UGC NET CS 2017 Nov paper3
Question 1 
In 8085 microprocessor which of the following flag(s) is (are) affected by an arithmetic operation ?
AC flag Only  
CY flag Only  
Z flag Only  
AC, CY, Z flags 
Question 1 Explanation:
AC Flag: This flag is set if thee is a carry to fifth bit from the fourth bits if two numbers.
Question 2 
In 8085 microprocessor the address bus is of __________ bits.
4  
8  
16  
32 
Question 2 Explanation:
In 8085 microprocessor the address bus is of 16 bits and data bus is 8 bits.
Question 3 
In the architecture of 8085 microprocessor match the following:
(a)(iv), (b)(i), (c)(ii)  
(a)(iii), (b)(iv), (c)(ii)  
(a)(ii), (b)(iii), (c)(i)  
(a)(i), (b)(ii), (c)(iv) 
Question 3 Explanation:
Processing unit→ ALU
Arithmetic logic unit(ALU) is a fundamental building block of many types of computing circuits, including the central processing unit (CPU) of computers, FPUs, and graphics processing units (GPUs).
Instruction unit →Timing and control
Timing and Control unit is responsible for hardwired control unit and Microprogrammed control unit
Storage and Interface unit → General purpose Register
The general purpose register can store a data (or) a memory location address.
Arithmetic logic unit(ALU) is a fundamental building block of many types of computing circuits, including the central processing unit (CPU) of computers, FPUs, and graphics processing units (GPUs).
Instruction unit →Timing and control
Timing and Control unit is responsible for hardwired control unit and Microprogrammed control unit
Storage and Interface unit → General purpose Register
The general purpose register can store a data (or) a memory location address.
Question 4 
Which of the following addressing mode is best suited to access elements of an array of contiguous memory locations ?
Indexed addressing mode  
Base Register addressing mode  
Relative address mode  
Displacement mode 
Question 4 Explanation:
Index mode:
The address of the operand is obtained by adding to the contents of the general register (called index register) a constant value. The number of the index register and the constant value are included in the instruction code.
→ Index Mode is used to access an array whose elements are in successive memory locations (or) contiguous memory locations.
The address of the operand is obtained by adding to the contents of the general register (called index register) a constant value. The number of the index register and the constant value are included in the instruction code.
→ Index Mode is used to access an array whose elements are in successive memory locations (or) contiguous memory locations.
Question 5 
Which of the following is correct statement ?
In memory  mapped I/O, the CPU can manipulate I/O data residing in interface registers that are not used to manipulate memory words.  
The isolated I/O method isolates memory and I/O addresses so that memory address range is not affected by interface address assignment.  
In asynchronous serial transfer of data the two units share a common clock.  
In synchronous serial transmission of data the two units have different clocks. 
Question 5 Explanation:
FALSE: In memory  mapped I/O, the CPU can manipulate I/O data residing in interface registers that are used to manipulate memory words.
TRUE: The isolated I/O method isolates memory and I/O addresses so that memory address range is not affected by interface address assignment.
FALSE: In asynchronous serial transfer of data the two units share a different clock.
FALSE: In synchronous serial transmission of data the two units have common clocks.
TRUE: The isolated I/O method isolates memory and I/O addresses so that memory address range is not affected by interface address assignment.
FALSE: In asynchronous serial transfer of data the two units share a different clock.
FALSE: In synchronous serial transmission of data the two units have common clocks.
Question 6 
A microinstruction format has microops field which is divided into three subfields F_{1}, F_{2}, F_{3} each having seven distinct microoperations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of microinstruction is:
17  
20  
24  
32 
Question 6 Explanation:
Given data,
 Microops field F_{1} = 7 distinct micro operations
 Microops field F_{2} = 7 distinct micro operations
 Microops field F_{3} = 7 distinct micro operations
 Condition field(CD) = 4 status bits
 Branch Field(BR) = 4 options
 Address space =128 memory locations
 Size of micro operation=?
Given microinstruction format
 Microops field F_{2} = 7 distinct micro operations
 Microops field F_{3} = 7 distinct micro operations
 Condition field(CD) = 4 status bits
 Branch Field(BR) = 4 options
 Address space =128 memory locations
 Size of micro operation=?
Given microinstruction format
Question 7 
Consider the following four schedules due to three transactions (indicated by the subscript) using read and write on a data item X, denoted by r(X) and w(X) respectively. Which one of them is conflict serializable ?
S1: r_{1}(X); r_{2}(X); w_{1}(X); r_{3}(X); w_{2}(X)
S2: r_{2}(X); r_{1}(X); w_{2}(X); r_{3}(X); w_{1}(X)
S3: r_{3}(X); r_{2}(X); r_{1}(X); w_{2}(X); w_{1}(X)
S4: r_{2}(X); w_{2}(X); r_{3}(X); r_{1}(X); w_{1}(X)
S1: r_{1}(X); r_{2}(X); w_{1}(X); r_{3}(X); w_{2}(X)
S2: r_{2}(X); r_{1}(X); w_{2}(X); r_{3}(X); w_{1}(X)
S3: r_{3}(X); r_{2}(X); r_{1}(X); w_{2}(X); w_{1}(X)
S4: r_{2}(X); w_{2}(X); r_{3}(X); r_{1}(X); w_{1}(X)
S_{1}  
S_{2}  
S_{3}  
S_{4} 
Question 7 Explanation:
Question 8 
Suppose a database schedule S involves transactions T1, T2, .............,Tn. Consider the precedence graph of S with vertices representing the transactions and edges representing the conflicts. If S is serializable, which one of the following orderings of the vertices of the precedence graph is guaranteed to yield a serial schedule ?
Topological order  
Depth  first order  
Breadth  first order  
Ascending order of transaction indices 
Question 8 Explanation:
If a schedule is conflict serializable then no cycle in precedence graph should be present.
But BFS and DFS are also possible for cyclic graphs.
And topological sort is not possible for cyclic graph.
Moreover option (D) is also wrong because in a transaction with more indices might come before lower one.
But BFS and DFS are also possible for cyclic graphs.
And topological sort is not possible for cyclic graph.
Moreover option (D) is also wrong because in a transaction with more indices might come before lower one.
Question 9 
If every nonkey attribute is functionally dependent on the primary key, then the relation is in __________ .
First normal form  
Second normal form  
Third normal form  
Fourth normal form 
Question 9 Explanation:
1NF : A relation is in 1NF if it does not contain multivalue attributes and composite attributes.
2NF: A relation is in 2NF is there is no partial dependency exist in the relation. i.e.
primary_key → nonkey attribute
prime_attribute → nonkey attribute.
3NF : A relation is in 3NF if there is no transitive dependency or we can say that a relation is in 3NF if either LHS of a functional dependency is a super key or the RHS of the functional dependency is a prime key attribute.
4NF : A relation is in 4NF if there is no multivalued dependency exist in the relation.
Hence the correct answer is option (C)
NOTE: A prime key attribute is a attribute which is the part of primary key attributes. For example: If ABC is the the primary key then A,B,C, AB, AC, BC are the prime key attributes.
