UGC NET CS 2014 June-paper-2
Question 1 |
Infrared signals can be used for short range communication in a closed area using _______ propagation.
ground | |
sky | |
line of sight | |
space |
Question 1 Explanation:
Propagation methods are 3 types
1. Ground
2. Sky
3. Line of sight
1. Ground
2. Sky
3. Line of sight
Question 2 |
A bridge has access to _______ address in the same network.
Physical | |
Network | |
Datalink | |
Application |
Question 2 Explanation:
A bridge use MAC addresses (DLL layer) and router uses IP addresses (network layer). A bridge has access to physical address in the same network.
Question 3 |
The minimum frame length for 10 Mbps Ethernet is _______ bytes and maximum is _______ bytes.
64 & 128 | |
128 & 1518 | |
1518 & 3036 | |
64 & 1518 |
Question 3 Explanation:
10Base-T is the Ethernet wiring standard for 10 Mbps (megabits per second) for a maximum distance of approximately 100 meters per segment over unshielded twisted pair cables terminated with RJ-45 connectors.
Note: 100Base-T (Fast Ethernet) and 100Base-T (Gigabit Ethernet)
Note: 100Base-T (Fast Ethernet) and 100Base-T (Gigabit Ethernet)
Question 4 |
The bit rate of a signal is 3000 bps. If each signal unit carries 6 bits, the baud rate of the signal is _______.
500 baud/sec | |
1000 baud/sec | |
3000 baud/sec | |
18000 baud/sec. |
Question 4 Explanation:
→ Bit rate is nothing but number of bits transmitted per second
→ Baud rate is nothing but number of signals units transmitted per unit time.
Given data,
-- Bit rate of a signal=3000 bps
-- Each signal unit carries=6 bits
-- Baud rate=?
Step-1: Baud rate= Bit rate of signal / baud rate
= 3000/6
= 500 baud/sec
→ Baud rate is nothing but number of signals units transmitted per unit time.
Given data,
-- Bit rate of a signal=3000 bps
-- Each signal unit carries=6 bits
-- Baud rate=?
Step-1: Baud rate= Bit rate of signal / baud rate
= 3000/6
= 500 baud/sec
Question 5 |
Match the following :
a-iv, b-v, c-ii, d-iii, e-i | |
a-v, b-iv, c-i, d-ii, e-iii | |
a-i, b-iii, c-ii, d-v, e-iv | |
a-i, b-ii, c-iv, d-iii, e-v |
Question 5 Explanation:
Physical layer→ Transmission of bit stream
Data link layer→ Formation of frames
Network layer→ Move packets from one destination to other
Transport layer→ Process to process message delivery
Application Layer→ Allow resources to network access
Data link layer→ Formation of frames
Network layer→ Move packets from one destination to other
Transport layer→ Process to process message delivery
Application Layer→ Allow resources to network access
Question 6 |
A grammar G is LL(1) if and only if the following conditions hold for two distinct productions
A → α | β
- First (α) ∩ First (β) ≠ {a} where a is some terminal symbol of the grammar.
- First (α) ∩ First (β) ≠ λ
I and II | |
I and III | |
II and III | |
I, II and III |
Question 6 Explanation:
A grammar G is LL(1) if and only if the following conditions hold for two distinct productions:
A → α | β
1. First (α) and First (β) must be disjoint if none of α and β contains NULL move.
2. At most one of the strings α or β can drive NULL move i.e. α → NULL(since First (α) and First (β) are disjoint). In this case, First (β) and Follow(A) must be disjoint.
Hence the answer is option(D).
A → α | β
1. First (α) and First (β) must be disjoint if none of α and β contains NULL move.
2. At most one of the strings α or β can drive NULL move i.e. α → NULL(since First (α) and First (β) are disjoint). In this case, First (β) and Follow(A) must be disjoint.
Hence the answer is option(D).
Question 7 |
Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar ?
Removing left recursion alone | |
Removing the grammar alone | |
Removing left recursion and factoring the grammar | |
None of the above |
Question 7 Explanation:
→ Left recursion removing (or) factoring the given grammar are not sufficient to convert an arbitrary CFG to an LL(1) grammar.
→ To convert an arbitrary CFG to an LL(1) grammar we need to remove the left recursion and as well as left factoring without that we cannot convert.
→ To convert an arbitrary CFG to an LL(1) grammar we need to remove the left recursion and as well as left factoring without that we cannot convert.
Question 8 |
A shift reduce parser suffers from
shift reduce conflict only | |
reduce reduce conflict only | |
both shift reduce conflict and reduce reduce conflict | |
shift handle and reduce handle conflicts |
Question 8 Explanation:
→ A shift-reduce parser scans and parses the input text in one forward pass over the text, without backing up. (That forward direction is generally left-to-right within a line, and top-to-bottom for multi-line inputs.) The parser builds up the parse tree incrementally, bottom up, and left to right, without guessing or backtracking.
A shift-reduce parser works by doing some combination of Shift steps and Reduce steps, hence the name.
→ A Shift step advances in the input stream by one symbol. That shifted symbol becomes a new single-node parse tree.
→ A Reduce step applies a completed grammar rule to some of the recent parse trees, joining them together as one tree with a new root symbol.
