## UGC NET CS 2013 June-paper-2

 Question 1
COCOMO stands for
 A COmposite COst MOdel B COnstructive COst MOdel C COnstructive Composite MOdel D COmprehensive Construction MOdel
Software-Engineering       COCOMO-Model
Question 1 Explanation:
COCOMO stands for COnstructive COst MOdel. It is a procedural software cost estimation model but not software process model. Linear sequential,prototype and spiral models are software process models.
The basic COCOMO equations take the form
1. Effort Applied (E) = a(KLOC)b [ man-months ]
2. Development Time (D) = c(Effort Applied)d [months]
3. People required (P) = Effort Applied / Development Time [count]
 Question 2
Match the following : A a-iii, b-ii, c-iv, d-i B a-ii, b-iii, c-iv, d-i C a-i, b-ii, c-iv, d-iii D a-i, b-ii, c-iii, d-iv
Question 2 Explanation:
Good quality→ Meets both functional and non-functional requirements
Correctness→ Meets the functional requirements
Predictable→ Process is under statistical control
Reliable→ Program does not fail for a specified time in a given environment
 Question 3
While estimating the cost of software, Lines Of Code(LOC) and Function Points(FP) are used to measure which one of the following ?
 A Length of code B Size of software C Functionality of software D None of the above
Software-Engineering       Software-requirements
Question 3 Explanation:
→ A function point is a "unit of measurement" to express the amount of business functionality an information system (as a product) provides to a user. Function points are used to compute a functional size measurement (FSM) of software. The cost (in dollars or hours) of a single unit is calculated from past projects.
→ Lines of code (LOC) is a software metric used to measure the size of a computer program by counting the number of lines in the text of the program's source code.
 Question 4
A good software design must have
 A High module coupling, High module cohesion B High module coupling, Low module cohesion C Low module coupling, High module cohesion D Low module coupling, Low module cohesion
Software-Engineering       Software-design
Question 4 Explanation:
→ A good software design must have low module coupling and high module cohesion.
→ Coupling and Cohesion are used in software design. Cohesion measures strength of a module while coupling measures interdependency between modules.
 Question 5
Cyclometric complexity of a flow graph G with n vertices and e edges is
 A V(G) = e+n–2 B V(G) = e–n+2 C V(G) = e+n+2 D V(G) = e–n–2
Engineering-Mathematics       Graph-Theory
Question 5 Explanation:
Cyclomatic complexity uses 3 formulas:
1. The number of regions corresponds to the cyclomatic complexity
2. V(G),Flow graph is defined as V(G)=E-N+2 where E is the number of flow graph edges, and N is the number of flow graph nodes.
3. V(G),Flow graph is defined as V(G)=P+1 where p is the number of predicate nodes contained in the flow graph G.
 Question 6
When the following code is executed what will be the value of x and y ? int x = 1, y = 0;      y = x++;
 A 2,1 B 2,2 C 1,1 D 1,2
Programming       Operator
Question 6 Explanation:
Given post increment for x value. In post increment the value assign first and update after assigning.
So, x=2 and y=1
 Question 7
How many values can be held by an array A(–1,m;1,m) ?
 A m B m2 C m(m+1) D m(m+2)
Programming       Arrays
Question 7 Explanation:
int A[n] means "n" elements and index starts from 0 to n-1.
→ Similarly, int A[n][n] means both column and rows has "n" elements and total elements are n*n nothing but n2 elements.
In the given example ,an array A(-1,m;1,m)
→ In which rows indexes are -1,0,1,2, and so on up to m then total rows are "m+2".
→ columns consist of 1,2,3 and so on up to m then total elements in columns are "m".
Total elements are (m+2)*m
 Question 8
What is the result of the expression (1&2)+(3/4) ?
