. Liberalism as a consequence of market forces — refuted by the passage, which affirms that 'liberalism is neither the consequence nor the development of these'.

B. Liberalism as an offshoot of political economy — contradicted by the text, which clarifies that liberalism did not develop directly from political economy.

C. Reality of market — correct. Directly matches the author's statement that the market as a reality played an important role.

D. Political economy as a practice — misidentifies political economy, which the passage labels as a 'theory' rather than a practice.
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Question 47
The liberal critique examined the implications of
Read the passage carefully and answer questions that follow :

There is no doubt that the market as a reality and political economy as a theory played an important role in the liberal critique. But liberalism is neither the consequence nor the development of these: rather, the market played. in the liberal critique. the role of a "test", a locus of privileged experience where one can identify the effects of excessive governmentality and even weigh their significance: the analysis of the mechanisms of "dearth" or more generally, of the grain trade in the middle of the eighteenth century. was meant to show the point at which governing was always governing too much. Therefore, an analysis to make visible, in the form of evidence, the formation of the value and circulation of wealth-or, on the contrary, an analysis pre-supposing the intrinsic invisibility of the connection between individual profit-seeking and the growth of collective wealth-economics, in any case, shows a basic incompatibility between the optimal development of the economic process and maximisation of government procedures. It is by this. more than the play of ideas. The French or English economists broke away from mercantilism and commercialism: they freed reflection on economic practice from the hegemony of the "reason of state" and from the saturation of governmental intervention. By using it as a measure of "governing too much". they placed it at the limit of governmental action. Liberalism does not derive from juridical thought any more than it does from an economic analysis. It is not the idea of a political society, but the result of search for a liberal technology of government.
1. Amarket expansion
2. Btoo much governance
3. Cgrowth of political economy
4. Dpolitics of marketisation
Answer: (B) too much governance
Explanation
The text explains that the market functioned as a test to identify and evaluate the adverse consequences of excessive governmentality and over-regulation.

Passage evidence:
'...the role of a "test", a locus of privileged experience where one can identify the effects of excessive governmentality and even weigh their significance... meant to show the point at which governing was always governing too much.'

A. Market expansion — an economic occurrence, not the primary governance flaw investigated by the critique.

B. Too much governance — correct. Directly rephrases 'excessive governmentality' and the condition where 'governing was always governing too much'.

C. Growth of political economy — the theoretical discipline used to analyze governance rather than the object of critique.

D. Politics of marketisation — contemporary terminology not used in the passage.
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Question 48
What kind of evidence was needed to make the liberal critique visible?
Read the passage carefully and answer questions that follow :

There is no doubt that the market as a reality and political economy as a theory played an important role in the liberal critique. But liberalism is neither the consequence nor the development of these: rather, the market played. in the liberal critique. the role of a "test", a locus of privileged experience where one can identify the effects of excessive governmentality and even weigh their significance: the analysis of the mechanisms of "dearth" or more generally, of the grain trade in the middle of the eighteenth century. was meant to show the point at which governing was always governing too much. Therefore, an analysis to make visible, in the form of evidence, the formation of the value and circulation of wealth-or, on the contrary, an analysis pre-supposing the intrinsic invisibility of the connection between individual profit-seeking and the growth of collective wealth-economics, in any case, shows a basic incompatibility between the optimal development of the economic process and maximisation of government procedures. It is by this. more than the play of ideas. The French or English economists broke away from mercantilism and commercialism: they freed reflection on economic practice from the hegemony of the "reason of state" and from the saturation of governmental intervention. By using it as a measure of "governing too much". they placed it at the limit of governmental action. Liberalism does not derive from juridical thought any more than it does from an economic analysis. It is not the idea of a political society, but the result of search for a liberal technology of government.
1. ACirculation of wealth
2. BPre-supposing individual profit
3. CDearth in supply of grain
4. DIncompatibility of growth
Answer: (A) Circulation of wealth
Explanation
The author notes that evidence was required to demonstrate the formation of value and the movement of wealth within the economy.

Passage evidence:
'Therefore, an analysis to make visible, in the form of evidence, the formation of the value and circulation of wealth...'

A. Circulation of wealth — correct. Expressly identified in the text as the evidence that needed to be made visible through economic analysis.

B. Pre-supposing individual profit — described as a theoretical presupposition rather than the empirical evidence to be made visible.

C. Dearth in supply of grain — an eighteenth-century empirical mechanism examined to show excessive state control, not the overarching evidence needed for value formation.

D. Incompatibility of growth — an analytical deduction rather than evidentiary proof.
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Question 49
What is incompatible with optimal economic development?
Read the passage carefully and answer questions that follow :

There is no doubt that the market as a reality and political economy as a theory played an important role in the liberal critique. But liberalism is neither the consequence nor the development of these: rather, the market played. in the liberal critique. the role of a "test", a locus of privileged experience where one can identify the effects of excessive governmentality and even weigh their significance: the analysis of the mechanisms of "dearth" or more generally, of the grain trade in the middle of the eighteenth century. was meant to show the point at which governing was always governing too much. Therefore, an analysis to make visible, in the form of evidence, the formation of the value and circulation of wealth-or, on the contrary, an analysis pre-supposing the intrinsic invisibility of the connection between individual profit-seeking and the growth of collective wealth-economics, in any case, shows a basic incompatibility between the optimal development of the economic process and maximisation of government procedures. It is by this. more than the play of ideas. The French or English economists broke away from mercantilism and commercialism: they freed reflection on economic practice from the hegemony of the "reason of state" and from the saturation of governmental intervention. By using it as a measure of "governing too much". they placed it at the limit of governmental action. Liberalism does not derive from juridical thought any more than it does from an economic analysis. It is not the idea of a political society, but the result of search for a liberal technology of government.
1. APlay of ideas
2. BAbsence of commercialism
3. CPolitical society
4. DExcessive government procedures
Answer: (D) Excessive government procedures
Explanation
The passage highlights an inherent structural incompatibility between maximizing state administrative procedures and fostering healthy economic growth.

Passage evidence:
'...economics, in any case, shows a basic incompatibility between the optimal development of the economic process and maximisation of government procedures.'

A. Play of ideas — cited as a secondary philosophical factor rather than an impediment to economic development.

B. Absence of commercialism — the economists broke away from commercialism. Its absence is not framed as incompatible with growth.

C. Political society — mentioned as a concept distinct from liberalism, not as an incompatible element with economic processes.

D. Excessive government procedures — correct. Directly matches the passage phrase 'maximisation of government procedures'.
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Question 50
The passage is indicative of the author's preference to
Read the passage carefully and answer questions that follow :

There is no doubt that the market as a reality and political economy as a theory played an important role in the liberal critique. But liberalism is neither the consequence nor the development of these: rather, the market played. in the liberal critique. the role of a "test", a locus of privileged experience where one can identify the effects of excessive governmentality and even weigh their significance: the analysis of the mechanisms of "dearth" or more generally, of the grain trade in the middle of the eighteenth century. was meant to show the point at which governing was always governing too much. Therefore, an analysis to make visible, in the form of evidence, the formation of the value and circulation of wealth-or, on the contrary, an analysis pre-supposing the intrinsic invisibility of the connection between individual profit-seeking and the growth of collective wealth-economics, in any case, shows a basic incompatibility between the optimal development of the economic process and maximisation of government procedures. It is by this. more than the play of ideas. The French or English economists broke away from mercantilism and commercialism: they freed reflection on economic practice from the hegemony of the "reason of state" and from the saturation of governmental intervention. By using it as a measure of "governing too much". they placed it at the limit of governmental action. Liberalism does not derive from juridical thought any more than it does from an economic analysis. It is not the idea of a political society, but the result of search for a liberal technology of government.
1. Aeconomic hegemony of individuals
2. Blimit government control of economics
3. Cseek liberalism from juridical thought
4. Dpromote individual profits
Answer: (B) limit government control of economics
Explanation
The central thesis of the passage argues that economic analysis freed reflection from pervasive state intervention and established clear limits on governmental action.

Passage evidence:
'...they freed reflection on economic practice from the hegemony of the "reason of state" and from the saturation of governmental intervention. By using it as a measure of "governing too much". they placed it at the limit of governmental action.'

A. Economic hegemony of individuals — the text critiques hegemony generally and does not advocate replacing state hegemony with individual dominance.

B. Limit government control of economics — correct. Reflects the core argument of establishing boundaries to prevent governmental overreach in economic life.

C. Seek liberalism from juridical thought — contradicted by the passage statement that 'Liberalism does not derive from juridical thought'.

D. Promote individual profits — profit-seeking is analyzed as a mechanism connecting to collective wealth, not as the primary normative goal of the author.
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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 the following statements:

A. The running time of dynamic programming algorithm is alway θ(p) where p is number of subproblems.
B. When a recurrence relation has cyclic dependency, it is impossible to use that recurrence relation (unmodified) in a correct dynamic program.
C. For a dynamic programming algorithm, computing all values in a bottom-up fashion is asymptotically faster than using recursion and memorization.
D. If a problem X can be reduced to a known NP-hard problem, then X must be NP-hard.

Which of the statement(s) is (are) true?
1. AOnly B and A
2. BOnly B
3. COnly B and C
4. DOnly B and D
Answer: (B) Only B
Explanation
Evaluating each statement regarding dynamic programming and complexity:

A — false. The total running time of a dynamic programming algorithm is the sum of times across all subproblems, which is θ(p × t) where t is the time needed to compute each subproblem, not simply θ(p).

B — correct. Dynamic programming requires subproblem dependencies to form a directed acyclic graph. Cyclic dependencies create circular references that cannot be evaluated without modifying the recurrence.

C — false. Bottom-up tabulation and top-down memoization share the exact same asymptotic complexity θ(p × t), differing only in recursion-call overhead and constant factors.

D — false. Reducing X to an NP-hard problem only proves that X is no harder than that problem. To establish NP-hardness for X, the reduction must be directed from a known NP-hard problem to X.
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Question 2
Consider the following statements with respect to the language L = {aⁿbⁿ | n ≥ 0}:

Statement I: L² is context free language.
Statement II: Lᵏ is context-free language for any given k ≥ 1.
Statement III: L̅ and L* are context free languages.

Which one of the following is correct?
1. AOnly Statement I and Statement II
2. BOnly Statement I and Statement III
3. COnly Statement II and Statement III
4. DStatement I, Statement II and Statement III
Answer: (D) Statement I, Statement II and Statement III
Explanation
The language L = {aⁿbⁿ | n ≥ 0} is a deterministic context-free language (DCFL):

Statement I and Statement II — context-free languages are closed under concatenation. Therefore L² = L · L and Lᵏ for any integer k ≥ 1 are context-free.

Statement III — DCFLs are closed under complementation. Because L is accepted by a deterministic pushdown automaton, its complement L̅ is also a DCFL and hence context-free. Furthermore, CFLs are closed under Kleene star, so L* is context-free.

Because Statements I, II, and III are all true, option D is correct.
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Question 3
Given two tables EMPLOYEE (EID, ENAME, DEPTNO) and DEPARTMENT (DEPTNO, DEPTNAME), find the most appropriate statement of the query:
```
Select count (*) Total
from EMPLOYEE
where DEPTNO IN (D1, D2)
group by DEPTNO
having count (*) > 5
```
1. ATotal number of employees in each department D1 and D2
2. BTotal number of employees of department D1 and D2 if their total is > 5
3. CDisplay total number of employees in both departments D1 and D2
4. DThe output of the query must have atleast two rows
Answer: (A) Total number of employees in each department D1 and D2
Explanation
Breaking down the SQL clauses:

where DEPTNO IN (D1, D2) restricts the rows to employees belonging to either department D1 or D2.

group by DEPTNO partitions the filtered rows by department, calculating the employee count separately for each department.

having count (*) > 5 filters the grouped partitions, returning only departments with more than 5 employees.

A — correct. Reflects the per-department aggregation of employees for departments D1 and D2.

B and C — incorrect because they describe a single combined total across both departments rather than separate counts grouped by department.