*** A shift reduce parser suffers from both shift reduce conflict and reduce reduce conflict.
A shift-reduce parser works by doing some combination of Shift steps and Reduce steps, hence the name.
→ A Shift step advances in the input stream by one symbol. That shifted symbol becomes a new single-node parse tree.
→ A Reduce step applies a completed grammar rule to some of the recent parse trees, joining them together as one tree with a new root symbol.
*** A shift reduce parser suffers from both shift reduce conflict and reduce reduce conflict.
Question 9 |
The context free grammar for the language L = {a^{n}b^{m}c^{k} | k = |n – m|,
n ≥ 0, m ≥ 0, k ≥ 0} is
S → S_{1}S_{3}, S_{1} → aS_{1}c | S_{2}| λ,
S_{2} → aS_{2}b|λ, S_{3 }→ aS_{3}b| S_{4} | λ,
S_{4} → bS_{4}c|λ | |
S → S_{1}S_{3}, S_{1}→ aS_{1}S_{2}c | λ,
S_{2} → aS_{2}b|λ, S_{3} → aS_{3}b| S_{4} |λ,
S_{4} → bS_{4}c|λ | |
S → S_{1}|S_{2}, S_{1}→ aS_{1}S_{2}c | λ,
S_{2} → aS_{2}b | λ, S_{3} → aS_{3}b | S_{4} |λ,
S_{4} → bS_{4}c|λ | |
S → S_{1} | S_{3}, S_{1}→ aS_{1}c|S_{2} | λ,
S_{2} → aS_{2}b | λ, S_{3} → a S_{3}b| S_{4} | λ,
S_{4} → bS_{4}c | λ |
Question 9 Explanation:
L = { λ, ab, ac, bc, aabb,aabc,aac, bbc, ........}
Option(A): Option(A) will generate the string "acab" which does not belongs to the sequence a^{n}b^{m}c^{k} . So, it is not the context free grammar for the language L.
Option(B): Option(B) will generate the string "acab" which does not belongs to the sequence a^{n}b^{m}c^{k}. So, it is not the context free grammar for the language L. Option(C): In this option production S_{3} and S_{4} are unreachable so it is not the context free grammar for the language L. Option (D): The grammar given in this option can generate { λ, ab, ac, bc, aabb,aabc,aac, bbc, ........}. So it is the context free grammar for the language L.
Option(A): Option(A) will generate the string "acab" which does not belongs to the sequence a^{n}b^{m}c^{k} . So, it is not the context free grammar for the language L.
Option(B): Option(B) will generate the string "acab" which does not belongs to the sequence a^{n}b^{m}c^{k}. So, it is not the context free grammar for the language L. Option(C): In this option production S_{3} and S_{4} are unreachable so it is not the context free grammar for the language L. Option (D): The grammar given in this option can generate { λ, ab, ac, bc, aabb,aabc,aac, bbc, ........}. So it is the context free grammar for the language L.
Question 10 |
The regular grammar for the language
L = {w|na(w) and nb(w) are both even,
w ∈ {a, b}*} is given by :
(Assume, p, q, r and s are states)
L = {w|na(w) and nb(w) are both even,
w ∈ {a, b}*} is given by :
(Assume, p, q, r and s are states)
p → aq | br | λ, q → bs | ap r → as | bp, s → ar | bq, p and s are initial and final states. | |
p → aq | br, q → bs | ap r → as | bp, s → ar | bq, p and s are initial and final states. | |
p → aq | br | λ, q → bs | ap r → as | bp, s → ar | bq p is both initial and final states. | |
p → aq | br, q → bs | ap r → as | bp, s → ar | bq p is both initial and final states |
Question 10 Explanation:
L={ λ. ab, λab, λaabb, aabb, ..........}
Option (A): The grammar given in this option does not contain a state which is accepting a terminal "λ" , having zero number of "a" and "b". Since the grammar is not accepting λ so it not the regular grammar for the given language L.
Option (B): The grammar given in this option is not generating any string "λ" which is having zero number of "a" and "b". Since the grammar is not generating λ so it not the regular grammar for the given language L.
Option (C): Here state "p" is the initial and final state and the grammar given here is generating string "λ" , having zero number of "a" and "b". Since the grammar is accepting λ so it is the regular grammar for the given language L.
Option (D): In this option (p) is the initial and final state but the grammar is not generating any string "λ" which also belongs to the given language L. So this grammar is not the regular grammar for the given language L.
Option (A): The grammar given in this option does not contain a state which is accepting a terminal "λ" , having zero number of "a" and "b". Since the grammar is not accepting λ so it not the regular grammar for the given language L.
Option (B): The grammar given in this option is not generating any string "λ" which is having zero number of "a" and "b". Since the grammar is not generating λ so it not the regular grammar for the given language L.
Option (C): Here state "p" is the initial and final state and the grammar given here is generating string "λ" , having zero number of "a" and "b". Since the grammar is accepting λ so it is the regular grammar for the given language L.
Option (D): In this option (p) is the initial and final state but the grammar is not generating any string "λ" which also belongs to the given language L. So this grammar is not the regular grammar for the given language L.
Question 11 |
KPA in CMM stands for
Key Process Area | |
Key Product Area | |
Key Principal Area | |
Key Performance Area |
Question 11 Explanation:
KPA(Key process area) in Capability Maturity Model(CMM):
A Process Area is a cluster of related practices in an area that, when implemented collectively, satisfy a set of goals considered important for making significant improvement in that area. All CMMI process areas are common to both continuous and staged representations.