 A 1 B 2 C 3 D 0
Programming       Operator
Question 8 Explanation:
Given statement, (1&2)+(3/4)
(1&2) means (1 AND 2)
(¾) means (3 divided by 4)
Step-1: First consider (1&2) Question 9
How many times the word ‘print’ shall be printed by the following program segment ?
for (i=1, i≤2,i++)
for (j=1,j≤2,j++)
for(k=1,k≤2,k++)
printf("print/n")
 A 1 B 3 C 6 D 8
Programming       Control-Statement
Question 9 Explanation:
Iteration-1: In iteration-1 of ith loop will execute once and jth loop 2 times.
Each jth loop will execute kth loop 2 times.
Finally, iteration-1 will display 4 times of “print/n” word.
Iteration-2: In iteration-2 of ith loop will execute once and jth loop 2 times.
Each jth loop will execute kth loop 2 times.
Finally, iteration-2 will display 4 times of “print/n” word.
Above code segment will print 8 times.
Better understanding purpose given modified code segment:
#include
int main()
{
int i,j,k,jj=0,kk=0,ii=0;
for(i=1;i<=2;i++)
{
ii++;
for(j=1;j<=2;j++)
{
jj++;
for(k=1;k<=2;k++)
{
kk++;
printf("print/n");
}
}
}
printf("ii=%d \t jj=%d \t kk=%d",ii,jj,kk);
return 0;
}
Output:
print/n print/n print/n print/n print/n print/n print/n print/n
ii=2 jj=4 kk=8
 Question 10
Which of the following is not a type of Database Management System ?
 A Hierarchical B Network C Relational D Sequential
Database-Management-System       Types-of-DBMS
Question 10 Explanation:
Most common types of database management system types are
1. Hierarchical databases
2. Network databases
3. Relational databases
4. Object-oriented databases
5. Graph databases
6. ER model databases
7. Document databases
 Question 11
Manager’s salary details are to be hidden from Employee Table. This Technique is called as
 A Conceptual level Datahiding B Physical level Datahiding C External level Datahiding D Logical level Datahiding
Question 11 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. Question 12
A Network Schema
 A restricts to one to many relationship B permits many to many relationship C stores Data in a Database D stores Data in a Relation
Database-Management-System       ER-Model
Question 12 Explanation:
→ The network model expands upon the hierarchical structure, allowing many-to-many relationships in a tree-like structure that allows multiple parents. It was most popular before being replaced by the relational model, and is defined by the CODASYL specification.
→ The network model organizes data using two fundamental concepts, called records and sets.
→ Records contain fields (which may be organized hierarchically, as in the programming language COBOL).
→ Sets (not to be confused with mathematical sets) define one-to-many relationships between records: one owner, many members.
→ A record may be an owner in any number of sets, and a member in any number of sets. Question 13
Which normal form is considered as adequate for usual database design ?
 A 2NF B 3NF C 4NF D 5NF
Database-Management-System       Normalization
Question 13 Explanation:
3NF normal form is considered as adequate for usual database design. BCNF doesn't guarantee dependency preservation.
 Question 14
If D1,D2, .. ..Dn are domains in a relational model, then the relation is a table, which is a subset of
 A D1+D2+ … +Dn B D1×D2× … ×Dn C D1∪D2∪ … ∪Dn D D1–D2– … –Dn
Database-Management-System       Relational-databases
Question 14 Explanation:
In D1,D2,..,Dn are domains in a relational model, then the relation is a table, which s a subset of D1xD2x..xDn
 Question 15
Which of the following addresses is used to deliver a message to the correct application program running on a host ?
 A Port B IP C Logical D Physical
Computer-Networks       Application-Layer-Protocol
Question 15 Explanation:
Port addresses is used to deliver a message to the correct application program running on a host.
 Question 16
In ________ substitution, a character in the plaintext is always changed to the same character in the ciphertext, regardless of its position in the text.
 A polyalphabetic B monoalphabetic C transpositional D multi alphabetic
Computer-Networks       Network-Security
Question 16 Explanation:
Monoalphabetic Substitution: The relationship between a character in the plaintext and a character in the ciphertext is always one-to-one
Polyalphabetic Substitution: This is an improvement over the Caesar cipher. In polyalphabetic substitution, each occurrence of a character may have a different substitute. Here the relationship between a character in the plaintext and a character in the ciphertext is always one-to-many.