2NF: A relation is in 2NF is there is no partial dependency exist in the relation. i.e.
primary_key → nonkey attribute
prime_attribute → nonkey attribute.
3NF : A relation is in 3NF if there is no transitive dependency or we can say that a relation is in 3NF if either LHS of a functional dependency is a super key or the RHS of the functional dependency is a prime key attribute.
4NF : A relation is in 4NF if there is no multivalued dependency exist in the relation.
Hence the correct answer is option (C)
NOTE: A prime key attribute is a attribute which is the part of primary key attributes. For example: If ABC is the the primary key then A,B,C, AB, AC, BC are the prime key attributes.
Question 10 
Consider a relation R (A, B, C, D, E, F, G, H), where each attribute is atomic, and following functional dependencies exist.
CH → G
A → BC
B → CFH
E → A
F → EG
The relation R is __________ .
CH → G
A → BC
B → CFH
E → A
F → EG
The relation R is __________ .
in 1NF but not in 2NF  
in 2NF but not in 3NF
 
in 3NF but not in BCNF  
in BCNF 
Question 10 Explanation:
The attribute D is not part of any FD's. So, D can be a candidate key or it may be part of the candidate key.
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD+ = {ABCDEFGH}
BD+ = {ABCDEFGH}
ED+ = {ABCDEFGH}
FD+= {ABCDEFGH}
But CD+, GD+ and HD+ does not give all the attributes hence CD, GD and HD are not candidate keys.
Here Candidate keys are AD, BD, ED and FD.
A → BC, B → CFH and F → EG etc are partial dependencies.
So given relation is in 1NF, but not in 2NF.
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD+ = {ABCDEFGH}
BD+ = {ABCDEFGH}
ED+ = {ABCDEFGH}
FD+= {ABCDEFGH}
But CD+, GD+ and HD+ does not give all the attributes hence CD, GD and HD are not candidate keys.
Here Candidate keys are AD, BD, ED and FD.
A → BC, B → CFH and F → EG etc are partial dependencies.
So given relation is in 1NF, but not in 2NF.
Question 11 
Given two relations R1(A, B) and R2(C, D), the result of following query
Select distinct A, B
from R1, R2
is guaranteed to be same as R1 provided one of the following condition is satisfied.
R1 has no duplicates and R2 is empty.  
R1 has no duplicates and R2 is non  empty.  
Both R1 and R2 have no duplicates.  
R2 has no duplicates and R1 is non  empty. 
Question 11 Explanation:
Option(A): Cartesian product with a empty table will result in zero tuple because we can't have any ordered pair with a empty table. Hence option(A) is incorrect option.
Option(B):
Option(B):
Question 12 
Consider a schema R(A, B, C, D) and following functional dependencies.
A → B
B → C
C → D
D → B
Then decomposition of R into R1(A, B), R2(B, C) and R3(B, D) is __________ .
A → B
B → C
C → D
D → B
Then decomposition of R into R1(A, B), R2(B, C) and R3(B, D) is __________ .
Dependency preserving and lossless join.  
Lossless join but not dependency preserving.  
Dependency preserving but not lossless join.  
Not dependency preserving and not lossless join. 
Question 12 Explanation:
(A, B) (B, C)  common attribute is (B) and due to B→C, B is a key for (B, C) and hence ABC can be losslessly decomposed into (A, B)and (B, C).
(A, B, C) (B, D)  common attributes is B and B→D is a FD (via B→C, C→D), and hence, B is a key for (B, D). So, decomposition of (A, B, C, D) into (A, B, C) (B, D) is lossless.
Thus the given decomposition is lossless.
The given decomposition is also dependency preserving as the dependencies A→B is present in (A, B), B→C is present in (B, C), D→B is present in (B, D) and C→D is indirectly present via C→B in (B, C) and B→D in (B, D).
(A, B, C) (B, D)  common attributes is B and B→D is a FD (via B→C, C→D), and hence, B is a key for (B, D). So, decomposition of (A, B, C, D) into (A, B, C) (B, D) is lossless.
Thus the given decomposition is lossless.
The given decomposition is also dependency preserving as the dependencies A→B is present in (A, B), B→C is present in (B, C), D→B is present in (B, D) and C→D is indirectly present via C→B in (B, C) and B→D in (B, D).
Question 13 
Which of the following is not a component of Memory tube display ?
Flooding gun  
Collector  
Ground  
Liquid Crystal 
Question 13 Explanation:
Memory tube display components:
1. Connector Pins
2. Electron Gun
3. Base
4. Focusing System
5. Control Grid Voltage
6. X/Y Deflect
7. Phosphor
8. Collector
9. Ground
Note: Liquid crystal is not a component of memory tube display
.
1. Connector Pins
2. Electron Gun
3. Base
4. Focusing System
5. Control Grid Voltage
6. X/Y Deflect
7. Phosphor
8. Collector
9. Ground
Note: Liquid crystal is not a component of memory tube display
.
Question 14 
Which of the following is not true in case of Oblique Projections?
Parallel projection rays are not perpendicular to the viewing plane.  
Parallel lines in space appear parallel on the final projected image.  
Used exclusively for pictorial purposes rather than formal working drawings.  
Projectors are always perpendicular to the plane of projection. 
Question 15 
With respect to CRT, the horizontal retrace is defined as:
The path an electron beam takes when returning to the left side of the CRT.  
The path an electron beam takes when returning to the right side of the CRT.  
The technique of turning the electron beam off while retracing.  
The technique of turning the electron beam on/off while retracing. 
Question 16 
Find the equation of the circle x^{2}+y^{2}=1 in terms of x'y' coordinates, assuming that the xy coordinate system results from a scaling of 3 units in the x' direction and 4 units in the y' direction.
3(x')^{2} + 4(y')^{2} = 1  
(x'/3)^{2} + (y'/4)^{2} = 1  
(3x')^{2} + 4(y')^{2} = 1  
1/3(x')^{2} + 1/4(y')^{2} = 1 
Question 17 
Find the normalization transformation that maps a window whose lower left corner is at (1, 1) and upper right corner is at (3, 5) onto a viewport that is the entire normalized device screen.
Question 18 
The three aspects of Quantization, programmers generally concerned with are:
Coding error, Sampling rate and Amplification  
Sampling rate, Coding error and Conditioning  
Sampling rate, Aperture time and Coding error  
Aperture time, Coding error and Strobing 
Question 19 
The logic of pumping lemma is an example of __________.
iteration  
recursion  
the divide and conquer principle  
the pigeonhole principle 
Question 19 Explanation:
→ A pumping lemma (or) pumping argument states that, for a particular language to be a member of a language class, any sufficiently long string in the language contains a section, or sections, that can be removed, or repeated any number of times, with the resulting string remaining in that language.
→ The proofs of these lemmas typically require counting arguments such as the pigeonhole principle.
→ Hence, the logic of pumping lemma is a good example of the pigeonhole principle.
→ The proofs of these lemmas typically require counting arguments such as the pigeonhole principle.
→ Hence, the logic of pumping lemma is a good example of the pigeonhole principle.
Question 20 
Heap allocation is required for languages that
use dynamic scope rules  
support dynamic data structures  
support recursion  
support recursion and dynamic data structures 
Question 20 Explanation:
→ Heap allocation is required for languages that support dynamic data structures. The heap is managed via calls to new, delete, callac, realloc, malloc, free, etc.
