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|Computer Science & Applications · June 2023

UGC NET Computer Science & Applications June 2023 Question PaperPaper 1 & Paper 2 with Answer Key

The complete UGC NET June 2023 question paper for Computer Science & Applications, covering Paper 1 (General Aptitude) and Paper 2 (Computer Science & Applications). Every question below is shown with its options and the correct answer, and a detailed explanation you can unlock by signing in. All 150 questions are free to practise.

UGC NET June 2023 at a glance

Exam	UGC NET June 2023
Papers	Paper 1 (General Aptitude) and Paper 2 (Computer Science & Applications)
Exam date(s)	June 17, 2023
Shift(s)	Shift 1
Paper 1 questions	50 questions · 60 minutes · 100 marks
Paper 2 questions	100 questions · 120 minutes · 200 marks
Total	150 questions
Marking scheme	+2 per correct answer · No negative marking

Paper 1

General Aptitude

UGC NET June 2023 · Computer Science & Applications sitting

50 questions60 min100 marks

Practice Now

Paper 2

Computer Science & Applications

UGC NET June 2023

100 questions120 min200 marks

Practice Now

Paper 1 — General Aptitude

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50 questions with the answer key and explanations.

Question 1
If in Delhi, the number of employees who work in Accounts is 40% more than the number who work in HR, what is the number of employees who work in IT in Delhi?
The following table shows the number of employees in an organization in five cities. The total number of employees is 4,260. Percentages show distribution across HR, Finance, IT and Accounts. Some values are missing and may be calculated when required. Each employee works in only one department.

City Employees HR Finance IT Accounts
Delhi 900 - 18% - 28%
Mumbai 760 15% - 30% -
Bengaluru - 18% 20% - 32%
Hyderabad - - 25% 18% 35%
Gurgaon 700 20% 22% - 20%
1. A266
2. B306
3. C352
4. D294
Answer: (B) 306
Explanation
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Question 2
What is the total number of employees in Mumbai and Gurgaon who work in the IT department?
The following table shows the number of employees in an organization in five cities. The total number of employees is 4,260. Percentages show distribution across HR, Finance, IT and Accounts. Some values are missing and may be calculated when required. Each employee works in only one department.

City Employees HR Finance IT Accounts
Delhi 900 - 18% - 28%
Mumbai 760 15% - 30% -
Bengaluru - 18% 20% - 32%
Hyderabad - - 25% 18% 35%
Gurgaon 700 20% 22% - 20%
1. A482
2. B452
3. C570
4. D494
Answer: (D) 494
Explanation
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Question 3
If the number of employees in Bengaluru is 10% less than the number in Hyderabad, what is the difference between the total number of employees working in the Accounts department and the total number working in the Finance department across these two cities?
The following table shows the number of employees in an organization in five cities. The total number of employees is 4,260. Percentages show distribution across HR, Finance, IT and Accounts. Some values are missing and may be calculated when required. Each employee works in only one department.

City Employees HR Finance IT Accounts
Delhi 900 - 18% - 28%
Mumbai 760 15% - 30% -
Bengaluru - 18% 20% - 32%
Hyderabad - - 25% 18% 35%
Gurgaon 700 20% 22% - 20%
1. A244
2. B208
3. C194
4. D270
Answer: (B) 208
Explanation
The total number of employees across all five cities is 4260.
Employees in Delhi + Mumbai + Gurgaon = 900 + 760 + 700 = 2360.
Employees in Bengaluru + Hyderabad = 4260 - 2360 = 1900.
Let the number of employees in Hyderabad be H. The number in Bengaluru is 0.9H.
0.9H + H = 1900 -> 1.9H = 1900 -> H = 1000. So, Bengaluru has 900 employees.
In Bengaluru: Finance = 20% of 900 = 180; Accounts = 32% of 900 = 288.
In Hyderabad: Finance = 25% of 1000 = 250; Accounts = 35% of 1000 = 350.
Total Finance = 180 + 250 = 430.
Total Accounts = 288 + 350 = 638.
Difference = 638 - 430 = 208.
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Question 4
If the number of employees who work in Accounts in Mumbai is 38 more than the number who work in IT in Gurgaon, what is the number of employees who work in Finance in Mumbai?
The following table shows the number of employees in an organization in five cities. The total number of employees is 4,260. Percentages show distribution across HR, Finance, IT and Accounts. Some values are missing and may be calculated when required. Each employee works in only one department.

City Employees HR Finance IT Accounts
Delhi 900 - 18% - 28%
Mumbai 760 15% - 30% -
Bengaluru - 18% 20% - 32%
Hyderabad - - 25% 18% 35%
Gurgaon 700 20% 22% - 20%
1. A114
2. B120
3. C98
4. D124
Answer: (A) 114
Explanation
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Question 5
If the number of employees in Bengaluru is 10% less than the number in Hyderabad, the total number of employees who work in Finance and Accounts is more than 415 in exactly ______ cities.
The following table shows the number of employees in an organization in five cities. The total number of employees is 4,260. Percentages show distribution across HR, Finance, IT and Accounts. Some values are missing and may be calculated when required. Each employee works in only one department.

City Employees HR Finance IT Accounts
Delhi 900 - 18% - 28%
Mumbai 760 15% - 30% -
Bengaluru - 18% 20% - 32%
Hyderabad - - 25% 18% 35%
Gurgaon 700 20% 22% - 20%
1. ATwo
2. BThree
3. CFour
4. DFive
Answer: (B) Three
Explanation
Using the data from question 3, Bengaluru = 900 and Hyderabad = 1000 employees.
Let's calculate Finance + Accounts for each city:
Delhi: (18% + 28%) = 46% of 900 = 414.
Mumbai: We know HR (15%) + IT (30%) = 45%, so Finance + Accounts must be 55%. 55% of 760 = 418.
Bengaluru: Finance (20%) + Accounts (32%) = 52% of 900 = 468.
Hyderabad: Finance (25%) + Accounts (35%) = 60% of 1000 = 600.
Gurgaon: Finance (22%) + Accounts (20%) = 42% of 700 = 294.
The total exceeds 415 in exactly three cities: Mumbai, Bengaluru, and Hyderabad.
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Question 6
When you sign up for something like a free e-mail account, you may be presented with a CAPTCHA. CAPTCHA stands for:
1. ACoverage Analysis and Planning Tool for Computers and Humans Apart
2. BCompletely Automated Public Turing Test to Tell Computers and Humans Apart
3. CCompletely Applied Parallel Technologies for Communication and Hardware Apart
4. DComparable Additional Protected Turing Test to Tell Computers and Humans Apart
Answer: (B) Completely Automated Public Turing Test to Tell Computers and Humans Apart
Explanation
CAPTCHA stands for 'Completely Automated Public Turing test to tell Computers and Humans Apart'. It is named after the Turing Test, proposed by Alan Turing in 1950, which tests whether a machine can exhibit behaviour indistinguishable from a human. A CAPTCHA reverses this concept — it is a test designed by a computer to determine whether the respondent is a human or an automated bot. Common forms include distorted text, image recognition challenges, and checkbox challenges. It is widely used to prevent spam, automated account creation, and brute-force attacks.
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Question 7
According to Abhinavagupta, the Rasa theory of communication provided an opportunity for the audience's:
1. ALiterary taste
2. BMaterialistic welfare
3. CSocialisation
4. DMoral cultivation
Answer: (D) Moral cultivation
Explanation
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Question 8
Which of the following arguments commits the fallacy of contradicted middle such that the middle term, instead of being pervaded by the presence of the major term, is pervaded by the absence of the major term?
1. AAll things are non-eternal because they are knowable.
2. BSound is a quality because it is visible.
3. CFire is cold because it is a substance.
4. DSound is eternal because it is produced.
Answer: (D) Sound is eternal because it is produced.
Explanation
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Question 9
A photovoltaic array is:
1. AA single solar cell
2. BA single solar module
3. CA single solar panel
4. DA combination of solar panels
Answer: (D) A combination of solar panels
Explanation
Solar energy generation is organised in a clear hierarchy:

Solar cell: The smallest unit; a single semiconductor device (typically silicon) that converts sunlight into electricity via the photovoltaic effect.

Solar module / panel: Multiple solar cells electrically wired together and encapsulated in a protective frame.

Photovoltaic array: Multiple solar panels (modules) connected together — in series, parallel, or a combination — to form a complete power-generating installation. Arrays can range from a rooftop setup to a utility-scale solar farm. An array is therefore never a single cell, module, or panel.
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Question 10

Match List I with List II.

List I - Security / data-loss issue	List II - Recovery / protection
A. Data loss caused by hard-disk crash	I. Anti-spyware software
B. Self-replicating software that can cause data loss	II. Anti-virus software
C. Preventing reading of illegally accessed documents	III. Backup files
D. Software that secretly records all key presses	IV. Encryption

AA-III, B-IV, C-II, D-I
BA-I, B-II, C-IV, D-III
CA-III, B-II, C-IV, D-I
DA-II, B-III, C-IV, D-I

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

Explanation

Question 11
Given below are two statements:
Statement I: The risk of rejecting the null hypothesis when it should be confirmed is more at the 0.05 p-level than at the 0.01 p-level.
Statement II: The risk of committing a Type-I error is lower at the 0.05 p-level than at the 0.01 p-level.
1. ABoth Statement I and Statement II are true
2. BBoth Statement I and Statement II are false
3. CStatement I is true but Statement II is false
4. DStatement I is false but Statement II is true
Answer: (C) Statement I is true but Statement II is false
Explanation
A Type I error is rejecting the null hypothesis when it is actually true (a 'false positive'). The probability of committing a Type I error equals the significance level (α).

At α = 0.05, there is a 5% chance of a Type I error.

At α = 0.01, there is only a 1% chance of a Type I error.

Therefore, the risk of a Type I error is higher at 0.05 than at 0.01, making Statement I true. Statement II says the risk is lower at 0.05, which directly contradicts the definition — so Statement II is false. A stricter (smaller) significance level reduces the probability of rejecting a true null hypothesis.
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Question 12
Given below are two statements:
Statement I: There is an argument that the emergence of cable and television audiences has brought in a more personalised and interactive media culture.
Statement II: Within an integrated communication system, messages do not acquire communicability and socialisation.
1. ABoth Statement I and Statement II are true
2. BBoth Statement I and Statement II are false
3. CStatement I is true but Statement II is false
4. DStatement I is false but Statement II is true
Answer: (C) Statement I is true but Statement II is false
Explanation
Statement I is correct. The rise of cable television, satellite channels, and later internet-based streaming fragmented the mass broadcast audience into numerous niche audiences, enabling more personalised, on-demand, and interactive media consumption — a shift scholars call the 'second media age' or the era of narrowcasting.

Statement II is false. An integrated communication system is specifically designed to maximise communicability — the capacity of messages to be understood — and to facilitate socialisation by coordinating multiple channels (text, audio, video, social media) into a unified framework. Messages in such a system acquire greater communicability, not less.
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Question 13
Given below are two statements:
Statement I: A cohort is a population whose members have some specific characteristics in common.
Statement II: In a prospective study, the researcher relies on people's memories and recollections of the past.
1. ABoth Statement I and Statement II are true
2. BBoth Statement I and Statement II are false
3. CStatement I is true but Statement II is false
4. DStatement I is false but Statement II is true
Answer: (C) Statement I is true but Statement II is false
Explanation
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Question 14
Media interactivity is experienced at different levels in different ______ relations.
1. APolitical
2. BEconomic
3. CTechno-social
4. DTechno-ethical
Answer: (C) Techno-social
Explanation
Media interactivity refers to the degree to which users can influence the form and content of a mediated environment. This interactivity does not operate uniformly; it is shaped by the intersection of technological infrastructure and the social norms, habits, and relationships of users — collectively called techno-social relations. Different techno-social contexts (e.g., a rural community with limited broadband vs. an urban professional network) experience and exercise media interactivity differently. The concept bridges purely technical affordances with the social practices that determine how those affordances are actually used.
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Question 15
Which of the following are the interrelated phases created by John Keller for stimulating and sustaining learners' motivation in the teaching and learning process?
A. Attention
B. Authority
C. Relevance
D. Confidence
E. Satisfaction
1. AA, C, D and E only
2. BB, D and E only
3. CA, C and D only
4. DB and C only
Answer: (A) A, C, D and E only
Explanation
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Question 16
Find the chronological order of the following open universities in India.
A. Karnataka State Open University
B. Netaji Subhas Open University
C. UP Rajarshi Tandon Open University
D. Madhya Pradesh Bhoj Open University
E. Nalanda Open University
1. AA, B, D, C, E
2. BE, D, A, B, C
3. CB, C, E, D, A
4. DD, E, A, B, C
Answer: (B) E, D, A, B, C
Explanation
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Question 17
Determine the next term in the number series: 3, 15, 35, 63, ?
1. A89
2. B91
3. C99
4. D106
Answer: (C) 99
Explanation
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Question 18
The Education Despatch of 1854 suggested that the medium of instruction in Indian universities should be:
1. APersian
2. BSanskrit
3. CHindi
4. DEnglish
Answer: (D) English
Explanation
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Question 19
Given below are two statements:
Statement I: A dichotomous variable is one which has more than two categories.
Statement II: A nominal variable can have more than two categories.
1. ABoth Statement I and Statement II are true
2. BBoth Statement I and Statement II are false
3. CStatement I is true but Statement II is false
4. DStatement I is false but Statement II is true
Answer: (D) Statement I is false but Statement II is true
Explanation
A dichotomous variable has exactly two mutually exclusive categories — for example, Yes/No, Employed/Unemployed, or Heads/Tails. It cannot have more than two categories, making Statement I false.

A nominal variable classifies observations into named categories without any intrinsic order or ranking. It can have two categories (making it dichotomous) or many more — for example, type of religion (Hindu, Muslim, Christian, Buddhist, …), blood group (A, B, AB, O), or country of birth. Statement II is therefore true.
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Question 20
Identify the correct order of the terms A-E to complete the paragraph:
A MAC address is a media access ______ address. A network device has a ______ MAC address that can help ______ the device in the network. An IP address is an Internet ______ address. An IP address can be static or ______.
A. Protocol
B. Control
C. Dynamic
D. Identify
E. Unique
1. AB, C, D, A, E
2. BC, E, D, A, B
3. CA, B, E, D, C
4. DB, E, D, A, C
Answer: (D) B, E, D, A, C
Explanation
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Question 21
The Union Government's mandate for the Humayun Kabir Committee was to explore the possibility of:
1. APrivatisation of public universities
2. BSetting up of new university
3. CEstablishing correspondence courses
4. DFunding private universities
Answer: (B) Setting up of new university
Explanation
The Humayun Kabir Committee (1964) was appointed by the Union Government to review the growing pressure on India's higher education system. Its specific mandate was to explore the feasibility and necessity of establishing new universities to meet regional demand, rather than expanding existing ones indefinitely.

The other options are incorrect because:

Privatisation of public universities (A) was specifically recommended against by the committee, which emphasised affordable public access.

Correspondence courses (C) were championed by separate initiatives; the committee's focus was on conventional new universities.

Funding private universities (D) was outside the committee's primary scope.
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Question 22
A passenger train running at 75 km/hr leaves a railway station 9 hours after a goods train leaves and overtakes it in 6 hours. The speed of the goods train is:
1. A35 km/hr
2. B32 km/hr
3. C30 km/hr
4. D28 km/hr
Answer: (C) 30 km/hr
Explanation
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Question 23
In a certain coding scheme, 'WOMEN' is coded as '70'. How will 'CHILDREN' be coded in this scheme?
1. A178
2. B143
3. C116
4. D73
Answer: (D) 73
Explanation
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Question 24
According to UGC Regulations 2018 on plagiarism, common knowledge or coincidental terms up to how many consecutive words shall be excluded while checking similarity for detection of plagiarism?
1. A8
2. B12
3. C14
4. D18
Answer: (C) 14
Explanation
The UGC (Promotion of Academic Integrity and Prevention of Plagiarism in Higher Educational Institutions) Regulations, 2018 specify that the following shall be excluded from similarity detection:

Properly attributed quotations

References and bibliography

Standard symbols, equations, and formulae

Table of contents, preface, and acknowledgements

Common knowledge or coincidental terms up to 14 consecutive words

The 14-word threshold prevents routine phrases and widely-shared knowledge from inflating similarity scores. This exclusion applies per matched segment — any run of identical text shorter than 14 words is not flagged as plagiarism.
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Question 25
If x = 0.25 and y = 0.16, find the value of the expression: y/x + √x × √y - xy
1. A0.6
2. B0.8
3. C0.45
4. D0.58
Answer: (B) 0.8
Explanation
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Question 26
Given below are two statements:
Statement I: Cultural eutrophication is caused by increased nutrient input into a water body mainly due to human activities.
Statement II: Cultural eutrophication decelerates the 'aging' of a water body enormously over natural rates.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (C) Statement I is correct but Statement II is incorrect
Explanation
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Question 27
You and your friends decided to go to Goa after you finish your finals. Your parents set a couple of conditions (see the spreadsheet below).

Category of condition Minimum results needed Actual results Allowed (Yes/No)
Hours spent on homework weekly 2 3
Midterm score 70 68
Final exam score 75 77

Imagine your parents care only about the Midterm and Final exam conditions. Which formula can be used to determine whether you are allowed to join your friends?
1. A=IF(C3>=B3,IF(C4>B4,"Yes","No"),"No")
2. B=IF(AND(C3>=B3,C4>=B4),"Yes","No")
3. C=IF(C3:C4>=B3:B4,"Yes","No")
4. D=IF(OR(C3>=B3,C4>=B4),"Yes","No")
Answer: (B) =IF(AND(C3>=B3,C4>=B4),"Yes","No")
Explanation
Since both the Midterm and Final exam conditions must be satisfied simultaneously, a logical AND is required — the student is allowed only if both results meet the minimums.

Option B — =IF(AND(C3>=B3,C4>=B4),"Yes","No") — correctly checks both conditions with >= (greater than or equal to) for each.