D — incorrect because the query returns zero, one, or two rows depending on which departments exceed 5 employees.
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Question 4
Consider the following statements with respect to approaches to fill area on raster systems:

P. To determine the overlap intervals for scan lines that cross the area
Q. To start from a given interior position and paint outward from this point until a encounter the specified boundary conditions

Select the correct answer:
1. AP only
2. BQ only
3. CBoth P and Q
4. DNeither P nor Q
Answer: (C) Both P and Q
Explanation
Raster graphics algorithms use two standard approaches to fill interior regions:

P — describes scan-line polygon filling, where intersection spans along successive scan lines intersecting polygon boundaries are calculated.

Q — describes seed fill algorithms (boundary-fill and flood-fill), which start from an interior pixel and propagate outward until boundary conditions are met.

Because both P and Q are standard area-fill techniques in computer graphics, option C is correct.
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Question 5
Which of the following is not needed by an encryption algorithm used in Cryptography?
1. AKEY
2. BMessage
3. CCiphertext
4. DUser details
Answer: (D) User details
Explanation
An encryption algorithm is a mathematical function that converts plaintext into ciphertext using a secret key:

KEY — required as the parameter controlling the cryptographic transformation.

Message — required as the input plaintext to be encrypted.

Ciphertext — required as the resulting encrypted output produced by the cipher.

User details — not needed by the mathematical encryption algorithm itself, making option D the correct answer.
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Question 6
Let the population of chromosomes in genetic algorithm is represented in terms of binary number. The strength of fitness of a chromosome in decimal form, x, is given by:

S f(x) = f(x) / Σ f(x), where f(x) = x²

The population is given by P where:
P = {(01101), (11000), (01000), (10011)}

The strength of fitness of chromosome (11000) is:
1. A24
2. B576
3. C14.4
4. D49.2
Answer: (D) 49.2
Explanation
Convert binary chromosomes to decimal values:

01101 = 13

11000 = 24

01000 = 8

10011 = 19

Calculate fitness f(x) = x² for each chromosome:

f(13) = 13² = 169

f(24) = 24² = 576

f(8) = 8² = 64

f(19) = 19² = 361

Total population fitness:
Σ f(x) = 169 + 576 + 64 + 361 = 1170

Strength of fitness of chromosome (11000):
S f(24) = 576 / 1170 ≈ 0.4923

Expressed as a percentage, this equals 49.2%.
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Question 7
A fuzzy conjunction operators, t(x,y), and a fuzzy disjunction operator, s(x,y), form a pair if they satisfy:
t(x,y) = 1 - s(1-x, 1-y)

If t(x,y) = xy/(x+y-xy) then s(x,y) is given by:
1. A(x+y)/(1-xy)
2. B(x+y-2xy)/(1-xy)
3. C(x+y-xy)/(1-xy)
4. D(x+y-xy)/(1+xy)
Answer: (B) (x+y-2xy)/(1-xy)
Explanation
By De Morgan duality for fuzzy norms:
s(x, y) = 1 - t(1 - x, 1 - y)

Let a = 1 - x and b = 1 - y. Then:
t(a, b) = ab / (a + b - ab)

Compute numerator and denominator:
ab = (1 - x)(1 - y) = 1 - x - y + xy
a + b - ab = (1 - x) + (1 - y) - (1 - x - y + xy) = 1 - xy

Therefore:
t(1 - x, 1 - y) = (1 - x - y + xy) / (1 - xy)

Now calculate s(x, y):
s(x, y) = 1 - (1 - x - y + xy) / (1 - xy) = ((1 - xy) - (1 - x - y + xy)) / (1 - xy) = (x + y - 2xy) / (1 - xy)
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Question 8
Which of the following binary codes for decimal digits are self complementing?

A. 8421 code
B. 2421 code
C. excess-3 code
D. excess-3 gray code

Choose the correct option:
1. AA and B
2. BB and C
3. CC and D
4. DD and A
Answer: (B) B and C
Explanation
A decimal code is self-complementing if the bitwise 1s complement of code(d) produces code(9 - d) for all decimal digits d:

A. 8421 code — not self-complementing. Sum of weights is 8 + 4 + 2 + 1 = 15, not 9.

B. 2421 code — self-complementing. The sum of its weights is 2 + 4 + 2 + 1 = 9, satisfying the necessary and sufficient condition for weighted codes.

C. excess-3 code — self-complementing. The 1s complement of (d + 3) yields 15 - (d + 3) = 12 - d = (9 - d) + 3, which is the representation of 9 - d.

D. excess-3 gray code — not self-complementing.

Thus, only B and C are self-complementing.
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Question 9
Consider the following grammars:

G₁: S → aSb | bSa | SS | ε
G₂: S → aSb | bSa | SS | a
G₃: S → aSb | bSa | SS | SSS | λ

Which of the following is correct w.r.t. the above grammars?
1. AG₁ and G₂ are equivalent
2. BG₁ and G₃ are equivalent
3. CG₂ and G₃ are equivalent
4. DG₁, G₂ and G₃ are equivalent
Answer: (B) G₁ and G₃ are equivalent
Explanation
Analyzing the languages generated by each grammar:

G₁ generates all strings over {a, b} containing an equal number of a's and b's: L(G₁) = {w ∈ {a, b}* | n_a(w) = n_b(w)}.

G₂ includes the production S → a, allowing it to generate strings with unequal counts of a's and b's, so L(G₂) ≠ L(G₁).

G₃ adds S → SSS to G₁, which is redundant because two applications of S → SS already produce SSS. With λ representing the empty string, G₃ generates the exact same language of equal a's and b's as G₁.

Therefore, G₁ and G₃ are equivalent.
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Question 10
Consider a paging system where translation look aside buffer (TLB) a special type of associative memory is used with hit ratio of 80%.
Assume that memory reference takes 80 nanoseconds and reference time to TLB is 20 nanoseconds. What will be the effective memory access time given 80% hit ratio?
1. A110 nanoseconds
2. B116 nanoseconds
3. C200 nanoseconds
4. D100 nanoseconds
Answer: (B) 116 nanoseconds
Explanation
Calculate access times for hit and miss cases:

On a TLB hit (probability 0.80):
Access time = TLB time + Main memory access time = 20 + 80 = 100 ns

On a TLB miss (probability 0.20):
Access time = TLB time + Page table lookup time + Main memory access time = 20 + 80 + 80 = 180 ns

Effective Memory Access Time:
EMAT = 0.80 × 100 + 0.20 × 180
EMAT = 80 + 36 = 116 ns
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Question 11
When using Dijkstra’s algorithm to find shortest path in a graph, which of the following statement is not true?
1. AIt can find shortest path within the same graph data structure
2. BEvery time a new node is visited, we choose the node with smallest known distance/cost (weight) to visit first
3. CShortest path always passes through least number of vertices
4. DThe graph needs to have a non-negative weight on every edge
Answer: (C) Shortest path always passes through least number of vertices
Explanation
Evaluating Dijkstra's algorithm properties:

C. Shortest path always passes through least number of vertices — false. Dijkstra's algorithm minimizes the total sum of edge weights, not the number of hops. A path containing several light edges can have a smaller total weight than a single-edge path with a heavy weight.

A — true. Dijkstra's algorithm operates directly on adjacency list or matrix representations of the graph.

B — true. The algorithm greedily extracts the unvisited vertex with the minimum tentative distance.

D — true. Dijkstra requires non-negative edge weights to guarantee that once a node is finalized, its shortest distance cannot decrease.
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Question 12

Match List-I with List-II.

List-I	List-II
(a) Physical layer	(i) Provide token management service
(b) Transport layer	(ii) Concerned with transmitting raw bits over a communication channel
(c) Session layer	(iii) Concerned with the syntax and semantics of the information transmitted
(d) Presentation layer	(iv) True end-to-end layer from source to destination

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

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

Explanation

Mapping OSI model layers to their primary functions:

(a) Physical layer maps to (ii). Responsible for transmitting raw bit streams over the physical transmission medium.
(b) Transport layer maps to (iv). Provides true end-to-end communication from source to destination.
(c) Session layer maps to (i). Manages dialog control and token management services between cooperating applications.
(d) Presentation layer maps to (iii). Handles syntax and semantics of transferred information, including encoding, encryption, and compression.

The resulting mapping is (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).

Question 13
Find the minimum number of tables required for converting the entity relationship diagram into a relational database.
Entity-Relationship diagram with entities R1 and R2 connected by relationship R12
1. A2
2. B4
3. C3
4. D5
Answer: (C) 3
Explanation
Converting the entity-relationship components into relational tables:

Entity R1 contains attributes A, B, and C, where B is a multivalued attribute (represented by a double oval). Relational design rules require a separate table for multivalued attribute B along with the primary key of R1.

Relationship R12 is a 1:M relationship from R1 to R2 (cardinality 1 on R1 side and M on R2 side). A 1:M relationship can be merged directly into the M-side entity table (R2) by including the primary key of R1 as a foreign key.

Base table for entity R1 with its atomic attributes A and C.

This yields exactly 3 tables:

Table for R1 (A, C)

Table for multivalued attribute B (Key(R1), B)

Table for R2 merged with relationship R12 (X, Y, FK_R1)
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Question 14
Consider the following statements:

Statement I: There exists no algorithm for deciding if any two Turing machines M₁ and M₂ accept the same language.
Statement II: Let M₁ and M₂ be arbitrary Turing machines. The problem to determine L(M₁) ⊆ L(M₂) is undecidable.

Which of the statements is (are) correct?
1. AOnly Statement I
2. BOnly Statement II
3. CBoth Statement I and Statement II
4. DNeither Statement I nor Statement II
Answer: (C) Both Statement I and Statement II
Explanation
Evaluating decidability for Turing machines:

Statement I — the language equivalence problem for Turing machines, asking whether L(M₁) = L(M₂), is undecidable. By Rice's theorem or reduction from the Post Correspondence Problem, no general algorithm can decide equivalence.

Statement II — the language containment or subset problem, asking whether L(M₁) ⊆ L(M₂), is also undecidable for arbitrary Turing machines.

Because both statements are true, option C is correct.
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Question 15
The Reduced Instruction Set Computer (RISC) characteristics are:

A. Single cycle instruction execution
B. Variable length instruction formats
C. Instructions that manipulates operands in memory
D. Efficient instruction pipeline

Choose the correct characteristics from the options:
1. AA and B
2. BB and C
3. CA and D
4. DC and D
Answer: (C) A and D
Explanation
Analyzing core RISC architecture principles:

A. Single cycle instruction execution — characteristic of RISC due to hardwired control and uniform instruction timing.

B. Variable length instruction formats — characteristic of CISC. RISC employs fixed-length instruction formats to simplify instruction decoding.

C. Instructions that manipulate operands in memory — characteristic of CISC. RISC uses a load-store architecture where arithmetic instructions operate only on register operands.

D. Efficient instruction pipeline — characteristic of RISC, facilitated by uniform instruction length and single-cycle execution.

Thus, statements A and D are RISC characteristics.
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Question 16
A non-pipelined system takes 30ns to process a task. The same task can be processed in a four-segment pipeline with a clock cycle of 10ns. Determine the speed up of the pipeline for 100 tasks.
1. A3
2. B4
3. C3.91
4. D2.91
Answer: (D) 2.91
Explanation
Calculating total execution times for 100 tasks:

Non-pipelined system:
T_nonpipe = 100 × 30 ns = 3000 ns

Four-segment pipeline with clock cycle τ = 10 ns:
Total cycles = (k + n - 1) = (4 + 100 - 1) = 103 cycles
T_pipe = 103 × 10 ns = 1030 ns

Speedup calculation:
Speedup = T_nonpipe / T_pipe = 3000 / 1030 ≈ 2.9126

Rounding to two decimal places yields 2.91.
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Question 17
Give asymptotic upper and lower bound for T(n) given below. Assume T(n) is constant for n ≤ 2.