→ Stack allocation is required for local variables. Space on the stack is reserved for local variables when they are declared.
→ Stack allocation is required for local variables. Space on the stack is reserved for local variables when they are declared.
Question 21 
Pumping lemma for regular language is generally used for proving:
whether two given regular expressions are equivalent  
a given grammar is ambiguous  
a given grammar is regular  
a given grammar is not regular 
Question 21 Explanation:
Pumping lemma for regular language is generally used for proving a given grammar is not regular.
Question 22 
Which of the following problems is undecidable?
To determine if two finite automata are equivalent  
Membership problem for context free grammar  
Finiteness problem for finite automata  
Ambiguity problem for context free grammar 
Question 23 
Finite state machine can recognize language generated by __________.
Only context free grammar  
Only context sensitive grammar  
Only regular grammar  
any unambiguous grammar 
Question 24 
The language L = {a^{i} b c^{i} │ i >= 0} over the alphabet {a, b, c} is:
a regular language.  
not a deterministic context free language but a context free language.  
recursive and is a deterministic context free language.  
not recursive.

Question 24 Explanation:
→ The given language is Context free language(CFL) because we can push a's onto the top of stack as a's come as input and when a "b" come as input don't push it on the top of stack just change the state and after that when c's come come as input pop one "a" from top of stack for each "c".
→ In this way the given language can be accepted by the deterministic pushdown automata and hence the language is deterministic context free language. And when a language is CFL it is also a recursive language.
Hence the correct answer is option (C)
→ In this way the given language can be accepted by the deterministic pushdown automata and hence the language is deterministic context free language. And when a language is CFL it is also a recursive language.
Hence the correct answer is option (C)
Question 25 
Suppose we want to download text documents at the rate of 100 pages per second. Assume that a page consists of an average of 24 lines with 80 characters in each line. What is the required bit rate of the channel?
192 kbps  
512 kbps  
1.248 Mbps  
1.536 Mbps 
Question 25 Explanation:
Given data,
 Number of Pages(P)=100 per second
 Number of Lines(L)= 24
 Number of characters(C) =80
 One character(O) =8 bits
 Bit rate of channel=?
Step1: Bit rate of channel= P*L*C*O
= 100*24*80*8 bits per second
= 1536000
Step2: Given options in Megabits per seconds.
= 1.536 Mbps
 Number of Pages(P)=100 per second
 Number of Lines(L)= 24
 Number of characters(C) =80
 One character(O) =8 bits
 Bit rate of channel=?
Step1: Bit rate of channel= P*L*C*O
= 100*24*80*8 bits per second
= 1536000
Step2: Given options in Megabits per seconds.
= 1.536 Mbps
Question 26 
Quadrature Amplitude Modulation means changing both:
Frequency and phase of the carrier.  
Frequency and Amplitude of the carrier.  
Amplitude and phase of the carrier.  
Amplitude and Wavelength of the carrier. 
Question 26 Explanation:
Quadrature Amplitude Modulation means changing both amplitude and phase of the carrier.
Question 27 
If a file consisting of 50,000 characters takes 40 seconds to send, then the data rate is __________.
1 kbps  
1.25 kbps  
2 kbps  
10 kbps 
Question 27 Explanation:
Given data,
 Total number of characters=50,000
 One character=8 bits
 Total time to send 50,000 characters= 40 sec
 Data rate=?
Step1: Data rate= (Total number of characters* one character bits)/ Total time
= (50,000*8)/40
= 4,00,000/40
= 10,000 bits per seconds
Step2: We calculated bits per second but given options in Kbps.
= 10 Mbps
 Total number of characters=50,000
 One character=8 bits
 Total time to send 50,000 characters= 40 sec
 Data rate=?
Step1: Data rate= (Total number of characters* one character bits)/ Total time
= (50,000*8)/40
= 4,00,000/40
= 10,000 bits per seconds
Step2: We calculated bits per second but given options in Kbps.
= 10 Mbps
Question 28 
Match the following:
(a)(iv), (b)(iii), (c)(i), (d)(ii)  
(a)(iii), (b)(iv), (c)(ii), (d)(i)  
(a)(iv), (b)(ii), (c)(iii), (d)(i)  
(a)(iv), (b)(iii), (c)(ii), (d)(i) 
Question 28 Explanation:
Data link layer→ Framing
Network layer→ Routing
Transport layer→ Connection control
Presentation layer→ Encryption
Network layer→ Routing
Transport layer→ Connection control
Presentation layer→ Encryption
Question 29 
The address of a class B host is to be split into subnets with a 6  bit subnet number. What is the maximum number of subnets and maximum number of hosts in each subnet?
62 subnets and 1022 hosts  
64 subnets and 1024 hosts  
62 subnets and 254 hosts  
64 subnets and 256 hosts 
Question 29 Explanation:
It is a class B address, so there 16bits for NID and 16bits for HID.
From HID, we took 6bits for subnetting.
Then total subnets possible = ( 2^{6} )  2 = 64
Total hosts possible for each subnet = (2^{10})  2 = 1022
From HID, we took 6bits for subnetting.
Then total subnets possible = ( 2^{6} )  2 = 64
Total hosts possible for each subnet = (2^{10})  2 = 1022
Question 30 
Which of the following statements are true?
(a) The fragmentation fields in the base header section of IPv4 have moved to the fragmentation extension header in IPv6.
(b) The authentication extension header is new in IPv6.
(c) The record route option is not implemented in IPv6.
(a) The fragmentation fields in the base header section of IPv4 have moved to the fragmentation extension header in IPv6.
(b) The authentication extension header is new in IPv6.
(c) The record route option is not implemented in IPv6.
(a) and (b) Only  
(b) and (c) Only  
(a) and (c) Only  
(a), (b) and (c) 
Question 30 Explanation:
Question 31 
Consider a full binary tree with n internal nodes, internal path length i, and external path length e. The internal path length of a full binary tree is the sum, taken over all nodes of the tree, of the depth of each node. Similarly, the external path length is the sum, taken over all leaves of the tree, of the depth of each leaf. Which of the following is correct for the full binary tree?
e = i+n  
e = i+2n  
e = 2i+n  
e = 2^{n}+i 
Question 31 Explanation:
→ A node's path length is the number of links (or branches) required to get back to the root.
→ The root has path length zero and the maximum path length in a tree is called the tree's height.
→ The sum of the path lengths of a tree's internal nodes is called the internal path length and the sum of the path lengths of a tree's external nodes is called the external path length.
External Path Length:
The sum over all external (square) nodes of the lengths of the paths from the root of an extended binary tree to each node. For example, in the tree above, the external path length is 25 (Knuth 1997, pp. 399400). The internal and external path lengths are related by
E = I + 2n,
where n is the number of internal nodes.
→ The root has path length zero and the maximum path length in a tree is called the tree's height.
→ The sum of the path lengths of a tree's internal nodes is called the internal path length and the sum of the path lengths of a tree's external nodes is called the external path length.