Why other options fail:

Option A uses C4>B4 (strict greater-than) for the final exam. If the student scores exactly the minimum (e.g., 75 vs 75 needed), Option A would return 'No' even though the condition is met — a subtle but critical bug.

Option C uses an array range (C3:C4>=B3:B4) which does not work as a single TRUE/FALSE in a standard IF formula.

Option D uses OR, meaning the student passes if either condition is met — but both must be met per the stated rule.
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Question 28
The process of comparing a student's performance against his/her previous performance is known as:
1. ADiagnostic evaluation
2. BSummative assessment
3. CIpsative assessment
4. DInformal assessment
Answer: (C) Ipsative assessment
Explanation
Assessment can be measured against three reference points:

Norm-referenced: A student's performance is compared to the group average (e.g., class rank).

Criterion-referenced / Summative: Performance is measured against a fixed external standard (e.g., scoring ≥ 50% to pass).

Ipsative assessment: Performance is compared to the student's own previous performance. The term comes from the Latin 'ipse' meaning 'self'. It focuses on personal growth and improvement over time, regardless of external benchmarks or peer comparison.

Ipsative assessment is widely used in sport, rehabilitation, and personalised learning contexts where individual progress matters more than peer ranking.
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Question 29
When participants drop out of a study, leading to a change in the nature of the sample, it would cause which of the following threats to internal validity?
1. ASelection threat
2. BAttrition threat
3. CHistory threat
4. DTesting threat
Answer: (B) Attrition threat
Explanation
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Question 30
Given below are two statements:
Statement I: A valid deductive argument that also has all true premises is called a "sound" argument.
Statement II: A strong inductive argument that has all true premises is called a "cogent" argument.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (A) Both Statement I and Statement II are correct
Explanation
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Question 31
When people metacommunicate, they are sending messages about:
1. ASenders
2. BReceivers
3. CChannels
4. DMessages
Answer: (D) Messages
Explanation
Communication theorist Gregory Bateson introduced the concept of metacommunication — communication about communication itself. When people metacommunicate, they are sending messages that frame, qualify, or comment on other messages in the interaction.

For example, saying "I was only joking" after a statement is metacommunication — it is a message about the preceding message. Tone of voice, facial expression, and framing devices like "to be serious for a moment…" are all forms of metacommunication. It operates at a higher logical level than the message content itself, shaping how the receiver should interpret the message.
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Question 32
Given below are two statements:
Statement I: The electronic space of the Internet has failed to displace institutional space in a society even an iota.
Statement II: Internet communication has broken down the hierarchies of race, ethnicity and gender.
1. ABoth Statement I and Statement II are true
2. BBoth Statement I and Statement II are false
3. CStatement I is true but Statement II is false
4. DStatement I is false but Statement II is true
Answer: (D) Statement I is false but Statement II is true
Explanation
Statement I is false. The Internet has significantly displaced and disrupted traditional institutional spaces — journalism, retail, education, banking, and government services have all been profoundly restructured by digital and online platforms. Claiming the Internet has not displaced institutional space 'even an iota' is empirically indefensible.

Statement II is considered true within the context of media and communication theory: the Internet has created democratised spaces where individuals can publish, organise, and interact regardless of race, ethnicity, or gender in ways that traditional hierarchical institutional spaces did not permit. While discrimination persists online, the structural barriers to voice and participation are considerably lower than in pre-internet institutional contexts.
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Question 33
Given below are two statements:
Statement I: A binary number that ends in 1, e.g. 110011, is always an odd decimal number.
Statement II: HTML is a set of rules that must be obeyed when transferring files across the Internet.
1. ABoth Statement I and Statement II are true
2. BBoth Statement I and Statement II are false
3. CStatement I is true but Statement II is false
4. DStatement I is false but Statement II is true
Answer: (C) Statement I is true but Statement II is false
Explanation
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Question 34
Given below are two statements:
Statement I: Metacognition is the understanding about one's own use of language.
Statement II: Mnemonics is the knowledge about our own thinking processes.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (B) Both Statement I and Statement II are incorrect
Explanation
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Question 35
The teaching that provides all students with challenging instruction and uses supports when needed but removes those supports as students become able to handle more on their own is called:
1. ARecessive teaching
2. BActive teaching
3. CAdaptive teaching
4. DProblem-oriented teaching
Answer: (C) Adaptive teaching
Explanation
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Question 36
Given below are two statements:
Statement I: Reputed scholars like Panini and Chanakya had their education from Nalanda University.
Statement II: Takshashila was a university known for education in Indian arts and sciences, especially medicine.
1. ABoth Statement I and Statement II are true
2. BBoth Statement I and Statement II are false
3. CStatement I is true but Statement II is false
4. DStatement I is false but Statement II is true
Answer: (D) Statement I is false but Statement II is true
Explanation
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Question 37
What is the correct increasing order of contribution to global warming by the following gases?
A. Methane (CH4)
B. Carbon dioxide (CO2)
C. Nitrous oxide (N2O)
1. AA, B, C
2. BC, A, B
3. CB, A, C
4. DB, C, A
Answer: (B) C, A, B
Explanation
When measuring total contribution to global warming, what matters is the total radiative forcing each gas exerts on the atmosphere — which depends on both its warming potency per molecule and its actual atmospheric concentration.

CO₂: Low warming potential per molecule (GWP = 1) but present in massive quantities; responsible for roughly 66–76% of total anthropogenic warming.

CH₄ (Methane): Much higher GWP per molecule (~28–84×) but far less abundant; contributes roughly 16–18%.

N₂O (Nitrous oxide): Very high GWP per molecule (~265–298×) but extremely low concentration; contributes roughly 6%.

Increasing order of total contribution: N₂O (C) < CH₄ (A) < CO₂ (B) → answer option B (C, A, B).
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Question 38
If A, B and C can complete a piece of work in 3, 5 and 6 days respectively, they together will complete the work in:
1. A1 3/7 days
2. B2 4/7 days
3. C1 1/2 days
4. D2 1/3 days
Answer: (A) 1 3/7 days
Explanation
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Question 39
Given below are two statements:
Statement I: Due to climate change, Arctic sea ice is significantly thinner now than about a century ago.
Statement II: Due to climate change, tropical regions have warmed much faster than the rest of the world.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (C) Statement I is correct but Statement II is incorrect
Explanation
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Question 40
Which informal fallacy is committed in the following statement: "American Indians are disappearing. That man is an American Indian. Therefore that man is disappearing"?
1. ARed Herring
2. BHasty Generalisation
3. CBegging the question
4. DFallacy of division
Answer: (D) Fallacy of division
Explanation
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Question 41

Match List I (Online teaching tool) with List II (Advantage for teachers).

List I - Online teaching tool	List II - Advantage for teachers
A. Learning Management System (LMS)	I. Encouraging student engagement and collaboration; fostering critical thinking and problem solving
B. Video conferencing	II. Creating and administering assessments quickly and easily; automating grading and feedback
C. Discussion forums	III. Streamlining course administration; managing student records and grades
D. Online assessment platform	IV. Delivering engaging and interactive class sessions; connecting with remote learners

AA-IV, B-III, C-I, D-II
BA-I, B-IV, C-II, D-III
CA-II, B-III, C-IV, D-I
DA-III, B-IV, C-I, D-II

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

Explanation

Question 42
As per the recommendation of the National Education Policy (2020), Indian standards of skills will be aligned with the International Standard Classification of Occupations of:
1. AUNDP
2. BUNESCO
3. CILO
4. DITU
Answer: (C) ILO
Explanation
The National Education Policy 2020 (NEP 2020) sets out to map Indian skill qualifications onto a globally recognisable framework. It specifies alignment with the International Standard Classification of Occupations (ISCO), which is maintained by the International Labour Organization (ILO) — the UN agency responsible for labour standards.

ISCO provides a hierarchical system for classifying and organising all occupations, enabling comparison of skill levels and job families across countries. This alignment allows Indian vocational and higher-education qualifications to be benchmarked and recognised internationally. The other options (UNDP — development finance; UNESCO — education/culture; ITU — telecommunications) do not maintain occupational classification standards.
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Question 43
Arrange the following group of terms in order of increasing intension:
A. Tiger
B. Mammal
C. Feline
D. Animal
1. AA, C, B, D
2. BD, A, C, B
3. CD, B, C, A
4. DA, B, C, D
Answer: (C) D, B, C, A
Explanation
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Question 44
Identify the formal fallacy committed in the following argument: "Some good actors are not powerful athletes. All professional wrestlers are powerful athletes. Therefore all professional wrestlers are good actors."
1. AExclusive premises
2. BExistential fallacy
3. CAffirmative conclusion from negative premises
4. DAffirming the consequent
Answer: (C) Affirmative conclusion from negative premises
Explanation
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Question 45
United Nations Framework Convention on Climate Change (UNFCCC) was the outcome of:
1. AParis Agreement
2. BMontreal Protocol
3. CConvention on Biodiversity
4. DEarth Summit
Answer: (D) Earth Summit
Explanation
The United Nations Framework Convention on Climate Change (UNFCCC) was one of the landmark agreements produced at the Earth Summit — formally the United Nations Conference on Environment and Development (UNCED) — held in Rio de Janeiro, Brazil, in June 1992. The Earth Summit also produced the Convention on Biological Diversity (CBD) and Agenda 21.

The UNFCCC entered into force in 1994 and provides the legal framework under which subsequent climate agreements operate, including the Kyoto Protocol (1997) and the Paris Agreement (2015).

The other options are incorrect because:

The Paris Agreement (A) is itself a product of the UNFCCC process, not the other way around.

The Montreal Protocol (B) addresses ozone-depleting substances, not greenhouse gas emissions.

The Convention on Biodiversity (C) was signed at the same Earth Summit but is a separate treaty.
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Question 46
"Black spots" are areas:
The transport department is planning to start a road safety campaign that will involve local resident welfare associations, market stakeholders, school children and society representatives. As part of the campaign, the district magistrates will form plans to address the increasing number of accidents in consensus with the locals.

The decision was taken after it was found during a government study that road sections passing through pedestrian-heavy areas having schools, metro stations and commercial sites were more prone to accidents.

A senior official said that funds worth crores, which were meant for ensuring road safety, were not being utilised. "The focus of the campaign will be to identify and rectify black spots, revisit speed limits, form road safety clubs, install AI cameras at strategic locations, encourage helmet use, check speed in high-risk locations and regulate overloaded commercial vehicles," the official explained.

The number of deaths due to road accidents in Delhi totalled 1,239 in 2021, an increase from 1,196 fatalities in the previous year, according to Delhi Crash Report 2021, which was prepared by the city's traffic police. Pedestrian deaths comprised more than 40% of the total fatalities.
1. AThat have been blackened
2. BThat have no light
3. CWhere a lot of accidents happen
4. DThat are busy
Answer: (C) Where a lot of accidents happen
Explanation
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Question 47
Read the following two statements:
Statement I: The focus of the passage is on making Delhi roads safer.
Statement II: Funds worth crores meant for ensuring road safety have been properly utilised.
The transport department is planning to start a road safety campaign that will involve local resident welfare associations, market stakeholders, school children and society representatives. As part of the campaign, the district magistrates will form plans to address the increasing number of accidents in consensus with the locals.

The decision was taken after it was found during a government study that road sections passing through pedestrian-heavy areas having schools, metro stations and commercial sites were more prone to accidents.

A senior official said that funds worth crores, which were meant for ensuring road safety, were not being utilised. "The focus of the campaign will be to identify and rectify black spots, revisit speed limits, form road safety clubs, install AI cameras at strategic locations, encourage helmet use, check speed in high-risk locations and regulate overloaded commercial vehicles," the official explained.

The number of deaths due to road accidents in Delhi totalled 1,239 in 2021, an increase from 1,196 fatalities in the previous year, according to Delhi Crash Report 2021, which was prepared by the city's traffic police. Pedestrian deaths comprised more than 40% of the total fatalities.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (C) Statement I is correct but Statement II is incorrect
Explanation
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Question 48
"Revisit speed limits" means:
The transport department is planning to start a road safety campaign that will involve local resident welfare associations, market stakeholders, school children and society representatives. As part of the campaign, the district magistrates will form plans to address the increasing number of accidents in consensus with the locals.

The decision was taken after it was found during a government study that road sections passing through pedestrian-heavy areas having schools, metro stations and commercial sites were more prone to accidents.

A senior official said that funds worth crores, which were meant for ensuring road safety, were not being utilised. "The focus of the campaign will be to identify and rectify black spots, revisit speed limits, form road safety clubs, install AI cameras at strategic locations, encourage helmet use, check speed in high-risk locations and regulate overloaded commercial vehicles," the official explained.

The number of deaths due to road accidents in Delhi totalled 1,239 in 2021, an increase from 1,196 fatalities in the previous year, according to Delhi Crash Report 2021, which was prepared by the city's traffic police. Pedestrian deaths comprised more than 40% of the total fatalities.
1. ASpeed limits will be revised
2. BSpeed limits will be re-assessed
3. CSpeed limits will be reduced
4. DSpeed limits will be increased
Answer: (B) Speed limits will be re-assessed
Explanation
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Question 49
The government study found that areas that were more accident-prone were:
The transport department is planning to start a road safety campaign that will involve local resident welfare associations, market stakeholders, school children and society representatives. As part of the campaign, the district magistrates will form plans to address the increasing number of accidents in consensus with the locals.

The decision was taken after it was found during a government study that road sections passing through pedestrian-heavy areas having schools, metro stations and commercial sites were more prone to accidents.

A senior official said that funds worth crores, which were meant for ensuring road safety, were not being utilised. "The focus of the campaign will be to identify and rectify black spots, revisit speed limits, form road safety clubs, install AI cameras at strategic locations, encourage helmet use, check speed in high-risk locations and regulate overloaded commercial vehicles," the official explained.

The number of deaths due to road accidents in Delhi totalled 1,239 in 2021, an increase from 1,196 fatalities in the previous year, according to Delhi Crash Report 2021, which was prepared by the city's traffic police. Pedestrian deaths comprised more than 40% of the total fatalities.
1. AMain roads
2. BPeripheral roads
3. CHighways
4. DPedestrian-heavy areas
Answer: (D) Pedestrian-heavy areas
Explanation
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Question 50
Fill in the blank with the appropriate option:
The number of deaths due to road accidents ______ from 2020 to 2021.
The transport department is planning to start a road safety campaign that will involve local resident welfare associations, market stakeholders, school children and society representatives. As part of the campaign, the district magistrates will form plans to address the increasing number of accidents in consensus with the locals.

The decision was taken after it was found during a government study that road sections passing through pedestrian-heavy areas having schools, metro stations and commercial sites were more prone to accidents.

A senior official said that funds worth crores, which were meant for ensuring road safety, were not being utilised. "The focus of the campaign will be to identify and rectify black spots, revisit speed limits, form road safety clubs, install AI cameras at strategic locations, encourage helmet use, check speed in high-risk locations and regulate overloaded commercial vehicles," the official explained.

The number of deaths due to road accidents in Delhi totalled 1,239 in 2021, an increase from 1,196 fatalities in the previous year, according to Delhi Crash Report 2021, which was prepared by the city's traffic police. Pedestrian deaths comprised more than 40% of the total fatalities.
1. ADeclined
2. BRemained the same
3. CIncreased
4. DDecreased
Answer: (C) Increased
Explanation
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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
Which of the following is not a solution representation in a genetic algorithm?
1. ABinary valued
2. BReal valued
3. CPermutation
4. DCombinations
Answer: (D) Combinations
Explanation
Genetic algorithms encode candidate solutions using a small set of standard representation schemes chosen to match the problem structure.

A. Binary valued — a standard GA encoding where each solution is a bit string, used for classic optimization problems like the knapsack problem.

B. Real valued — used when parameters are continuous, such as tuning weights in a neural network, so each gene stores a floating-point value directly.

C. Permutation — used for ordering problems such as the travelling salesman problem, where a solution is a sequence of distinct elements.

D. Combinations — correct. There is no standard GA chromosome scheme called combination encoding, unlike binary, real-valued, and permutation encoding, which are all textbook representation types.
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Question 2

Match List I with List II on algorithm running times.

List I	List II
A. Parallel FFT	I. Θ(n²)
B. Iterative FFT	II. Θ(n)
C. Evaluation of polynomial at n points by Horner method	III. Θ(lg n)
D. Product of two polynomials that are represented in point value form	IV. Θ(n lg n)

List I to List II matching for FFT and polynomial algorithm running times

AA-III, B-I, C-II, D-IV
BA-II, B-I, C-III, D-IV
CA-III, B-IV, C-I, D-II
DA-II, B-III, C-IV, D-I

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

Explanation

Each pairing depends on a distinct algorithmic complexity result.

A-III, B-IV, C-I, D-II

Parallel FFT runs in Θ(lg n) time because the butterfly network of the FFT computation can be evaluated across O(lg n) parallel stages.
Iterative FFT, the standard sequential divide-and-conquer FFT, runs in Θ(n lg n) time.
Evaluating a degree n-1 polynomial at n points using Horner's method costs Θ(n) per point, so evaluating at all n points costs Θ(n²).
Multiplying two polynomials already given in point-value form only needs a pointwise multiplication of the y-values, which is Θ(n).

Question 3
A software system experienced five failures during testing. The time elapsed since the previous failure, in hours, is shown in the table.

Error number 1 2 3 4 5
Time since last failure (hours) 6 4 8 5 6

Using the exponential reliability model with the failure rate estimated from the mean time between these five failures, what is the reliability of the software running for the next one hour without failing?
1. Ae^(−9/29)
2. Be^(−7/29)
3. Ce^(−5/29)
4. De^(−3/29)
Answer: (C) e^(−5/29)
Explanation
The exponential software-reliability model estimates the failure rate as the number of observed failures divided by the total observed operating time, then applies R(t) = e^(-λt) for future operating time t.

Total elapsed time across the five recorded failures = 6 + 4 + 8 + 5 + 6 = 29 hours.

Failure rate λ = (number of failures observed) / (total time) = 5/29 per hour.