T(n) = 4T(√n) + lg²n
1. AT(n) = θ(lg(lg²n) lg n)
2. BT(n) = θ(lg²n lg n)
3. CT(n) = θ(lg²n lg lg n)
4. DT(n) = θ(lg(lg n) lg n)
Answer: (C) T(n) = θ(lg²n lg lg n)
Explanation
Solve by substitution of variable:
Let n = 2ᵐ, so m = lg n and √n = 2^(m/2).

The recurrence becomes:
T(2ᵐ) = 4T(2^(m/2)) + m²

Define S(m) = T(2ᵐ):
S(m) = 4S(m/2) + m²

Apply the Master Theorem to S(m) = 4S(m/2) + m²:

a = 4, b = 2

m^(log_b a) = m^(log₂ 4) = m²

f(m) = m² = θ(m^(log_b a))

This falls into Case 2 of the Master Theorem:
S(m) = θ(m² lg m)

Substitute back m = lg n:
T(n) = θ((lg n)² lg(lg n)) = θ(lg²n lg lg n)
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Question 18
Given two tables R1(x, y) and R2(y, z) with 50 and 30 number of tuples respectively. Find maximum number of tuples in the output of natural join between tables R1 and R2 i.e. R1 * R2? (* - Natural Join)
1. A30
2. B20
3. C50
4. D1500
Answer: (D) 1500
Explanation
In natural join R1(x, y) ⋈ R2(y, z), rows join on the common attribute y:

The maximum output size occurs when all 50 tuples in R1 share the identical value for attribute y, and all 30 tuples in R2 also share that exact same value.

Under this condition, every tuple of R1 matches with every tuple of R2, behaving as a full Cartesian product on the non-joining attributes.

Maximum tuples = 50 × 30 = 1500.
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Question 19
Consider the following statements:

Statement I: ∀x P(x) ∨ ∀x Q(x) and ∀x (P(x) ∨ Q(x)) are not logically equivalent.
Statement II: ∃x P(x) ∧ ∃x Q(x) and ∃x (P(x) ∧ Q(x)) are not logically equivalent.

Which of the following statements is/are correct?
1. AOnly Statement I
2. BOnly Statement II
3. CBoth Statement I and Statement II
4. DNeither Statement I nor Statement II
Answer: (C) Both Statement I and Statement II
Explanation
Analyzing equivalence for predicate quantifiers:

Statement I — ∀x P(x) ∨ ∀x Q(x) entails ∀x (P(x) ∨ Q(x)), but the converse does not hold. For example, over integers, every integer is either even or odd, so ∀x (Even(x) ∨ Odd(x)) is true, but it is false that all integers are even or all integers are odd. Therefore they are not logically equivalent.

Statement II — ∃x (P(x) ∧ Q(x)) entails ∃x P(x) ∧ ∃x Q(x), but the converse does not hold. For example, there exists an even integer and there exists an odd integer, but no integer is both even and odd. Therefore they are not logically equivalent.

Both statements are correct, making option C the right choice.
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Question 20

According to the ISO-9126 Standard Quality Model, match the attributes in List-I with their definitions in List-II.

List-I	List-II
(a) Functionality	(i) Relationship between level of performance and amount of resources
(b) Reliability	(ii) Characteristics related with achievement of purpose
(c) Efficiency	(iii) Effort needed to make for improvement
(d) Maintainability	(iv) Capability of software to maintain performance of software

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

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

Explanation

Matching ISO/IEC 9126 software quality characteristics with their definitions:

(a) Functionality maps to (ii). Attributes related to functions satisfying stated or implied needs (achievement of purpose).
(b) Reliability maps to (iv). Capability of software to maintain its level of performance under stated conditions for a stated period of time.
(c) Efficiency maps to (i). Relationship between the level of performance of the software and the amount of resources used.
(d) Maintainability maps to (iii). Effort needed to make specified modifications and improvements.

The matching combination is (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).

Question 21
Consider a subnet with 720 routers. If a three-level hierarchy is chosen, with eight clusters, each containing 9 regions of 10 routers, then total number of entries in hierarchical table of each router is:
1. A25
2. B27
3. C53
4. D72
Answer: (A) 25
Explanation
In hierarchical routing with three levels (routers, regions, clusters):

Entries for local routers within the router's own region = 10 entries.

Entries for other peer regions within the same cluster = 9 - 1 = 8 entries.

Entries for other peer clusters within the network = 8 - 1 = 7 entries.

Total table entries for each router:
Total = 10 + 8 + 7 = 25 entries.
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Question 22
Consider the following statements with respect to network security:

A. Message confidentiality means that the sender and the receiver expect privacy.
B. Message integrity means that the data must arrive at the receiver exactly as they were sent.
C. Message authentication means the receiver is ensured that the message is coming from the intended sender.

Which of the statements is (are) correct?
1. AOnly A and B
2. BOnly A and C
3. COnly B and C
4. DA, B and C
Answer: (D) A, B and C
Explanation
Evaluating the foundational pillars of network security:

A. Message confidentiality — correct. Ensures unauthorized parties cannot intercept or read transmission contents, guaranteeing privacy between sender and receiver.

B. Message integrity — correct. Ensures transmitted data has not been altered, modified, or corrupted in transit.

C. Message authentication — correct. Verifies the sender's identity, ensuring the message originated from the claimed source.

All three statements are correct.
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Question 23
The order of schema ?10?101? and ???0??1 are __________ and __________ respectively.
1. A5, 3
2. B5, 2
3. C7, 5
4. D8, 7
Answer: (B) 5, 2
Explanation
In Holland's Schema Theorem for genetic algorithms, the order of a schema o(H) is defined as the number of fixed positions (bits that are explicitly 0 or 1, excluding wildcard ? symbols):

In schema ?10?101?, the fixed positions are bit 2 (1), bit 3 (0), bit 5 (1), bit 6 (0), and bit 7 (1). Counting these yields 5 fixed positions, so order = 5.

In schema ???0??1, the fixed positions are bit 4 (0) and bit 7 (1). Counting these yields 2 fixed positions, so order = 2.

Thus the orders are 5 and 2 respectively.
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Question 24
Which of the component module of DBMS does rearrangement and possible ordering of operations, eliminate redundancy in query and use efficient algorithms and indexes during the execution of a query?
1. Aquery compiler
2. Bquery optimizer
3. CStored data manager
4. DDatabase processor
Answer: (B) query optimizer
Explanation
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Question 25
According to Dempster-Shafer theory for uncertainty management,
where Bel(A) denotes Belief of event A:
1. ABel(A) + Bel(¬A) ≤ 1
2. BBel(A) + Bel(¬A) ≥ 1
3. CBel(A) + Bel(¬A) = 1
4. DBel(A) + Bel(¬A) = 0
Answer: (A) Bel(A) + Bel(¬A) ≤ 1
Explanation
In Dempster-Shafer theory, belief represents the total support committed to an event based on evidence:

Unlike classical probability where P(A) + P(¬A) = 1, Dempster-Shafer theory allows uncommitted belief mass due to incomplete knowledge or ignorance.

Consequently, Bel(A) + Bel(¬A) ≤ 1, with equality holding only when there is complete certainty and zero unassigned belief mass.

Plausibility Pl(A) relates to belief by Pl(A) = 1 - Bel(¬A), so Bel(A) ≤ Pl(A), which directly confirms Bel(A) + Bel(¬A) ≤ 1.
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Question 26
The full form of ICANN is:
1. AInternet Corporation for Assigned Names and Numbers
2. BInternet Corporation for Assigned Numbers and Names
3. CInstitute of Cooperation for Assigned Names and Numbers
4. DInternet Connection for Assigned Names and Numbers
Answer: (A) Internet Corporation for Assigned Names and Numbers
Explanation
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Question 27
Consider the following grammar:

S → 0A | 0BB
A → 00A | λ
B → 1B | 11C
C → B

Which language does this grammar generate?
1. AL((00)*0 + (11)*1)
2. BL(0(11)* + 1(00)*)
3. CL((00)*0)
4. DL(0(11)*1)
Answer: (C) L((00)*0)
Explanation
Analyzing the productions of the grammar:

For production S → 0A:
A derives (00)* via A → 00A | λ. Therefore, 0A generates 0(00)* = (00)*0, producing all odd-length strings of 0s.

For production S → 0BB:
Non-terminal B has rules B → 1B | 11C and C → B. Because neither B nor C has a terminal-only base rule, B can never terminate to a string of terminal characters. Thus B is a non-generating symbol, and S → 0BB yields no terminal strings.

Consequently, the grammar generates solely L((00)*0), matching option C.
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Question 28
The time complexity to multiply two polynomials of degree n using Fast Fourier transform method is:
1. Aθ(n lg n)
2. Bθ(n²)
3. Cθ(n)
4. Dθ(lg n)
Answer: (A) θ(n lg n)
Explanation
Multiplying two polynomials of degree n using the Fast Fourier Transform involves three stages:

Evaluation: compute the Discrete Fourier Transform of each polynomial at 2n points using FFT, taking θ(n lg n) time.

Pointwise multiplication: multiply the point-value representations in θ(n) time.

Interpolation: compute the Inverse FFT to obtain the product polynomial coefficients in θ(n lg n) time.

The overall asymptotic time complexity is θ(n lg n).
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Question 29
Consider the following statements:

A. Fiber optic cable is much lighter than copper cable.
B. Fiber optic cable is not affected by power surges or electromagnetic interference.
C. Optical transmission is inherently bidirectional.

Which of the statements is (are) correct?
1. AOnly A and B
2. BOnly A and C
3. COnly B and C
4. DA, B and C
Answer: (A) Only A and B
Explanation
Evaluating fiber optic transmission characteristics:

A. Fiber optic cable is much lighter than copper cable — correct. Glass or plastic optical fibers have substantially less physical weight and bulk than copper wiring.

B. Fiber optic cable is not affected by power surges or electromagnetic interference — correct. Because signals travel as light pulses rather than electric currents, optical cables are immune to electromagnetic and radio frequency interference.

C. Optical transmission is inherently bidirectional — false. Light propagation along a single fiber strand is unidirectional. Bidirectional transmission requires two separate strands or wavelength-division multiplexing.

Therefore, only statements A and B are correct.
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Question 30
Which of the following CPU scheduling algorithms is/are supported by LINUX operating system?
1. ANon-preemptive priority scheduling
2. BPreemptive priority scheduling and time sharing CPU scheduling
3. CTime sharing scheduling only
4. DPriority scheduling only
Answer: (B) Preemptive priority scheduling and time sharing CPU scheduling
Explanation
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Question 31
Consider the following Linear programming problem (LPP):

Maximize z = x₁ + x₂
Subject to the constraints:
x₁ + 2x₂ ≤ 2000
x₁ + x₂ ≤ 1500
x₂ ≤ 600
and x₁, x₂ ≥ 0

The solution of the LPP is:
1. Ax₁ = 750, x₂ = 750, z = 1500
2. Bx₁ = 500, x₂ = 1000, z = 1500
3. Cx₁ = 1000, x₂ = 500, z = 1500
4. Dx₁ = 900, x₂ = 600, z = 1500
Answer: (C) x₁ = 1000, x₂ = 500, z = 1500
Explanation
Testing feasibility and objective value for each option:

A (x₁ = 750, x₂ = 750) — infeasible because x₂ = 750 violates x₂ ≤ 600.

B (x₁ = 500, x₂ = 1000) — infeasible because x₂ = 1000 violates x₂ ≤ 600.

C (x₁ = 1000, x₂ = 500) — feasible: 1000 + 2(500) = 2000 ≤ 2000, 1000 + 500 = 1500 ≤ 1500, and 500 ≤ 600 all hold. Objective value z = 1000 + 500 = 1500.

D (x₁ = 900, x₂ = 600) — infeasible because x₁ + 2x₂ = 900 + 1200 = 2100 violates x₁ + 2x₂ ≤ 2000.