External Path Length:
The sum over all external (square) nodes of the lengths of the paths from the root of an extended binary tree to each node. For example, in the tree above, the external path length is 25 (Knuth 1997, pp. 399400). The internal and external path lengths are related by
E = I + 2n,
where n is the number of internal nodes.
Question 32 
You are given a sequence of n elements to sort. The input sequence consists of n/k subsequences,each containing k elements. The elements in a given subsequence are all smaller than the elements in the succeeding subsequence and larger than the elements in the preceding subsequence. Thus, all that is needed to sort the whole sequence of length n is to sort the k elements in each of the n/k subsequences. The lower bound on the number of comparisons needed to solve this variant of the sorting problem is
Ω(n)  
Ω(n/k)  
Ω(nlogk )  
Ω(n/klogn/k) 
Question 32 Explanation:
There are n/k subsequences and each can be ordered in k! ways. This makes a (k!)^{n/k} outputs. We can use the same reasoning:
(k!)^{n/k} ≤ 2^{h}
Taking the logarithm of both sides, we get:
h ≥ lg(k!)^{n/k}
= (n/k)lg(k!)
≥ (n/k)(k/2)lg(k/2)
= (1/2)*(nlogk)(1/2)*n
= Ω(nlogk)
(k!)^{n/k} ≤ 2^{h}
Taking the logarithm of both sides, we get:
h ≥ lg(k!)^{n/k}
= (n/k)lg(k!)
≥ (n/k)(k/2)lg(k/2)
= (1/2)*(nlogk)(1/2)*n
= Ω(nlogk)
Question 33 
T(n)= 8T (n/2)+Cn, if n>1 = b, if n=1
Consider the recurrence relation:
Where b and c are constants. The order of the algorithm corresponding to above recurrence relation is:
n  
n^{2}  
n log n  
n^{3} 
Question 33 Explanation:
The above recurrence is in the form of masters theorem.
a=8,b=2,k=0 and p=0
Case1: a>b^{k} = 8>2^{0}
T(n)=O(n^{logb^a})
= O(n^{3})
a=8,b=2,k=0 and p=0
Case1: a>b^{k} = 8>2^{0}
T(n)=O(n^{logb^a})
= O(n^{3})
Question 34 
Consider the following two sequences :
X = < B, C, D, C, A, B, C > and
Y = < C, A, D, B, C, B >
The length of longest common subsequence of X and Y is :
X = < B, C, D, C, A, B, C > and
Y = < C, A, D, B, C, B >
The length of longest common subsequence of X and Y is :
5  
3  
4  
2 
Question 34 Explanation:
Question 35 
A text is made up of the characters a, b, c, d, e each occurring with the probability 0.11, 0.40, 0.16, 0.09 and 0.24 respectively. The optimal Huffman coding technique will have the average length of:
2.40  
2.16  
2.26  
2.15 
Question 35 Explanation:
Question 36 
An undirected graph G (V, E) contains n (n > 2) nodes named v_{1} , v_{2} ,...,v_{n}. Two nodes v_{i} and v_{j} are connected if and only if 0 <│i−j│≤2. Each edge (v_{i }, v_{j} ) is assigned a weight i+j. The cost of the minimum spanning tree of such a graph with 10 nodes is:
88  
91  
49  
21 
Question 36 Explanation:
Method1:
Question 37 
An XML document that adheres to syntax rules specified by XML 1.0 specification in that it must satisfy both physical and logical structured, is called:
Well  formed  
Reasonable  
Valid  
Sophisticated 
Question 37 Explanation:
A wellformed document in XML is a document that "adheres to the syntax rules specified by the XML 1.0 specification in that it must satisfy both physical and logical structures".
Well formed documents requirements:
1. Content be defined.
2. Content be delimited with a beginning and end tag
3. Content be properly nested (parents within roots, children within parents)
Well formed documents requirements:
1. Content be defined.
2. Content be delimited with a beginning and end tag
3. Content be properly nested (parents within roots, children within parents)
Question 38 
Which of the following statement(s) is/are TRUE regarding Java Servlets ?
(a) A Java Servlet is a serverside component that runs on the web server and extends the capabilities of a server.
(b) A Servlet can use the user interface classes like AWT or Swing.
(a) A Java Servlet is a serverside component that runs on the web server and extends the capabilities of a server.
(b) A Servlet can use the user interface classes like AWT or Swing.
Only (a) is TRUE.  
Only (b) is TRUE.  
Both (a) and (b) are TRUE.  
Neither (a) nor (b) is TRUE. 
Question 38 Explanation:
TRUE: A Java Servlet is a serverside component that runs on the web server and extends the capabilities of a server.
FALSE: A Servlet can not use the user interface classes like AWT or Swing.
→ Servlets can respond to any types of requests, they most commonly implement web containers for hosting web applications on web servers and thus qualify as a serverside servlet web API.
FALSE: A Servlet can not use the user interface classes like AWT or Swing.
→ Servlets can respond to any types of requests, they most commonly implement web containers for hosting web applications on web servers and thus qualify as a serverside servlet web API.
Question 39 
Consider the following HTML table definition:
<table border = 1>
<tr>
<td> colspan = 2> Text A </td>
</tr>
<tr>
<td> Text B </td>
<td> Text C </td*gt;
</tr>
<tr>
<td rowspan = 2> Text D </td>
</tr>
<tr>
<td> Text E </td>
<tr>
<td> Text F </td>
</tr>
</table>
The above HTML code would render on screen as:
<table border = 1>
<tr>
<td> colspan = 2> Text A </td>
</tr>
<tr>
<td> Text B </td>
<td> Text C </td*gt;
</tr>
<tr>
<td rowspan = 2> Text D </td>
</tr>
<tr>
<td> Text E </td>
<tr>
<td> Text F </td>
</tr>
</table>
The above HTML code would render on screen as:
Question 39 Explanation:
Question 40 
Which of the following statements is/are TRUE ?
(a) In HTML, character entities are used to incorporate external content into a web page, such as images.
(b) Once a web server returns a cookie to a browser, the cookie will be included in all future requests from the browser to the same server.
(a) In HTML, character entities are used to incorporate external content into a web page, such as images.
(b) Once a web server returns a cookie to a browser, the cookie will be included in all future requests from the browser to the same server.
Only (a) is TRUE.  
Only (b) is TRUE.  
Both (a) and (b) are TRUE.  
Neither (a) nor (b) is TRUE. 
Question 40 Explanation:
FALSE: Entities are used to display characters that have special HTML meaning, such as “<” and “>”.
FALSE: Once a web server returns a cookie to a browser, the cookie will be included in all future requests from the browser to the same server.
FALSE: Once a web server returns a cookie to a browser, the cookie will be included in all future requests from the browser to the same server.
Question 41 
Which of the following statements is/are TRUE regarding JAVA ?
(a) Constants that cannot be changed are declared using the ‘static’ keyword.
(b) A class can only inherit one class but can implement multiple interfaces.
Only (a) is TRUE.  
Only (b) is TRUE.  
Both (a) and (b) are TRUE.  
Neither (a) nor (b) are TRUE. 
Question 41 Explanation:
FALSE: Constants that cannot be changed are declared using the ‘static’ keyword.