Reliability of no failure in the next hour, R(1) = e^(-λ×1) = e^(-5/29).

A. e^(−9/29) — does not correspond to the ratio of failures to total observed time.

B. e^(−7/29) — not obtainable from the given failure count or total time.

C. e^(−5/29) — correct. It uses the correct failure rate of 5 failures over 29 hours.

D. e^(−3/29) — understates the failure count used in the rate estimate.
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Question 4
Which of the following concurrency control protocols ensures both conflict serializability and freedom from deadlock?

Statement I: Two phase locking
Statement II: Timestamp ordering
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (D) Statement I is incorrect but Statement II is correct
Explanation
A concurrency control protocol must guarantee conflict serializability and also avoid deadlock to satisfy both required properties.

Statement I: Two phase locking — 2PL guarantees conflict serializability because every transaction's lock and unlock phases enforce a serializable order, but basic or strict 2PL can still deadlock when two transactions each hold a lock the other is waiting for.

Statement II: Timestamp ordering — this protocol assigns each transaction a fixed timestamp and orders conflicting operations by timestamp, guaranteeing conflict serializability, and because it never uses locks, transactions never wait on each other, so deadlock cannot occur.

Statement I is incorrect because 2PL does not eliminate deadlock, and Statement II is correct because timestamp ordering achieves both properties simultaneously.
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Question 5
The clipping process in computer graphics is used for
1. AAdding graphics
2. BCopying
3. CZooming
4. DRemoving objects and lines
Answer: (D) Removing objects and lines
Explanation
Clipping identifies which parts of a scene lie outside a defined window or viewport and removes them before rendering, so only the visible portion is drawn.

A. Adding graphics — clipping never creates new content, it only discards what falls outside the visible region.

B. Copying — copying duplicates content, unrelated to boundary testing against a window.

C. Zooming — zooming changes scale through a transformation such as scaling, not through removing out-of-bound geometry.

D. Removing objects and lines — correct. Clipping algorithms such as Cohen-Sutherland test each line or object against the clip boundary and discard the portions outside it.
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Question 6
Assertion (A): I/O protection is ensured by a hardware trap.
Reason (R): I/O interrupt is caused by conditions like I/O completion and device malfunction occurring within the I/O devices.
1. ABoth A and R are true and R is the correct explanation of A
2. BBoth A and R are true but R is NOT the correct explanation of A
3. CA is true but R is false
4. DA is false but R is true
Answer: (B) Both A and R are true but R is NOT the correct explanation of A
Explanation
I/O protection and I/O interrupts are related but distinct hardware mechanisms in operating systems.

Assertion A is true because I/O instructions are privileged, so any attempt by a user program to execute one directly causes a hardware trap into the operating system, which then validates and performs the operation on the program's behalf.

Reason R is also true as a general statement, because I/O interrupts are indeed signals raised by a device to report events such as completion of a transfer or a malfunction.

R does not explain A because the trap mechanism that enforces I/O protection is triggered by a user program's attempt to run a privileged instruction, not by the device-generated completion or malfunction interrupts that R describes. These are two separate mechanisms serving different purposes, so R is true but not the reason A holds.
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Question 7
Let R = {x : x ∈ N, x is multiple of 3 and x ≤ 100} and S = {x : x ∈ N, x is a multiple of 5 and x < 100}. What is the number of elements in (R ∩ S) × (S ∩ R)?
1. A36
2. B33
3. C20
4. D6
Answer: (A) 36
Explanation
R ∩ S contains numbers that are multiples of both 3 and 5, so they must be multiples of 15, and the bound x < 100 from S also restricts the intersection.

R ∩ S = S ∩ R = {15, 30, 45, 60, 75, 90}, which has 6 elements, since 105 is excluded because it fails x < 100.

|(R ∩ S) × (S ∩ R)| = |R ∩ S| × |S ∩ R| = 6 × 6 = 36.

A. 36 — correct, matches 6 × 6.

B. 33 — this is |R| alone, the count of multiples of 3 up to 100, not the intersection.

C. 20 — close to |S|, the count of multiples of 5 below 100, not the required product.

D. 6 — this is the size of R ∩ S itself, not the size of its Cartesian product with itself.
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Question 8
Statement I: Fuzzifier is a part of a fuzzy system.
Statement II: Inference engine is a part of a fuzzy system.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (A) Both Statement I and Statement II are correct
Explanation
A standard fuzzy inference system is built from four components working in sequence, a fuzzifier, a rule base, an inference engine, and a defuzzifier.

Statement I is correct because the fuzzifier converts crisp numeric inputs into fuzzy membership values before any rule is applied.

Statement II is correct because the inference engine applies the fuzzy rule base to these membership values to produce a fuzzy output.

Since both components are standard parts of a fuzzy system, both statements hold.
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Question 9
What will be the output of the following code?
```
#include <stdio.h>
int main(){
 int a, b, c;
 a = 0x10;
 b = 010;
 c = a + b;
 printf("%d", c);
 return 0;
}
```
1. A20
2. B24
3. CGarbage
4. Derror
Answer: (B) 24
Explanation
C interprets a leading 0x as a hexadecimal literal and a leading 0 as an octal literal, so both variables here hold different decimal values than their digits suggest.

a = 0x10 is hexadecimal 10, which equals 16 in decimal.

b = 010 is octal 10, which equals 8 in decimal.

c = a + b = 16 + 8 = 24, and printf("%d", c) prints this decimal value directly.

A. 20 — would only follow from treating both literals as decimal, ignoring the 0x and leading-zero prefixes.

B. 24 — correct, from 16 + 8.

C. Garbage — the code has no undefined behavior, both variables are assigned valid literal values.

D. error — the program is syntactically valid and compiles without any error.
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Question 10
What is the output of the following code?
```
int main() {
 struct s1 {
  char *z;
  int i;
  struct s1 *p;
 };
 static struct s1 a[] = {
  {"Nagpur", 1, a + 1},
  {"Raipur", 2, a + 2},
  {"Kanpur", 3, a}
 };
 struct s1 *ptr = a;
 printf("%s %s %s\n", a[0].z, ptr->z, a[2].p->z);
 return 0;
}
```
1. ANagpur Raipur Kanpur
2. BNagpur Nagpur Nagpur
3. CKanpur Kanpur Kanpur
4. DError
Answer: (B) Nagpur Nagpur Nagpur
Explanation
Each element of array a is a struct containing a name, an integer, and a pointer to another struct in the same array, so tracing the pointers requires following which array index each p field actually points to.

a[0].z is "Nagpur", the string stored directly in the first element.

ptr is initialized to a, which is the same address as a[0], so ptr->z is also a[0].z, which is "Nagpur".

a[2].p is initialized to a, the address of a[0], so a[2].p->z dereferences back to a[0].z, again "Nagpur".

A. Nagpur Raipur Kanpur — would only hold if each printed value came from a different array index, but ptr and a[2].p both resolve back to a[0].

B. Nagpur Nagpur Nagpur — correct, because a[0].z, ptr->z, and a[2].p->z all trace back to the same struct, a[0].

C. Kanpur Kanpur Kanpur — would require all three expressions to resolve to a[2], but only a[2].z itself is "Kanpur", not the pointer-dereferenced values used here.

D. Error — the program is syntactically valid, and the self-referential array initialization is legal because a is already declared by the time its own elements are initialized with pointers into it.
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Question 11

Match List I with List II on Fibonacci computation running times.

List I	List II
A. The running time of the straightforward recursive method to compute the nth Fibonacci number Fn	I. O(n²)
B. The running time to compute Fn using memoization	II. O(lg n)
C. The running time to compute a Fibonacci number Fn using only integer addition and multiplication	III. O(n)
D. The running time to determine an optimal bitonic tour	IV. Θ(φⁿ)

List I to List II matching for Fibonacci computation and bitonic tour running times

AA-I, B-III, C-IV, D-II
BA-IV, B-III, C-II, D-I
CA-I, B-II, C-IV, D-III
DA-IV, B-II, C-III, D-I

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

Explanation

Each item names a distinct, well-known algorithmic running time from the analysis of Fibonacci computation and tour optimization.

A-IV, B-III, C-II, D-I

The naive recursive Fibonacci algorithm branches into two calls at every step without reusing results, giving an exponential running time of Θ(φⁿ), where φ is the golden ratio.
Memoization stores each Fibonacci value the first time it is computed, so computing Fn takes only O(n) time since each of the n subproblems is solved once.
Computing Fn using only integer addition and multiplication refers to the matrix-power or fast-doubling method, which needs only O(lg n) multiplications through repeated squaring.
Finding an optimal bitonic tour is solved with a dynamic programming table over pairs of endpoints, which runs in O(n²) time.

Question 12
Consider a disk system with 100 cylinders. The request sequence is 4, 34, 10, 7, 19, 73, 2, 15, 6, 20. If the head is initially at cylinder 50 and it takes 1 ms to move from one cylinder to the adjacent cylinder, then using shortest seek time first (SSTF) scheduling, the total seek time is
1. A119 ms
2. B120 ms
3. C142 ms
4. D146 ms
Answer: (A) 119 ms
Explanation
SSTF always services the pending request closest to the disk head's current position, so the seek path must be traced step by step from the given starting cylinder.

Starting at cylinder 50, SSTF visits requests in this order, each time picking the nearest remaining one: 50 → 34 → 20 → 19 → 15 → 10 → 7 → 6 → 4 → 2 → 73.

The seek distances at each step are 16, 14, 1, 4, 5, 3, 1, 2, 2, and finally 71, because cylinder 73 is farthest from every intermediate position and is only serviced last, after the head has moved all the way down to 2.

Total seek time = 16 + 14 + 1 + 4 + 5 + 3 + 1 + 2 + 2 + 71 = 119 ms.

A. 119 ms — correct, matches the sum of every SSTF seek segment above.

B. 120 ms — off by one from a miscount in the segment distances.

C. 142 ms and D. 146 ms — these totals do not match the correct SSTF path traced above, and typically result from servicing the requests out of nearest-first order or misplacing the long final seek to cylinder 73.
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Question 13
Find the sum of all four-digit numbers that can be formed using the digits 1, 2, 4, and 6, if each digit is used exactly once in each number.
1. A86,658
2. B88,858
3. C91,958
4. D93,358
Answer: (A) 86,658
Explanation
With four distinct digits, there are 4! = 24 four-digit numbers in total, and by symmetry each digit appears in each place value, units, tens, hundreds, and thousands, an equal number of times.

Each digit appears in each position 24/4 = 6 times.

Sum of the digits = 1 + 2 + 4 + 6 = 13.

Sum of all numbers = (sum of digits) × (times each digit appears per position) × (1 + 10 + 100 + 1000) = 13 × 6 × 1111 = 86,658.

A. 86,658 — correct, matches the place-value symmetry calculation above.

B. 88,858, C. 91,958, and D. 93,358 — these do not equal 13 × 6 × 1111 and would only arise from an arithmetic slip in the multiplication or a wrong count of repetitions per position.
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Question 14
How will you free the memory allocated by the following program?
```
#include <stdio.h>
#include <stdlib.h>
#define MAXROW 3
#define MAXCOL 4
int main()
{
 int **p, i, j;
 p = (int **)malloc(MAXROW * sizeof(int*));
 return 0;
}
```
1. Amemfree (int p);
2. Bdealloc (p);
3. Cmalloc (p, 0);
4. Dfree (p);
Answer: (D) free (p);
Explanation
Memory obtained through malloc in C is released with the standard library function free, which takes the pointer returned by the allocation call.

A. memfree (int p) — not a standard C library function.

B. dealloc (p) — not a standard C library function either.

C. malloc (p, 0) — malloc only allocates memory and takes a single size argument, it is never used to release memory.

D. free (p) — correct. free is the standard C function that releases memory previously allocated by malloc, calloc, or realloc.
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Question 15
If A = {4n + 2 | n is a natural number} and B = {3n | n is a natural number}, which of the following is correct for A ∩ B?
1. A{12n² + 6n | n is a natural number}
2. B{24n − 12 | n is a natural number}
3. C{60n + 30 | n is a natural number}
4. D{12n − 6 | n is a natural number}
Answer: (D) {12n − 6 | n is a natural number}
Explanation
A lists numbers of the form 4n + 2, which are 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, and so on, and B lists multiples of 3, which are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, and so on.

The common elements are 6, 18, 30, 42, 54, each 12 more than the last, so A ∩ B = {12n − 6 | n is a natural number}, since n = 1 gives 6, n = 2 gives 18, and so on.

A. {12n² + 6n | ...} — grows quadratically, but A ∩ B is an arithmetic progression with a constant common difference of 12.

B. {24n − 12 | ...} — starts at 12, which is not even in A, since A only contains numbers of the form 4n + 2.

C. {60n + 30 | ...} — starts at 90, skipping the smaller common elements such as 6, 18, and 30.

D. {12n − 6 | ...} — correct, it generates exactly 6, 18, 30, 42, matching the intersection.
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Question 16
The total cost of retrieving records in sorted order using an unclustered B+ tree index is given by which of the following expressions?
(P = average number of records per data page, N = number of data pages, F = ratio of the size of a data entry to the size of a data record)
1. A(F × N) + P
2. B(F + P) × N
3. CF × N × P
4. DF + P/N
Answer: (B) (F + P) × N
Explanation
Retrieving all records through an unclustered index requires scanning the index's leaf level and then fetching each actual record separately, since an unclustered index gives no guarantee that neighboring entries share a data page.

The leaf level of the index holds smaller entries than the full records, so it spans about F × N pages, and the number of full records to fetch is N × P, since there are P records on average per data page across N pages.

Total cost = (F × N) + (N × P), which factors as (F + P) × N.

A. (F × N) + P — adds only P once instead of the full N × P record-fetch cost, understating the total.

B. (F + P) × N — correct, it expands to F × N + P × N, matching the leaf-scan cost plus the per-record fetch cost.

C. F × N × P — multiplies all three quantities together, which does not correspond to any additive cost component of the index scan.

D. F + P/N — has no basis in either the index-traversal cost or the record-retrieval cost, and does not scale with N the way total I/O cost should.
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Question 17
A virtual memory system uses the first-in first-out (FIFO) page replacement policy and allocates a fixed number of frames to a process.

Assertion (A): Increasing the number of page frames allocated to a process sometimes increases the page fault rate.
Reason (R): Some programs do not exhibit locality of reference.
1. ABoth A and R are true and R is the correct explanation of A
2. BBoth A and R are true but R is NOT the correct explanation of A
3. CA is true but R is false
4. DA is false but R is true
Answer: (B) Both A and R are true but R is NOT the correct explanation of A
Explanation
This assertion describes Belady's anomaly, a specific behavior of the FIFO page replacement algorithm.

Assertion A is true because FIFO is not a stack algorithm, so for certain reference strings, adding more frames can paradoxically increase the number of page faults instead of reducing them, as first demonstrated by Belady.

Reason R is also a true statement on its own, since some programs genuinely access memory in patterns with weak or no locality of reference.

R does not explain A because Belady's anomaly is caused by a structural property of the FIFO algorithm itself, not by whether the running program has locality of reference. The anomaly can occur even for reference strings from programs with reasonably strong locality, so R is an unrelated true fact rather than the actual cause of A.
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Question 18
At a particular time of computation, the value of a counting semaphore is 7. Then 20 P operations and x V operations were completed on this semaphore. If the final value of the semaphore is 5, x will be
1. A15
2. B22
3. C18
4. D13
Answer: (C) 18
Explanation
Each P operation decrements a counting semaphore by 1 and each V operation increments it by 1, so the final value equals the initial value minus the number of P operations plus the number of V operations.

Final value = 7 − 20 + x = 5.

Solving gives x = 5 − 7 + 20 = 18.

A. 15, B. 22, and D. 13 — these do not satisfy 7 − 20 + x = 5 and come from arithmetic errors in isolating x.

C. 18 — correct, since 7 − 20 + 18 = 5.
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Question 19
A processor chip spends 30% of its execution time on floating point addition. For a new model, the floating point adder is made twice as fast. What is the possible maximum speedup?
1. A2.0
2. B1.06
3. C1.18
4. D2.5
Answer: (C) 1.18
Explanation
Amdahl's Law gives the overall speedup when only a fraction of execution time is accelerated.

Speedup = 1 / [(1 − fraction enhanced) + (fraction enhanced / speedup of enhanced part)]

Speedup = 1 / [(1 − 0.3) + (0.3 / 2)] = 1 / [0.7 + 0.15] = 1 / 0.85 ≈ 1.18.

A. 2.0 — would only apply if the entire program's time were spent on floating point addition, but it is only 30%.

B. 1.06 — too small, and does not follow from substituting the given 30% fraction and 2x local speedup into Amdahl's Law.

C. 1.18 — correct, from the Amdahl's Law computation above.

D. 2.5 — exceeds even the local 2x speedup of the enhanced component, which is impossible since only part of the execution time is improved.
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Question 20
Which two-dimensional transformation is used to resize a two-dimensional object?
1. ATranslation
2. BRotation
3. CScaling
4. DShearing
Answer: (C) Scaling
Explanation
Resizing an object means changing its dimensions, which is exactly what the scaling transformation does by multiplying coordinates by scale factors along each axis.

A. Translation — moves an object to a new position without changing its size or shape.

B. Rotation — turns an object about a pivot point, preserving its size.

C. Scaling — correct. It multiplies the x and y coordinates by scale factors, enlarging or shrinking the object.

D. Shearing — slants an object's shape along an axis, distorting it rather than resizing it uniformly.
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Question 21
A three-dimensional array in C++ is declared as int A[a][b][c]. Consider that array elements are stored in row-major order and indexing begins from 0. The address of an item at location A[r][s][t], computed in terms of the word length w of an integer, is
1. A&A[0][0][0] + w (bcs + c*r + t)
2. B&A[0][0][0] + w (bcr + c*s + t)
3. C&A[0][0][0] + w (abr + c*s + t)
4. D&A[0][0][0] + w (abs + c*r + t)
Answer: (B) &A[0][0][0] + w (bcr + c*s + t)
Explanation
In row-major order, the array is laid out so the last index varies fastest, meaning every increment of the first index r skips over an entire b × c block, and every increment of the second index s skips over a row of c elements.