Transposition Cipher: The transposition cipher, the characters remain unchanged but their positions are changed to create the ciphertext. A transposition cipher does not substitute one symbol for another, instead it changes the location of the symbols. A transposition cipher reorders symbols.
 Question 17
 A Class A B Class B C Class C D Class D
Question 17 Explanation:
→ Class A addresses only include IP starting from 1.x.x.x to 126.x.x.x only. The IP range 127.x.x.x is reserved for loopback IP addresses.
→ Class B IP Addresses range from 128.0.x.x to 191.255.x.x. The default subnet mask for Class B is 255.255.x.x.
→ Class C IP addresses range from 192.0.0.x to 223.255.255.x. The default subnet mask for Class C is 255.255.255.x.
 Question 18
In hierarchical routing with 4800 routers, what region and cluster sizes should be chosen to minimize the size of the routing table for a three layer hierarchy ?
 A 10 clusters, 24 regions and 20 routers B 12 clusters, 20 regions and 20 routers C 16 clusters, 12 regions and 25 routers D 15 clusters, 16 regions and 20 routers
Computer-Networks       Routing
Question 18 Explanation:
Given data, -- Hierarchical routing with 4800 routers.
Hierarchical routing = cluster size * Regions * Routers
-- Cluster size=?
-- Regions=?
-- Routers=?
Step-1: Minimize size= (Clusters-1) + (Regions-1) + Routers
Option-A: 10 clusters, 24 regions and 20 routers
= 9 + 23 + 20
= 52
Option-B: 12 clusters, 20 regions and 20 routers
= 11 + 19 + 20
= 50
Option-C: 16 clusters, 12 regions and 25 routers
= 15 + 11 + 25
= 51
Option-D: 15 clusters, 16 regions and 20 routers
= 14 + 15 + 20
= 49
 Question 19
In IPv4 header, the ______ field is needed to allow the destination host to determine which datagram a newly arrived fragments belongs to.
 A identification B fragment offset C time to live D header checksum
Question 19 Explanation:
→ In IPv4 header, the identification field is needed to allow the destination host to determine which datagram a newly arrived fragments belongs to.
→ Identification field is primarily used for uniquely identifying the group of fragments of a single IP datagram.
→ Some experimental work has suggested using the ID field for other purposes, such as for adding packet-tracing information to help trace datagrams with spoofed source addresses.
 Question 20
Given L1=L(a*baa*) and L2=L(ab*). The regular expression corresponding to language L3 = L1/L2 (right quotient) is given by
 A a*b B a*baa* C a*ba* D None of the above
Theory-of-Computation       Regular-Expression
Question 20 Explanation:
L1=L(a*baa*) and L2=L(ab*). The regular expression corresponding to language L3 = L1/L2
L1={ba, baa, baaa, .........................aba, aaba, aaaba,................abaa, abaaa, ....................}
L2={a, ab, abb, abbb, ..........................}
Take "a" from L1:
L1/a ={ b, ba, baa, .................ab, aab, aaab, ....................aba, abaa, .....................}
So we get : a*ba*
If you take "ab" from L2
Like:
L1/ab = you will not get any string i,e. empty set will be in result in case of L1/ab
as no string in L1 end with "ab"
 Question 21
Given the production rules of a grammar G1 as
S1 → AB | aaB
A → a | Aa
B → b
and the production rules of a grammar G2 as
S2 → aS2bS2 | bS2aS2 | λ
Which of the following is correct statement ?
 A G1 is ambiguous and G2 is not ambiguous. B G1 is ambiguous and G2 is ambiguous. C G1 is not ambiguous and G2 is ambiguous. D G1 is not ambiguous and G2 is not ambiguous.