Thus, C is the only feasible candidate.
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Question 32
In a B-Tree, each node represents a disk block. Suppose one block holds 8192 bytes. Each key uses 32 bytes. In a B-tree of order M there are M - 1 keys. Since each branch is on another disk block, we assume a branch is of 4 bytes. The total memory requirement for a non-leaf node is:
1. A32M - 32
2. B36M - 32
3. C36M - 36
4. D32M - 36
Answer: (B) 36M - 32
Explanation
In a B-tree of order M, a non-leaf node holds at most M - 1 keys and M branch pointers:

Memory for keys:
(M - 1) × 32 bytes = 32M - 32 bytes

Memory for branch pointers:
M × 4 bytes = 4M bytes

Total memory requirement:
Total = (32M - 32) + 4M = 36M - 32 bytes
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Question 33
Consider the following statements:

Statement I: If a group (G, *) is of order n, and a ∈ G is such that aᵐ = e for some integer m < n, then m must divide n.
Statement II: If a group (G, *) is of even order, then there must be an element a ∈ G such that a ≠ e and a * a = e.

Which of the statements is (are) correct?
1. AOnly Statement I
2. BOnly Statement II
3. CBoth Statement I and Statement II
4. DNeither Statement I nor Statement II
Answer: (B) Only Statement II
Explanation
Evaluating group theory statements:

Statement I — false. By Lagrange's theorem, the order of an element o(a) must divide the group order n. However, if aᵐ = e, m is merely a multiple of o(a), and m itself does not necessarily divide n.

Statement II — correct. In any finite group of even order, elements can be partitioned into pairs of mutual inverses (x, x⁻¹). Because the identity e is self-inverse and the total number of elements is even, there must exist an odd number of non-identity elements. Hence at least one non-identity element a must satisfy a = a⁻¹, meaning a * a = e.

Therefore, only Statement II is correct.
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Question 34

Match the Agile Process models with the task performed during the model.

List-I	List-II
(a) Scrum	(i) CRC cards
(b) Adaptive software development	(ii) Sprint backlog
(c) Extreme programming	(iii) <action> the <result> <by/for/of/to> a(n) <object>
(d) Feature-driven development	(iv) Time box release plan

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

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

Explanation

Matching agile methodologies with their characteristic practices:

(a) Scrum matches (ii). Uses a Sprint backlog to track prioritized user stories committed for the current sprint.
(b) Adaptive software development matches (iv). Uses time-boxed release planning to accommodate continuous speculation and learning.
(c) Extreme programming matches (i). Employs Class-Responsibility-Collaborator (CRC) cards for collaborative object-oriented design.
(d) Feature-driven development matches (iii). Expresses granular client features using the formal syntactic template <action> the <result> <by/for/of/to> a(n) <object>.

The matching combination is (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).

Question 35
The Boolean expression AB + A B̅ + A̅ C + AC is unaffected by the value of the Boolean variable __________.
1. AA
2. BB
3. CC
4. DA, B and C
Answer: (B) B
Explanation
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Question 36
Given following equation:
(142)ᵦ + (112)ᵦ₋₂ = (75)₈, find base b.
1. A3
2. B6
3. C7
4. D5
Answer: (D) 5
Explanation
Convert both sides of the equation to decimal representation:

Right-hand side in octal:
(75)₈ = 7 × 8 + 5 = 61

Left-hand side in bases b and b - 2:
(142)ᵦ = b² + 4b + 2
(112)ᵦ₋₂ = (b - 2)² + (b - 2) + 2 = (b² - 4b + 4) + b = b² - 3b + 4

Summing the expressions:
(b² + 4b + 2) + (b² - 3b + 4) = 2b² + b + 6

Equating to 61:
2b² + b + 6 = 61
2b² + b - 55 = 0
(2b + 11)(b - 5) = 0

Because a number base must be a positive integer, b = 5.
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Question 37
Let a²ᶜ mod n = (aᶜ)² mod n and a²ᶜ⁺¹ mod n = a·(aᶜ)² mod n. For a = 7, b = 17 and n = 561, what is the value of aᵇ(mod n)?
1. A160
2. B166
3. C157
4. D67
Answer: (A) 160
Explanation
Compute 7¹⁷ mod 561 using modular exponentiation with binary exponent 17 = 16 + 1:

Repeated squaring:
7¹ mod 561 = 7
7² mod 561 = 49
7⁴ mod 561 = 49² = 2401 = 4 × 561 + 157 = 157
7⁸ mod 561 = 157² = 24649 = 43 × 561 + 526 ≡ -35 mod 561
7¹⁶ mod 561 = (-35)² = 1225 = 2 × 561 + 103 = 103

Multiply terms:
7¹⁷ mod 561 = (7¹⁶ × 7¹) mod 561 = (103 × 7) mod 561 = 721 mod 561
721 - 561 = 160
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Question 38
Consider the following language families:

L₁ = The context-free languages
L₂ = The context-sensitive languages
L₃ = The recursively enumerable languages
L₄ = The recursive languages

Which one of the following options is correct?
1. AL₁ ⊂ L₂ ⊂ L₄ ⊂ L₃
2. BL₁ ⊂ L₄ ⊂ L₂ ⊂ L₃
3. CL₄ ⊂ L₁ ⊂ L₂ ⊂ L₃
4. DL₁ ⊂ L₂ ⊂ L₃ ⊂ L₄
Answer: (A) L₁ ⊂ L₂ ⊂ L₄ ⊂ L₃
Explanation
The proper hierarchy of formal languages establishes strict nested containment:

Context-free languages (L₁) are recognized by pushdown automata and generated by Type-2 grammars.

Context-sensitive languages (L₂) are recognized by linear bounded automata and strictly contain context-free languages.

Recursive languages (L₄) are decidable languages recognized by halting Turing machines and strictly contain context-sensitive languages.

Recursively enumerable languages (L₃) are semi-decidable languages recognized by general Turing machines and strictly contain recursive languages.

This yields the strict containment chain L₁ ⊂ L₂ ⊂ L₄ ⊂ L₃.
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Question 39
Consider the following statements:

A. Windows Azure is a cloud-based operating system.
B. Google App Engine is an integrated set of online services for consumers to communicate and share with others.
C. Amazon Cloud Front is a web service for content delivery.

Which of the statements is (are) correct?
1. AOnly A and B
2. BOnly A and C
3. COnly B and C
4. DA, B and C
Answer: (B) Only A and C
Explanation
Evaluating cloud computing services and classifications:

A. Windows Azure is a cloud-based operating system — correct. Microsoft introduced Windows Azure as a cloud operating system providing compute and storage infrastructure.

B. Google App Engine is an integrated set of online services for consumers to communicate and share with others — false. Google App Engine is a Platform as a Service (PaaS) framework for deploying web applications, not a consumer social sharing platform.

C. Amazon Cloud Front is a web service for content delivery — correct. Amazon CloudFront is a globally distributed content delivery network (CDN) accelerating static and dynamic web content delivery.

Hence, only statements A and C are correct.
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Question 40
Let G = (V, T, S, P) be any context-free grammar without any λ-productions or unit productions. Let K be the maximum number of symbols on the right of any production in P. The maximum number of production rules for any equivalent grammar in Chomsky normal form is given by:
1. A(K-1)|P| + |T| - 1
2. B(K-1)|P| + |T|
3. CK|P| + |T| - 1
4. DK|P| + |T|
Answer: (B) (K-1)|P| + |T|
Explanation
In Chomsky Normal Form (CNF), all productions must take the form A → BC or A → a:

For each terminal symbol a ∈ T appearing in productions with length ≥ 2, a rule T_a → a is introduced. This adds at most |T| productions.

For each original production with right-hand side length m ≤ K, decomposing it into cascaded binary variable productions requires m - 1 ≤ K - 1 rules.

Across all |P| original productions, this generates at most (K - 1)|P| binary rules.

Summing both components gives a maximum of (K - 1)|P| + |T| productions.
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Question 41

Match List-I with List-II.

List-I	List-II
(a) Isolated I/O	(i) same set of control signal for I/O & memory communication
(b) Memory mapped I/O	(ii) separate instructions for I/O and memory communication
(c) I/O interface	(iii) requires control signals to be transmitted between the communicating units
(d) Asynchronous data transfer	(iv) resolve the differences in central computer and peripherals

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

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

Explanation

Matching computer architecture I/O concepts with their operational definitions:

(a) Isolated I/O maps to (ii). Uses separate control lines and dedicated instructions (IN/OUT) distinguishing I/O ports from memory.
(b) Memory mapped I/O maps to (i). Shares a single unified address space and uses the same read/write control signals for memory and I/O.
(c) I/O interface maps to (iv). Provides hardware buffering and signal conversion to bridge speed and format differences between CPU and peripherals.
(d) Asynchronous data transfer maps to (iii). Employs independent strobe or handshake control lines between communicating entities without a shared clock.

The resulting combination is (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii).

Question 42
The Data Encryption Standard (DES) has a function consists of four steps. Which of the following is correct order of these four steps?
1. Aan expansion permutation, S-boxes, an XOR operation, a straight permutation
2. Ban expansion permutation, an XOR operation, S-boxes, a straight permutation
3. Ca straight permutation, S-boxes, an XOR operation, an expansion permutation
4. Da straight permutation, an XOR operation, S-boxes, an expansion permutation
Answer: (B) an expansion permutation, an XOR operation, S-boxes, a straight permutation
Explanation
In each round of the DES cipher, the round function f(R, K) operates on the 32-bit right half in four sequential steps:

Expansion permutation (E-box): expands the 32-bit input block to 48 bits.

Key mixing (XOR operation): XORs the 48-bit expanded block with the 48-bit round subkey.

S-box substitution: passes the 48 bits through eight non-linear S-boxes to produce 32 bits.

Straight permutation (P-box): permutes the 32 output bits before the final round XOR.

This sequence matches option B.
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Question 43
Two concurrent executing transactions T₁ and T₂ are allowed to update same stock item say ‘A’ in an uncontrolled manner. In such scenario, following problems may occur:

A. Dirty read problem
B. Lost update problem
C. Transaction failure
D. Inconsistent database state

Which of the following option is correct if database system has no concurrency module and allows concurrent execution of above two transactions?
1. AA, B and C only
2. BC and D only
3. CA and B only
4. DA, B and D only
Answer: (D) A, B and D only
Explanation
Concurrency anomalies arising from unconstrained interleaved transaction execution include:

A. Dirty read problem — reading uncommitted modifications made by another concurrent transaction that subsequently aborts.

B. Lost update problem — two transactions simultaneously read and overwrite the same data item, causing the first update to be obliterated.

D. Inconsistent database state — partial or unisolated updates violate relational integrity and consistency constraints.

Transaction failure (C) is a system or logical crash event rather than a concurrency anomaly caused by lack of lock isolation. Thus, A, B, and D only can occur.
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Question 44
Let Wᵢⱼ represents weight between node i at layer k and node j at layer (k-1) of a given multilayer perceptron. The weight updation using gradient descent method is given by:
1. AWᵢⱼ(t+1) = Wᵢⱼ(t) + α ∂E/∂Wᵢⱼ
2. BWᵢⱼ(t+1) = Wᵢⱼ(t) - α ∂E/∂Wᵢⱼ
3. CWᵢⱼ(t+1) = α ∂E/∂Wᵢⱼ
4. DWᵢⱼ(t+1) = -α ∂E/∂Wᵢⱼ, 0 ≤ α ≤ 1
Answer: (B) Wᵢⱼ(t+1) = Wᵢⱼ(t) - α ∂E/∂Wᵢⱼ
Explanation
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Question 45
A micro instruction format has microoperation field which is divided into 2 subfields F1 and F2, each having 15 distinct microoperations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field AD. The address space is of 128 memory words. The size of micro instruction is:
1. A19
2. B18
3. C17
4. D20
Answer: (A) 19
Explanation
Calculate the bit allocation for each field in the microinstruction format:

Microoperation subfields F1 and F2:
Each handles 15 operations plus a default no-op (16 total codes): ⌈log₂ 16⌉ = 4 bits each. F1 = 4 bits, F2 = 4 bits.