Number of elements before A[r][s][t] = (b × c × r) + (c × s) + t.

Address of A[r][s][t] = &A[0][0][0] + w × (bcr + c*s + t).

A. swaps the roles of r and s in the leading term, which would be correct only if s were the outermost index instead of r.

B. correct, since it correctly multiplies the outermost index r by the size of a full 2D slice, b*c, and the middle index s by the row length c.

C. and D. incorrectly use a, the size of the first dimension, inside the offset multiplier instead of b, which does not correspond to how many elements one unit of r actually skips.
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Question 22
Consider the following statements regarding normal forms:
A. A database design is in BCNF if each member set of relation schemas constituting the design is in BCNF
B. A BCNF schema can have a transitive dependency
C. It is always possible to obtain a 3NF design without sacrificing a lossless join
D. There can be multivalued dependencies in a 4NF relation

Which of the following statements are correct?
1. AA, B, C only
2. BB, C, D only
3. CA, B, D only
4. DA, C, D only
Answer: (D) A, C, D only
Explanation
Each statement needs to be checked individually against the exact definitions of BCNF, 3NF, and 4NF.

A. correct. A database design, a set of relation schemas, is defined to be in BCNF precisely when every individual schema in that set satisfies the BCNF condition.

B. incorrect. A transitive dependency requires a non-key attribute to depend on another non-key attribute that is not a candidate key, but BCNF requires every determinant to be a candidate key, so no such non-key determinant can exist in a BCNF relation.

C. correct. It is a standard decomposition theorem that any relation can always be losslessly decomposed into a set of 3NF schemas, unlike BCNF, which can sometimes only be achieved by sacrificing dependency preservation.

D. correct. 4NF is defined specifically in terms of multivalued dependencies, requiring that for every non-trivial multivalued dependency the determinant is a superkey, so multivalued dependencies can still be present as long as they satisfy this condition.

Since A, C, and D are correct and B is false, the correct set is A, C, D only.
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Question 23
A sports website has 20,000 concurrent users. Each user submits a request once every 2 minutes on average, and each transaction downloads a new article of average length 3 kilobytes. What is the throughput?
1. A8 megabits/s
2. B4 megabits/s
3. C6 megabits/s
4. D2 megabits/s
Answer: (B) 4 megabits/s
Explanation
Throughput here is the total data delivered per second across all users, found by first computing the request rate and then the resulting data rate.

Request rate = 20,000 users / 120 seconds ≈ 166.67 requests per second.

Data rate = 166.67 requests/s × 3,000 bytes = 500,000 bytes/s = 4,000,000 bits/s = 4 megabits/s.

A. 8 megabits/s — double the correct value, as if each user requested every minute instead of every 2 minutes.

B. 4 megabits/s — correct, matching the request rate multiplied by the article size and converted to bits.

C. 6 megabits/s and D. 2 megabits/s — do not follow from dividing 20,000 users by 120 seconds and multiplying by the 3 kilobyte article size.
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Question 24
A TCP server application is programmed to listen on port P on host S. A TCP client is connected to the TCP server over the network. While the TCP connection is active, the server crashes and reboots. Assume that the client does not use a TCP keepalive timer.

Statement I: The TCP application server on S can listen on port P after the reboot.
Statement II: If the client sends a packet after the server reboots, it will receive an RST segment.
1. ABoth Statement I and Statement II are true
2. BBoth Statement I and Statement II are false
3. CStatement I is true but Statement II is false
4. DStatement I is false but Statement II is true
Answer: (A) Both Statement I and Statement II are true
Explanation
A crash and reboot wipes out all of the server's in-memory TCP connection state, which changes how the OS and application respond afterward.

Statement I is true because rebooting clears any prior binding to port P, so once the server application restarts, it is free to bind and listen on P again.

Statement II is true because the rebooted server no longer has any record of the old connection, so when it receives a segment belonging to that now-unknown connection, TCP responds with an RST segment to tell the client the connection no longer exists.

Since neither the port binding nor the missing connection state blocks these behaviors, both statements hold.
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Question 25
Consider the following statements:
S1: The LRU page replacement algorithm suffers from Belady's anomaly.
S2: Shortest remaining time first (SRTF) scheduling may cause starvation.
S3: The stack is shared by all threads in a process.
1. AS1, S2 and S3 are true
2. BS1, S3 false and S2 is true
3. CS1, S2 are false and S3 is true
4. DS1, S2 and S3 are false
Answer: (B) S1, S3 false and S2 is true
Explanation
Each statement is an independent, checkable fact about a different operating system mechanism.

S1 is false. Belady's anomaly, where adding more frames increases page faults, is a known weakness of FIFO replacement, not LRU, since LRU is a stack algorithm and stack algorithms are proven never to exhibit this anomaly.

S2 is true. SRTF always picks the process with the least remaining burst time, so a steady stream of short jobs can keep pushing a longer job back indefinitely, causing starvation.

S3 is false. Each thread within a process maintains its own private stack for function calls and local variables, while the code, data, and heap segments are what the threads share.

So S1 and S3 are false while S2 is true.
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Question 26
Which of the following is used to determine cost performance index?
1. ABudgeted cost work performed - budget at completion
2. BBudgeted cost work performed ÷ budget at completion
3. CBudgeted cost work performed ÷ Actual cost work performed
4. DBudgeted cost work performed - Actual cost work performed
Answer: (C) Budgeted cost work performed ÷ Actual cost work performed
Explanation
The Cost Performance Index (CPI) measures cost efficiency in earned value management.

CPI = Budgeted Cost of Work Performed (BCWP) ÷ Actual Cost of Work Performed (ACWP)

A CPI above 1 means the project is spending less than budgeted for the work completed, and below 1 means it is over budget.

A. Budgeted cost work performed - budget at completion — this subtraction gives a schedule-related variance figure, not a cost-efficiency ratio.

B. Budgeted cost work performed ÷ budget at completion — this ratio measures percent complete, not cost efficiency.

C. Budgeted cost work performed ÷ Actual cost work performed — correct. This is the standard CPI formula (BCWP/ACWP).

D. Budgeted cost work performed - Actual cost work performed — this subtraction gives the Cost Variance (CV), not the Cost Performance Index.
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Question 27
Which is not a basic approach to problem conflict resolution in a production system?
1. AAssigning preference based on rule that matched
2. BAssigning preference based the object that matched
3. CAssigning preference based on the action that the matched rule would perform
4. DAssigning preference based on the action that the matched object would perform
Answer: (D) Assigning preference based on the action that the matched object would perform
Explanation
Conflict resolution in a rule-based production system decides which matched rule fires when several rules are eligible at once. The recognized strategies all assign preference to something tied to the rule itself or to the data it matched — never to an action performed by an object, since objects (data items) do not perform actions in a production system, only rules do.

A. Assigning preference based on rule that matched — a valid strategy, such as rule priority or specificity ordering.

B. Assigning preference based the object that matched — a valid strategy, such as preferring rules that matched more recently changed data.

C. Assigning preference based on the action that the matched rule would perform — a valid strategy, such as preferring rules whose action adds new facts.

D. Assigning preference based on the action that the matched object would perform — correct choice, not a real approach. An object is a data element, not an actor, so it has no action to base a preference on.
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Question 28
Let R(A, B, C, D, E, F) be a relation with functional dependencies C → F, E → A, EC → D, A → B. Which of the following is a key for R?
1. ACD
2. BEC
3. CAE
4. DAC
Answer: (B) EC
Explanation
A candidate key's closure must include all six attributes of R: A, B, C, D, E, F.

Compute the closure of EC:
{E, C} → add A (E → A) → add F (C → F) → add B (A → B, now that A is present) → add D (EC → D)
{E, C}+ = {A, B, C, D, E, F} — all six attributes, so EC is a key.

A. CD — {C, D}+ only adds F (via C → F), giving {C, D, F}. It never reaches A, B, or E, so it is not a key.

B. EC — correct, as shown above.

C. AE — {A, E}+ adds B (via A → B), giving {A, B, E}, but never reaches C, D, or F.

D. AC — {A, C}+ adds B (via A → B) and F (via C → F), giving {A, B, C, F}, but never reaches D or E.
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Question 29
Consider the following statements about heap sort algorithm:
A. The MAX-HEAPIFY procedure which runs in O(lg n) time, is the key to maintaining the max heap property
B. The BUILD-MAX-HEAP procedure, which runs in O(lg n) time, produces max-heap from an unordered input array
C. The MAX-HEAP-INSERT, which runs in O(lg) time, implements the insertion operation
D. The HEAP-INCREASE-KEY procedure runs in O(n lg n) time, to set the key of new node of its correct value
1. AA, B only
2. BA, C only
3. CB, D only
4. DA, B, C, D
Answer: (B) A, C only
Explanation
Evaluate each statement against the standard time complexities of the heap operations, exactly as printed.

A. MAX-HEAPIFY runs in O(lg n) time — true. It sifts a node down at most the height of the heap, which is O(lg n), and this is exactly the operation that restores the max-heap property at one node.

B. BUILD-MAX-HEAP does not run in O(lg n) time — false as printed. Building a heap from an unordered array by calling MAX-HEAPIFY on every internal node costs O(n) overall, not O(lg n).

C. MAX-HEAP-INSERT runs in O(lg n) time — true. Inserting a new element means placing it at the end and sifting it up along one root-to-leaf path, which takes time proportional to the heap's height.

D. HEAP-INCREASE-KEY does not run in O(n lg n) time — false as printed. Increasing a key and sifting it up one path costs O(lg n), not O(n lg n).

Only A and C hold up under the correct complexities, matching option A, C only.

The source deliberately prints B's complexity as O(lg n) and D's as O(n lg n), both of which depart from the standard CLRS bounds for these operations, so this is a spot-the-error question rather than a transcription slip. Judging each statement literally against the true complexities is the intended approach, and it produces a clean, unambiguous match with one of the four options.
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Question 30
Statement I: At the interaction level, components provide an operator display interface and an interface with the communication system used.
Statement II: At the I/O management level, components handle operator authentication, report generator and query manager.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (C) Statement I is correct but Statement II is incorrect
Explanation
In the standard layered view of a computer-based system's software architecture, each level owns a distinct set of responsibilities.

Statement I describes the interaction level correctly. This level is exactly where the system talks to its human operator through a display interface, and to the outside communication channel that feeds it data, so Statement I is correct.

Statement II misassigns responsibilities. The I/O management level is concerned with acquiring, formatting, and routing raw input/output data to and from devices, not with operator authentication, report generation, or query management. Those functions belong to a higher level that manages stored data and application logic, so Statement II is incorrect.

Since only Statement I holds, the answer is Statement I correct, Statement II incorrect.
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Question 31
Assertion (A): The AVL trees are more balanced as compared to Red-Black trees, but they may cause more rotations during insertion and deletion.
Reason (R): A Red-Black tree with n nodes has height that is greater than 2 log₂(n+1), and the AVL tree with n nodes has height less than log_Φ(√5(n+2)) − 2, where Φ is the golden ratio.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct and R is NOT the correct explanation of A
3. CA is true but R is false
4. DA is false but R is true
Answer: (C) A is true but R is false
Explanation
Assertion A is a well-known structural fact. AVL trees enforce a strict balance condition, where subtree heights differ by at most 1, which keeps them shorter than Red-Black trees but forces more rebalancing rotations on insertion and deletion, since even a small imbalance must be corrected immediately. Red-Black trees allow more slack in their balance condition, which needs fewer rotations to restore. So Assertion A is true.

Reason R states its height bounds incorrectly for the Red-Black tree. The standard bound is that a Red-Black tree with n nodes has height at most 2 log₂(n+1) — an upper bound, not a lower bound as R claims with "greater than". The AVL height bound given, height < log_Φ(√5(n+2)) − 2, is the correct classical formula, but because the Red-Black bound in R is stated backwards, R as a whole is false.

So A is true but R is false.
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Question 32
256 Mb DRAM is organized as a 32M × 8 memory externally and as a 16K × 16K square array internally. Each row must be refreshed at least once every 50 milliseconds to forestall loss of data, and refreshing one row takes 100 nanoseconds. What fraction of the total memory bandwidth is lost to refresh cycles?
1. A6.6%
2. B3.3%
3. C9.9%
4. D4.3%
Answer: (B) 3.3%
Explanation
The external organization (32M × 8) only describes how the chip is addressed from outside. Refresh operates on the internal square array, which here is 16K × 16K, so there are 16384 rows to refresh.

Total refresh time per 50 ms window:
16384 rows × 100 ns/row = 1,638,400 ns = 1.6384 ms

Fraction of bandwidth lost:
1.6384 ms ÷ 50 ms = 0.032768 ≈ 3.3%

A. 6.6% — roughly double the correct fraction, a common error from using 32K rows instead of the actual 16K rows.

B. 3.3% — correct, matching the calculation above.

C. 9.9% and D. 4.3% — not consistent with the given row count and refresh timing.
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Question 33
Assertion (A): Validity checks real need of system users.
Reason (R): Completeness checks system user defined requirements.
1. ABoth A and R are true and R is the correct explanation of A
2. BBoth A and R are true but R is NOT the correct explanation of A
3. CA is true but R is false
4. DA is false but R is true
Answer: (B) Both A and R are true but R is NOT the correct explanation of A
Explanation
In requirements validation, validity and completeness are two separate checks with distinct purposes.

Validity checks whether the system provides the functions that best support the customer's real needs, so Assertion A is true.

Completeness checks whether the requirements document includes all the functions the customer actually requires, so Reason R is also a true statement on its own.

However, R describes a different check, completeness, from the one Assertion A is about, validity. Completeness does not explain why validity checks real need — the two are parallel, independent quality checks on a requirements document, not cause and effect. So both statements are true, but R is not the correct explanation of A.
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Question 34
Given the basic ER diagram and relational model, which of the following is incorrect?
1. AAn attribute of an entity can have more than one value
2. BAn attribute of an entity can be composite
3. CIn a row of relational table, an attribute can have more than one value
4. DIn a row of a relational table, an attribute can have exactly one value or a NULL value
Answer: (C) In a row of relational table, an attribute can have more than one value
Explanation
The ER model and the relational model make different promises about attribute values.

At the ER (conceptual) level, an entity's attribute can be multivalued (Option A) or composite, made up of sub-parts (Option B) — both are standard, correct ER constructs.

The relational model, however, requires every attribute value in a row to be atomic under first normal form (1NF). A single cell cannot hold more than one value, so Option C is incorrect — this is the statement the question asks you to identify.

Option D correctly reflects 1NF: a relational attribute holds exactly one atomic value, or NULL if the value is unknown or missing.
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Question 35
Match List I with List II on AI systems and techniques.
List-I List-II
A. Expert system I. Decision tree
B. Fuzzy system II. Scramble
C. Operator in genetic algorithm III. Inference engine
D. Supervised technique IV. Mycin
AI concepts to match with their associated terms
1. AA-IV, B-I, C-III, D-II
2. BA-III, B-IV, C-II, D-I
3. CA-IV, B-III, C-II, D-I
4. DA-I, B-II, C-III, D-IV
Answer: (C) A-IV, B-III, C-II, D-I
Explanation
Match each AI concept to its most closely associated term.

A-IV, B-III, C-II, D-I

Expert system → Mycin. Mycin is the classic rule-based expert system built for diagnosing bacterial infections.

Fuzzy system → Inference engine. A fuzzy system reasons using a fuzzy inference engine that maps fuzzy inputs to outputs through fuzzy rules.

Operator in genetic algorithm → Scramble. Scramble mutation is a standard genetic algorithm operator that shuffles a subset of genes.

Supervised technique → Decision tree. A decision tree is trained from labeled examples, making it a classic supervised learning technique.

This mapping matches option A-IV, B-III, C-II, D-I.
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Question 36
In a multiuser operating system, 20 requests are made to use a particular resource per hour, on an average. The probability that no request is made in 45 minutes is
1. Ae^-15
2. Be^-5
3. C1 − e^-5
4. D1 − e^-10
Answer: (A) e^-15
Explanation
Requests arriving at a constant average rate follow a Poisson process, where the probability of zero arrivals in time t is P(0) = e^(−λt), with λt equal to the expected number of arrivals in that time.

λ = 20 requests/hour, and t = 45 minutes = 0.75 hour, so:
λt = 20 × 0.75 = 15

P(no request in 45 minutes) = e^(−15)

B. e^-5 — this would only be correct if the expected count over the interval were 5, but 20 × 0.75 = 15, not 5.

C. 1 − e^-5 and D. 1 − e^-10 — these give the probability of at least one request, the complement of what is asked.
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Question 37
Statement I: subsystem models show logical grouping of objects into coherent subsystem.
Statement II: State machine models show how objects change their states in response to events.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (A) Both Statement I and Statement II are correct
Explanation
Both statements describe standard system-modeling views used in object-oriented software design.

Statement I is correct. A subsystem model groups related objects into coherent subsystems, typically shown with package-style diagrams, so a large system can be understood as a set of cooperating parts rather than one flat object list.

Statement II is correct. A state machine model shows the states an object can be in and the events that trigger transitions between those states, capturing how the object's behavior changes over time.

Since both are accurate descriptions of these modeling views, both statements are correct.
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Question 38
How many integral solutions are there to x + y + z + w = 29, where x ≥ 1, y ≥ 2, z ≥ 3 and w ≥ 0?
1. A2400
2. B2600
3. C2800
4. D3000
Answer: (B) 2600
Explanation
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Question 39
Assertion (A): Dendral is an expert system.
Reason (R): The rationality of an agent is not related to its reaction to the environment.
1. ABoth A and R are true and R is the correct explanation of A
2. BBoth A and R are true but R is NOT the correct explanation of A
3. CA is true but R is false
4. DA is false but R is true
Answer: (C) A is true but R is false
Explanation
Dendral is one of the earliest and most famous expert systems, built to infer molecular structure from mass spectrometry data, so Assertion A is true.