Compiler-Design       Ambiguous-and-Unambiguous-Grammar
Question 21 Explanation: Question 22
Given a grammar : S1 → Sc, S → SA | A, A → aSb | ab, there is a rightmost derivation
S1 ⇒ Sc ⇒ SAC ⇒ SaSbc
Thus, SaSbc is a right sentential form, and its handle is
 A SaS B bc C Sbc D aSb
Compiler-Design       Handles
Question 22 Explanation:
A “handle” of a string is a substring that matches the RHS of a production and whose reduction to the non-terminal (on the LHS of the production) represents one step along the reverse of a rightmost derivation toward reducing to the start symbol.
And in above question aSb is a handle because it's reduction to the LHS of production A → aSb represents one step along the reverse of a rightmost derivation toward reducing to the start symbol.
 Question 23
The equivalent production rules corresponding to the production rules S → Sα1 | Sα2 | β1 | β2 is
 A S → β1 | β2, A → α1A | α2A | λ B S → β1| β2 | β1A | β2A, A → α1A | α2A C S → β1 | β2, A → α1A | α2A D S → β1 | β2 | β1A | β2A, A → α1A | α2A | λ
Compiler-Design       Grammars
Question 23 Explanation:
Given grammar can generate { β1, β2 , β2α2, , β2 α1, , β1α2, , β1α1 , β1 α1α2..................}
Option A can generate only {β1, β2} so it is not a correct option.
Option B is not correct because it have no terminating point strings containing {α1 , α2}
Option C is not correct because it can generate only {β1, β2}
Option D is correct answer because it can generate all the strings generated by given grammar.
 Question 24
Given a Non-deterministic Finite Automation (NFA) with states p and r as initial and final states respectively and transition table as given below : The minimum number of states required in Deterministic Finite Automation (DFA) equivalent to NFA is
 A 5 B 4 C 3 D 2
Theory-of-Computation       Finite-Automata
Question 24 Explanation: Question 25
Which is the correct statement(s) for Non Recursive predictive parser ?
S1 : First(α) = { t | α⇒ *tβ for some string β}⇒ *tβ
S2 : Follow(X) = { a | S⇒ *αXaβ for some strings α and β}
 A Both statements S1 and S2 are incorrect. B S1 is incorrect and S2 is correct. C S1 is correct and S2 is incorrect. D Both statements S1 and S2 are correct.
Compiler-Design       Parsers
Question 25 Explanation:
Statement-1: → See the symbol (⇒ *) means after some step , here * represent an arbitrary number of steps. * is not part of terminal.
→ So if alpha after some step has t as first symbol (terminal) in some sentential form , then first(alpha) must be {t}
Statement-2:
α→∗tβ Here First(α) = {*} .
S→∗αXaβ
Follow(X) = {a}
So Statement S2 is correct.
 Question 26
Given an open address hash table with load factor α < 1, the expected number of probes in a successful search is
 A At most (1/α) ln ((1–α)/α) B At most (1/α) ln (1/(1–α)) C At least (1/α) ln (1/(1– α)) D At least (1/α) ln (α/(1– α))
Data-Structures       Hashing
Question 26 Explanation:
→ Given an open-address hash table with load factor α = n/m < 1, the expected number of probes in an unsuccessful search is at most 1/(1 − α), assuming uniform hashing.
→ Given an open-address hash table with load factor α = n/m < 1, the expected number of probes in a successful search is at most (1/α) ln (1/(1 − α)), assuming uniform hashing and assuming that each key in the table is equally likely to be searched for.
 Question 27
For a B-tree of height h and degree t, the total CPU time used to insert a node is
 A O(h log t) B O(h log t) C O(t2h) D O(th)
Database-Management-System       B-and-B+-Trees
Question 27 Explanation:
→ For a B-tree of height h and degree t, the total CPU time used to insert a node is O(th).
→ The number of disk pages accessed by B-tree search is therefore O(h) = O(logt n), where ‘h’ is the height of the B-tree and ‘n’ is the number of keys in the B-tree.