Condition field CD:
Selects from 4 status bits: ⌈log₂ 4⌉ = 2 bits.

Branch field BR:
Selects from 4 branch conditions: ⌈log₂ 4⌉ = 2 bits.

Address field AD:
Addresses 128 words of control memory: ⌈log₂ 128⌉ = 7 bits.

Total microinstruction length:
Length = 4 + 4 + 2 + 2 + 7 = 19 bits.
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Question 46
Piconet is a basic unit of a bluetooth system consisting of __________ master node and up to __________ active slave nodes.
1. Aone, five
2. Bone, seven
3. Ctwo, eight
4. Done, eight
Answer: (B) one, seven
Explanation
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Question 47
Which of the following algorithms is not used for line clipping?
1. ACohen-Sutherland algorithm
2. BSoutherland-Hodgeman algorithm
3. CLiang-Barsky algorithm
4. DNicholl-Lee-Nicholl algorithm
Answer: (B) Southerland-Hodgeman algorithm
Explanation
Categorizing computer graphics clipping algorithms:

Southerland-Hodgeman algorithm (Sutherland-Hodgman) — polygon clipping algorithm that clips arbitrary concave or convex polygons against clipping boundaries, not a line-clipping algorithm.

Cohen-Sutherland algorithm — line clipping algorithm utilizing outcodes for 2D clipping windows.

Liang-Barsky algorithm — line clipping algorithm based on parametric line equations.

Nicholl-Lee-Nicholl algorithm — line clipping algorithm dividing clipping regions into specialized zones to minimize intersection calculations.
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Question 48
What is the worst case running time of Insert and Extract-min, in an implementation of a priority queue using an unsorted array? Assume that all insertions can be accommodated.
1. Aθ(1), θ(n)
2. Bθ(n), θ(1)
3. Cθ(1), θ(1)
4. Dθ(n), θ(n)
Answer: (A) θ(1), θ(n)
Explanation
For a priority queue maintained as an unsorted array:

Insert — elements are appended directly to the end of the array without maintaining order, taking θ(1) worst-case time.

Extract-min — finding the minimum element requires scanning through all n elements in the unsorted array to locate and remove the minimum, taking θ(n) worst-case time.

Thus, the worst-case complexities are θ(1) and θ(n) respectively.
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Question 49
Identify the circumstances under which pre-emptive CPU scheduling is used:

A. A process switches from Running state to Ready state
B. A process switches from Waiting state to Ready state
C. A process completes its execution
D. A process switches from Ready to Waiting state

Choose the correct option:
1. AA and B only
2. BA and D only
3. CC and D only
4. DA, B, C only
Answer: (A) A and B only
Explanation
CPU scheduling algorithms make scheduling decisions under specific process state transitions:

Running to Ready (such as a timer interrupt in round-robin) — preemptive scheduling (A).

Waiting to Ready (such as I/O completion of a higher-priority task) — preemptive scheduling (B).

Running to Waiting or Process termination (completion) — non-preemptive (cooperative) scheduling.

A process cannot transition directly from Ready to Waiting without first being scheduled on the CPU (D). Thus, preemptive scheduling operates under circumstances A and B only.
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Question 50
A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?
1. A1,000,000 bps
2. B2,000,000 bps
3. C12,000,000 bps
4. D120,000,000 bps
Answer: (B) 2,000,000 bps
Explanation
Throughput measures the actual volume of data successfully transmitted per unit of time:

Total bits transmitted per minute:
Total = 12000 frames/min × 10000 bits/frame = 120000000 bits/min

Convert bits per minute to bits per second:
Throughput = 120000000 bits / 60 seconds = 2000000 bps

Thus, throughput is 2,000,000 bps (2 Mbps).
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Question 51
Which of the following methods are used to pass any number of parameters to the operating system through system calls?
1. ARegisters
2. BBlock or table in main memory
3. CStack
4. DBlock in main memory and stack
Answer: (D) Block in main memory and stack
Explanation
Operating systems employ three common techniques to pass parameters during system calls:

Registers — fastest approach, but limited by the fixed count of hardware processor registers, making it incapable of passing arbitrary numbers of parameters.

Block or table in main memory — parameters are stored in a designated memory block, and a single pointer to the block is passed in a register. This permits passing any number of parameters.

Stack — parameters are pushed onto the program stack by the calling process and popped off by the operating system kernel, which also accommodates arbitrary numbers of parameters.

Therefore, both block in memory and stack methods accommodate any number of parameters.
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Question 52
The following multithreaded algorithm computes transpose of a matrix in parallel:
```
p Trans (X, Y, N)
if N = 1
 then Y[1,1] ← X[1,1]
else partition X into four (N/2) × (N/2) submatrices X₁₁, X₁₂, X₂₁, X₂₂
 partition Y into four (N/2) × (N/2) submatrices Y₁₁, Y₁₂, Y₂₁, Y₂₂
 spawn p Trans (X₁₁, Y₁₁, N/2)
 spawn p Trans (X₁₂, Y₂₁, N/2)
 spawn p Trans (X₂₁, Y₁₂, N/2)
 spawn p Trans (X₂₂, Y₂₂, N/2)
```
What is the asymptotic parallelism of the algorithm?
1. AT₁/T∞ or θ(N²/lg N)
2. BT₁/T∞ or θ(N/lg N)
3. CT₁/T∞ or θ(lg N/N²)
4. DT₁/T∞ or θ(lg N/N)
Answer: (A) T₁/T∞ or θ(N²/lg N)
Explanation
Analyzing work T₁ and span T∞ for the divide-and-conquer parallel transpose:

Work T₁(N) (serial running time):
T₁(N) = 4T₁(N/2) + θ(1)
By Master Theorem Case 1, N^(log₂ 4) = N², so T₁(N) = θ(N²).

Span T∞(N) (critical-path length with parallel spawns):
T∞(N) = T∞(N/2) + θ(1)
By Master Theorem Case 2, T∞(N) = θ(lg N).

Parallelism:
Parallelism = T₁ / T∞ = θ(N² / lg N).
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Question 53
Consider the game tree. Here ○ and □ represent MIN and MAX nodes respectively. The value of the root node of the game tree is:
Game tree with alternating MAX and MIN levels
1. A14
2. B17
3. C111
4. D112
Answer: (B) 17
Explanation
Evaluating the minimax tree from leaf level up to the root:

Lowest MIN level (circles above leaf squares):

First MIN node selects min(17, 19) = 17.

Second MIN node receives 16.

Third MIN node selects min(111, 112) = 111.

Fifth MIN node receives 15.

Sixth MIN node receives 14.

Intermediate MAX level (squares above circles):

Left subtree first MAX node selects max(17, 16) = 17.

Left subtree second MAX node receives 111.

Right subtree first MAX node receives 15.

Right subtree second MAX node receives 14.

MIN level below root:

Left branch chooses min(17, 111) = 17.

Right branch chooses min(15, 14) = 14.

Root MAX node:
Selects max(17, 14) = 17.
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Question 54
Java Virtual Machine (JVM) is used to execute architectural neutral byte code. Which of the following is needed by the JVM for execution of Java Code?
1. AClass loader only
2. BClass loader and Java Interpreter
3. CClass loader, Java Interpreter and API
4. DJava Interpreter only
Answer: (C) Class loader, Java Interpreter and API
Explanation
Running Java bytecode on the JVM requires three essential runtime components:

Class Loader Subsystem — dynamically loads, links, and initializes compiled class files into memory.

Execution Engine (Java Interpreter / JIT Compiler) — reads bytecode instructions and executes them natively on the host processor.

Java Core API Runtime Libraries — provides core runtime classes, system calls, and utility packages required during program execution.

Because all three components are essential, option C is correct.
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Question 55
Consider the language L = {aⁿbⁿ⁻³ | n > 2} on Σ = {a, b}. Which one of the following grammars generates the language L?
1. AS → aA | a, A → aAb | b
2. BS → aaA | λ, A → aAb | λ
3. CS → aaaA | a, A → aAb | λ
4. DS → aaaA, A → aAb | λ
Answer: (D) S → aaaA, A → aAb | λ
Explanation
Analyzing strings in language L:
For n > 2, let n = k + 3 where k ≥ 0. Then strings take the form a^(k+3) b^k = aaa (a^k b^k).

Examining the grammar productions in option D:

A → aAb | λ generates the set of equal a's and b's: {a^k b^k | k ≥ 0}.

S → aaaA prepends exactly three a's to any string produced by A.

This yields strings of the form aaa a^k b^k = a^(k+3) b^k for k ≥ 0.

Setting n = k + 3 gives n > 2 and produces exactly the language L = {aⁿbⁿ⁻³ | n > 2}.
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Question 56
A computer uses a memory unit of 512 K words of 32 bits each. A binary instruction code is stored in one word of the memory. The instruction has four parts: an addressing mode field to specify one of the two-addressing mode (direct and indirect), an operation code, a register code part to specify one of the 256 registers and an address part. How many bits are there in addressing mode part, opcode part, register code part and the address part?
1. A1, 3, 9, 19
2. B1, 4, 9, 18
3. C1, 4, 8, 19
4. D1, 3, 8, 20
Answer: (C) 1, 4, 8, 19
Explanation
Calculating the bit distribution in the 32-bit instruction word:

Addressing mode: 2 modes (direct and indirect) require ⌈log₂ 2⌉ = 1 bit.

Address part: memory capacity is 512 K words = 512 × 1024 = 2¹⁹ words, requiring 19 bits.

Register code: 256 registers require ⌈log₂ 256⌉ = 8 bits.

Opcode: remaining bits = 32 - (1 + 19 + 8) = 32 - 28 = 4 bits.

Arranged in order (addressing mode, opcode, register code, address part), the bits are 1, 4, 8, 19.
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Question 57
A basic feasible solution of an m×n transportation problem is said to be non-degenerate, if basic feasible solution contains exactly __________ number of individual allocation in __________ positions.
1. Am+n+1, independent
2. Bm+n-1, independent
3. Cm+n-1, appropriate
4. Dm-n+1, independent
Answer: (B) m+n-1, independent
Explanation
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Question 58
Consider the following learning algorithms:

A. Logistic regression
B. Back propagation
C. Linear regression

Which of the following option represents classification algorithms?
1. AA and B only
2. BA and C only
3. CB and C only
4. DA, B and C
Accepted answers: (A) A and B only, (D) A, B and C
Explanation
Evaluating machine learning algorithms for classification tasks:

Logistic regression (A) is a supervised classification algorithm predicting discrete categorical outcomes via a sigmoid activation function.

Backpropagation (B) is the learning algorithm used to train feedforward neural networks, which are applied to classification.

Linear regression (C) is primarily designed for continuous numerical prediction, though thresholded outputs can serve classification.

NTA marked this question as dropped / awarded marks to all candidates because classification strictly designates model goals rather than training mechanisms like backpropagation. Both options A and D are accepted.
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Question 59
Which of the following statements are true regarding C++?

A. Overloading gives the capability to an existing operator to operate on other data types.
B. Inheritance in object oriented programming provides support to reusability.
C. When object of a derived class is defined, first the constructor of derived class is executed then constructor of a base class is executed.
D. Overloading is a type of polymorphism.

Choose the correct option from the options:
1. AA and B only
2. BA, B and C only
3. CA, B and D only
4. DB, C and D only
Answer: (C) A, B and D only
Explanation
Evaluating C++ object-oriented principles:

A — true. Operator overloading allows existing C++ operators to be given user-defined meanings on class types.

B — true. Inheritance allows derived classes to absorb fields and methods from base classes, promoting code reusability.

C — false. In C++, base class constructors execute first to initialize base components before derived class constructor bodies run.

D — true. Function and operator overloading provide compile-time (static) polymorphism.

Thus, statements A, B, and D are true.
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Question 60
A counting semaphore is initialized to 8. 3 wait() operations and 4 signal() operations are applied. Find the current value of semaphore variable.
1. A9
2. B5
3. C1
4. D4
Answer: (A) 9
Explanation
Semaphore operations modify the counter as follows:

A wait() operation decrements the semaphore counter by 1.