TRUE: A class can only inherit one class but can implement multiple interfaces.
TRUE: A class can only inherit one class but can implement multiple interfaces.
Question 42 
What is the output of the following JAVA program ?
Class Test {
public static void main(String[] args) {
Test obj = new Test();
obj.start();
}
void start() {
String stra = ”do”;
String strb = method(stra);
System.out.print(“: ”+stra + strb);
}
String method(String stra) {
stra = stra + ”good”;
System.out.print(stra);
return“ good”;
}
}
Class Test {
public static void main(String[] args) {
Test obj = new Test();
obj.start();
}
void start() {
String stra = ”do”;
String strb = method(stra);
System.out.print(“: ”+stra + strb);
}
String method(String stra) {
stra = stra + ”good”;
System.out.print(stra);
return“ good”;
}
}
dogood : dogoodgood  
dogood : gooddogood  
dogood : dodogood  
dogood : dogood 
Question 43 
Statistical software quality assurance in software engineering involves __________ .
using sampling in place of exhaustive testing of software.  
surveying customers to find out their opinions about product quality.  
tracing each defect to its underlying cause, isolating the vital few causes, and moving to correct them.  
tracing each defect to its underlying causes, and using the Pareto principle to correct each problem found. 
Question 43 Explanation:
→ Statistical software quality assurance in software engineering involves tracing each defect to its underlying cause, isolating the vital few causes, and moving to correct them.
→ Statistical software quality assurance uses pareto principle to identify vital causes (80% of defects can be traced to 20% of causes) and moves to correct the problems that have caused the defects.
→ Statistical software quality assurance uses pareto principle to identify vital causes (80% of defects can be traced to 20% of causes) and moves to correct the problems that have caused the defects.
Question 44 
Which of the following statements is/are FALSE with respect to software testing ?
S1 : Whitebox tests are based on specifications; better at telling whether program meets specification, better at finding errors of omission.
S2 : Black box tests are based on code; better for finding crashes, out of bounds errors, file not closed errors.
S3 : Alpha testing is conducted at the developer’s site by a team of highly skilled testers for software that is developed as a product to be used by many customers.
S1 : Whitebox tests are based on specifications; better at telling whether program meets specification, better at finding errors of omission.
S2 : Black box tests are based on code; better for finding crashes, out of bounds errors, file not closed errors.
S3 : Alpha testing is conducted at the developer’s site by a team of highly skilled testers for software that is developed as a product to be used by many customers.
Only S1 and S2 are FALSE.  
Only S1 and S3 are FALSE.  
Only S2 and S3 are FALSE.  
All of S1, S2, and S3 are FALSE. 
Question 44 Explanation:
FALSE: Black box tests are based on specifications; better at telling whether program meets specification, better at finding errors of omission.
FALSE: White box tests are based on code; better for finding crashes, out of bounds errors, file not closed errors.
TRUE: Alpha testing is conducted at the developer’s site by a team of highly skilled testers for software that is developed as a product to be used by many customers.
FALSE: White box tests are based on code; better for finding crashes, out of bounds errors, file not closed errors.
TRUE: Alpha testing is conducted at the developer’s site by a team of highly skilled testers for software that is developed as a product to be used by many customers.
Question 45 
A signal processor software is expected to operate for 91.25 days after repair, and the mean software repair time is expected to be 5 minutes. Then, the availability of the software is:
96.9862%  
97.9862%  
98.9962%  
99.9962% 
Question 45 Explanation:
Given data,
 Software expected to operate after repair(MTBF) = 91.25 days
= 91.25 * 24 hours
= 2190 hours
 Mean software repair time(MTTR) = 5 mins
= 5/60 hrs
= 0.083 hrs
 Software Availability=?
Step1: Software Availability =MTBF / (MTBF + MTTR)
= [2190 / (2190 + 0.083)]*100
= (2190 / 2190.083)*100
= 0.9996211481 * 100
= 99.9962%
 Software expected to operate after repair(MTBF) = 91.25 days
= 91.25 * 24 hours
= 2190 hours
 Mean software repair time(MTTR) = 5 mins
= 5/60 hrs
= 0.083 hrs
 Software Availability=?
Step1: Software Availability =MTBF / (MTBF + MTTR)
= [2190 / (2190 + 0.083)]*100
= (2190 / 2190.083)*100
= 0.9996211481 * 100
= 99.9962%
Question 46 
Consider the method mcq ( ):
int mcq(boolean a, boolean b, boolean c, boolean d)
{
int ans=1;
if(a) { ans = 2; }
else if (b) { ans = 3; }
else if (c) {
if (d) { ans = 4; }
return ans;
}
if
M1 = Number of tests to exhaustively test mcq( );
M2 = Minimum number of tests to achieve full statement coverage for mcq( ); and
M3 = Minimum number of tests to achieve full branch coverage for mcq( );
then (M1, M2, M3) = __________.
int mcq(boolean a, boolean b, boolean c, boolean d)
{
int ans=1;
if(a) { ans = 2; }
else if (b) { ans = 3; }
else if (c) {
if (d) { ans = 4; }
return ans;
}
if
M1 = Number of tests to exhaustively test mcq( );
M2 = Minimum number of tests to achieve full statement coverage for mcq( ); and
M3 = Minimum number of tests to achieve full branch coverage for mcq( );
then (M1, M2, M3) = __________.
(16, 3, 5)  
(8, 5, 3)  
(8, 3, 5)  
(16, 4, 4) 
Question 46 Explanation:
Question 47 
A simple stand  alone software utility is to be developed in ‘C’ programming by a team of software experts for a computer running Linux and the overall size of this software is estimated to be 20,000 lines of code. Considering (a, b) = (2.4, 1.05) as multiplicative and exponention factor for the basic COCOMO effort estimation equation and (c, d) = (2.5, 0.38) as multiplicative and exponention factor for the basic COCOMO development time estimation equation, approximately how long does the software project take to complete ?
10.52 months  
11.52 months  
12.52 months  
14.52 months 
Question 47 Explanation:
Given data,
 lines of code(LOC)= 20,000 (or 20K)
 Multiplicative value ‘a’=2.4
 Exponential ‘b’=1.05
 Basic COCOMO formula= a*(LOC in K)^{b}
 Multiplicative value ‘c’=2.5
 Exponential ‘d’=0.38
 Development=c*(basic COCOMO result)^{d}
Step1: Basic COCOMO formula= a*(LOC in K)^{b}
= 2.4*(20)^{1.05}
= 55.756
Step2: Development= c*(basic COCOMO result)^{d}
= 2.5(55.756)^{0.38}
= 11.52 months
 lines of code(LOC)= 20,000 (or 20K)
 Multiplicative value ‘a’=2.4
 Exponential ‘b’=1.05
 Basic COCOMO formula= a*(LOC in K)^{b}
 Multiplicative value ‘c’=2.5
 Exponential ‘d’=0.38
 Development=c*(basic COCOMO result)^{d}
Step1: Basic COCOMO formula= a*(LOC in K)^{b}
= 2.4*(20)^{1.05}
= 55.756
Step2: Development= c*(basic COCOMO result)^{d}
= 2.5(55.756)^{0.38}
= 11.52 months
Question 48 
In Software Configuration Management (SCM), which of the following is a usecase supported by standard version control systems ?