→ The time taken by each node is O(t), and the total CPU time is O(th) = O(t logt n).
 Question 28
The time complexity to build a heap with a list of n numbers is
 A O(log n) B O(n) C O(n logn) D O(n2)
Data-Structures       Heap-Tree
Question 28 Explanation:
Build-Max-Heap(A)
// Input: A: an (unsorted) array
// Output: A modified to represent a heap.
// Running Time: O(n) where n = length[A]
1 heap-size[A] ← length[A]
2 for i ← Floor(length[A]/2) downto 1
3 Max-Heapify(A, i)
The Build-Max-Heap function that follows, converts an array A which stores a complete binary tree with n nodes to a max-heap by repeatedly using Max-Heapify in a bottom up manner. It is based on the observation that the array elements indexed by floor(n/2) + 1, floor(n/2) + 2, ..., n are all leaves for the tree (assuming that indices start at 1), thus each is a one-element heap.
→ Build-Max-Heap runs Max-Heapify on each of the remaining tree nodes.
 Question 29
The value of postfix expression :
8 3 4 + – 3 8 2 / + * 2 \$ 3 + is
 A 17 B 131 C 64 D 52
Data-Structures       Queues-and-Stacks
Question 29 Explanation: Question 30
Consider the following statements for priority queue :
S1 : It is a data structure in which the intrinsic ordering of the elements does determine the result of its basic operations.
S2 : The elements of a priority queue may be complex structures that are ordered on one or several fields.
Which of the following is correct ?
 A Both S1 and S2 are incorrect. B S1 is correct and S2 is incorrect. C S1 is incorrect and S2 is correct. D Both S1 and S2 are correct
Data-Structures       Queues-and-Stacks
Question 30 Explanation:
S1: TRUE: It is a data structure in which the intrinsic ordering of the elements does determine the result of its basic operations.
S2: TRUE: The elements of a priority queue may be complex structures that are ordered on one or several fields.
 Question 31
Repository of information gathered from multiple sources, storing under unified scheme at a single site is called as
 A Data mining B Meta data C Data warehousing D Database
Question 31 Explanation:
→ Repository of information gathered from multiple sources, storing under unified scheme at a single site is called as Data Warehousing.
→ Data warehousing will support OLAP and OLTP.
→ Business intelligence and Data warehousing is used for forecasting and Data mining.
→ Business intelligence and Data warehousing is used for analysis of large volumes of sales data.
→ The Data warehouse Hadoop cluster at Facebook has become the largest known Hadoop storage cluster in the world.
 Question 32
The task of correcting and preprocessing data is called as
 A Data streaming B Data cleaning C Data mining D Data storming
Question 32 Explanation:
→ Data cleansing or data cleaning is the process of detecting and correcting (or removing) corrupt or inaccurate records from a record set, table, or database and refers to identifying incomplete, incorrect, inaccurate or irrelevant parts of the data and then replacing, modifying, or deleting the dirty or coarse data.
→ Data cleansing may be performed interactively with data wrangling tools, or as batch processing through scripting.
 Question 33
Using data p=3, q=11, n=pq, d=7 in RSA algorithm find the cipher text of the given plain text SUZANNE
 A BUTAEEZ B SUZANNE C XYZABCD D ABCDXYZ
Computer-Networks       Network-Security
Question 33 Explanation:
n = p*q = 33,
Φ = (p-1)*(q-1) = 20,
d = 7 and e = 3 (solve by e*d = 1 mod 20)
Message M= SUZANNE
Here take A = 1, B = 2, C = 3 ….. Z = 26, A = 27, …...
Cipher text = (Me mod n)
⇒ S = 19, Cipher = 193 mod 33 = 28 = B
⇒ U = 21, Cipher = 213 mod 33 = 21 = U
and so on ..
 Question 34
The relation “divides” on a set of positive integers is ________.
 A Symmetric and transitive B Anti symmetric and transitive C Symmetric only D Transitive only
Engineering-Mathematics       Set-Theory
Question 34 Explanation:
The relation “divides” on a set of positive integers is Anti symmetric and Transitive.