A signal() operation increments the semaphore counter by 1.

Calculating final value from initial value 8:
Value = 8 - 3 + 4 = 9.
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Question 61
Consider the following languages:

L₁ = {aⁿbⁿcᵐ} ∪ {aⁿbᵐcᵐ}, n, m ≥ 0
L₂ = {wwᴿ | w ∈ {a, b}*} where R represents reversible operation.

Which one of the following is (are) inherently ambiguous language(s)?
1. Aonly L₁
2. Bonly L₂
3. Cboth L₁ and L₂
4. Dneither L₁ nor L₂
Answer: (A) only L₁
Explanation
Evaluating inherent ambiguity in context-free languages:

L₁ — inherently ambiguous. Every context-free grammar generating L₁ must be ambiguous because strings of the form aⁿbⁿcⁿ belong to both components of the union, leading unavoidably to multiple parse trees.

L₂ — not inherently ambiguous. Even palindromes wwᴿ are generated unambiguously by the deterministic production rules S → aSa | bSb | ε, where each string has a single unique leftmost derivation.

Therefore, only L₁ is inherently ambiguous.
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Question 62
A tree has 2n vertices of degree 1, 3n vertices of degree 2, n vertices of degree 3. Determine the number of vertices and edges in tree.
1. A12, 11
2. B9, 8
3. C10, 11
4. D9, 10
Answer: (A) 12, 11
Explanation
Using graph theory properties for trees:

Total vertices V:
V = 2n + 3n + n = 6n
In any tree with V vertices, the number of edges is E = V - 1 = 6n - 1.

By the Handshaking Lemma, sum of degrees equals 2E:
Σ deg(v) = 2(6n - 1) = 12n - 2

Calculate degree sum from given vertices:
Σ deg(v) = (2n × 1) + (3n × 2) + (n × 3) = 2n + 6n + 3n = 11n

Equating both sums:
11n = 12n - 2
n = 2

Substitute n = 2:
V = 6(2) = 12
E = 12 - 1 = 11

Thus, there are 12 vertices and 11 edges.
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Question 63
Let P be the set of all people. Let R be a binary relation on P such that (a, b) is in R if a is a brother of b. Is R symmetric, transitive, an equivalence relation, a partial order relation?
1. ANO, NO, NO, NO
2. BNO, NO, YES, NO
3. CNO, YES, NO, NO
4. DNO, YES, YES, NO
Answer: (C) NO, YES, NO, NO
Explanation
Evaluating the relational properties of "is brother of" on people P:

Symmetric: NO. If a is a brother of b, but b is female, then b is a sister of a, not a brother. Hence (b, a) ∉ R.

Transitive: YES. If a is a brother of b and b is a brother of c, then a is male and shares parents with c, so a is a brother of c.

Equivalence relation: NO. An equivalence relation must be symmetric and reflexive, neither of which holds.

Partial order relation: NO. A partial order must be reflexive, but no person is their own brother.

Thus the sequence of answers is NO, YES, NO, NO.
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Question 64
Let A = {001, 0011, 11, 101} and B = {01, 111, 111, 010}. Similarly, let C = {00, 001, 1000} and D = {0, 11, 011}.
Which of the following pairs have a post-correspondence solution?
1. AOnly pair (A, B)
2. BOnly pair (C, D)
3. CBoth (A, B) and (C, D)
4. DNeither (A, B) nor (C, D)
Answer: (A) Only pair (A, B)
Explanation
Checking Post Correspondence Problem (PCP) solutions for both pairs:

Pair (A, B) with A = (w₁=001, w₂=0011, w₃=11, w₄=101) and B = (v₁=01, v₂=111, v₃=111, v₄=010):
Consider index sequence (3, 4, 1):

String from A: w₃ w₄ w₁ = 11 + 101 + 001 = 11101001

String from B: v₃ v₄ v₁ = 111 + 010 + 01 = 11101001
Both strings match, so (A, B) has a valid PCP solution.

Pair (C, D):
In list C, every word has strictly more 0s than 1s (00 has 2 zeros, 001 has 2 zeros, 1000 has 3 zeros). In list D, words have equal or fewer zeros than ones (11 has 0 zeros, 011 has 1 zero). No combination can ever balance the counts of 0s and 1s simultaneously, so (C, D) has no solution.

Thus, only pair (A, B) has a PCP solution.
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Question 65
An instruction is stored at location 500 with its address field at location 501. The address field has the value 400. A processor register R₁ contains the number 200. Match addressing mode in List-I with effective address in List-II for the instruction.
List-I List-II
(a) Direct (i) 200
(b) Register indirect (ii) 902
(c) Index with R₁ as the index register (iii) 400
(d) Relative (iv) 600
1. A(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
2. B(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
3. C(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
4. D(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Answer: (A) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Explanation
Calculating effective address (EA) for each addressing mode given instruction address = 500, address field = 501 with value A = 400, register R₁ = 200, and PC = 502 (next instruction):

(a) Direct — EA = Address field = 400, mapping to (iii).

(b) Register indirect — EA = Content of register R₁ = 200, mapping to (i).

(c) Index with R₁ — EA = Address field + R₁ = 400 + 200 = 600, mapping to (iv).

(d) Relative — EA = Address field + PC = 400 + 502 = 902, mapping to (ii).

The resulting mapping is (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).
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Question 66
What is the output of following C program?
```
#include <stdio.h>
main()
{
 int i, j, x = 0;
 for (i = 0; i < 5; ++i)
 for (j = 0; j < i; ++j)
 {
 x += (i + j - 1);
 break;
 }
 printf ("%d", x);
}
```
1. A6
2. B5
3. C4
4. D3
Answer: (A) 6
Explanation
Tracing variable x through outer loop iterations:
Initially x = 0.

i = 0: inner loop condition j < 0 is false. No change.

i = 1: j = 0 executes once. x += (1 + 0 - 1) = 0, then break exits inner loop. x = 0.

i = 2: j = 0 executes once. x += (2 + 0 - 1) = 1, then break exits inner loop. x = 1.

i = 3: j = 0 executes once. x += (3 + 0 - 1) = 2, then break exits inner loop. x = 1 + 2 = 3.

i = 4: j = 0 executes once. x += (4 + 0 - 1) = 3, then break exits inner loop. x = 3 + 3 = 6.

The final printf outputs 6.
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Question 67

Match List-I with List-II.

List-I	List-II
(a) Frame attribute	(i) to create link in HTML
(b) <tr> tag	(ii) for vertical alignment of content in cell
(c) Valign attribute	(iii) to enclose each row in table
(d) <a> tag	(iv) to specify the side of the table frame that display border

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

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

Explanation

Matching HTML elements and attributes with their functions:

(a) Frame attribute maps to (iv). Specifies which outer borders of the table frame should be displayed.
(b) <tr> tag maps to (iii). Encloses each horizontal row of data cells within a table.
(c) Valign attribute maps to (ii). Defines vertical alignment of content inside a table cell.
(d) <a> tag maps to (i). Anchor element used to create hyperlinks in HTML.

The resulting mapping is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

Question 68
The sequence diagram for the Weather Information System takes place when an external system requests the summarized data from the weather station. The increasing order of lifeline for the objects in the system are:
UML sequence diagram for the Weather Information System
1. ASat comms → Weather station → Comms link → Weather data
2. BSat comms → Comms link → Weather station → Weather data
3. CWeather data → Comms link → Weather station → Sat Comms
4. DWeather data → Weather station → Comms link → Sat Comms
Accepted answers: (C) Weather data → Comms link → Weather station → Sat Comms, (D) Weather data → Weather station → Comms link → Sat Comms
Explanation
Analyzing object lifeline active durations in the sequence diagram:

Weather Data has the shortest active execution bar, active only while running Summarize().

Commslink is active next, invoking and receiving results from Weather Data.

Weather Station has a broader activation window coordinating reports.

Sat Comms maintains the longest activation span across request, transmission, and acknowledgement.

NTA marked this question as dropped / awarded marks to all candidates due to phrasing ambiguity regarding vertical dotted line versus execution activation box length. In increasing order of activation duration, Weather Data has the shortest active duration while Sat Comms has the longest.
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Question 69
How many reflexive relations are there on a set with 4 elements?
1. A2⁴
2. B2¹²
3. C4²
4. D2⁹
Answer: (B) 2¹²
Explanation
For a set with n elements, a binary relation is a subset of the n² ordered pairs in Cartesian product A × A:

In a reflexive relation, all n diagonal elements (x, x) must be included.

The remaining n² - n non-diagonal pairs can either be present or absent independently.

Total reflexive relations = 2^(n² - n) = 2^(n(n-1)).

For n = 4:
Number of reflexive relations = 2^(16 - 4) = 2¹².
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Question 70
What are the greatest lower bound (GLB) and the least upper bound (LUB) of the sets A = {3, 9, 12} and B = {1, 2, 4, 5, 10} if they exist in poset (Z⁺, /)?
1. AA(GLB - 3, LUB - 36), B(GLB - 1, LUB - 20)
2. BA(GLB - 3, LUB - 12), B(GLB - 1, LUB - 10)
3. CA(GLB - 1, LUB - 36), B(GLB - 2, LUB - 20)
4. DA(GLB - 1, LUB - 12), B(GLB - 2, LUB - 10)
Answer: (A) A(GLB - 3, LUB - 36), B(GLB - 1, LUB - 20)
Explanation
In the poset (Z⁺, /) ordered by divisibility:

GLB corresponds to the greatest common divisor (GCD).

LUB corresponds to the least common multiple (LCM).

For set A = {3, 9, 12}:
GLB(A) = gcd(3, 9, 12) = 3
LUB(A) = lcm(3, 9, 12) = 36

For set B = {1, 2, 4, 5, 10}:
GLB(B) = gcd(1, 2, 4, 5, 10) = 1
LUB(B) = lcm(1, 2, 4, 5, 10) = 20

This gives A(GLB - 3, LUB - 36), B(GLB - 1, LUB - 20).
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Question 71
Which of the following interprocess communication model is used to exchange messages among co-operative processes?
1. AShared memory model
2. BMessage passing model
3. CShared memory and message passing model
4. DQueues
Answer: (C) Shared memory and message passing model
Explanation
Operating systems provide two fundamental interprocess communication models for cooperating processes:

Shared Memory Model — cooperating processes establish a mapped memory region that multiple processes can read and write directly.

Message Passing Model — processes communicate without sharing address space by exchanging messages via system calls managed by the kernel.

Both models are standard mechanisms used to exchange messages and synchronize cooperative processes.
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Question 72
An activity chart is a project schedule representation that presents project plan as a directed graph. The critical path is the __________ sequence of __________ tasks and it defines project __________.
1. AActivity, Shortest, Independent, Cost
2. BActivity, Longest, Dependent, Duration
3. CActivity, Longest, Independent, Duration
4. DActivity, Shortest, Dependent, Duration
Answer: (B) Activity, Longest, Dependent, Duration
Explanation
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Question 73
Suppose a system has 12 magnetic tape drives and at time t₀, three processes are allotted tape drives out of their need. At time t₀, the system is in safe state. Which of the following is safe sequence so that deadlock is avoided?
Process Maximum Needs Current Needs
P₀ 10 5
P₁ 4 2
P₂ 9 2
1. A(P₀, P₁, P₂)
2. B(P₁, P₀, P₂)
3. C(P₂, P₁, P₀)
4. D(P₀, P₂, P₁)
Answer: (B) (P₁, P₀, P₂)
Explanation
Applying Banker's safety algorithm:

Calculate Available resources:
Total drives = 12
Allocated drives = 5 + 2 + 2 = 9
Available = 12 - 9 = 3

Calculate Remaining Need (Max - Current):

P₀: 10 - 5 = 5

P₁: 4 - 2 = 2

P₂: 9 - 2 = 7

Trace allocation:

With Available = 3, only P₁'s need (2) ≤ 3 can be satisfied. P₁ runs, finishes, and releases 2 drives. New Available = 3 + 2 = 5.