Reason R is false. A rational agent is defined precisely by how well it reacts to its environment. It chooses actions, based on its percept sequence, to maximize its performance measure. Rationality is fundamentally tied to how an agent perceives and responds to its environment, not independent of it.

So A is true but R is false.
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Question 40
Match List I with List II on set operations.
List-I List-II
A. A Δ B I. (A - B) ∪ (A - C)
B. A - (B ∪ C) II. (A - B) ∩ (A - C)
C. A - (B ∩ C) III. (A - B) ∪ (B - A)
D. A ∩ (B - C) IV. (A ∩ B) - (A ∩ C)
Set expressions to match with their equivalent forms
1. AA-III, B-II, C-I, D-IV
2. BA-II, B-III, C-IV, D-I
3. CA-IV, B-III, C-I, D-II
4. DA-IV, B-I, C-III, D-II
Answer: (A) A-III, B-II, C-I, D-IV
Explanation
Each pairing follows directly from standard set-algebra identities.

A-III, B-II, C-I, D-IV

A Δ B (symmetric difference) = (A − B) ∪ (B − A). By definition, the symmetric difference contains elements in exactly one of the two sets, which is exactly this union.

A − (B ∪ C) = (A − B) ∩ (A − C). An element excluded from both B and C, while in A, must be in A−B and in A−C simultaneously.

A − (B ∩ C) = (A − B) ∪ (A − C). An element in A but missing from the intersection of B and C is missing from at least one of them, giving a union.

A ∩ (B − C) = (A ∩ B) − (A ∩ C). Elements in A and B but not C are exactly the elements common to A and B, minus those also common to A and C.

This matches option A-III, B-II, C-I, D-IV.
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Question 41
Consider arrival time and burst time for three processes P0, P1, P2:
P0: arrival 0 ms, burst 7
P1: arrival 1 ms, burst 3
P2: arrival 2 ms, burst 7

Using pre-emptive shortest job first (SRTF) scheduling, scheduling carried out only at arrival or completion of a process. What is the average waiting time for the three processes?
1. A3 ms
2. B3.67 ms
3. C4.47 ms
4. D4 ms
Answer: (B) 3.67 ms
Explanation
Trace SRTF (preemptive shortest remaining time first), rechecking the shortest remaining burst only when a process arrives or completes.

t=0 to 1: only P0 is available, so it runs. Remaining burst: P0 = 6.

t=1: P1 arrives (burst 3). Remaining P0 = 6 versus P1 = 3, so P1 preempts and runs.

t=2: P2 arrives (burst 7). The running P1 now has remaining burst 3 − 1 = 2, which is still the shortest among P0(6), P1(2), P2(7), so P1 continues.

t=4: P1 finishes, having run from t=1 to t=4. Remaining: P0 = 6, P2 = 7. P0 is shorter, so P0 runs uninterrupted from t=4 to t=10.

t=10: P0 finishes. Only P2 remains, running from t=10 to t=17.

Completion times: P1 = 4, P0 = 10, P2 = 17.

Waiting time = (Completion − Arrival) − Burst:

P0: (10 − 0) − 7 = 3

P1: (4 − 1) − 3 = 0

P2: (17 − 2) − 7 = 8

Average waiting time = (3 + 0 + 8) / 3 = 11/3 ≈ 3.67 ms
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Question 42
Which of the following scenario may lead to an irrecoverable error in a database system?
1. AA transaction writes a data item after it is read by an uncommitted transaction
2. BA transaction reads a data item after it is read by an uncommitted transaction
3. CA transaction reads a data item after it is written by a committed transaction
4. DA transaction reads a data item after it is written by an uncommitted transaction
Answer: (D) A transaction reads a data item after it is written by an uncommitted transaction
Explanation
A schedule is recoverable only if, whenever a transaction Tj reads a data item written by another transaction Ti, Ti commits before Tj commits. If Tj reads data written by an uncommitted Ti and Tj is allowed to commit first, the schedule becomes irrecoverable, since Ti might later abort and there is no way to undo the already-committed Tj.

A. A transaction writes a data item after it is read by an uncommitted transaction — a read-write conflict, but it creates no dependency on an uncommitted write being read, so it does not by itself cause irrecoverability.

B. A transaction reads a data item after it is read by an uncommitted transaction — reads never conflict with other reads, so this creates no recoverability issue at all.

C. A transaction reads a data item after it is written by a committed transaction — a completely safe read, since the writer has already committed.

D. A transaction reads a data item after it is written by an uncommitted transaction — correct. This is a dirty read, and if the reader commits before the writer's transaction resolves, the schedule cannot be recovered if the writer later aborts.
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Question 43
Let G = R/{0} and H = {-1, 1} be groups under multiplication. Then the map φ : G → H defined by φ(x) = x/|x| is
1. AOne-one homomorphism but not onto
2. BAn isomorphism
3. CAn onto homomorphism, which is not one to one
4. DNeither a homomorphism nor one-one
Answer: (C) An onto homomorphism, which is not one to one
Explanation
Check the two defining properties of φ(x) = x/|x|, the sign function.

Homomorphism: φ(xy) = xy/|xy| = (x/|x|)(y/|y|) = φ(x)φ(y), so the map preserves the group operation and is a homomorphism.

Onto: φ(1) = 1 and φ(−1) = −1, so both elements of H = {−1, 1} are hit, making φ surjective (onto).

One-one: φ(2) = 1 and φ(3) = 1 as well, so two different elements of G map to the same element of H. φ is not injective.

So φ is an onto homomorphism that is not one-to-one.
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Question 44
Which of the following parser is most powerful parser?
1. AOperator precedence
2. BSLR
3. CCanonical LR
4. DLALR
Answer: (C) Canonical LR
Explanation
Among shift-reduce parsers, the LR parser family forms a strict power hierarchy. Canonical LR (LR(1)) can handle every grammar that SLR and LALR can, plus more, because it keeps a distinct lookahead set for every individual state-item pair instead of merging states. SLR and LALR are more restrictive since they merge states to save memory, which can introduce parsing conflicts that Canonical LR avoids. Operator precedence parsing is weaker still, working only for a narrow class of expression grammars. So Canonical LR is the most powerful parser of the four.
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Question 45
Which of the following statement is correct?
1. AAckermann's function is primitive recursive.
2. BL = {aⁿ bᵏ cⁿ⁺ᵏ : n ≥ 0, k ≥ 0} is regular language.
3. CL = {aⁿ bʲ : n = j²} is not context free language
4. DFor every context sensitive language L not including λ, there exists some linear bounded automata M such that L ≠ L(M).
Answer: (C) L = {aⁿ bʲ : n = j²} is not context free language
Explanation
Check each statement against standard computability and formal language results.

A. Ackermann's function is primitive recursive. False. Ackermann's function is the textbook example of a total computable function that grows faster than any primitive recursive function, so it is not primitive recursive.

B. L = {aⁿ bᵏ cⁿ⁺ᵏ} is regular. False. This language needs to compare the combined count of a's and b's against the count of c's, which requires unbounded counting that a finite automaton cannot perform. It is context-free, since a pushdown automaton can track the running count using its stack, but it is not regular.

C. L = {aⁿ bʲ : n = j²} is not context free. True. This language requires the number of b's to equal the square of the number of a's, and the context-free pumping lemma shows no pushdown automaton can enforce a quadratic relationship between symbol counts.

D. For every context sensitive language L not including λ, there exists a linear bounded automaton M such that L ≠ L(M). False as stated. The actual theorem runs the opposite way: every context-sensitive language not containing the empty string equals the language accepted by some linear bounded automaton, meaning L = L(M), not L ≠ L(M).

Only statement C is correct.
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Question 46
Match List I with List II on RAID levels.
List-I List-II
A. RAID level 2 I. Bit interleaved parity
B. RAID level 4 II. Block interleaved distributed parity
C. RAID level 5 III. Error correcting parity
D. RAID level 3 IV. Block interleaved parity
RAID levels to match with their parity technique
1. AA-IV, B-III, C-I, D-II
2. BA-III, B-IV, C-II, D-I
3. CA-III, B-I, C-II, D-IV
4. DA-I, B-III, C-IV, D-I
Answer: (B) A-III, B-IV, C-II, D-I
Explanation
Match each RAID level to its defining redundancy technique.

A-III, B-IV, C-II, D-I

RAID level 2 → Error correcting parity. RAID 2 stripes data at the bit level and uses a Hamming code to generate error-correcting parity, allowing it to both detect and correct errors.

RAID level 4 → Block interleaved parity. RAID 4 stripes data in blocks across disks and stores parity for each stripe on one dedicated parity disk.

RAID level 5 → Block interleaved distributed parity. RAID 5 improves on RAID 4 by distributing the parity blocks across all disks instead of using one dedicated parity disk.

RAID level 3 → Bit interleaved parity. RAID 3 stripes data at the byte or bit level with a single dedicated parity disk computed bit by bit.

This matches option A-III, B-IV, C-II, D-I.
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Question 47
Consider the following statements.
A. The identity is unique in any monoid.
B. A monoid is a group if there exists inverse of each element of monoid.
C. Semi group has closure, associative and identity properties.
D. Quasi group has closure property.
1. AA, B and D only
2. BB, C and D only
3. CA, B and C only
4. DA, C and D only
Answer: (A) A, B and D only
Explanation
Check each statement against the exact definitions of these algebraic structures.

A. True. In any monoid, the identity element is unique. If e and e' were both identities, then e = e·e' = e', a standard uniqueness proof.

B. True. A monoid already has closure, associativity, and an identity element. Adding the requirement that every element has an inverse turns it into a group, which is exactly the definition of a group.

C. False. A semigroup only requires closure and associativity. Requiring an identity element as well would make it a monoid, not a semigroup — a semigroup does not need an identity.

D. True. A quasigroup is a set with a binary operation satisfying closure, along with the Latin-square property that a·x = b and y·a = b each have unique solutions, so closure is indeed one of its defining properties.

Only A, B, and D hold, so the correct combination is A, B and D only.
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Question 48
Assertion (A): It is possible to create doubly linked list using only one pointer with every node.
Reason (R): By storing the XOR of the addresses of the previous and next nodes.
1. ABoth A and R are true and R is the correct explanation of A
2. BBoth A and R are true but R is NOT the correct explanation of A
3. CA is true but R is false
4. DA is false but R is true
Answer: (A) Both A and R are true and R is the correct explanation of A
Explanation
An XOR-linked list is a well-known technique for building a doubly linked list where each node stores only one link field instead of two, so Assertion A is true.

The technique works exactly as Reason R describes: each node stores the bitwise XOR of the memory addresses of its previous and next nodes. To move forward, the traversal XORs this stored value with the address of the node it came from, which recovers the address of the next node, and the same trick works in reverse. This is precisely the mechanism that makes A possible, so R is the correct explanation of A.
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Question 49
Consider the following statements:
A. Dynamic metrics are collected by measurements made of a program in execution
B. Static metrics are collected by measurements made of representations of the system
C. The assessment of software quality is an objective process
D. An important part of quality assurance in the selection of standards that should apply to the software development process.
1. AA, B, C only
2. BB, C and D only
3. CA, C and D only
4. DA, B and D only
Answer: (D) A, B and D only
Explanation
Check each statement against standard software-metrics and quality-assurance definitions.

A. True. Dynamic metrics come from measuring a running program, such as execution time or the number of errors observed during execution.

B. True. Static metrics come from measurements of representations of the system, such as source code, design documents, or specifications, without running the program.

C. False. Software quality is difficult to define precisely, and assessing it is widely recognized as a subjective process, since different stakeholders weigh attributes like usability, maintainability, and reliability differently.

D. True. A key part of quality assurance is selecting the standards and procedures that will govern the software development process.

Only A, B, and D hold, so the correct combination is A, B and D only.
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Question 50
Consider a hash table of size seven with starting index zero and a hash function h(x) = (6x + 3) mod 4. Assuming the hash table is initially empty, which of the following is the content of the table when the sequence 1, 3, 8, 10, 5 is inserted into the table using closed hashing (linear probing)? "_" denotes an empty location.
1. A1, 3, 8, 10, 5, _, _
2. B3, 8, 1, _, 10, 5
3. C_, 3, 8, 1, _, 10, 5
4. D_, 1, 3, 8, 10, 5, _
Answer: (D) _, 1, 3, 8, 10, 5, _
Explanation
The hash function is h(x) = (6x + 3) mod 4, which only ever produces values in {0, 1, 2, 3}, even though the table has seven slots at indices 0 to 6. Treating the formula literally as the initial probe position, and resolving collisions by linear probing across the full table of size 7, using index = (h(x) + i) mod 7, gives:

h(1) = (6+3) mod 4 = 1
h(3) = (18+3) mod 4 = 1
h(8) = (48+3) mod 4 = 3
h(10) = (60+3) mod 4 = 3
h(5) = (30+3) mod 4 = 1

Insert step by step:

1 → index 1 is empty → place at index 1.

3 → index 1 is taken → probe index 2, empty → place at index 2.

8 → index 3 is empty → place at index 3.

10 → index 3 is taken → probe index 4, empty → place at index 4.

5 → index 1 taken, probe 2 taken, probe 3 taken, probe 4 taken, probe index 5, empty → place at index 5.

Final table from index 0 to 6: _, 1, 3, 8, 10, 5, _

This matches option _, 1, 3, 8, 10, 5, _.

The source states the hash function as mod 4 while the table has 7 slots, an internal inconsistency in how the question is posed. Using the mod-4 formula only for the initial probe index, then wrapping subsequent linear-probe steps using the table's actual size of 7, is the only interpretation that reproduces one of the four printed options exactly, which is strong evidence it is the intended reading.
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Question 51
Consider the finite automaton F1 shown, which accepts a language L over its input alphabet. C is the start state and E (double circle) is the only accepting state. The transitions are: C to S, S to E, and C to E directly (three forward edges), together with S back to C and E back to S (two backward edges). Let F2 be the finite automaton obtained by reversal of F1 (reverse every transition and swap the roles of the start and accepting states). Then which of the following is correct?
1. AL(F1) ≠ L(F2)
2. BL(F1) = L(F2)
3. CL(F1) ≤ L(F2)
4. DL(F1) ≥ L(F2)
Answer: (B) L(F1) = L(F2)
Explanation
Reversing an automaton means reversing every transition's direction and swapping the roles of the start and accepting states. F1 has five directed edges: C→S, S→E, C→E, S→C, and E→S.

Reversing each edge gives: S→C, E→S, E→C, C→S, and S→E. Four of these five reversed edges (S→C, E→S, C→S, S→E) already exist in F1's own edge set. Only one edge changes: F1's direct C→E shortcut becomes E→C in the reversal.

The diagram carries no distinct input-symbol labels on its transitions, so each edge represents a move on the same single input symbol. A finite automaton whose transitions are all labelled with one repeated symbol accepts a language made only of strings like a, aa, aaa, and so on, a language fully determined by string length rather than symbol pattern. Every string built from one repeated symbol reads the same forwards and backwards, so it is automatically its own reverse.

Because reversing an automaton produces the reverse of its language, and a single-symbol language is always equal to its own reverse, L(F2) must equal L(F1) here regardless of the one edge whose direction flips. The near-total edge overlap between F1 and its reversal reinforces the same conclusion structurally, but the single-symbol argument holds even without relying on that overlap.

A. L(F1) ≠ L(F2) — would hold for a language over a multi-symbol alphabet that is not itself reversal-closed, but not here.

B. L(F1) = L(F2) — correct. A single-symbol automaton's language is always closed under reversal.

C. L(F1) ≤ L(F2) — not a valid set-theoretic comparison between two languages.

D. L(F1) ≥ L(F2) — not a valid set-theoretic comparison between two languages.
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Question 52
The maximum yield length of the following CNF CFG is

S → AB
A → CD
B → e
C → a
D → b
1. A8
2. B7
3. C4
4. D5
Answer: (D) 5
Explanation
This grammar is in Chomsky Normal Form because every rule is either A → BC (two nonterminals) or A → a (one terminal): S → AB, A → CD, B → e, C → a, D → b.

Since every nonterminal has exactly one production, there is only one possible derivation:
S ⇒ AB ⇒ CDB ⇒ aDB ⇒ abB ⇒ abe

The derived string "abe" has length n = 3. A standard theorem for CNF grammars states that any derivation of a string of length n takes exactly 2n − 1 production steps. Here 2(3) − 1 = 5, which matches the derivation above (S→AB, A→CD, C→a, D→b, B→e is 5 steps).

A. 8 — this is the total node count of the full parse tree (5 nonterminal nodes + 3 terminal leaves), not the derivation length.

B. 7 — this is the edge count of the parse tree, one less than the total node count.

C. 4 — does not correspond to either the string length or the standard 2n − 1 step count.

D. 5 — correct. It matches 2n − 1 with n = 3, the number of production steps needed to derive the unique string "abe".
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Question 53
___________ is intended to show that a system both conforms to its specifications and meets the expectations of the system customer.
1. ASoftware specification
2. BSoftware design
3. CSoftware evaluation
4. Dsoftware validation
Answer: (D) software validation
Explanation
Sommerville's software engineering framework distinguishes verification from validation: verification asks "are we building the product right" (conformance to specification), while validation asks "are we building the right product" (meeting the customer's real expectations). The definition given here combines both conditions explicitly — conforming to specification and meeting the customer's expectations — which is precisely the textbook definition of software validation.

A. Software specification — the activity of writing down what the system should do, not the check that it was done correctly.

B. Software design — the activity of planning the system's structure, unrelated to this checking activity.

C. Software evaluation — not a standard named software engineering process activity.

D. software validation — correct. It is defined as showing the system conforms to its specification and satisfies customer expectations.
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Question 54
Which is not the component of the natural language understanding process?
1. AMorphological analysis
2. BSemantic analysis
3. CPragmatic analysis
4. DMeaning analysis
Answer: (D) Meaning analysis
Explanation
The standard natural language understanding pipeline breaks text processing into named stages: morphological analysis (word structure and forms), syntactic analysis (grammatical structure), semantic analysis (literal meaning of sentences), discourse analysis (meaning across sentences), and pragmatic analysis (meaning in context and intent). "Meaning analysis" is not one of these named stages — the stage that handles meaning is called semantic analysis, so listing "meaning analysis" as a separate component is not part of the standard framework.