(a) Managing several versions or releases of a software
(b) Filing bug reports and tracking their progress
(c) Allowing team members to work in parallel
(d) Identifying when and where a regression occurred
(a) Managing several versions or releases of a software
(b) Filing bug reports and tracking their progress
(c) Allowing team members to work in parallel
(d) Identifying when and where a regression occurred
Only (a), (c), and (d)  
Only (a), (b), and (c)  
Only (a), (b), and (d)  
Only (b), (c), and (d) 
Question 48 Explanation:
Software Configuration Management (SCM) is a usecase supported by standard version control systems:
1. Develop the next version of a piece of software while fixing problems with the current one.
2. Share code with other team members in a controlled way, allowing you to develop code in parallel with others and join with the current state of the codeline.
3. Identify what versions of code went into a particular component.
4. Analyze where change happened in the history of a component's development.
1. Develop the next version of a piece of software while fixing problems with the current one.
2. Share code with other team members in a controlled way, allowing you to develop code in parallel with others and join with the current state of the codeline.
3. Identify what versions of code went into a particular component.
4. Analyze where change happened in the history of a component's development.
Question 49 
Consider the following four processes with the arrival time and length of CPU burst given in milliseconds :
The average waiting time for preemptive SJF scheduling algorithm is __________.
The average waiting time for preemptive SJF scheduling algorithm is __________.
6.5  
7.5  
6.75  
7.75 
Question 49 Explanation:
Question 50 
Consider a virtual page reference string 7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2, 1, 2, 0, 1, 7, 0, 1 Suppose a demand paged virtual memory system running on a computer system such that the main memory has 3 page frames. Then __________ page replacement algorithm has minimum number of page faults.
FIFO  
LIFO  
LRU  
Optimal 
Question 50 Explanation:
Question 51 
User level threads are threads that are visible to the programmer and are unknown to the kernel. The operating system kernel supports and manages kernel level threads. Three different types of models relate user and kernel level threads. Which of the following statements is/are true ?
(a) (i) The Many  to  one model maps many user threads to one kernel thread
(ii) The one  to  one model maps one user thread to one kernel thread
(iii) The many  to  many model maps many user threads to smaller or equal kernel threads
(b) (i) Many  to  one model maps many kernel threads to one user thread
(ii) One  to  one model maps one kernel thread to one user thread
(iii) Many  to  many model maps many kernel threads to smaller or equal user threads
(a) (i) The Many  to  one model maps many user threads to one kernel thread
(ii) The one  to  one model maps one user thread to one kernel thread
(iii) The many  to  many model maps many user threads to smaller or equal kernel threads
(b) (i) Many  to  one model maps many kernel threads to one user thread
(ii) One  to  one model maps one kernel thread to one user thread
(iii) Many  to  many model maps many kernel threads to smaller or equal user threads
(a) is true; (b) is false  
(a) is false; (b) is true  
Both (a) and (b) are true  
Both (a) and (b) are false 
Question 51 Explanation:
Question 52 
Consider a system with five processes P0 through P4 and three resource types A, B and C. Resource type A has seven instances, resource type B has two instances and resource type C has six instances suppose at time T0we have the following allocation.
If we implement Deadlock detection algorithm we claim that system is ____________.
(1) Semaphore
(2) Deadlock state
(3) Circular wait
(4) Not in deadlock state
(1), (2) and (3)  
(1) and (3)  
(3) and (4)  
All are correct. 
Question 52 Explanation:
Question 53 
Consider a disk queue with requests for I/O to blocks on cylinders
98, 183, 37, 122, 14, 124, 65, 67
Suppose SSTF disk scheduling algorithm implemented to meet the requests then the total number of head movements are __________ if the disk head is initially at 53.
98, 183, 37, 122, 14, 124, 65, 67
Suppose SSTF disk scheduling algorithm implemented to meet the requests then the total number of head movements are __________ if the disk head is initially at 53.
224  
248  
236  
240 
Question 53 Explanation:
Question 54 
The Bounded buffer problem is also known as __________.
Producer  consumer problem  
Reader  writer problem  
Dining Philosophers problem  
Both (2) and (3) 
Question 54 Explanation:
→ The Bounded buffer problem is also known as producerconsumer problem.
→ The problem describes two processes, the producer and the consumer, who share a common, fixedsize buffer used as a queue.
→ The producer's job is to generate data, put it into the buffer, and start again. At the same time, the consumer is consuming the data (i.e., removing it from the buffer), one piece at a time.
→ The problem is to make sure that the producer won't try to add data into the buffer if it's full and that the consumer won't try to remove data from an empty buffer.
→ The problem describes two processes, the producer and the consumer, who share a common, fixedsize buffer used as a queue.
→ The producer's job is to generate data, put it into the buffer, and start again. At the same time, the consumer is consuming the data (i.e., removing it from the buffer), one piece at a time.
→ The problem is to make sure that the producer won't try to add data into the buffer if it's full and that the consumer won't try to remove data from an empty buffer.
Question 55 
In Artificial Intelligence (AI), which agent deals with happy and unhappy state?
Simple reflex agent  
Model based agent
 
Learning agent  
Utility based agent

Question 55 Explanation:
Simple reflex agents[edit]
Simple reflex agents act only on the basis of the current percept, ignoring the rest of the percept history. The agent function is based on the conditionaction rule: "if condition, then action".
Modelbased reflex agents: A modelbased agent can handle partially observable environments. Its current state is stored inside the agent maintaining some kind of structure which describes the part of the world which cannot be seen. This knowledge about "how the world works" is called a model of the world, hence the name "modelbased agent".
Utilitybased agents: Goalbased agents only distinguish between goal states and nongoal states. It is possible to define a measure of how desirable a particular state is. This measure can be obtained through the use of a utility function which maps a state to a measure of the utility of the state. A more general performance measure should allow a comparison of different world states according to exactly how happy they would make the agent. The term utility can be used to describe how "happy" the agent is.
Learning agents: Learning has the advantage that it allows the agents to initially operate in unknown environments and to become more competent than its initial knowledge alone might allow. The most important distinction is between the "learning element", which is responsible for making improvements, and the "performance element", which is responsible for selecting external actions.
Simple reflex agents act only on the basis of the current percept, ignoring the rest of the percept history. The agent function is based on the conditionaction rule: "if condition, then action".
Modelbased reflex agents: A modelbased agent can handle partially observable environments. Its current state is stored inside the agent maintaining some kind of structure which describes the part of the world which cannot be seen. This knowledge about "how the world works" is called a model of the world, hence the name "modelbased agent".
Utilitybased agents: Goalbased agents only distinguish between goal states and nongoal states. It is possible to define a measure of how desirable a particular state is. This measure can be obtained through the use of a utility function which maps a state to a measure of the utility of the state. A more general performance measure should allow a comparison of different world states according to exactly how happy they would make the agent. The term utility can be used to describe how "happy" the agent is.
Learning agents: Learning has the advantage that it allows the agents to initially operate in unknown environments and to become more competent than its initial knowledge alone might allow. The most important distinction is between the "learning element", which is responsible for making improvements, and the "performance element", which is responsible for selecting external actions.