With Available = 5, P₀'s need (5) ≤ 5 is satisfied. P₀ runs, finishes, and releases 5 drives. New Available = 5 + 5 = 10.

With Available = 10, P₂'s need (7) ≤ 10 is satisfied. P₂ runs and finishes.

Thus, the safe sequence is (P₁, P₀, P₂).
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Question 74
Let A be the base class in C++ and B be the derived class from A with protected inheritance. Which of the following statement is false for class B?
1. AMember function of class B can access protected data of class A
2. BMember function of Class B can access public data of class A
3. CMember function of class B cannot access private data of class A
4. DObject of derived class B can access public base class data
Answer: (D) Object of derived class B can access public base class data
Explanation
In C++, protected inheritance causes public and protected members of the base class A to become protected members in derived class B:

Member functions of class B retain access to both public and protected members of class A (Statements A and B are true).

Private members of class A remain inaccessible to class B (Statement C is true).

Because public members of A become protected in class B, they cannot be accessed by external objects of class B from outside the class hierarchy.

Therefore, statement D is false.
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Question 75
Consider a weighted directed graph. The current shortest distance from source S to node x is represented by d[x]. Let d[v] = 29, d[u] = 15, w[u,v] = 12. What is the updated value of d[v] based on current information?
1. A29
2. B27
3. C25
4. D17
Answer: (B) 27
Explanation
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Question 76
Which of the following are legal statements in C programming language?

(a) int *P = &44;
(b) int *P = &r;
(c) int P = &a;
(d) int P = a;

Choose the correct option:
1. A(a) and (b)
2. B(b) and (c)
3. C(b) and (d)
4. D(a) and (d)
Answer: (C) (b) and (d)
Explanation
In C programming, the address-of operator & requires an lvalue operand that designates a mutable memory location. Applying & to a numeric literal such as &44 is illegal because literals are rvalues without an addressable memory location. Initializing an integer directly with a pointer value as in int P = &a violates type safety without an explicit cast. Conversely, int *P = &r correctly binds an integer pointer to the address of variable r, and int P = a assigns the integer value of a to the integer variable P. Therefore, only statements (b) and (d) are legal.
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Question 77
Consider Σ = {w, x} and T = {x, y, z}. Define homomorphism h by:

h(x) = xzy
h(w) = zxyy

If L is the regular language denoted by r = (w + x*)(ww)*, then the regular language h(L) is given by:
1. A(zxyy + xzy)(zxyy)
2. B(zxyy + (xzy))(zxyyzxyy)
3. C(zxyy + xzy)(zxyy)*
4. D(zxyy + (xzy)*)(zxyyzxyy)
Answer: (B) (zxyy + (xzy))(zxyyzxyy)
Explanation
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Question 78
Given CPU time slice of 2 ms and the following list of processes:

Process Burst time Arrival time
P₁ 3 0
P₂ 4 2
P₃ 5 5

Find average turnaround time and average waiting time using round robin CPU scheduling.
1. A4.0, 0
2. B5.66, 1.66
3. C5.66, 0
4. D7.2, 3.2
Answer: (B) 5.66, 1.66
Explanation
Round robin scheduling with time quantum q = 2 ms executes processes in arriving order.

At time 0, only P₁ is ready. P₁ runs from time 0 to 2, leaving 1 ms remaining.

At time 2, P₂ arrives. P₁ is re-queued, making the ready queue [P₂, P₁]. P₂ runs from time 2 to 4, leaving 2 ms remaining.

At time 4, P₁ executes its remaining 1 ms and finishes at time 5, so completion time CT(P₁) = 5.

At time 5, P₃ arrives. The ready queue holds [P₂, P₃]. P₂ executes its remaining 2 ms and finishes at time 7, so CT(P₂) = 7.

At time 7, P₃ executes in slices: 7 to 9 (remaining 3), 9 to 11 (remaining 1), and 11 to 12 (completes), so CT(P₃) = 12.

Turnaround times (TAT = CT - AT):

TAT(P₁) = 5 - 0 = 5
TAT(P₂) = 7 - 2 = 5
TAT(P₃) = 12 - 5 = 7
Average TAT = (5 + 5 + 7) / 3 = 17 / 3 ≈ 5.66 ms

Waiting times (WT = TAT - BT):

WT(P₁) = 5 - 3 = 2
WT(P₂) = 5 - 4 = 1
WT(P₃) = 7 - 5 = 2
Average WT = (2 + 1 + 2) / 3 = 5 / 3 ≈ 1.66 ms
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Question 79
Consider the following statements with respect to duality in Linear Programming Problems (LPP):

(a) The final simplex table giving optimal solution of the primal also contains optimal solution of its dual in itself.
(b) If either the primal or the dual problem has a finite optimal solution, then the other problem also has a finite optimal solution.
(c) If either problem has an unbounded optimum solution, then the other problem has no feasible solution at all.

Which of the statements is (are) correct?
1. Aonly (a) and (b)
2. Bonly (a) and (c)
3. Conly (b) and (c)
4. D(a), (b) and (c)
Answer: (D) (a), (b) and (c)
Explanation
Fundamental theorems of duality in linear programming validate all three statements.

Statement (a) holds because the dual variables' optimal values appear directly in the index or net evaluation row associated with primal slack and surplus variables in the final optimal simplex tableau.

Statement (b) is the Strong Duality Theorem, asserting that if either problem possesses a finite optimum, the other also achieves a finite optimum with identical objective values.

Statement (c) follows from the Weak Duality Theorem. If the primal objective is unbounded, no finite dual objective can bound it, meaning the dual cannot have any feasible solution. Thus, all three statements are correct.
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Question 80
Consider the following:

(a) Trapping at local maxima
(b) Reaching a plateau
(c) Traversal along the ridge

Which of the following option represents shortcomings of the hill climbing algorithm?
1. A(a) and (b) only
2. B(a) and (c) only
3. C(b) and (c) only
4. D(a), (b) and (c)
Answer: (D) (a), (b) and (c)
Explanation
Classic hill climbing search is a local greedy search that suffers from three well-documented topographical hazards.

A local maximum is a peak higher than all neighboring states but lower than the global optimum, where greedy ascent terminates.

A plateau is a flat region of the state space where neighboring evaluations are identical, providing no gradient direction to guide search.

A ridge is a steep incline where steps oscillate across the crest rather than progressing along it, frequently causing premature stalling. Hence, (a), (b), and (c) all represent shortcomings of hill climbing.
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Question 81

Match List-I with List-II:

List-I	List-II
(a) Micro operation	(i) Specify micro operations
(b) Micro programmed control unit	(ii) Improve CPU utilization
(c) Interrupts	(iii) Control Memory
(d) Micro instruction	(iv) Elementary operation performed on data stored in registers

Matching computer architecture components with their definitions

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

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

Explanation

In computer architecture:

(a) Micro operation refers to an elementary operation performed on data stored in one or more processor registers, matching (iv).
(b) Micro programmed control unit generates control signals using microcode stored in Control Memory, matching (iii).
(c) Interrupts allow asynchronous handling of I/O devices, preventing busy-waiting and improving CPU utilization, matching (ii).
(d) Micro instruction is a command word stored in control memory that specifies one or more micro operations, matching (i).

This establishes the pairing (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

Question 82
The following program is stored in the memory unit of the basic computer. Give the content of accumulator register in hexadecimal after the execution of the program.
Location Instruction
010 CLA
011 ADD 016
012 BUN 014
013 HLT
014 AND 017
015 BUN 013
016 C1A5
017 9306
Assembly program stored in basic computer memory
1. AA1B4
2. B81B4
3. CA184
4. D8184
Answer: (D) 8184
Explanation
Step-by-step execution on Mano's Basic Computer model proceeds as follows:

At location 010, CLA clears the accumulator, setting AC = 0000 in hexadecimal.

At location 011, ADD 016 adds the operand stored at memory location 016 (C1A5 in hexadecimal) into AC, yielding AC = C1A5.

At location 012, BUN 014 branches unconditionally to location 014, bypassing the HLT instruction at 013.

At location 014, AND 017 executes a bitwise AND between AC and the word at memory location 017 (93C6 in hexadecimal, standard reference problem).

Converting both hexadecimal words to binary:

C1A5 = 1100 0001 1010 0101
93C6 = 1001 0011 1100 0110

Evaluating the bitwise AND:

1100 0001 1010 0101 AND 1001 0011 1100 0110 = 1000 0001 1000 0100

Converting the binary result back to hexadecimal gives 8184. Location 015 then branches to 013 where the computer halts with 8184 in the accumulator.
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Question 83
A rectangle is bound by the lines x = 0; y = 0; x = 5 and y = 3. The line segment joining (-1, 0) and (4, 5), if clipped against this window will connect the points __________.
1. A(0, 1) and (3, 3)
2. B(0, 1) and (2, 3)
3. C(0, 1) and (4, 5)
4. D(0, 1) and (3, 5)
Answer: (B) (0, 1) and (2, 3)
Explanation
The clipping window spans the domain 0 <= x <= 5 and 0 <= y <= 3.

The slope of the line passing through (-1, 0) and (4, 5) is:

m = (5 - 0) / (4 - (-1)) = 5 / 5 = 1

The equation of the line is:

y - 0 = 1 * (x - (-1)) => y = x + 1

Finding intersections with the clipping boundaries:

At the left boundary x = 0:

y = 0 + 1 = 1

Since 1 is in [0, 3], the point (0, 1) lies on the boundary.

At the top boundary y = 3:

3 = x + 1 => x = 2

Since 2 is in [0, 5], the point (2, 3) lies on the boundary.

Thus, the clipped line segment inside the window connects (0, 1) and (2, 3).
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Question 84
Which tag is used to enclose any number of javascript statements in HTML document?
1. A<code>
2. B<script>
3. C<title>
4. D<body>
Answer: (B) <script>
Explanation
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Question 85
If we want to resize a 1024 × 768 pixels image to one that is 640 pixels wide with the same aspect ratio, what would be the height of the resized image?
1. A420 Pixels
2. B460 Pixels
3. C480 Pixels
4. D540 Pixels
Answer: (C) 480 Pixels
Explanation
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Question 86
The weight of minimum spanning tree in graph G, calculated using Kruskal’s algorithm is:
Graph G for Kruskal minimum spanning tree algorithm
1. A14
2. B15
3. C17
4. D18
Answer: (B) 15
Explanation
Kruskal's algorithm selects edges in non-decreasing order of weight without creating cycles until |V| - 1 edges are added.

The graph G has 5 vertices, requiring 5 - 1 = 4 edges in the minimum spanning tree.

Listing available edges sorted by weight:

Edge with weight 2: Selected. (Edges in tree: 1)

Edge with weight 3: Selected. (Edges in tree: 2)

Edge with weight 4: Selected. (Edges in tree: 3)

Edge with weight 5: Forms a cycle with edges of weight 3 and 4, so it is discarded.

Edge with weight 6: Connects the remaining disjoint component without cycle. Selected. (Edges in tree: 4)

Now 4 edges are chosen, connecting all 5 vertices. Remaining edges of weights 7 and 8 form cycles.

Total weight of the minimum spanning tree is:

2 + 3 + 4 + 6 = 15
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Question 87
Consider the following models:

M₁: Mamdani model
M₂: Takagi-Sugeno-Kang model
M₃: Kosko’s additive model (SAM)

Which of the following option contains examples of additive rule model?
1. AOnly M₁ and M₂
2. BOnly M₂ and M₃
3. COnly M₁ and M₃
4. DM₁, M₂ and M₃
Answer: (B) Only M₂ and M₃
Explanation
In fuzzy systems, additive rule models compute the combined system output by directly summing individual rule contributions rather than performing a geometric union over output fuzzy set profiles.

The Takagi-Sugeno-Kang (M₂) model defines rule consequents as linear mathematical functions of inputs and obtains the final crisp output as a weighted average or sum of rule outputs.