A Process Area is a cluster of related practices in an area that, when implemented collectively, satisfy a set of goals considered important for making significant improvement in that area. All CMMI process areas are common to both continuous and staged representations.
Question 12 |
Which one of the following is not a risk management technique for managing the risk due to unrealistic schedules and budgets ?
Detailed multi source cost and schedule estimation. | |
Design cost | |
Incremental development | |
Information hiding |
Question 12 Explanation:
Information hiding is not a risk management technique for managing the risk due to unrealistic schedules and budgets.
Question 13 |
_______ of a system is the structure or structures of the system which comprise software elements, the externally visible properties of these elements and the relationship amongst them.
Software construction | |
Software evolution | |
Software architecture | |
Software reuse |
Question 13 Explanation:
Software architecture of a system is the structure or structures of the system which comprise software elements, the externally visible properties of these elements and the relationship amongst them.
Question 14 |
In function point analysis, the number of complexity adjustment factors is
10 | |
12 | |
14 | |
20 |
Question 14 Explanation:
→ In function point analysis, the number of complexity adjustment factors is 14.
→ The adjusted function point denoted by FP is given by the formula:
FP = total UFP * (0.65 + (0.01 * Total complexity adjustment value)) or
FP = total UFP * (Complexity adjustment factor)
→ Total complexity adjustment value is counted based on responses to questions called complexity weighting factors in the table below. Each complexity weighting factor is assigned a value (complexity adjustment value) that ranges between 0 (not important) to 5 (absolutely essential).
Table Adjusted Function Points
→ The adjusted function point denoted by FP is given by the formula:
FP = total UFP * (0.65 + (0.01 * Total complexity adjustment value)) or
FP = total UFP * (Complexity adjustment factor)
→ Total complexity adjustment value is counted based on responses to questions called complexity weighting factors in the table below. Each complexity weighting factor is assigned a value (complexity adjustment value) that ranges between 0 (not important) to 5 (absolutely essential).
Table Adjusted Function Points
Question 15 |
Regression testing is primarily related to
Functional testing | |
Development testing | |
Data flow testing | |
Maintenance testing |
Question 15 Explanation:
→ Regression testing is re-running functional and non-functional tests to ensure that previously developed and tested software still performs after a change. If not, that would be called a regression.
→ Changes that may require regression testing include bug fixes, software enhancements, configuration changes, and even substitution of electronic components.
→ As regression test suites tend to grow with each found defect, test automation is frequently involved.
→ Regression testing is primarily related to Maintenance testing.
→ Changes that may require regression testing include bug fixes, software enhancements, configuration changes, and even substitution of electronic components.
→ As regression test suites tend to grow with each found defect, test automation is frequently involved.
→ Regression testing is primarily related to Maintenance testing.
Question 16 |
How many different truth tables of the compound propositions are there that
involve the propositions p & q ?
2 | |
4 | |
8 | |
16 |
Question 16 Explanation:
→ The total number of possibilities are 2^4=16. Here, number of different functions will be the number of ways this 4 rows will be assigned values.
Question 17 |
A Boolean function F is called self-dual if and only if
F(x_{1}, x_{2}, … x_{n}) = F(͞x_{1},͞x_{2}, …͞x_{n})
How many Boolean functions of degree n are self-dual ?
2^{n} | |
(2)^{2^n} | |
(2)^{n^2} | |
(2)^{(2^(n-1))} |
Question 17 Explanation:
→ Number of possible minterms = 2^{n}.
→ Number of mutually exclusive pairs of minterms = 2^{n-1}.
→ There are 2 choices for each pair i.e., we can choose one of the two minterms from each pair of minterms for the function.
Therefore number of functions = 2*2* …. 2^{n-1} times.
= 2^{(2^(n-1)) }
→ Number of mutually exclusive pairs of minterms = 2^{n-1}.
→ There are 2 choices for each pair i.e., we can choose one of the two minterms from each pair of minterms for the function.
Therefore number of functions = 2*2* …. 2^{n-1} times.
= 2^{(2^(n-1)) }
Question 18 |
Which of the following statement(s) is (are) not correct ?
i. The 2’s complement of 0 is 0.
ii. In 2’s complement, the left most bit cannot be used to express a quantity.
iii. For an n-bit word (2’s complement) which includes the sign bit, there are 2^{n–1} positive
integers, 2^{n+1} negative integers and one 0 for a total of 2n unique states.
iv. In 2’s complement the significant information is contained in the 1’s of positive numbers and 0’s of the negative numbers.
i. The 2’s complement of 0 is 0.
ii. In 2’s complement, the left most bit cannot be used to express a quantity.
iii. For an n-bit word (2’s complement) which includes the sign bit, there are 2^{n–1} positive
integers, 2^{n+1} negative integers and one 0 for a total of 2n unique states.
iv. In 2’s complement the significant information is contained in the 1’s of positive numbers and 0’s of the negative numbers.
i & iv | |
i & ii | |
iii | |
iv |
Question 18 Explanation:
With n-bit 2's complement number we can represent
2^{n–1} -1 positive integers,
2^{n-1} negative integers
and
zero.