Proof:
Let assume a,b two set elements.
The relation is antisymmetric if and only if for every a,b in the set.
IF(a∣b AND b∣a), then it must follow that a=b.
→ If it's FALSE then both a∣b AND b∣a, then it's perfectly consistent to have a≠b.
→ The only time a∣b AND b∣a is exactly when a=b, since then we have a∣b ⟺ a∣a is TRUE for all a.
Hence the relation is antisymmetric.
 Question 35
Give as good a big–O estimate as possible for the following functions : (nlogn+n2)(n3+2) and (n!+2n)(n3+log(n2+1))
 A O(n5+2n2) & O(n3*n!) B O(n5) & O(n3*2n) C O(n5) & O(n3* n!) D O(n5+2n2) & O(n3*2n)
Algorithms       Asymptotic-Complexity
Question 35 Explanation:
Step-1: Consider first function (nlogn+n2)(n3+2).
Here, It is performing multiplication of 2 polynomials.
(nlogn+n2) → Raising value is n2. We can write asymptotically O(n2)
(n3+2) → Raising value is n3. We can write asymptotically O(n3)
= O(n2)*O(n3)
= O(n5)
Step-2: Second function (n!+2n)(n3+log(n2+1))
Here, It is performing multiplication of 2 polynomials.
(n!+2n) → Raising value is n!. We can write asymptotically O(n!)
(n3+log(n2+1)) → Raising value is n3. We can write asymptotically O(n3)
= O(n!)*O(n3)
= O(n3* n!)
 Question 36
A test contains 100 true/false questions. How many different ways can a student answer the questions on the test, if the answer may be left blank also.
 A 100P2 B 100C2 C 2100 D 3100
Engineering-Mathematics       Combinatorics
Question 36 Explanation:
Given data,
-- Total number of questions=100
-- Total possibilities for each question=3 (Note: BLANK or TRUE or FALSE)
Step-1: 1 question= 3 possibilities. Question 37
Which of the following connected simple graph has exactly one spanning tree ?
 A Complete graph B Hamiltonian graph C Euler graph D None of the above
Algorithms       Minimum-Spanning-Tree
Question 37 Explanation:
→ Complete graph have nn-2 spanning trees.
→ Hamiltonian graph will get more than one spanning tree.
→ Euler graph will get more than one spanning tree.
 Question 38
How many edges must be removed to produce the spanning forest of a graph with N vertices, M edges and C connected components ?
 A M+N–C B M–N–C C M–N+C D M+N+C
Engineering-Mathematics       Graph-Theory
Question 38 Explanation:
→ Spanning forest is nothing but a set of spanning trees, one for each connected component in the graph
→ In the spanning forest the components are represented by a set of edges, rather than a set of vertices. Any vertices in the graph that are entirely unconnected will not appear in the spanning forest.
→ In spanning forest, we will use multiple trees to represent a path through the graph.
→ Consider the below example Number of components(C)=3
Number of vertices(N)=9
Number of Edges(M)=8
By removing two edges, we can make spanning forest in the above graph.
The number of edges can be removed to make spanning forest is M-N+C=8-9+3=2
So option C is correct.
 Question 39
Which of the following shall be a compound proposition involving the propositions p, q and r, that is true when exactly two of the p, q and r are true and is false otherwise
 A (p ∨ q ∧ ⌐r) ∨ ( p ∧ q ∧ r) ∧ (⌐p ∧ q ∨ r) B (p ∧ q ∨ r) ∧ ( p ∧ q ∧ r) ∨ (⌐q ∧ ⌐p∧ ⌐r) C (p ∧ q ∧⌐ r) ∨ ( p ∧ ⌐q ∧ r) ∨ (⌐ p ∧ q ∧ r) D (p ∨ r ∧ q) ∨ ( p ∧ q ∧ r) ∨ (⌐p ∧ q ∧ r)
Engineering-Mathematics       Propositional-Logic
Question 39 Explanation:
From the question, the propositions consists of exactly two of the variables should true and final proposition should be true.