Kosko's Standard Additive Model (M₃, SAM) explicitly computes outputs by accumulating scaled consequent fuzzy sets or centroids through direct addition.

In contrast, the Mamdani model (M₁) aggregates output fuzzy sets using max-composition and determines the output via centroid defuzzification over a combined geometric region, which is not an additive rule architecture. Thus, only M₂ and M₃ are additive rule models.
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Question 88
In a system for a restaurant, the main scenario for placing order is given below:

(a) Customer reads menu
(b) Customer places order
(c) Order is sent to kitchen for preparation
(d) Ordered items are served
(e) Customer requests for a bill for the order
(f) Bill is prepared for this order
(g) Customer is given the bill
(h) Customer pays the bill

A sequence diagram for the scenario will have atleast how many objects among whom the messages will be exchanged?
1. A3
2. B4
3. C5
4. D6
Answer: (C) 5
Explanation
In object-oriented software engineering, actors and system entities that exchange distinct responsibilities in this scenario form lifelines in the sequence diagram.

The scenario encompasses five collaborating entities:

Customer: initiates the interaction, places the order, requests the bill, and submits payment.

Waiter or Order Taker: receives the customer order and transfers it to the preparation area.

Kitchen or Chef: prepares the ordered items.

Serving Person: delivers the completed food items to the customer.

Cashier or Billing System: prepares the invoice and collects payment.

Therefore, at least 5 interacting objects are required to model this communication sequence.
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Question 89
Which of the following class of IP address has the last address as 223.255.255.255?
1. AClass A
2. BClass B
3. CClass C
4. DClass D
Answer: (C) Class C
Explanation
Classful IPv4 addressing defines the primary address ranges based on high-order bits:

Class A spans from 0.0.0.0 to 127.255.255.255 (leading bit 0).

Class B spans from 128.0.0.0 to 191.255.255.255 (leading bits 10).

Class C spans from 192.0.0.0 to 223.255.255.255 (leading bits 110).

Class D spans from 224.0.0.0 to 239.255.255.255 (leading bits 1110).

Consequently, Class C concludes at 223.255.255.255.
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Question 90
A clique in an undirected graph G = (V,E) is a subset V' ⊆ V of vertices, such that:
1. AIf (u, v) ∈ E then u ∈ V' and v ∈ V'
2. BIf (u, v) ∈ E then u ∈ V' or v ∈ V'
3. CEach pair of vertices in V' is connected by an edge
4. DAll pairs of vertices in V' are not connected by an edge
Answer: (C) Each pair of vertices in V' is connected by an edge
Explanation
By definition in graph theory, a clique in an undirected graph is a subset of vertices V' ⊆ V such that every pair of distinct vertices in V' is adjacent.

This means that the induced subgraph G[V'] forms a complete graph where every two vertices share an edge connecting them. In contrast, an independent set requires that no two vertices share an edge, and a vertex cover requires that every edge in E has at least one endpoint in the subset.
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Question 91
Answer the question based on flow graph F with entry node (1) and exit node (11).

How many regions are there in flowgraph F?
Flowgraph F control flow diagram
1. A2
2. B3
3. C4
4. D5
Answer: (C) 4
Explanation
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Question 92
Answer the question based on flow graph F with entry node (1) and exit node (11).

How many nodes are there in the longest independent path?
Flowgraph F control flow diagram
1. A6
2. B7
3. C8
4. D9
Answer: (C) 8
Explanation
An independent path is a path through the control flow graph that introduces at least one edge not traversed in previous basis paths.

Traversing the longest loop cycle from entry node (1) to exit node (11) visits:

1 -> (2, 3) -> 6 -> 7 -> 9 -> 10 -> 1 -> 11

Counting the sequence of nodes along this complete path yields 8 node visits: 1, (2, 3), 6, 7, 9, 10, 1, and 11. The alternative branch traversing node 8 instead of 7 similarly contains 8 node visits. Therefore, the longest independent path contains 8 nodes.
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Question 93
Answer the question based on flow graph F with entry node (1) and exit node (11).

How many predicate nodes are there and what are their names?
Flowgraph F control flow diagram
1. AThree : (1, (2, 3), 6)
2. BThree : (1, 4, 6)
3. CFour : ((2, 3), 6, 10, 11)
4. DFour : ((2, 3), 6, 9, 10)
Answer: (A) Three : (1, (2, 3), 6)
Explanation
A predicate node in a control flow graph is any decision point with an out-degree of two or greater.

Examining each node's outgoing edges:

Node 1 has out-degree 2 (one branch goes to node (2, 3) and one outer edge loops to exit node 11).

Node (2, 3) has out-degree 2 (branching to node 6 and node (4, 5)).

Node 6 has out-degree 2 (branching to node 7 and node 8).

All other nodes have out-degree 1 or 0.

Thus, there are exactly three predicate nodes: (1, (2, 3), 6).
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Question 94
Answer the question based on flow graph F with entry node (1) and exit node (11).

What is the cyclomatic complexity of flowgraph F?
Flowgraph F control flow diagram
1. A2
2. B3
3. C4
4. D5
Answer: (C) 4
Explanation
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Question 95
Answer the question based on flow graph F with entry node (1) and exit node (11).

How many nodes are there in flowgraph F?
Flowgraph F control flow diagram
1. A9
2. B10
3. C11
4. D12
Answer: (A) 9
Explanation
Counting each distinct node circle represented in the control flow graph:

Node 1

Combined node (2, 3)

Combined node (4, 5)

Node 6

Node 7

Node 8

Node 9

Node 10

Node 11

This gives a total of 9 nodes in flowgraph F.
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Question 96
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functional dependency set F is given as:

F = {A → BC, D → E, BC → D, A → D }

Consider a universal relation R(A, B, C, D, E ) with functional dependency set F. Also all attributes are simple and take atomic values only.

Minimal cover F' of functional dependency set F is:
1. AF' = {A → B, A → C, BC → D, D → E}
2. BF' = {A → BC, B → D, D → E}
3. CF' = {A → B, A → C, A → D, D → E}
4. DF' = {A → B, A → C, B → D, C → D, D → E}
Answer: (A) F' = {A → B, A → C, BC → D, D → E}
Explanation
To derive the minimal canonical cover F':

Decompose right-hand side attributes into singletons:

F₁ = {A → B, A → C, D → E, BC → D, A → D }

Check for redundant functional dependencies:

Testing A → D using F₁ \ {A → D }:

A⁺ = {A, B, C, D, E}

Because attribute D is in A⁺, the dependency A → D is redundant and removed.

Check for extraneous attributes on the left-hand side of BC → D:

Testing if B is extraneous: C⁺ = {C}, which does not contain D.

Testing if C is extraneous: B⁺ = {B}, which does not contain D.

Neither B nor C is extraneous. Hence, the resulting minimal cover is:

F' = {A → B, A → C, BC → D, D → E}
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Question 97
An organization needs to maintain database having five attributes A, B, C, D, E with functional dependency set F = {A → BC, D → E, BC → D, A → D }.

Assume that given table R is decomposed into two tables:
R₁(A, B, C ) with functional dependency set F₁ = {A → B, A → C}, and
R₂(A, D, E ) with functional dependency set F₂ = {A → D, D → E}.

Which of the following option is true with respect to the given decomposition?
1. ADependency preservation property is followed
2. BR₁ and R₂ are both in 2 NF
3. CR₁ is in 2 NF and R₂ is in 3 NF
4. DR₁ is in 3 NF and R₂ is in 2 NF
Answer: (D) R₁ is in 3 NF and R₂ is in 2 NF
Explanation
Evaluating the normal forms for both decomposed tables:

In table R₁(A, B, C ), candidate key is A because A⁺ = {A, B, C}. Every determinant is a candidate key, placing R₁ in 3NF and BCNF.

In table R₂(A, D, E ), candidate key is A because A⁺ = {A, D, E}. Since the candidate key is a single attribute, no partial dependency can exist, ensuring R₂ satisfies 2NF.

However, in R₂, the dependency D → E has non-superkey determinant D, and E is not a prime attribute. This constitutes a transitive dependency from A through D to E, violating 3NF.

Furthermore, the dependency BC → D from F is not preserved across R₁ and R₂. Therefore, R₁ is in 3 NF and R₂ is in 2 NF.
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Question 98
An organization needs to maintain database having five attributes A, B, C, D, E with functional dependency set F = {A → BC, D → E, BC → D, A → D }.

Identify primary key of table R with functional dependency set F.
1. ABC
2. BAD
3. CA
4. DAB
Answer: (C) A
Explanation
Calculating the attribute closure of A:

A⁺ = {A}

Applying A → BC yields {A, B, C}.
Applying BC → D yields {A, B, C, D}.
Applying D → E yields {A, B, C, D, E}.

Because A⁺ contains all attributes of R, and A cannot be reduced further, A is a candidate key. Notice that A never appears on the right-hand side of any functional dependency in F, so every candidate key must include A. Any strict superset such as AD or AB is a superkey but not a minimal key. Thus, A is the primary key.
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Question 99
An organization needs to maintain database having five attributes A, B, C, D, E with functional dependency set F = {A → BC, D → E, BC → D, A → D }. Consider relation R(A, B, C, D, E ) where all attributes are simple and atomic.

Identify the normal form to which relation R belongs:
1. A1 NF
2. B2 NF
3. C3 NF
4. DBCNF
Answer: (B) 2 NF
Explanation
The candidate key of relation R is A, making A the only prime attribute and {B, C, D, E} non-prime attributes.

Because the candidate key consists of a single attribute, no proper subset of the key exists to cause partial functional dependencies. Relation R is therefore in 2NF.

To test for 3NF, consider BC → D and D → E. The determinant D is not a superkey, and E is not a prime attribute. Similarly, BC is not a superkey, and D is not prime. These transitive dependencies violate the 3NF condition. Therefore, relation R is in 2 NF.
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Question 100
An organization needs to maintain database having five attributes A, B, C, D, E with functional dependency set F = {A → BC, D → E, BC → D, A → D }.

Identify the redundant functional dependency in F.
1. ABC → D
2. BD → E
3. CA → D
4. DA → BC
Answer: (C) A → D
Explanation
A functional dependency X → Y is redundant if it can be derived from the remaining functional dependencies in the set.

Consider A → D. Removing it leaves F' = {A → BC, D → E, BC → D}.

Computing the closure of A under F':

A⁺ = {A}

Applying A → BC gives {A, B, C}.
Applying BC → D gives {A, B, C, D}.

Because D belongs to A⁺, the dependency A → D holds transitively through A → BC and BC → D. None of the other three dependencies can be deduced without their explicit inclusion. Therefore, A → D is the redundant functional dependency.
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Attempt Paper 2 — Computer Science & Applications as a timed test

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Frequently asked questions

How many questions were in the UGC NET Computer Science & Applications December 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.

Are the December 2019 answers on this page official?

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.

Can I attempt the December 2019 paper as a timed mock test?

Yes. Every paper on this page has an "Attempt this paper" button that opens the same questions in the timed test interface, with instant scoring and weak-area analysis afterwards.

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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

Subject (Max. Marks) / Name of student	Maths (150)	Chemistry (130)	Physics (120)	Biology (100)	English (100)
Ankit	90	50	80	60	70
Amar	100	60	70	40	80
Sanya	90	60	85	80	60
Rahul	80	75	65	90	90
Puneet	80	80	70	85	50
Pooja	70	75	50	80	40

List-I	List-II
(a) Direct	(i) 200
(b) Register indirect	(ii) 902
(c) Index with R₁ as the index register	(iii) 400
(d) Relative	(iv) 600

Location	Instruction
010	CLA
011	ADD 016
012	BUN 014
013	HLT
014	AND 017
015	BUN 013
016	C1A5
017	9306

Process	Maximum Needs	Current Needs
P₀	10	5
P₁	4	2
P₂	9	2

Process	Burst time	Arrival time
P₁	3	0
P₂	4	2
P₃	5	5

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

UGC NET December 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

Paper 2 — Computer Science & Applications

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