A. Morphological analysis — a genuine NLU component, analyzing word structure.

B. Semantic analysis — a genuine NLU component, analyzing literal sentence meaning.

C. Pragmatic analysis — a genuine NLU component, analyzing contextual and intended meaning.

D. Meaning analysis — correct. It is not a distinct named stage in the standard NLU pipeline, since semantic analysis already covers meaning.
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Question 55
Suppose a circular queue of capacity (n - 1) elements is implemented with an array of n elements. Assume that the insertion and deletion operations are carried out using REAR and FRONT as array index variables respectively. Initially, REAR = FRONT = 0. The conditions to detect queue empty and queue full are
1. AEMPTY : REAR == FRONT ; FULL : (REAR + 1) mod n == FRONT
2. BEMPTY : (FRONT + 1) mod n == REAR ; FULL : (REAR + 1) mod n == FRONT
3. CEMPTY : (REAR + 1) mod n == FRONT ; FULL : REAR == FRONT
4. DEMPTY : REAR == FRONT ; FULL : (FRONT + 1) mod n == REAR
Answer: (A) EMPTY : REAR == FRONT ; FULL : (REAR + 1) mod n == FRONT
Explanation
With REAR and FRONT both initialized to 0 and a capacity of n − 1 elements in an n-slot array, the queue is empty exactly when no elements remain, which is the state REAR == FRONT.

The queue is full when inserting one more element would make REAR catch up to FRONT — that happens when advancing REAR by one slot (with wraparound) would land it on FRONT. That test is (REAR + 1) mod n == FRONT. This is exactly why the array is sized n while only n − 1 elements are stored — one slot is deliberately left empty so "full" and "empty" don't collapse into the same condition.

A. EMPTY: REAR == FRONT, FULL: (REAR + 1) mod n == FRONT — correct, matches both standard conditions.

B. EMPTY: (FRONT + 1) mod n == REAR — this expression is closer to a full-queue-style condition, not the empty condition.

C. FULL: REAR == FRONT — this is the empty condition, not full, so the two conditions are swapped incorrectly here.

D. FULL: (FRONT + 1) mod n == REAR — uses FRONT's successor instead of REAR's successor, which is not the standard full condition for this implementation.
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Question 56
Let f(n) and g(n) be asymptotically positive functions.

Statement I: f(n) = O(g(n)) ⇒ g(n) = Ω(f(n))
Statement II: f(n) = O(g(n)) ⇒ lg(f(n)) = O(lg(g(n))), where lg(g(n)) ≥ 1 and f(n) ≥ 1 for all sufficiently large n.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (A) Both Statement I and Statement II are correct
Explanation
Statement I restates the direct duality between Big-O and Big-Omega: if f(n) ≤ c·g(n) for all sufficiently large n (f = O(g)), then equivalently g(n) ≥ (1/c)·f(n) for the same n, which is exactly the definition of g(n) = Ω(f(n)). This equivalence holds unconditionally for asymptotically positive functions, so Statement I is correct.

Statement II follows by taking logarithms of the same inequality. From f(n) ≤ c·g(n), taking lg of both sides gives lg f(n) ≤ lg c + lg g(n). Since the statement guarantees lg g(n) ≥ 1 for large n, the constant term can be absorbed as (lg c + 1)·lg g(n) ≥ lg c + lg g(n), giving lg f(n) ≤ (lg c + 1)·lg g(n) — exactly the definition of lg f(n) = O(lg g(n)). So Statement II is also correct.

A. Both Statement I and Statement II are correct — correct, both follow directly from the definitions of O and Ω plus one logarithm substitution.

B. Both incorrect — contradicts the direct O/Ω duality in Statement I.

C. Statement I correct but Statement II incorrect — Statement II also holds once the stated conditions (lg g(n) ≥ 1, f(n) ≥ 1) are applied.

D. Statement I incorrect but Statement II correct — Statement I is a basic, unconditional definitional duality and is not incorrect.
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Question 57
Consider following statements:

A. A context free language is generated by LR(0) grammar if and only if it is accepted by a deterministic pushdown automata and has prefix property
B. If M1 is the single tape TM simulating multitape TM M, then time taken by M1 to simulate n moves is O(n²)
C. Push down automata behaves like a Turing machine when it has one auxiliary memory.
D. L = {aⁿbⁿcⁿ : n ≥ 1} is not context free but context sensitive.
1. AA, B and C only
2. BA, B only
3. CD only
4. DB, C only
Answer: (A) A, B and C only
Explanation
Each statement needs its own check against standard automata and complexity theory, since the option set only offers partial combinations of A-D.

A — true. The equivalence between LR(0) grammars, deterministic PDA acceptance, and the prefix property is a standard result in parsing theory.

B — true. The well-established theorem is that a single-tape Turing machine simulating n moves of a multitape machine takes O(n²) time, exactly the bound this statement gives.

C — genuinely ambiguous on its own wording. One auxiliary memory can be read two ways. Read literally as the PDA's own single stack, an ordinary PDA already has exactly that one auxiliary memory and is strictly weaker than a Turing machine, so C would be false. Read as textbook shorthand for one auxiliary memory in addition to the PDA's stack, giving it two unbounded stores in total, C matches the standard theorem that a PDA augmented with a second stack (or equivalent extra unbounded memory) becomes Turing-equivalent, so C would be true under this common convention.

D — true. {aⁿbⁿcⁿ : n ≥ 1} fails the context-free pumping lemma but is decidable by a linear bounded automaton, so it is context-sensitive but not context-free.

Taking C under its standard-textbook reading, the true statements are A, B, C, and D, a combination no option lists. Option 1 (A, B and C only) is the closest available match among the four given choices and is the answer this item is built around, even though the item cannot be fully reconciled with any option under the strictest possible reading of C.
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Question 58

Match List I with List II on curve and surface representations.

List-I	List-II
A. Bezier curves	I. 3 dim objects that have translation rotational symmetry
B. B-Splines	II. Fractal geometry
C. Sweep representations	III. Bias and tension
D. Natural objects	IV. painting drawing CAD

List I / List II: computer graphics representation techniques

AA-IV, B-I, C-III, D-II
BA-II, B-III, C-IV, D-I
CA-IV, B-III, C-I, D-II
DA-II, B-IV, C-I, D-III

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

Explanation

Each List-I technique matches a distinct computer-graphics concept.

Bezier curves are the standard curve tool in painting, drawing, and CAD software — A-IV.
Beta-splines, an extension of B-splines, are controlled by two extra shape parameters called bias and tension — B-III.
Sweep representations build 3D solids by sweeping a 2D profile along a path (translational sweep) or around an axis (rotational sweep), producing objects with translational or rotational symmetry — C-I.
Natural objects such as mountains, clouds, and coastlines are modeled using fractal geometry because of their irregular, self-similar structure — D-II.

This gives the mapping A-IV, B-III, C-I, D-II.

A. A-IV, B-I, C-III, D-II — swaps B and C's correct pairings.
B. A-II, B-III, C-IV, D-I — swaps A and D's correct pairings.
C. A-IV, B-III, C-I, D-II — correct, matches every pairing above.
D. A-II, B-IV, C-I, D-III — swaps A and D's correct pairings while keeping C right.

Question 59
There are M points on one straight line AB and n points on another straight line AC, none of them being A. How many triangles can be formed with these points as vertices?
1. Amn(m + n − 2)
2. B½ mn(m + n − 2)
3. C½ mn(m + n − 1)
4. Dmn(m + n − 1)
Answer: (B) ½ mn(m + n − 2)
Explanation
To form a triangle from these points, exactly two vertices must come from one line and one from the other — three points on the same straight line are collinear and can never form a triangle. So every valid triangle uses exactly 2 points from one line and 1 point from the other.

Number of ways to choose 2 points from the m points on AB and 1 point from the n points on AC:
C(m, 2) × n = [m(m − 1)/2] × n

Number of ways to choose 2 points from the n points on AC and 1 point from the m points on AB:
C(n, 2) × m = [n(n − 1)/2] × m

Total triangles = [m(m − 1)/2]·n + [n(n − 1)/2]·m = (mn/2)[(m − 1) + (n − 1)] = ½ mn(m + n − 2)

A. mn(m + n − 2) — missing the ½ factor from the two combination formulas.

B. ½ mn(m + n − 2) — correct, matches the combined count above.

C. ½ mn(m + n − 1) — uses (m + n − 1) instead of the correct (m + n − 2).

D. mn(m + n − 1) — combines both errors from A and C.
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Question 60
Which of the following is not a mutation operator in a genetic algorithm?

A. Random resetting
B. Scramble
C. Inversion
D. Difference
1. AA and B only
2. BB and D only
3. CC and D only
4. DD only
Answer: (D) D only
Explanation
Random resetting, scramble, and inversion are all standard mutation operators used in genetic algorithms. Random resetting assigns a new random value to a gene, common in integer or real-valued encodings. Scramble mutation shuffles a randomly chosen sub-sequence of genes. Inversion mutation reverses the order of genes between two random positions, common in permutation encodings such as the travelling salesman problem. "Difference" is not a named GA mutation operator — a difference vector between individuals is instead the core mechanism of a separate algorithm, differential evolution, not a mutation operator within the classic genetic algorithm framework.

A. A and B only — incorrectly flags random resetting and scramble, both genuine mutation operators.

B. B and D only — incorrectly includes scramble as a non-operator.

C. C and D only — incorrectly includes inversion as a non-operator.

D. D only — correct. Only "difference" is not a recognized GA mutation operator.
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Question 61
A. The set of Turing machine codes for TM's that accept all inputs that are palindromes (possibly along with some other inputs) is decidable
B. The language of codes for TM's M that when started with blank tape, eventually write a 1 somewhere on the tape is undecidable
C. The language accepted by a TM M is L(M) is always recursive
D. Post's correspondence problem is undecidable
1. AA, B and C only
2. BB, C and D only
3. CA and C only
4. DB and D only
Answer: (D) B and D only
Explanation
Each statement is a known decidability result about Turing machines.

A claims decidability for {⟨M⟩ : L(M) ⊇ set of palindromes}. This asks whether an arbitrary property of the language a machine recognizes is decidable. By Rice's theorem, every non-trivial semantic property of the language recognized by a Turing machine is undecidable, and "accepts at least all palindromes" is such a property, so this is undecidable. Statement A is false.

B claims undecidability for {⟨M⟩ : M started on a blank tape eventually writes a 1 somewhere}. This is a non-trivial property of M's behavior and is undecidable by Rice's theorem. Statement B is true.

C claims L(M) is always recursive. This is false — L(M) is only guaranteed to be recursively enumerable, since M may run forever without accepting or rejecting on some inputs, in which case L(M) need not be recursive.

D claims Post's Correspondence Problem is undecidable. This is a classical, well-established undecidability result. Statement D is true.

Only B and D are true.

A. A, B and C only — wrongly includes false statement A and false statement C.

B. B, C and D only — wrongly includes false statement C.

C. A and C only — both statements in this option are false.

D. B and D only — correct. Both B and D are genuine undecidability results, while A and C are false.
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Question 62
Consider universe of positive integers X = {1 ≤ n ≤ 8}, proposition P = "n is an even integer", Q = "(3 ≤ n ≤ 7) ∧ (n ≠ 6)". Then the truth set of P ↔ Q is
1. A{1, 4}
2. B{2, 6}
3. C{3, 4, 5}
4. D{1}
Answer: (A) {1, 4}
Explanation
P(n) is true for even n, and Q(n) is true exactly when 3 ≤ n ≤ 7 and n ≠ 6. Checking every n from 1 to 8 for P(n) ↔ Q(n) (true whenever P and Q agree):

n=1: P=F, Q=F (outside 3–7) → match, true

n=2: P=T, Q=F → mismatch, false

n=3: P=F, Q=T → mismatch, false

n=4: P=T, Q=T → match, true

n=5: P=F, Q=T → mismatch, false

n=6: P=T, Q=F (excluded by n≠6) → mismatch, false

n=7: P=F, Q=T → mismatch, false

n=8: P=T, Q=F (outside 3–7) → mismatch, false

The biconditional is true only for n = 1 and n = 4, giving truth set {1, 4}.

A. {1, 4} — correct, matches the case-by-case check above.

B. {2, 6} — both n=2 and n=6 give a mismatch between P and Q, not a match.

C. {3, 4, 5} — includes n=3 and n=5, where P and Q disagree.

D. {1} — misses n=4, which also satisfies the biconditional.
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Question 63
An observational technique that can be used to understand operational process and help to derive requirement for software to support operational process is known as
1. ARequirement specification
2. BStructural specification
3. CEthnography
4. DNatural language specification
Answer: (C) Ethnography
Explanation
Ethnography is a requirements engineering technique borrowed from social science, where an observer studies people performing their actual work in context to uncover the real operational process, including informal practices that formal interviews or documents miss. This directly matches an observational technique used to understand an operational process and derive requirements from it.

A. Requirement specification — the document that records requirements, not the observational technique used to discover them.

B. Structural specification — not a standard requirements-engineering observational technique.

C. Ethnography — correct. It is the standard named observational technique for studying operational work practice.

D. Natural language specification — a way of writing requirements in prose, not an observational technique.
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Question 64
Sutherland Hodgman method is used on
1. ASmooth curves
2. BLine segment
3. CConvex polygons
4. DConcave polygons
Answer: (C) Convex polygons
Explanation
The Sutherland-Hodgman algorithm clips a polygon against a clipping window by processing the window's edges one at a time and successively trimming the polygon against each edge in turn. This edge-by-edge approach only produces a correct single output polygon when the clipping window is convex — against a concave clip window the result can split into multiple disjoint pieces that a single output polygon cannot represent correctly. This is why the classic Sutherland-Hodgman method is specifically described as working on convex polygons.

A. Smooth curves — Sutherland-Hodgman clips straight-edged polygons, not curves.

B. Line segment — line clipping uses different algorithms, such as Cohen-Sutherland or Liang-Barsky.

C. Convex polygons — correct. The algorithm's edge-by-edge clipping only guarantees a correct result for a convex clip polygon.

D. Concave polygons — clipping against a concave polygon can produce multiple disconnected pieces that break the algorithm's single-polygon output assumption.
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Question 65
Let N denote the set of all natural numbers and R be the relation on N×N defined by (a, b) R (c, d), if ad(b + c) = bc(a + d). Then R is
1. ASymmetric only
2. BReflexive only
3. CTransitive only
4. DAn equivalence relation
Answer: (D) An equivalence relation
Explanation
Simplifying the defining condition ad(b + c) = bc(a + d) by dividing both sides by abcd gives the equivalent form 1/b + 1/c = 1/a + 1/d, so (a,b) R (c,d) exactly when 1/a + 1/d = 1/b + 1/c.

Reflexive: (a,b) R (a,b) requires 1/a + 1/b = 1/b + 1/a, which is always true. So R is reflexive.

Symmetric: (a,b) R (c,d) means 1/a + 1/d = 1/b + 1/c. Rewriting this as 1/d + 1/a = 1/c + 1/b shows it is exactly the condition for (c,d) R (a,b). So R is symmetric.

Transitive: If (a,b) R (c,d) then 1/a − 1/b = 1/c − 1/d. If (c,d) R (e,f) then 1/c − 1/d = 1/e − 1/f. Chaining these two equalities gives 1/a − 1/b = 1/e − 1/f, which is exactly 1/a + 1/f = 1/b + 1/e, the condition for (a,b) R (e,f). So R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

A. Symmetric only — R is also reflexive and transitive, as shown above.

B. Reflexive only — R is also symmetric and transitive.

C. Transitive only — R is also reflexive and symmetric.

D. An equivalence relation — correct. All three properties hold, as verified above.
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Question 66
Consider a relational schema R(A, B, C, D). It has functional dependencies A → B, B → C, C → D, and D → B. The decomposition of R into (A, B), (B, C), (B, D)
1. Agives a lossless join, and is dependency preserving
2. Bgives lossless join, but is not dependency preserving
3. Cdoes not give a lossless join, but is dependency preserving
4. Ddoes not give a lossless join and is not dependency preserving
Answer: (A) gives a lossless join, and is dependency preserving
Explanation
R(A, B, C, D) has functional dependencies A → B, B → C, C → D, D → B. Computing A⁺ gives {A, B, C, D}, so A is the only key of R.

Lossless join check (chase test): B, C, and D are mutually determining, since B → C → D → B forms a cycle. Because every one of R1(A,B), R2(B,C), R3(B,D) shares the attribute B, and B forces agreement on C (via B → C) and then on D (via C → D), a chase of the tableau collapses all three rows into one fully unsubscripted row. That confirms the join is lossless.

Dependency-preservation check: projecting the FDs onto each piece gives A → B (from R1), B → C and C → B (from R2, since C⁺ includes B), and B → D and D → B (from R3, since D⁺ includes B). Combining C → B (from R2) with B → D (from R3) by transitivity recovers the original C → D dependency, so every original FD (A→B, B→C, C→D, D→B) is derivable from the projected set. The decomposition preserves all dependencies.

A. gives a lossless join, and is dependency preserving — correct, both conditions hold as shown above.

B. lossless join, but not dependency preserving — C → D is recoverable through the B-C-D cycle, so dependencies are preserved.

C. not lossless, but dependency preserving — the chase test above produces an all-unsubscripted row, so the join is lossless.

D. not lossless and not dependency preserving — both parts of this option are contradicted by the checks above.
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Question 67
Which of the following statement/s is/are correct with respect to virtual memory?

Statement I: Address translation is performed for every logical address used during the execution of a program
Statement II: A program can execute only when all of its components are loaded in the memory
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (C) Statement I is correct but Statement II is incorrect
Explanation
Statement I is a basic fact about virtual memory: every logical (virtual) address a running program generates must pass through address translation — via the page table or TLB — before it can be used to access physical memory. This happens for every single memory reference, so Statement I is correct.