Question 56 
If b is the branching factor and m is the maximum depth of the search tree, what is the space complexity of greedy search?
O(b+m)  
O(bm)  
O(b^{m})  
O(m^{m}) 
Question 56 Explanation:
Question 57 
Let P, Q, R and S be Propositions. Assume that the equivalences P ⇔ (Q ∨ ¬ Q) and Q ⇔ R hold. Then the truth value of the formula (P ∧ Q) ⇒ ((P ∧ R) ∨ S) is always:
True  
False  
Same as truth table of Q  
Same as truth table of S 
Question 57 Explanation:
Given data,
 Equivalence condition(Both RHS and LHS should be TRUE)= P ⇔ (Q V ~Q)
 Holding condition(Both RHS and LHS may TRUE/FALSE)= Q ⇔ R
 Equivalence condition(Both RHS and LHS should be TRUE)= P ⇔ (Q V ~Q)
 Holding condition(Both RHS and LHS may TRUE/FALSE)= Q ⇔ R
Question 58 
“If X, then Y unless Z” is represented by which of the following formulae in propositional logic?
(X ∧ Y) → ¬ Z  
(X ∧ ¬ Z) → Y  
X → (Y ∧ ¬ Z)  
Y → (X ∧ ¬ Z) 
Question 58 Explanation:
"If X then Y unless Z" ⇒ ¬Z → (X→Y)
⇒ Z ∨ ¬X ∨ Y
⇒ ¬X ∨ Z ∨ Y
Option B: (X ∧ ¬Z) → Y
= ¬(X ∧ ¬Z ) ∨ Y
= ¬X ∨ Z ∨ Y
Hence, option (B) is correct.
⇒ Z ∨ ¬X ∨ Y
⇒ ¬X ∨ Z ∨ Y
Option B: (X ∧ ¬Z) → Y
= ¬(X ∧ ¬Z ) ∨ Y
= ¬X ∨ Z ∨ Y
Hence, option (B) is correct.
Question 59 
Consider the following two wellformed formulas in propositional logic.
F1 : P ⇒ ¬ P
F2 : (P ⇒ ¬ P) ∨ (¬ P ⇒ P)
Which of the following statements is correct?
F1 : P ⇒ ¬ P
F2 : (P ⇒ ¬ P) ∨ (¬ P ⇒ P)
Which of the following statements is correct?
F1 is Satisfiable, F2 is valid  
F1 is unsatisfiable, F2 is Satisfiable  
F1 is unsatisfiable, F2 is valid  
F1 and F2 both are Satisfiable 
Question 59 Explanation:
Question 60 
Standard planning algorithms assume environment to be __________.
Both deterministic and fully observable  
Neither deterministic nor fully observable  
Deterministic but not fully observable  
Not deterministic but fully observable 
Question 60 Explanation:
→ Classical planning environments that are fully observable, deterministic, finite, static and discrete (in time, action, objects and effects).
Question 61 
Which of the following statements is not correct?
Every recursive language is recursively enumerable.  
L = {0^{n}1^{n} 0^{n} │n=1, 2 , 3, ....} is recursively enumerable.  
Recursive languages are closed under intersection.  
Recursive languages are not closed under intersection. 
Question 61 Explanation:
TRUE: Every recursive language is recursively enumerable.
TRUE: L = {0^{n}1^{n} 0^{n} │n=1, 2 , 3, ....} is recursively enumerable.
TRUE: Recursive languages are closed under intersection.
→ Recursive closed under union,intersection,complementation,etc.., except Homomorphism and substitution.
FALSE: Recursive languages are not closed under intersection.
→ Recursively enumerable languages are closed under union,intersection,etc.., except set difference and complementation.
TRUE: L = {0^{n}1^{n} 0^{n} │n=1, 2 , 3, ....} is recursively enumerable.
TRUE: Recursive languages are closed under intersection.
→ Recursive closed under union,intersection,complementation,etc.., except Homomorphism and substitution.
FALSE: Recursive languages are not closed under intersection.
→ Recursively enumerable languages are closed under union,intersection,etc.., except set difference and complementation.
Question 62 
Context free grammar is not closed under:
Concatenation  
Complementation  
Kleene Star  
Union 
Question 62 Explanation:
→ Contextfree grammar are closed under Union,Concatenation and Kleene closure.
→ Contextfree grammar are not closed under set difference,Complementation and intersection.
→ Contextfree grammar are not closed under set difference,Complementation and intersection.
Question 63 
Consider the following languages:
L_{1} = {a^{m} b^{n} │ m ≠ n}
L_{2} = {a^{m} b^{n} │ m = 2n+1}
L_{3} = {a^{m} b^{m} │ m ≠ 2n}
Which one of the following statement is correct ?
Only L_{1} and L_{2} are context free languages  
Only L_{1} and L_{3} are context free languages  
Only L_{2} and L_{3} are context free languages  
L_{1}, L_{2} and L_{3} are context free languages 
Question 63 Explanation:
L_{1}={a^{m} b^{n}  m≠n }
→ It means m could be greater than or less than ‘n’ but m will not be equal to ‘n’.
→ Since for accepting the language L1 we need a storage so then no. of a’s can be stored in it and when b’s come as input a’s and b’s can be compared.
→ Since finite automata do not have a storage element hence the language is not regular.
→ Now state transition diagram to check whether L1 is CFL or not:
→ Hence L_{2} is also a CFL.
L3: First simply read one 'a', then push one 'a' in the stack after reading two a's and then pop all the a's by reading the b's. Since can be done by PDA hence CFL.
→ It means m could be greater than or less than ‘n’ but m will not be equal to ‘n’.
→ Since for accepting the language L1 we need a storage so then no. of a’s can be stored in it and when b’s come as input a’s and b’s can be compared.
→ Since finite automata do not have a storage element hence the language is not regular.
→ Now state transition diagram to check whether L1 is CFL or not:
→ Hence L_{2} is also a CFL.
L3: First simply read one 'a', then push one 'a' in the stack after reading two a's and then pop all the a's by reading the b's. Since can be done by PDA hence CFL.
Question 64 
A 4*4 DFT matrix is given by :
Where values of x and y are _____, _____ respectively.
Where values of x and y are _____, _____ respectively.
1, −1  
−1, 1  
−j, j  
j, −j 
Question 65 
Entropy of a discrete random variable with possible values {x_{1}, x_{2}, ..., x_{n}} and probability density function P(X) is :
The value of b gives the units of entropy. The unit for b=10 is :
The value of b gives the units of entropy. The unit for b=10 is :
bits  
bann  
nats  
deca 
Question 66 
For any binary (n, h) linear code with minimum distance (2t+1) or greater
2t+1  
t+1  
t  
t1 
Question 67 
Which of the following is a valid reason for causing degeneracy in a transportation problem ?
Here m is no. of rows and n is no. of columns in transportation table
Here m is no. of rows and n is no. of columns in transportation table
When the number of allocations is m+n−1.  
When two or more occupied cells become unoccupied simultaneously.  
When the number of allocations is less than m+n−1.  