Only option C consists of exactly two of the variables are true and all remaining variables deviating this rule.
In this proposition, (p ∧ q ∧⌐ r) ∨ ( p ∧ ⌐q ∧ r) ∨ (⌐ p ∧ q ∧ r)
(p ∧ q ∧⌐ r) → p and q are true
( p ∧ ⌐q ∧ r) → p and r are true
(⌐ p ∧ q ∧ r) → q and r are true
Finally the proposition consists of “V” operation , so the entire proposition is true.
 Question 40
The truth value of the statements : ∃!xP(x) → ∃xP(x) and ∃!x ⌐P(x) →⌐∀xP(x), (where the notation ∃!xP(x) denotes the proposition “There exists a unique x such that P(x) is true”) are :
 A True and False B False and True C False and False D True and True
Engineering-Mathematics       Propositional-Logic
Question 40 Explanation:
From the given question, → The symbol ∃ is call the existential quantifier and represents the phrase “there exists” or “for some”. The existential quantification of P(x) is the statement “P(x) for some values x in the universe”, or equivalently, “There exists a value for x such that P(x) is true”, which is written ∃xP(x). So the statement ∃!xP(x) → ∃xP(x) is true.
→ If P(x) is true for at least one element in the domain, then ∃xP(x) is true. Otherwise it is false.
→ Accordingly DeMorgan’s laws for quantifiers:the following statements are true.
¬∀xP(x) ≡ ∃x¬P(x)
¬∃xP(x) ≡ ∀x¬P(x) then the statement is ∃!x ⌐P(x) →⌐∀xP(x) is true.
 Question 41
How many different Boolean functions of degree 4 are there ?
 A 24 B 28 C 212 D 216
Digital-Logic-Design       Boolean-Function
Question 41 Explanation:
Boolean functions of n variables, there are 22n different Boolean functions on n Boolean variables. Question 42
A Boolean operator Ɵ is defined as follows :
1 Ɵ 1 = 1,
1 Ɵ 0 = 0,
0 Ɵ 1 = 0,
0 Ɵ 0 = 1
What will be the truth value of the expression (x Ɵ y) Ɵ z = x Ɵ (y Ɵ z) ?
 A Always false B Always true C Sometimes true D True when x, y, z are all true
Digital-Logic-Design       Boolean-Operator
Question 42 Explanation:
They given XNOR property for Ɵ. Question 43
Which one of the following is decimal value of a signed binary number 1101010, if it is in 2’s complement form ?
 A -42 B -22 C -21 D -106
Digital-Logic-Design       Number-Systems
Question 43 Explanation:
Step-1: 2's complement number are weighted number. So, 1101010= -1*(0.26) + 1*25 + 1*23 + 1*21 = -64 + 32 +8 +2 = -22 Hence (1101010)=(-22) in 2's complement form
 Question 44
A set of processors P1, P2, ……, Pk can execute in parallel if Bernstein’s conditions are satisfied on a pairwise basis; that is P1 || P2 || P3 || ….. || Pk if and only if :
 A Pi || Pj for all i ≠ j B Pi || Pj for all i = j+1 C Pi || Pj for all i ≤ j D Pi || Pj for all i ≥ j
Question 44 Explanation:
Bernstein’s conditions for parallelism:
Define:
Ii as the input set of a process Pi
Oi as the output set of a process Pi
P1 and P2 can execute in parallel (denoted as P1 || P2) under the condition:
I1 ∩ O2 = 0
I2 ∩ O1 = 0
O1 ∩ O2 = 0
Note that I1 ∩ I2 <> 0 does not prevent parallelism.