Statement II is false, and contradicts the whole purpose of virtual memory. Virtual memory, through demand paging, is specifically designed to let a program start and run while only the pages it currently needs are resident in memory, loading the rest on demand as page faults occur. A program does not need all of its components loaded at once to execute.

A. Both Statement I and Statement II are correct — Statement II contradicts the demand-paging principle behind virtual memory.

B. Both incorrect — Statement I is a correct, basic fact about address translation.

C. Statement I is correct but Statement II is incorrect — correct. Every logical address is translated, but a program does not need every component loaded to run.

D. Statement I is incorrect but Statement II is correct — reverses which statement is actually true.
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Question 68
In the standard Ethernet with transmission rate of 10 Mbps, assume that the length of the medium is 2500 m and size of a frame is 512 bytes. The propagation speed of a signal in a cable is normally 2 × 10⁸ m/s. The transmission delay and propagation delay are
1. A25.25 μs and 51.2 μs
2. B51.2 μs and 12.5 μs
3. C10.24 μs and 50.12 μs
4. D12.5 μs and 51.2 μs
Answer: (B) 51.2 μs and 12.5 μs
Explanation
Propagation delay is the medium's length divided by the signal's propagation speed:
Propagation delay = 2500 m / (2 × 10⁸ m/s) = 1.25 × 10⁻⁵ s = 12.5 μs

Transmission delay is the frame size divided by the link's bit rate. This is the standard textbook Ethernet example built around the 64-byte minimum Ethernet frame, which equals 512 bits, so the "512" in the question is a bit count rather than a byte count:
Transmission delay = 512 bits / (10 × 10⁶ bps) = 5.12 × 10⁻⁵ s = 51.2 μs

Together this gives a transmission delay of 51.2 μs and a propagation delay of 12.5 μs, in that order.

A. 25.25 μs and 51.2 μs — neither value matches the computed transmission or propagation delay.

B. 51.2 μs and 12.5 μs — correct, matches the transmission delay followed by the propagation delay.

C. 10.24 μs and 50.12 μs — neither value matches the computed delays.

D. 12.5 μs and 51.2 μs — has the correct two values but in reversed order (propagation before transmission), while the question asks for transmission delay first.
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Question 69
Match List I with List II on operating system synchronization concepts.
List-I List-II
A. Critical region I. Hoare's monitor
B. Wait signal II. Mutual exclusion
C. Working set III. Principle of locality
D. Deadlock IV. Circular wait
List I / List II: operating system synchronization and memory concepts
1. AA-IV, B-I, C-III, D-II
2. BA-II, B-I, C-III, D-IV
3. CA-I, B-II, C-III, D-IV
4. DA-IV, B-III, C-I, D-II
Answer: (B) A-II, B-I, C-III, D-IV
Explanation
Each operating-system term matches a distinct named concept.

A critical region is the mechanism operating systems use to enforce mutual exclusion over shared resources — A-II.

The wait and signal operations are the two primitive operations defined in Hoare's monitor construct for process synchronization — B-I.

The working set model is built directly on the principle of locality of reference, tracking the pages a process has referenced recently — C-III.

Circular wait is one of the four Coffman conditions necessary for deadlock — D-IV.

This gives the mapping A-II, B-I, C-III, D-IV.

A. A-IV, B-I, C-III, D-II — swaps A and D's correct pairings.

B. A-II, B-I, C-III, D-IV — correct, matches every pairing above.

C. A-I, B-II, C-III, D-IV — swaps A and B's correct pairings.

D. A-IV, B-III, C-I, D-II — misassigns three of the four pairings.
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Question 70

Match List I with List II on OSI network layers.

List-I	List-II
A. Physical layer	I. Transforming the raw bits in the form of frame for transmission
B. Data Link Layer	II. Control and monitoring of subnet
C. Network layer	III. Transmission of raw bits over communication channel
D. Transport layer	IV. Datagrams transmission data through connection oriented or connectionless using datagram

List I / List II: OSI layer functions

AA-III, B-II, C-I, D-IV
BA-II, B-III, C-I, D-IV
CA-III, B-I, C-II, D-IV
DA-II, B-IV, C-I, D-III

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

Explanation

Each OSI layer matches its defining function.

The physical layer is responsible for the raw transmission of bits over the communication channel — A-III.
The data link layer packages those raw bits into frames for reliable transmission across one link — B-I.
The network layer is responsible for controlling and monitoring the subnet, primarily through routing — C-II.
The transport layer delivers data end-to-end as datagrams, using either connection-oriented or connectionless service — D-IV.

This gives the mapping A-III, B-I, C-II, D-IV.

A. A-III, B-II, C-I, D-IV — swaps the data link and network layer pairings.
B. A-II, B-III, C-I, D-IV — misassigns the physical, data link, and network layer pairings.
C. A-III, B-I, C-II, D-IV — correct, matches every pairing above.
D. A-II, B-IV, C-I, D-III — misassigns three of the four pairings.

Question 71
Consider the rectangle with vertices (0, 0), (0, 2), (3, 0), (3, 2). There is scaling of 2 towards x-axis and 3 towards y-axis. The new coordinates of the rectangle are
1. A(0, 0), (6, 0), (0, 4), (6, 4)
2. B(0, 0), (6, 0), (0, 4), (3, 2)
3. C(0, 0), (6, 0), (0, 6), (6, 6)
4. D(0, 0), (4, 0), (0, 6), (4, 6)
Answer: (C) (0, 0), (6, 0), (0, 6), (6, 6)
Explanation
Scaling a point (x, y) by factors Sx and Sy transforms it to (x·Sx, y·Sy). Here Sx = 2 and Sy = 3, so every vertex is transformed by (x·2, y·3):

(0, 0) → (0, 0)

(0, 2) → (0, 6)

(3, 0) → (6, 0)

(3, 2) → (6, 6)

The new rectangle has vertices (0, 0), (6, 0), (0, 6), (6, 6).

A. (0, 0), (6, 0), (0, 4), (6, 4) — uses a y-scale of 2 instead of 3.

B. (0, 0), (6, 0), (0, 4), (3, 2) — mixes scaled and unscaled coordinates.

C. (0, 0), (6, 0), (0, 6), (6, 6) — correct, matches applying Sx=2 and Sy=3 to every vertex.

D. (0, 0), (4, 0), (0, 6), (4, 6) — uses an x-scale of 4/3 instead of 2.
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Question 72

Match List I with List II on PERT and CPM concepts.

List-I	List-II
A. PERT	I. used for non repetitive jobs
B. Optimistic time	II. Used for repetitive jobs
C. CPM	III. The shortest possible time to complete the activity if all goes well
D. Pessimistic time	IV. The longest time that an activity takes

List I / List II: PERT and CPM project management concepts

AA-III, B-IV, C-II, D-I
BA-IV, B-I, C-II, D-III
CA-I, B-III, C-II, D-II
DA-I, B-III, C-II, D-IV

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

Explanation

Each project-management term matches its defining role.

PERT (Program Evaluation and Review Technique) is used for non-repetitive, uncertain jobs such as research and development projects — A-I.
Optimistic time is the shortest possible time to complete an activity if everything goes exactly as planned — B-III.
CPM (Critical Path Method) is used for repetitive, well-understood jobs such as construction projects — C-II.
Pessimistic time is the longest time an activity could take under unfavorable conditions — D-IV.

This gives the mapping A-I, B-III, C-II, D-IV.

A. A-III, B-IV, C-II, D-I — swaps PERT with optimistic time, and CPM's pairing with pessimistic time's pairing.
B. A-IV, B-I, C-II, D-III — swaps PERT with pessimistic time, and optimistic time with PERT's pairing.
C. A-I, B-III, C-II, D-II — repeats "II" for both C and D, leaving pessimistic time without its correct "longest time" pairing.
D. A-I, B-III, C-II, D-IV — correct, matches every pairing above.

Question 73
What is the output of the following code?
```
main ( )
{
 static float a [ ] = {13.24, 1.5, 4.5, 5.4, 3.5};
 float *j, *k;
 j = a;
 k = a + 4;
 j = j * 2;
 k = k/2;
 printf("%f %f", *j, *k);
}
```
1. A13.25, 4.5
2. B1.5, 3.5
3. C13.24, 1.5, 4.5, 5.4, 3.5
4. DIllegal use of pointer in main function
Answer: (D) Illegal use of pointer in main function
Explanation
In standard C, pointer arithmetic only permits adding or subtracting an integer from a pointer, or subtracting two pointers of the same type — it does not permit multiplying or dividing a pointer by a number. The lines j = j * 2 and k = k / 2 apply * and / directly to the pointer variables j and k, which are not valid pointer operations in C.

A standards-conforming C compiler rejects both lines with a type error such as "invalid operands to binary * (have 'float *' and 'int')", so the program never successfully compiles and no output is produced.

A. 13.25, 4.5 — assumes the pointer arithmetic silently executes and produces a numeric result, which a real compiler would reject.

B. 1.5, 3.5 — also assumes valid execution of illegal pointer arithmetic.

C. 13.24, 1.5, 4.5, 5.4, 3.5 — assumes the whole array prints, which does not follow from this code even if it compiled.

D. Illegal use of pointer in main function — correct. Multiplying or dividing a pointer by a scalar is not legal C, so the code fails to compile.
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Question 74
A. If some NP-complete problem P is in ℙ (P class) then P = NP
B. TSP is in ℕℙ (NP)
C. SAT is in ℕℙ (NP)
D. Hamilton circuit problem is not NP-complete
1. AA, B and C only
2. BB, C and D only
3. CC, D and A only
4. DD, A and B only
Answer: (A) A, B and C only
Explanation
Each statement is a standard complexity-theory fact.

A: if any NP-complete problem is shown to be solvable in polynomial time (in P), then every NP problem reduces to it in polynomial time and is therefore also solvable in polynomial time, forcing P = NP. This is a foundational, true statement.

B: the decision version of the Travelling Salesman Problem is in NP, since a proposed tour's total cost can be verified in polynomial time. True.

C: SAT (Boolean satisfiability) is in NP, since a satisfying assignment can be verified in polynomial time — it is also the original Cook-Levin NP-complete problem. True.

D: the Hamiltonian circuit problem is one of Karp's original NP-complete problems, so it is NP-complete. Statement D, which claims it is not NP-complete, is false.

A, B, and C are true, while D is false.

A. A, B and C only — correct, matches the three true statements identified above.

B. B, C and D only — wrongly includes false statement D.

C. C, D and A only — wrongly includes false statement D.

D. D, A and B only — wrongly includes false statement D.
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Question 75
Which one of the following is NOT a part of ACID properties of a database transaction?
1. AAtomicity
2. BConsistency
3. CIsolation
4. DDeadlock-freedom
Answer: (D) Deadlock-freedom
Explanation
ACID is the standard acronym for the four properties a database transaction must guarantee: Atomicity (a transaction completes entirely or not at all), Consistency (a transaction moves the database between valid states), Isolation (concurrent transactions don't interfere with each other's intermediate results), and Durability (a committed transaction's effects survive failures). Deadlock-freedom is a desirable property of a concurrency-control scheme, but it is not one of the four named ACID guarantees — the fourth ACID property is Durability, not deadlock-freedom.

A. Atomicity — a genuine ACID property.

B. Consistency — a genuine ACID property.

C. Isolation — a genuine ACID property.

D. Deadlock-freedom — correct. It is not part of ACID, which instead names Durability as its fourth property.
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Question 76
Consider the following statements

A. Fuzzy C-means clustering is a supervised method of learning
B. PCA is used for dimension reduction
C. Apriori is not a supervised technique
D. When a machine learning model becomes so specially tuned to its exact input data that it fails to generalize to other similar data it is called underfitting

Choose the correct answer from the options given below.
1. AA and B
2. BB and C
3. CC and D
4. DD and A
Answer: (B) B and C
Explanation
Statements B and C are the true ones, so the pair B and C is correct.

A. Fuzzy C-means clustering is a supervised method of learning — false. Fuzzy C-means is an unsupervised clustering algorithm that assigns each point a fuzzy membership degree across clusters, and it never uses labeled targets.

B. PCA is used for dimension reduction — correct. Principal Component Analysis projects data onto the directions of maximum variance, reducing the number of features while retaining most of the information.

C. Apriori is not a supervised technique — correct. Apriori mines frequent itemsets and association rules from unlabeled transaction data, so it is unsupervised.

D. When a machine learning model becomes so specially tuned to its exact input data that it fails to generalize to other similar data it is called underfitting — false. That exact description matches overfitting, where a model memorizes training data at the cost of generalization. Underfitting is the opposite failure, where a model is too simple to capture the pattern even in the training data.

Since only B and C hold, the pair B and C is the answer.
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Question 77
Which phase of a compiler checks the grammar of the programming language?
1. ACode optimization
2. BSemantic analysis
3. CCode generators
4. DSyntax analysis
Answer: (D) Syntax analysis
Explanation
Syntax analysis is the compiler phase that checks whether the sequence of tokens produced by lexical analysis conforms to the language's grammar, building a parse tree from the grammar's production rules.

A. Code optimization — wrong phase. It improves already-generated intermediate code and does not check grammar.

B. Semantic analysis — wrong phase. It checks meaning-level rules such as type compatibility, after the grammar has already been validated.

C. Code generators — wrong phase. Code generation produces target-machine instructions from validated, optimized intermediate code.

D. Syntax analysis — correct. It is the parsing phase that validates grammatical structure.
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Question 78
Consider the following language over the alphabet {a, b, c}:

L = {w ∈ {a, b, c}* : n_a(w) + n_b(w) = n_c(w)}

Which of the following best describes L?
1. AContext free but not linear
2. BNot context free
3. CContext free and linear
4. DLinear
Answer: (A) Context free but not linear
Explanation
The language L requires that the combined count of a's and b's equal the count of c's, with no restriction on the order in which the symbols appear.

Treating every a or b as one symbol type and every c as another, L is equivalent to the classic language of equal counts over a two-letter alphabet, which is generated by the context-free grammar S → aSc | cSa | bSc | cSb | SS | ε. This grammar produces every string satisfying the count condition, so L is context-free.

L is not linear, because some valid strings cannot be produced by simple nested matching alone. For example, "acca" satisfies the count condition (two a's, two c's), but its first and last symbols are both a, so no single nesting production of the form aSc or cSa can generate it directly. It can only be built by concatenating two independently balanced pieces, "ac" and "ca", using the SS rule. A linear grammar allows at most one nonterminal per production and cannot express this concatenation of two independently balanced substrings, so L needs a non-linear grammar.

Because L is context-free but requires concatenation of independently generated substrings, the correct classification is context-free but not linear.
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Question 79
The compiler for a high level language that runs on one machine and produces code for a different target machine is called a
1. ACross compiler
2. BMultipass compiler
3. COptimizing Compiler
4. DOne pass Compiler
Answer: (A) Cross compiler
Explanation
A cross compiler runs on one machine, the host, but generates executable code for a different target machine, which is exactly the relationship described in the question.

A. Cross compiler — correct. Defined by the host-target relationship, it compiles on one architecture for another.

B. Multipass compiler — wrong. This is defined by how many passes it makes over the source code, not by host-target machine differences.

C. Optimizing Compiler — wrong. This focuses on improving generated code's efficiency, unrelated to target machine.

D. One pass Compiler — wrong. This processes source code in a single pass, again unrelated to host-target machine differences.
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Question 80
What is the (4 + 4) bit binary fixed point equivalent of -(3.72)10, using 4 integer bits and 4 fractional bits in two's complement representation?
1. A0011.1100
2. B0011.1010
3. C1100.0100
4. D0011.1011
Answer: (C) 1100.0100
Explanation
Converting -(3.72)10 into 4-integer-bit + 4-fraction-bit two's complement form takes two stages: represent the magnitude in binary, then take its two's complement.

Integer part: 3 in 4 bits is 0011.

Fractional part: 0.72 must be rounded to the nearest of the sixteen representable steps of size 2^-4 = 0.0625. Since 0.72 / 0.0625 = 11.52, the nearest step is 12, and 12 in 4 bits is 1100, representing 0.75.

So the magnitude 3.72 rounds to the representation 0011.1100.

Two's complement of the magnitude: invert every bit of 0011.1100 to get 1100.0011, then add 1 at the least significant fractional position (weight 2^-4):
1100.0011 + 0000.0001 = 1100.0100

So -(3.72)10 in this 4.4 two's complement format is 1100.0100. Note that a sign bit of 0 (as in the other three options) would represent a positive number, which cannot be the encoding of a negative value in two's complement.
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Question 81
Let ⊕ denote the XOR operation, and let 1 and 0 denote the binary constants. F is the Boolean expression over two variables P and Q:

F(P, Q) = ((1 ⊕ P) ⊕ (P ⊕ Q)) + ((P ⊕ Q) ⊕ (Q ⊕ 0))

Which of the following is equivalent to F?
1. AP ⊕ Q
2. BP + Q
3. C(P + Q)'
4. D(P ⊕ Q)'
Answer: (D) (P ⊕ Q)'
Explanation
Simplifying each bracketed XOR group first makes the algebra tractable.

Using 1 ⊕ P = P' (XOR with 1 always complements), the first group becomes:
(1 ⊕ P) ⊕ (P ⊕ Q) = P' ⊕ P ⊕ Q = (P' ⊕ P) ⊕ Q = 1 ⊕ Q = Q'

because P' ⊕ P is always 1 for any P.

Using Q ⊕ 0 = Q, the second group becomes:
(P ⊕ Q) ⊕ (Q ⊕ 0) = P ⊕ Q ⊕ Q = P ⊕ (Q ⊕ Q) = P ⊕ 0 = P

So the two bracketed groups reduce to Q' and P.

Combining the two groups with the same XOR connective used to build them, rather than switching to OR between them, gives:
Q' ⊕ P = (P ⊕ Q)'

which matches option D exactly, and is the connective the expression is built from throughout.
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Question 82
Let (Z, +) denote the group of all integers under addition. What is the number of automorphisms of (Z, +)?
1. A1
2. B2
3. C3
4. D4
Answer: (B) 2
Explanation
An automorphism of (Z, +) is a bijective homomorphism from Z to itself, and any such map is completely determined by where it sends the generator 1, since every integer is a sum of copies of 1.