Total=2^{n–1} -1+ 2^{n-1} +1= 2^{n} unique states for 2^{n} unique integers
2^{n–1} -1 positive integers,
2^{n-1} negative integers
and
zero.
Total=2^{n–1} -1+ 2^{n-1} +1= 2^{n} unique states for 2^{n} unique integers
Question 19 |
The notation ∃!xP(x) denotes the proposition “there exists a unique x such that P(x) is true”.
Give the truth values of the following statements :
I. ∃!xP(x) → ∃xP(x)
II. ∃!x ¬ P(x) → ¬∀xP(x)
Give the truth values of the following statements :
I. ∃!xP(x) → ∃xP(x)
II. ∃!x ¬ P(x) → ¬∀xP(x)
Both I & II are true. | |
Both I & II are false. | |
I – false, II – true | |
I – true, II – false |
Question 19 Explanation:
I: If there exists a unique x with P(x) true, then there exist an x with P(x) true. This is TRUE as exactly 1 is a subset of at least one.
II: If there exists a unique x with P(x) false, then there does not exist an x with P(x) true. This is FALSE (contradiction) as all except one x can are having P(x) true.
II: If there exists a unique x with P(x) false, then there does not exist an x with P(x) true. This is FALSE (contradiction) as all except one x can are having P(x) true.
Question 20 |
Give a compound proposition involving propositions p, q and r that is true when exactly two of p, q and r are true and is false otherwise.
(p∨q∧¬r) ∧ (p∧¬q∧r) ∧ (¬p∧q∧r) | |
(p∧q∧¬r) ∧ (p∨q∧¬r) ∧ (¬p∧q∧r) | |
(p∧q∧¬r) ∨ (p∧¬q∧r) ∧ (¬p∧q∧r) | |
(p∧q∧¬r) ∨ (p∧¬q∧r) ∨ (¬p∧q∧r) |
Question 20 Explanation:
Here, As per the given question, option-D is correct answer.
Question 21 |
Consider the graph given below as :
Which one of the following graph is isomorphic to the above graph ?
Which one of the following graph is isomorphic to the above graph ?
Question 21 Explanation:
If two graphs are isomorphic, they must have:
→The same number of vertices, edges and degrees for corresponding vertices.
→The same number of connected components,loops and parallel edges.
The above graph each vertex having degree 3 and total number of vertices are 8. Total number of edges are 12.
Option-A: Vertex v_{5} having degree 4. So, it is not isomorphic graph.
Option-B: Vertex v_{6} having degree 4. So, it is not isomorphic graph.
Option-C: Every vertex having degree 3 and total number of edges and vertices are 12 and 8.
So, it is isomorphic.
Option-D: Vertex v_{6} having degree 4. So, it is not isomorphic graph.
→The same number of vertices, edges and degrees for corresponding vertices.
→The same number of connected components,loops and parallel edges.
The above graph each vertex having degree 3 and total number of vertices are 8. Total number of edges are 12.
Option-A: Vertex v_{5} having degree 4. So, it is not isomorphic graph.
Option-B: Vertex v_{6} having degree 4. So, it is not isomorphic graph.
Option-C: Every vertex having degree 3 and total number of edges and vertices are 12 and 8.
So, it is isomorphic.
Option-D: Vertex v_{6} having degree 4. So, it is not isomorphic graph.
Question 22 |
The upper bound and lower bound for the number of leaves in a B-tree of degree K with height h is given by :
K^{h} and 2⌈K/2⌉^{h–1} | |
K*h and 2⌊K/2⌋^{h–1} | |
K^{h} and 2⌊K/2⌋^{h–1} | |
K*h and 2⌈K/2⌉^{h–1} |
Question 22 Explanation:
→ The upper bound for the number of leaves in a B-tree of degree K with height h is K^{h}
→ The lower bound for the number of leaves in a B-tree of degree K with height h is 2⌈K/2⌉^{h–1}
→ The lower bound for the number of leaves in a B-tree of degree K with height h is 2⌈K/2⌉^{h–1}
Question 23 |
Consider a complete bipartite graph km,n. For which values of m and n does
this, complete graph have a Hamilton circuit
m = 3, n = 2 | |
m = 2, n = 3 | |
m = n ≥ 2 | |
m = n ≥ 3 |
Question 23 Explanation:
Complete graph have a Hamilton circuit m=n ≥ 2. If G is connected simple graph with n vertices where n≥3, then G has a Hamilton circuit if the degree of each vertex is at least n/2.
Question 24 |
Big-O estimates for the factorial function and the logarithm of the factorial function i.e. n! and log n! is given by
O(n!) and O(n log n) | |
O(n^{n}) and O(n log n) | |
O(n!) and O(log n!) | |
O(n^{n}) and O(log n!) |
Question 24 Explanation:
→ Given factorial function is n!. we can also write into n^{n}. When we are writing into asymptotic order also the value remains same O(n^{n}).
→ Given logarithm of the factorial function log n!.
= log n! (or)
= log n^{n} (or)
= nlogn
When we are writing into asymptotic order also the value remains same O(nlogn).
→ Given logarithm of the factorial function log n!.
= log n! (or)
= log n^{n} (or)
= nlogn
When we are writing into asymptotic order also the value remains same O(nlogn).