When a loop cannot be drawn without using unoccupied cells, except the starting cell of the loop. 
Question 68 
Consider the following LPP :
Max Z = 15x_{1} + 10x_{2}
Subject to the constraints
4x_{1 }+ 6x_{2 }≤ 360
3x_{1 }+ 0x_{2} ≤ 180
0x_{1 }+ 5x_{2} ≤ 200
x_{1}, x_{2} / 0
The solution of the LPP using Graphical solution technique is :
Max Z = 15x_{1} + 10x_{2}
Subject to the constraints
4x_{1 }+ 6x_{2 }≤ 360
3x_{1 }+ 0x_{2} ≤ 180
0x_{1 }+ 5x_{2} ≤ 200
x_{1}, x_{2} / 0
The solution of the LPP using Graphical solution technique is :
x_{1}=60, x_{2}=0 and Z=900  
x_{1}=60, x_{2}=20 and Z=1100  
x_{1}=60, x_{2}=30 and Z=1200  
x_{1}=50, x_{2}=40 and Z=1150 
Question 69 
Consider the following LPP :
Min Z = 2x1 + x2 + 3x3
Subject to :
x_{1} − 2x_{2} + x_{3 }≥ 4
2x_{1} + x_{2} + x_{3} ≤ 8
x_{1} − x_{3} ≥ 0
x_{1}, x_{2}, x_{3} ≥ 0
The solution of this LPP using Dual Simplex Method is :
Min Z = 2x1 + x2 + 3x3
Subject to :
x_{1} − 2x_{2} + x_{3 }≥ 4
2x_{1} + x_{2} + x_{3} ≤ 8
x_{1} − x_{3} ≥ 0
x_{1}, x_{2}, x_{3} ≥ 0
The solution of this LPP using Dual Simplex Method is :
x_{1}=0, x_{2=0, x3=3 and Z=9}  
x_{1}=0, x_{2}=6, x_{2}=0 and Z=6  
x_{1}=4, x_{2}=0, x_{2}=0 and Z=8  
x_{1}=2, x_{2}=0, x_{2}=2 and Z=10 
Question 70 
Consider a Takagi  Sugeno  Kang (TSK) Model consisting rules of the form: If X_{i} is A_{i1} and... and x_{r} is A_{ir} THEN y = f_{i}(x_{1}, x_{2},.... x_{r}) = b_{i0} + b_{i1}x_{1} + b_{ir}x_{r} assume, a_{i} is the matching degree of rule i, then the total output of the model is given by:
Question 71 
Consider a single perceptron with sign activation function. The perceptron is represented by weight vector [0.4 −0.3 0.1]t and a bias θ = 0. If the input vector to the perceptron is X = [0.2 0.6 0.5] then the output of the perceptron is:
1  
0  
0.25  
1 
Question 72 
The Sigmoid activation function f(t) is defined as:
Question 73 
Consider the following statements :
(a) UNIX provides three types of permissions
* Read
* Write
* Execute
(b) UNIX provides three sets of permissions
* permission for owner
* permission for group
* permission for others
Which of the above statement/s is/are true ?
(a) UNIX provides three types of permissions
* Read
* Write
* Execute
(b) UNIX provides three sets of permissions
* permission for owner
* permission for group
* permission for others
Which of the above statement/s is/are true ?
only (a)  
only (b)  
Both (a) and (b)  
Neither (a) nor (b) 
Question 73 Explanation:
→ UNIX provides three types of permissions using octal numbers
1. Read(R or 4)
2. Write(W or 2)
3. Execute(X or 1)
→ UNIX provides three sets of permissions. Each set of permission we are having 3 types of permissions.
1. Permission for owner(RWX or 421)
2. Permission for group(RWX or 421)
3. Permission for others(RWX or 421)
→ We can change/modify permissions using chmod, chown,etc.., commands
→ We can see the permissions for every file using ls l command
1. Read(R or 4)
2. Write(W or 2)
3. Execute(X or 1)
→ UNIX provides three sets of permissions. Each set of permission we are having 3 types of permissions.
1. Permission for owner(RWX or 421)
2. Permission for group(RWX or 421)
3. Permission for others(RWX or 421)
→ We can change/modify permissions using chmod, chown,etc.., commands
→ We can see the permissions for every file using ls l command
Question 74 
Which of the following routing technique / techniques is / are used in distributed systems ?
(a) Fixed Routing
(b) Virtual Routing
(c) Dynamic Routing
(a) Fixed Routing
(b) Virtual Routing
(c) Dynamic Routing
(a) only  
(a) and (b) only  
(c) only  
All (a), (b), (C) 
Question 74 Explanation:
Distributed systems are using fixed routing only.
Fixed routing: A path from A to B is specified in advance; path changes only if a hardware failure disables it.
1. Since the shortest path is usually chosen, communication costs are minimized.
2. Fixed routing cannot adapt to load changes
3. Ensures that messages will be delivered in the order in which they were sent
Virtual routing: A path from A to B is fixed for the duration of one session.
1. Different sessions involving messages from A to B may have different paths
2. Partial remedy to adapting to load changes
3. Ensures that messages will be delivered in the order in which they were sent
Dynamic routing: The path used to send a message form site A to site B is chosen only when a message is sent
1. Usually a site sends a message to another site on the link least used at that particular time
2. Adapts to load changes by avoiding routing messages on heavily used path
3. Messages may arrive out of order; This problem can be remedied by appending a sequence number to each message
4. Most complex to set up
Fixed routing: A path from A to B is specified in advance; path changes only if a hardware failure disables it.
1. Since the shortest path is usually chosen, communication costs are minimized.
2. Fixed routing cannot adapt to load changes
3. Ensures that messages will be delivered in the order in which they were sent
Virtual routing: A path from A to B is fixed for the duration of one session.
1. Different sessions involving messages from A to B may have different paths
2. Partial remedy to adapting to load changes
3. Ensures that messages will be delivered in the order in which they were sent
Dynamic routing: The path used to send a message form site A to site B is chosen only when a message is sent
1. Usually a site sends a message to another site on the link least used at that particular time
2. Adapts to load changes by avoiding routing messages on heavily used path
3. Messages may arrive out of order; This problem can be remedied by appending a sequence number to each message
4. Most complex to set up
Question 75 
Match the following WINDOWS system calls and UNIX system calls with reference to process control and File manipulation.
(a)(iii), (b)(iv), (c)(i), (d)(ii)  
(a)(iv), (b)(iii), (c)(i), (d)(ii)  
(a)(iv), (b)(iii), (c)(ii), (d)(i)  
(a)(iii), (b)(iv), (c)(ii), (d)(i) 
Question 75 Explanation:
In UNIX Wait() system call is equivalent to windows system call is WaitForSingleObject() .
In UNIX Fork() system call is equivalent to windows system call is Createprocess().
In UNIX Create() system call is equivalent to windows system call is CreateFile().
In UNIX Close() system call is equivalent to windows system call is CloseHandle()
In UNIX Fork() system call is equivalent to windows system call is Createprocess().
In UNIX Create() system call is equivalent to windows system call is CreateFile().
In UNIX Close() system call is equivalent to windows system call is CloseHandle()
There are 75 questions to complete.