Input set: also called read set or domain of a process
Output set: also called write set or range of a process
A set of processes can execute in parallel if Bernstein’s conditions are satisfied on a pairwise basis; that is, P1 || P2 || … || PK if and only if Pi || Pj for all i<>j
The parallelism relation is commutative: Pi || Pj implies that Pj || Pi
The relation is not transitive: Pi || Pj and Pj || Pk do not necessarily mean Pi || Pk
Associativity: Pi || Pj || Pk implies that (Pi || Pj) || Pk = Pi || (Pj || Pk)
For n processes, there are 3n(n-1)/2 conditions; violation of any of them prohibits parallelism collectively or partially
Statements or processes which depend on run-time conditions are not transformed to parallelism. (IF or conditional branches)
The analysis of dependences can be conducted at code, subroutine, process, task, and program levels; higher-level dependence can be inferred from that of subordinate levels.
 Question 45
When a mobile telephone physically moves from one to another cell, the base station transfers ownership to the cell getting strongest signal. This process is known as _______.
 A handoff B mobile switching C mobile routing D cell switching
Question 45 Explanation:
→ When a mobile telephone physically moves from one to another cell, the base station transfers ownership to the cell getting strongest signal. This process is known as handoff.
Hard handover (or) Hard Handoff: Early systems used hard handoff. In a hard handoff, a mobile station only communicates with one base station. When the MS moves from one cell to another, communication must first be broken with the previous base station before communication can be established with the new one.
Soft handover (or) Soft Handoff: New systems use a soft Handoff. In this case, a mobile station can communicate with base stations at the same time. This means that, during handoff, a mobile station may continue with the new base station before breaking off from the old one.
 Question 46
A virtual memory based memory management algorithm partially swaps out a process. This is an example of
 A short term scheduling B long term scheduling C medium term scheduling D mutual exclusion
Operating-Systems       Process-Scheduling
Question 46 Explanation:
A virtual memory based memory management algorithm partially swaps out a process. This is an example of medium term scheduling.
Medium-term scheduling:The decision to add to the number of processes that are partially or fully in main memory.
Long-term scheduling:The decision to add to the pool of processes to be executed.
Short-term scheduling:The decision as to which available process will be executed by the processor.
 Question 47
Assuming that the disk head is located initially at 32, find the number of disk moves required with FCFS if the disk queue of I/O block requests are 98, 37, 14, 124, 65, 67 :
 A 310 B 324 C 320 D 321
Operating-Systems       Disk-Scheduling
Question 47 Explanation: = 66 + 61 + 23 + 110 + 59 + 2
= 321
 Question 48
Let the page fault service time be 10 millisecond(ms) in a computer with average memory access time being 20 nanosecond(ns). If one page fault is generated for every 106 memory accesses, what is the effective access time for memory ?
 A 21 ns B 23 ns C 30 ns D 35 ns
Operating-Systems       Memory-Management
Question 48 Explanation:
P=page fault rate
EA = p*page fault service time + (1-p) * Memory access time
=1/106*10*106+(1-1/106)*20
≅ 29.9 ns
 Question 49
Consider the following UNIX command :
sort <in> temp; head – 30 <temp; rm temp Which of the following functions shall be performed by this command ?
 A Sort, taking the input from“temp”, prints 30 lines from temp and delete the file temp B Sort the file “temp”, removes 30 lines from temp and delete the file temp C Sort, taking the input from “in” and writing the output to“temp” then prints 30 lines from temp on terminal. Finally “temp” is removed. D Sort, taking the input from “temp” and then prints 30 lines from “temp” on terminal. Finally “temp” is removed
Operating-Systems       UNIX-Operating-System
Question 49 Explanation:
sort temp; → Sort, taking the input from “in” and writing the output to“temp”
head – 30 rm temp→ “temp” is removed.
 Question 50
The ‘mv’ command changes
 A the inode B the inode-number C the directory entry D both the directory entry and the inode
Operating-Systems       UNIX-Operating-System
Question 50 Explanation:
mv (short for move) is a Unix command that moves one or more files or directories from one place to another. If both filenames are on the same filesystem, this results in a simple file rename; otherwise the file content is copied to the new location and the old file is removed.
There are 50 questions to complete.