Question 25 |
How many cards must be chosen from a deck to guarantee that at least
I. two aces of two kinds are chosen.
II. two aces are chosen.
III. two cards of the same kind are chosen.
IV. two cards of two different kinds are chosen.
50, 50, 14, 5 | |
51, 51, 15, 7 | |
52, 52, 14, 5 | |
51, 51, 14, 5 |
Question 26 |
Match the following with respect to the mobile computing technologies :
a-iii, b-iv, c-ii, d-i | |
a-iv, b-i, c-ii, d-iii | |
a-ii, b-iii, c-iv, d-i | |
a-ii, b-i, c-iv, d-iii |
Question 26 Explanation:
GPRS → An emerging wireless service that offers a mobile data
GSM → An integrated digital radio standard
UMTS → The Universal Mobile Telecommunications System (UMTS) is a third generation mobile cellular system for networks based on the GSM standard. Developed and maintained by the 3GPP (3rd Generation Partnership Project), UMTS is a component of the International Telecommunications Union IMT-2000 standard set and compares with the CDMA2000 standard set for networks based on the competing cdmaOne technology. UMTS uses wideband code division multiple access (W-CDMA) radio access technology to offer greater spectral efficiency and bandwidth to mobile network operators.
EDGE → Enhanced GPRS or EGPRS. It is a data system used on top of GSM networks. It provides nearly 3 times faster speeds than the outdated GPRS system. Nine different schemes for modulation and error correction.
GSM → An integrated digital radio standard
UMTS → The Universal Mobile Telecommunications System (UMTS) is a third generation mobile cellular system for networks based on the GSM standard. Developed and maintained by the 3GPP (3rd Generation Partnership Project), UMTS is a component of the International Telecommunications Union IMT-2000 standard set and compares with the CDMA2000 standard set for networks based on the competing cdmaOne technology. UMTS uses wideband code division multiple access (W-CDMA) radio access technology to offer greater spectral efficiency and bandwidth to mobile network operators.
EDGE → Enhanced GPRS or EGPRS. It is a data system used on top of GSM networks. It provides nearly 3 times faster speeds than the outdated GPRS system. Nine different schemes for modulation and error correction.
Question 27 |
Object Request Broker (ORB) is
I. A software program that runs on the client as well as on the application server.
II. A software program that runs on the client side only.
III. A software program that runs on the application server, where most of the components reside.I, II & III
I. A software program that runs on the client as well as on the application server.
II. A software program that runs on the client side only.
III. A software program that runs on the application server, where most of the components reside.I, II & III
I, II & III | |
I & II | |
II & III | |
I only |
Question 27 Explanation:
Object Request Broker (ORB) is a middleware which allows program calls to be made from one computer to another via a computer network, providing location transparency through remote procedure calls. ORBs promote interoperability of distributed object systems, enabling such systems to be built by piecing together objects from different vendors, while different parts communicate with each other via the ORB.
→ A software program that runs on the client as well as on the application server used in distributed systems.
→ A software program that runs on the client as well as on the application server used in distributed systems.
Question 28 |
A software agent is defined as
I. A software developed for accomplishing a given task.
II. A computer program which is capable of acting on behalf of the user in order to accomplish a given computational task.
III. An open source software for accomplishing a given task.
I. A software developed for accomplishing a given task.
II. A computer program which is capable of acting on behalf of the user in order to accomplish a given computational task.
III. An open source software for accomplishing a given task.
I | |
II | |
III | |
All of the above |
Question 28 Explanation:
→ A software agent is a computer program that acts for a user or other program in a relationship of agency, which derives from the Latin agere (to do): an agreement to act on one's behalf. Such "action on behalf of" implies the authority to decide which, if any, action is appropriate. Agents are colloquially known as bots, from robot.
→ Software agents interacting with people (e.g. chatbots, human-robot interaction environments) may possess human-like qualities such as natural language understanding and speech, personality or embody humanoid form.
→ Software agents interacting with people (e.g. chatbots, human-robot interaction environments) may possess human-like qualities such as natural language understanding and speech, personality or embody humanoid form.
Question 29 |
Match the following :
a-iii, b-iv, c-ii, d-i | |
a-iv, b-iii, c-i, d-ii | |
a-iii, b-iv, c-i, d-ii | |
a-iv, b-iii, c-ii, d-i |
Question 29 Explanation:
Classification → K-nearest neighbour(K-NN) algorithm. It is supervised learning.
Clustering→ K-means algorithm. It is unsupervised learning.
Feature Extraction→ Principal component analysis(PCA)
Feature Selection→ Branch and Bound
Clustering→ K-means algorithm. It is unsupervised learning.
Feature Extraction→ Principal component analysis(PCA)
Feature Selection→ Branch and Bound
Question 30 |
SET, an open encryption and security specification model that is designed for
protecting credit card transactions on the internet, stands for
Secure Electronic Transaction | |
Secular Enterprise for Transaction | |
Security Electronic Transmission | |
Secured Electronic Termination |
Question 30 Explanation:
Secure Electronic Transaction (SET) is a communications protocol standard for securing credit card transactions over networks, specifically, the Internet.
→ SET was not itself a payment system, but rather a set of security protocols and formats that enabled users to employ the existing credit card payment infrastructure on an open network in a secure fashion.