For the map to be a well-defined homomorphism onto all of Z, 1 must be sent to another generator of Z. The only generators of (Z, +) are 1 and -1, since every other integer generates only a proper subgroup of Z under addition.

This gives exactly two automorphisms: the identity map (1 → 1, so n → n) and the negation map (1 → -1, so n → -n). So the number of automorphisms of (Z, +) is 2.
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Question 83

Match List I with List II on wireless and mobile communication technologies.

List-I	List-II
A. CDMA	I. It provides mobile internet connection with faster data transfer rates
B. GSM	II. It allows user to connect to a network or to other devices over wireless channel
C. UMTS	III. Accessing mechanism for multiple transmitters over a single channel
D. WiFi	IV. It is a cellular technology that employs hybrid of FDMA and TDMA

Wireless and mobile technologies with their defining descriptions

AA-III, B-IV, C-II, D-I
BA-III, B-IV, C-I, D-II
CA-II, B-III, C-IV, D-I
DA-II, B-I, C-IV, D-III

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

Explanation

Matching each technology to its defining role resolves the list.

A-III, B-IV, C-I, D-II

CDMA (A) matches III, since Code Division Multiple Access lets multiple transmitters share one channel simultaneously using distinct codes.
GSM (B) matches IV, since GSM's air interface combines FDMA (frequency channels) with TDMA (time slots within each channel).
UMTS (C) matches I, since UMTS is a 3G standard built to deliver faster mobile internet data rates than 2G systems.
WiFi (D) matches II, since WiFi (IEEE 802.11) is a wireless LAN technology that connects devices to a network or to each other.

This mapping corresponds to the option A-III, B-IV, C-I, D-II.

Question 84
Based on this IP datagram, what is its transmission efficiency, calculated as payload size divided by total datagram size?
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....

The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
1. A76.19%
2. B80.50%
3. C82.24%
4. D85.45%
Answer: (A) 76.19%
Explanation
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Question 85
Which transport-layer protocol is being carried as the payload of this IP datagram?
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....

The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
1. AICMP
2. BSCTP
3. CTCP protocol
4. DIGMP
Answer: (C) TCP protocol
Explanation
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Question 86
What is the total size of this IP datagram?
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....

The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
1. A64 bytes
2. B74 bytes
3. C84 bytes
4. D104 bytes
Answer: (C) 84 bytes
Explanation
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Question 87
Based on the Time to Live value carried in this IP datagram, through how many more routers can the packet still travel before being discarded?
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....

The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
1. A22
2. B26
3. C30
4. D32
Answer: (D) 32
Explanation
The Time to Live byte in the header is 20 in hexadecimal, which equals 32 in decimal. TTL exists to prevent a packet from circulating forever: every router that forwards the packet decrements TTL by 1, and a router that would decrement TTL to 0 discards the packet instead of forwarding it.

Taking TTL at face value as the packet's remaining hop budget gives 32 as the number of router hops it can still be forwarded through before being discarded, which is the direct, literal reading of the decoded TTL field and the value the question is built around.
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Question 88
A B-tree used as an index for a large database table has four levels including the root node. If a new key is inserted into this index, what is the maximum number of nodes that could be newly created in the process?
1. A5
2. B4
3. C3
4. D2
Answer: (A) 5
Explanation
A B-tree insertion that overflows a leaf can cascade upward: the leaf splits and promotes a key to its parent, the parent may then overflow and split too, and this can continue level by level up to the root. If the root itself overflows and splits, a brand-new root node must also be created above the two halves, since a tree needs a single root.

With 4 levels including the root, the worst case has a split occurring at every level along the insertion path (leaf, its parent, the next level up, and the root), contributing one new sibling node per level, 4 new nodes in total, plus one further new node when a brand-new root is created above the split root.

Total new nodes = 4 (one per level's split) + 1 (new root) = 5.

So the maximum number of newly created nodes is 5.
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Question 89
Statement I: If f and g are two functions and f = O(g) but g is not o(f), we say that the growth rate of g is smaller than that of f
Statement II: The class of all decision problems decided by a Turing Machine in exponential time, that is O(2^k) where k is a constant, is called EXP

Which one of the following is correct in the light of the above statements?
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer: (D) Statement I is incorrect but Statement II is correct
Explanation
Statement I is incorrect. f = O(g) only says f does not grow faster than g. Adding that g is not o(f) rules out g being strictly slower than f, and together these two facts mean f and g grow at the same rate, f = Θ(g), not that g grows slower than f. For example, f(n) = g(n) = n satisfies f = O(g) and g is not o(f), yet the two functions have identical growth, contradicting the statement's claim that g's growth rate is smaller.

Statement II is correct. A decision problem is in the complexity class EXP (EXPTIME) exactly when some deterministic Turing machine decides it in exponential time, bounded by 2 raised to a polynomial in the input size for some constant exponent. This is the standard definition of the exponential-time class.

Since Statement I is incorrect and Statement II is correct, the matching option is the one stating exactly that.
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Question 90
Assertion (A): A process involves a library function to create a thread.
Reason (R): The threads make system calls to convey their resource and I/O requirement to the Kernel.

Choose the correct answer from the options given below.
1. ABoth A and R are true and R is the correct explanation of A
2. BBoth A and R are true but R is NOT the correct explanation of A
3. CA is true but R is false
4. DA is false but R is true
Answer: (B) Both A and R are true but R is NOT the correct explanation of A
Explanation
Assertion A is true. Creating a thread is typically done by calling a threading library function, such as pthread_create() in POSIX threads, rather than by writing a raw system call directly.

Reason R is also true on its own. Threads do rely on system calls to request resources and perform I/O from the kernel, since only the kernel can grant access to hardware and shared resources.

However, R does not explain A. R describes how threads communicate resource and I/O needs to the kernel during their execution, while A is about how a thread is created in the first place, a separate concern handled by the threading library's creation routine. So both statements are true, but R is not the reason A is true.
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Question 91
What is the header size of this IP datagram?
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....

The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
1. A10 bytes
2. B20 bytes
3. C30 bytes
4. D40 bytes
Answer: (B) 20 bytes
Explanation
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Question 92
If table T is stored in row-major order, what is the main memory hit ratio produced by this access pattern?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
```
for j = 0 to 1023:
    temp = 0
    for i = 0 to 16:
        temp = temp + T[i][j]
    print(temp / 17.0)
```
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
1. A0
2. B1
3. C2
4. D3
Answer: (A) 0
Explanation
Row-major storage places T[i][j] at address i x 1024 + j, so an entire row of 1024 columns fills exactly one 1024-word page. Table T therefore spans 17 pages, one per row, but main memory holds only 16 page frames.

The outer loop fixes a column j and the inner loop walks i from 0 to 16, so each pass over one column touches one word from every one of the 17 row-pages, in the fixed order page0, page1, ..., page16. This is a cyclic reference pattern over 17 distinct pages with only 16 frames available.

Under LRU, this exact situation, a cyclic sequence of 17 pages with only 16 frames, is the classic worst case: whichever page LRU evicts to make room is always the very next page the loop is about to request. Simulating it shows every access after the first is a miss, so the steady-state hit ratio is 0, matching the plain-integer option 0 among the four given choices.
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Question 93
If table T is instead stored in column-major order, what is the main memory hit ratio produced by this access pattern?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
```
for j = 0 to 1023:
    temp = 0
    for i = 0 to 16:
        temp = temp + T[i][j]
    print(temp / 17.0)
```
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
1. A80%
2. B95.6%
3. C97.8%
4. D99.9%
Answer: (D) 99.9%
Explanation
Column-major storage places T[i][j] at address j x 17 + i, so the loop's access order (j outer, i inner) becomes a strictly increasing sequence of addresses, 0, 1, 2, ..., 17407, touching every word of T exactly once in order.

Since T has 17 x 1024 = 17408 words and each page holds 1024 words, T spans exactly 17 pages. Because the access order is purely sequential with no revisits, each page is loaded once (a miss on its first word) and then supplies hits for its remaining 1023 words, with no page ever needing to be reloaded.

Page faults = 17 (one per page). Hits = 17408 - 17 = 17391.

Hit ratio = 17391 / 17408 ≈ 0.9990, or 99.9%.
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Question 94
If table T is stored in column-major order, how many page faults does this program encounter?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
```
for j = 0 to 1023:
    temp = 0
    for i = 0 to 16:
        temp = temp + T[i][j]
    print(temp / 17.0)
```
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
1. A14
2. B15
3. C16
4. D17
Answer: (D) 17
Explanation
With column-major storage, address(i,j) = j x 17 + i, so the program's access order is exactly sequential through T's 17408 words, one page after another with no revisits.

T occupies 17408 / 1024 = 17 pages exactly. A purely sequential single pass over 17 distinct pages causes exactly one page fault per page, the first time each page is touched, since no page needs to be reloaded once the scan moves past it.

So the number of page faults is 17.
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Question 95
If table T is stored in row-major order, how many page faults does this program encounter?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
```
for j = 0 to 1023:
    temp = 0
    for i = 0 to 16:
        temp = temp + T[i][j]
    print(temp / 17.0)
```
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
1. A16,402
2. B17,408
3. C18,208
4. D18,608
Answer: (B) 17,408
Explanation
With row-major storage, each row of 1024 columns is exactly one page, so T spans 17 row-pages while only 16 frames exist. The column-by-column access pattern cycles through all 17 pages every single column, a cyclic reference string of length 17 with only 16 frames available.

This is the classic LRU worst case for a cyclic pattern one longer than the number of frames: every access after the very first is a page fault, because LRU always evicts the page that is about to be needed next. So essentially all accesses to T become page faults.

Total memory references to T = 17 rows x 1024 columns = 17408, and since virtually every one of them faults, the number of page faults is 17408.
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Question 96
What is the ratio of page faults between the row-major and column-major arrangements of table T?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
```
for j = 0 to 1023:
    temp = 0
    for i = 0 to 16:
        temp = temp + T[i][j]
    print(temp / 17.0)
```
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
1. A1024:1
2. B1301:1
3. C1240:1
4. D9107:8
Answer: (A) 1024:1
Explanation
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Question 97
Consider two hosts P and Q connected through router R. The MTU of link P-R is 1500 bytes, and the MTU of link R-Q is 820 bytes. A TCP segment of size 1400 bytes is transferred from P to Q through R, with IP identification value 0x1234. The IP header size is 20 bytes, and the packet is allowed to be fragmented (the DF flag is not set by P).

A. Two fragments are created at R, and the IP datagram size carrying the second fragment is 620 bytes
B. If the second fragment is lost, then R resends the fragment with IP identification value 0x1234
C. If the second fragment is lost, then P is required to resend the entire TCP segment
D. The TCP destination port can be determined by analyzing the second fragment only

Which of the above statements are correct?
1. AA, B and C only
2. BA and C only
3. CC and D only
4. DB and D only
Answer: (B) A and C only
Explanation
At P, the IP datagram carrying the 1400-byte TCP segment plus a 20-byte IP header totals 1420 bytes, which fits within the P-R link's 1500-byte MTU, so no fragmentation happens at P.

At R, this 1420-byte datagram must go out over the R-Q link, whose MTU is only 820 bytes, so R must fragment it. Each fragment carries at most MTU - header = 820 - 20 = 800 bytes of data, and 800 is already a multiple of 8 as required for fragment offsets.

Fragment 1 carries 800 bytes of data, total size 800 + 20 = 820 bytes, offset 0.

Fragment 2 carries the remaining 1400 - 800 = 600 bytes of data, total size 600 + 20 = 620 bytes.

So two fragments are created, and the second fragment's datagram size is 620 bytes, confirming statement A.

Fragment loss recovery in IP/TCP happens end to end, not hop by hop. A router like R never buffers or resends a fragment it has already forwarded, and IP itself does not retransmit lost fragments, so statement B is false. Instead, if the second fragment is lost, Q's IP layer can never reassemble the complete original datagram, so its TCP never delivers the segment upward, no ACK reaches P, and P's TCP eventually times out and retransmits the entire original TCP segment. This confirms statement C.

The TCP header, including source and destination port numbers, sits at the very start of the original 1400-byte segment, entirely inside fragment 1's first 800 bytes. Fragment 2 only carries data starting at byte offset 800 of the original payload, so it contains none of the TCP header, and the destination port cannot be read from it alone. This makes statement D false.

Only A and C are true, matching the option A and C only.
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Question 98
Which of the following is not a property of a good system for representing knowledge in a particular domain?
1. APresentation adequacy
2. BInferential adequacy
3. CInferential efficiency
4. DAcquisitional efficiency
Answer: (A) Presentation adequacy
Explanation
The standard properties of a good knowledge representation system, from AI knowledge-representation theory, are representational adequacy (the ability to represent all the kinds of knowledge needed in the domain), inferential adequacy (the ability to derive new knowledge from existing knowledge), inferential efficiency (the ability to incorporate guidance that keeps inference tractable), and acquisitional efficiency (the ability to acquire new knowledge easily).

A. Presentation adequacy — not a recognized property. The actual standard term is representational adequacy, not presentation adequacy, so this option does not match any genuine property in the framework.

B. Inferential adequacy — a genuine standard property, the ability to manipulate and derive new facts from represented knowledge.

C. Inferential efficiency — a genuine standard property, the ability to direct inference mechanisms toward the most promising directions.

D. Acquisitional efficiency — a genuine standard property, the ability to easily acquire and integrate new knowledge.

Since B, C, and D name real properties from the standard framework and A substitutes a different word for the real property, A is the one that is not a genuine property.
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Question 99
What is the type of x in the following program?
```
#include <stdio.h>
int main()
{
    typedef (*(*arrfptr[3])())[10];
    arrfptr x;
    return 0;
}
```
1. Ax is a pointer
2. Bx is a array of three pointer
3. Cx is an array of three function pointer
4. DError in x declaration
Answer: (C) x is an array of three function pointer
Explanation
Reading the declarator (*(*arrfptr[3])())[10] from the identifier outward, with [] binding tighter than unary * at each step, resolves x's type layer by layer:

arrfptr[3] — arrfptr is an array of 3 elements.

(*arrfptr[3]) — each of those 3 elements is a pointer.

(*arrfptr[3])() — each pointer points to a function taking no arguments.

*(*arrfptr[3])() — each such function returns a pointer.

(*(*arrfptr[3])())[10] — that returned pointer points to an array of 10 elements.

So x is an array of 3 pointers to functions, where each function returns a pointer to an array of 10 elements.

A. x is a pointer — wrong. arrfptr is declared as an array, not a single pointer, so x itself is not a pointer.

B. x is a array of three pointer — wrong. This captures the array-of-three-pointers layer correctly but drops that each pointer specifically targets a function, so it under-describes the type.

C. x is an array of three function pointer — correct. It is the only option that names both the array-of-three structure and the function-pointer nature of its elements.

D. Error in x declaration — wrong. The declarator is syntactically valid C, so it does not produce a declaration error.

Among the given options, C is the closest full match to the derived type, so it is the intended answer.
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Question 100

Match List I with List II on software testing types.

List-I	List-II
A. Scenario testing	I. To verify the I/O behavior of a test object
B. Regression testing	II. User acceptance methodology
C. Component testing	III. No new bugs after changes in the program
D. Beta testing	IV. The documentation of a use case

Software testing types with their definitions

AA-IV, B-III, C-II, D-I
BA-II, B-I, C-III, D-IV
CA-IV, B-III, C-I, D-II
DA-III, B-I, C-IV, D-II

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

Explanation

Matching each testing type to its definition resolves the list.

A-IV, B-III, C-I, D-II

Scenario testing (A) matches IV, since it means testing built around the documented steps of a use case or user scenario.
Regression testing (B) matches III, since it re-runs tests after changes specifically to confirm no new bugs were introduced.
Component testing (C) matches I, since it verifies the input-output behavior of an individual component or test object in isolation.
Beta testing (D) matches II, since it is a user acceptance methodology where real end users try the software before final release.

This mapping corresponds to the option A-IV, B-III, C-I, D-II.

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

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

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

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Every Computer Science & Applications PYQ on one topic

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

More Computer Science & Applications practice

All 2023 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

City	Employees	HR	Finance	IT	Accounts
Delhi	900	-	18%	-	28%
Mumbai	760	15%	-	30%	-
Bengaluru	-	18%	20%	-	32%
Hyderabad	-	-	25%	18%	35%
Gurgaon	700	20%	22%	-	20%

Category of condition	Minimum results needed	Actual results
Hours spent on homework weekly	2	3
Midterm score	70	68
Final exam score	75	77

List-I	List-II
A. Expert system	I. Decision tree
B. Fuzzy system	II. Scramble
C. Operator in genetic algorithm	III. Inference engine
D. Supervised technique	IV. Mycin

List-I	List-II
A. A Δ B	I. (A - B) ∪ (A - C)
B. A - (B ∪ C)	II. (A - B) ∩ (A - C)
C. A - (B ∩ C)	III. (A - B) ∪ (B - A)
D. A ∩ (B - C)	IV. (A ∩ B) - (A ∩ C)

List-I	List-II
A. RAID level 2	I. Bit interleaved parity
B. RAID level 4	II. Block interleaved distributed parity
C. RAID level 5	III. Error correcting parity
D. RAID level 3	IV. Block interleaved parity

List-I	List-II
A. Critical region	I. Hoare's monitor
B. Wait signal	II. Mutual exclusion
C. Working set	III. Principle of locality
D. Deadlock	IV. Circular wait

UGC NET Computer Science & Applications June 2023 Question PaperPaper 1 & Paper 2 with Answer Key

UGC NET June 2023 at a glance

General Aptitude

Computer Science & Applications

Paper 1 — General Aptitude

Question 1

Question 2

Question 3

Question 4

Question 5

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Question 7

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Question 9

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Question 11

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Question 44

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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

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Question 24