→ SET was not itself a payment system, but rather a set of security protocols and formats that enabled users to employ the existing credit card payment infrastructure on an open network in a secure fashion.
Question 31 |
In a paged memory management algorithm, the hit ratio is 70%. If it takes 30 nanoseconds to search Translation Lookaside Buffer (TLB) and 100 nanoseconds (ns) to access memory, the effective memory access time is
91 ns | |
69 ns | |
200 ns | |
160 ns |
Question 31 Explanation:
Given data,
-- Hit ratio=70%
=70/100
=0.7
-- Miss ratio=(1-Hit ratio)
=(1-0.7)
=0.3
-- TLB search time=30ns
-- Access memory=100ns
-- Effective access memory=?
Step-1: [ Hit ratio*(TLB Search time+Access memory) +
Miss ratio*(TLB Search time+2*Access memory) ]
= 0.7*(30+100) + 0.3*(30+2*100)
= 0.7*(130) + 0.3*(30+200)
= 0.7*130 + 0.3*230
= 91 + 69
= 160
-- Hit ratio=70%
=70/100
=0.7
-- Miss ratio=(1-Hit ratio)
=(1-0.7)
=0.3
-- TLB search time=30ns
-- Access memory=100ns
-- Effective access memory=?
Step-1: [ Hit ratio*(TLB Search time+Access memory) +
Miss ratio*(TLB Search time+2*Access memory) ]
= 0.7*(30+100) + 0.3*(30+2*100)
= 0.7*(130) + 0.3*(30+200)
= 0.7*130 + 0.3*230
= 91 + 69
= 160
Question 32 |
Match the following :
a-i, b-iii, c-ii, d-iv | |
a-iv, b-iii, c-ii, d-i | |
a-iii, b-i, c-iv, d-i | |
a-ii, b-iii, c-iv, d-i |
Question 32 Explanation:
Multilevel feedback queue→ Criteria to move processes between queues
FCFS → Batch processing
Shortest process next → Exponential smoothening
Round robin scheduling → Time slicing
FCFS → Batch processing
Shortest process next → Exponential smoothening
Round robin scheduling → Time slicing
Question 33 |
Consider a system with five processes P_{0} through P_{4} and three resource typesR_{1}, R_{2 }and R_{3}. Resource type R_{1} has 10 instances, R_{2} has 5 instances and R_{3}has 7 instances. Suppose that at time T_{0}, the following snapshot of the system has been taken :
Assume that now the process P<sub>1</sub> requests one additional instance of type R<sub>1</sub> and two instances of resource type R<sub>3</sub>. The state resulting after this allocation will be
Allocation | |||
R1 | R2 | R3 | |
P0 | 0 | 1 | 0 |
P1 | 2 | 0 | 0 |
P2 | 3 | 0 | 2 |
P3 | 2 | 1 | 1 |
P4 | 0 | 2 | 2 |
Max | |||
R1 | R2 | R3 | |
P0 | 7 | 5 | 3 |
P1 | 3 | 2 | 2 |
P2 | 9 | 0 | 2 |
P3 | 2 | 2 | 2 |
P4 | 4 | 3 | 3 |
Available | ||
R1 | R2 | R3 |
3 | 3 | 2 |
Ready state | |
Safe state | |
Blocked state | |
Unsafe state |
Question 33 Explanation:
To find safe state, we need to find need matrix.
Need=Max-Allocation
Step-3: Request is less than availability. Then we are subtracting request to available
available= 3-1, 3-0, 2-2
= 2, 3, 0
Step-4: So, we are adding P_{1} allocation to availability
Availability=2+3, 0+3, 0+2
= 5, 3, 2
Step-5: Next we can serve to P_{3} then availability is 7 4 3
Step-6: Next. we can serve P_{4} then availability is 7 4 5
Step-7: Next, we can serve P_{0} then availability is 7 5 3
Step-8: Next, we can serve P_{2} then availability is 10 5 7
Hence, The state resulting after this allocation will be safe state.
Step-3: Request is less than availability. Then we are subtracting request to available
available= 3-1, 3-0, 2-2
= 2, 3, 0
Step-4: So, we are adding P_{1} allocation to availability
Availability=2+3, 0+3, 0+2
= 5, 3, 2
Step-5: Next we can serve to P_{3} then availability is 7 4 3
Step-6: Next. we can serve P_{4} then availability is 7 4 5
Step-7: Next, we can serve P_{0} then availability is 7 5 3
Step-8: Next, we can serve P_{2} then availability is 10 5 7
Hence, The state resulting after this allocation will be safe state.
Question 34 |
Match the following :
a-iii, b-iv, c-ii, d-i | |
a-iii, b-ii, c-iv, d-i | |
a-i, b-ii, c-iv, d-iii | |
a-i, b-iv, c-ii, d-iii |
Question 34 Explanation:
Contiguous allocation→ Number of disks required to access file is minimal.
Linked allocation→ This allocation technique supports only sequential files.
Indexed allocation→ This technique suffers from maximum wastage of space in storing pointers.
Multi-level indexed→ This scheme supports very large file sizes.
Linked allocation→ This allocation technique supports only sequential files.
Indexed allocation→ This technique suffers from maximum wastage of space in storing pointers.
Multi-level indexed→ This scheme supports very large file sizes.
Question 35 |
Which of the following commands will output “onetwothree” ?
for val; do echo-n $val; done < one two three | |
for one two three; do echo-n-; done | |
for n in one two three; do echo-n $n; done | |
for n in one two three {echo –n $ n} |
Question 35 Explanation:
for→ It will execute a statement will work like normal ‘C’ programming.
echo→ This statement will print the value
$n → Printing the value of ‘n’ variable.
; → Terminating the current statement.
According to question, Option-C is most suitable answer.
echo→ This statement will print the value
$n → Printing the value of ‘n’ variable.
; → Terminating the current statement.
According to question, Option-C is most suitable answer.
Question 36 |
Mergesort makes two recursive calls. Which statement is true after these two
recursive calls finish, but before the merge step ?
The array elements form a heap. | |
Elements in each half of the array are sorted amongst themselves. | |
Elements in the first half of the array are less than or equal to elements in second half of the array. | |
All of the above |
Question 36 Explanation:
A merge sort works as follows:
→Divide the unsorted list into n sublists, each containing one element (a list of one element is considered sorted).
→Repeatedly merge sublists to produce new sorted sublists until there is only one sublist remaining. This will be the sorted list.
As per the merge sort, The elements in each half of the array are sorted amongst themselves.
→Divide the unsorted list into n sublists, each containing one element (a list of one element is considered sorted).
→Repeatedly merge sublists to produce new sorted sublists until there is only one sublist remaining. This will be the sorted list.
As per the merge sort, The elements in each half of the array are sorted amongst themselves.
Question 37 |
A text is made up of the characters α,β,γ,δ and σ with the probability 0.12, 0.40, 0.15, 0.08 and 0.25 respectively. The optimal coding technique will have the average length of
1.7 | |
2.15 | |
3.4 | |
3.8 |
Question 37 Explanation:
Step-1: Sort in descending order according to their probabilities.
Step-2: Perform optimal merge pattern
0.40, 0.25, 0.15, 0.12, 0.08
Step-4: Multiply with length of the code
= (0.40*1) + (0.25*2) + (0.15*3) + (0.12*4) + (0.08*4)
= 2.15
Step-2: Perform optimal merge pattern
0.40, 0.25, 0.15, 0.12, 0.08
Step-4: Multiply with length of the code
= (0.40*1) + (0.25*2) + (0.15*3) + (0.12*4) + (0.08*4)
= 2.15
Question 38 |
Searching for an element in the hash table requires O(1) time for the _____ time, whereas for direct addressing it holds for the ________ time.
worst-case, average | |
worst-case, worst-case | |
average, worst-case | |
best, average |
Question 38 Explanation:
→ Hash tables are a simple and effective method to implement dictionaries. Average time to search for an element is O(1), while worst-case time is O(n)
Question 39 |
An algorithm is made up of 2 modules M1 and M2. If time complexity of modules M_{1} and M_{2} are h(n) and g(n) respectively, the time complexity of the algorithm is
min(h(n), g(n)) | |
max(h(n), g(n)) | |
h(n) + g(n) | |
h(n) * g(n) |
Question 39 Explanation:
→ The time complexity of modules M_{1} and M_{2} are h(n) and g(n).
Let take some constants c_{1}, c_{2}, n_{1}, n_{2} such that
T(n) ≤ c_{1}*h(n), for all n≥n_{1}.
T(n) ≤ c_{2} *h(n), for all n ≥ n_{2}.
N=max(n_{1}, n_{2}) and C=max(c_{1}, c_{2}).
So,
T(n)≤C * h(n) for all n≥N
T(n)≤C * g(n) for all n≥N
T(n)≤C/2 * (h(n)+g(n))
Without loss of generality, let max(h(n), g(n))=h(n) .
So, T(n)≤C/2 (h(n)+h(n))≤C*h(n) .
So, complexity of h(n) is max(h(n), g(n))
Let take some constants c_{1}, c_{2}, n_{1}, n_{2} such that
T(n) ≤ c_{1}*h(n), for all n≥n_{1}.
T(n) ≤ c_{2} *h(n), for all n ≥ n_{2}.
N=max(n_{1}, n_{2}) and C=max(c_{1}, c_{2}).
So,
T(n)≤C * h(n) for all n≥N
T(n)≤C * g(n) for all n≥N
T(n)≤C/2 * (h(n)+g(n))
Without loss of generality, let max(h(n), g(n))=h(n) .
So, T(n)≤C/2 (h(n)+h(n))≤C*h(n) .
So, complexity of h(n) is max(h(n), g(n))
Question 40 |
What is the maximum number of parenthesis that will appear on the stack at any one time for parenthesis expression given by ( ( ) ( ( ) ) ( ( ) ) )
2 | |
3 | |
4 | |
5 |
Question 40 Explanation:
Here, when parenthesis match, we are popping until we can push many expression.
i.e.., Maximum 3 parenthesis that will appear on the stack at any one time for parenthesis expression.
i.e.., Maximum 3 parenthesis that will appear on the stack at any one time for parenthesis expression.
Question 41 |
Match the following :
a-iii, b-iv, c-i, d-ii | |
a-iii, b-iv, c-ii, d-i | |
a-iv, b-iii, c-ii, d-i | |
a-iv, b-iii, c-i, d-ii |
Question 41 Explanation:
Automatic storage class→ Value stored in memory and local to the block in which the variable is defined.
Register storage class → Value stored in CPU registers rather than main memory.
Static storage class → Value of the variable persists between different function calls.
External storage class → Scope of the variable is global.
Register storage class → Value stored in CPU registers rather than main memory.
Static storage class → Value of the variable persists between different function calls.
External storage class → Scope of the variable is global.
Question 42 |
When we pass an array as an argument to a function, what actually gets passed ?
Address of the array | |
Values of the elements of the array | |
Base address of the array | |
Number of elements of the array |
Question 42 Explanation:
When we pass an array as an argument to a function it actually gets passed base address of the array.
Question 43 |
While(87) printf(“computer”); The above C statement will
print “computer” 87 times | |
print “computer” 0 times | |
print “computer” 1 times | |
print “computer” infinite times |
Question 43 Explanation:
When we are giving any positive number in while loop. It considered into TRUE value.
So, printf(“computer”); will execute infinite number of times.
while(1) or while(87) are same. It consider like TRUE only.
So, printf(“computer”); will execute infinite number of times.
while(1) or while(87) are same. It consider like TRUE only.
Question 44 |
A friend function can be used to
avoid arguments between classes. | |
allow access to classes whose source code is unavailable. | |
allow one class to access an unrelated class. | |
None of the above |
Question 44 Explanation:
A friend function of a class is defined outside that class scope but it has the right to access all private and protected members of the class. Even though the prototypes for friend functions appear in the class definition, friends are not member functions.
→A function can only be declared a friend by a class itself.
→It can have access to all members of the class, even private ones.
→ A friend function can be used to allow one class to access an unrelated class.
→A function can only be declared a friend by a class itself.
→It can have access to all members of the class, even private ones.
→ A friend function can be used to allow one class to access an unrelated class.
Question 45 |
Which of the following is the correct value returned to the operating system upon the successful completion of a program ?
0 | |
1 | |
-1 | |
Program do not return a value. |
Question 45 Explanation:
The operating system will return 0 when it’s successfully execute a program.
Question 46 |
Manager’s salary details are hidden from the employee. This is called as
Conceptual level data hiding | |
Physical level data hiding | |
External level data hiding | |
Local level data hiding |
Question 46 Explanation:
1. Physical (or) Internal view: is at the lowest level of abstraction, closest to the physical storage method used. It indicates how the data will be stored and describes the data structures and access methods to be used by the database. There is one internal view for the entire database.
2. Global or Conceptual View : At this level of abstraction all the database entities and the relationships among them are included. There is one conceptual view for the entire database.
3. External or User View: The external or user view is at the highest level of database abstraction where only those portions of the database concern to a user or application programme are included. Any number if external or user views may exists for a given global or conceptual view.
2. Global or Conceptual View : At this level of abstraction all the database entities and the relationships among them are included. There is one conceptual view for the entire database.
3. External or User View: The external or user view is at the highest level of database abstraction where only those portions of the database concern to a user or application programme are included. Any number if external or user views may exists for a given global or conceptual view.
Question 47 |
Which of the following statements is false ?
Any relation with two attributes is in BCNF. | |
A relation in which every key has only one attribute is in 2NF. | |
A prime attribute can be transitively dependent on a key in 3NF relation. | |
A prime attribute can be transitively dependent on a key in BCNF relation. |
Question 47 Explanation:
Rules for BCNF is:
i) It is in 3NF.
ii) For any dependency X→ Y
where X is a super key.
iii) Functional dependency has been removed.
Option D is false.
→ Because a prime attribute can’t be transitive dependent on a key in a BCNF relation.
i) It is in 3NF.
ii) For any dependency X→ Y
where X is a super key.
iii) Functional dependency has been removed.
Option D is false.
→ Because a prime attribute can’t be transitive dependent on a key in a BCNF relation.
Question 48 |
A clustering index is created when _______.
primary key is declared and ordered | |
no key ordered | |
foreign key ordered | |
there is no key and no order |
Question 48 Explanation:
→ A Clustering index is created when primary key is declared and ordered. Clustering index is defined on an ordered data file. The data file is ordered on a non-key field.
→ Ordered Indexing is of two types −Dense Index and Sparse Index.
→ Ordered Indexing is of two types −Dense Index and Sparse Index.
Question 49 |
Let R ={A, B, C, D, E, F} be a relation schema with the following dependencies
C → F, E → A, EC → D, A → B Which of the following is a key for R ?
CD | |
EC | |
AE | |
AC |
Question 49 Explanation:
Simple way to solve this question is,
→First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
→In right hand side, CE are missing. It means definitely EC should be there in candidate key.
(EC)+=(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
→First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
→In right hand side, CE are missing. It means definitely EC should be there in candidate key.
(EC)+=(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
Question 50 |
Match the following :
a-ii, b-i, c-iii, d-iv
| |
a-i, b-ii, c-iv, d-iii | |
a-iii, b-ii, c-i, d-iv | |
a-iv, b-i, c-ii, d-iii |
Question 50 Explanation:
DDL→ REVOKE
DML→ LOCK TABLE
TCL→ COMMIT
BINARY Operation→ Natural Difference
DML→ LOCK TABLE
TCL→ COMMIT
BINARY Operation→ Natural Difference
There are 50 questions to complete.