UGC NET CS 2013 Decpaper2
Question 1 
When data and acknowledgement are sent in the same frame, this is called as
Piggy packing  
Piggy backing  
Backpacking  
Good packing 
Question 1 Explanation:
→ In twoway communication, wherever a frame is received, the receiver waits and does not send the control frame (acknowledgement or ACK) back to the sender immediately.
→ The receiver waits until its network layer passes in the next data packet. The delayed acknowledgement is then attached to this outgoing data frame.
→ This technique of temporarily delaying the acknowledgement so that it can be hooked with next outgoing data frame is known as piggybacking.
→ The receiver waits until its network layer passes in the next data packet. The delayed acknowledgement is then attached to this outgoing data frame.
→ This technique of temporarily delaying the acknowledgement so that it can be hooked with next outgoing data frame is known as piggybacking.
Question 2 
Encryption and Decryption is the responsibility of _______ Layer.
Physical  
Network  
Application  
Datalink 
Question 2 Explanation:
Presentation layer is responsible for
1. Translation, authentication
2. Encryption and Decryption
3. Compression and decompression
But according application layer also can do these responsibilities. So, OptioC is the correct answer.
1. Translation, authentication
2. Encryption and Decryption
3. Compression and decompression
But according application layer also can do these responsibilities. So, OptioC is the correct answer.
Question 3 
An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, then baud rate and bit rate of the signal are and
4000 bauds \ sec & 1000 bps  
2000 bauds \ sec & 1000 bps  
1000 bauds \ sec & 500 bps  
1000 bauds \ sec & 4000 bps 
Question 3 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=4000 bps
 Each signal unit carries=4 bits
 Baud rate=?
Step1: Baud rate= Bit rate of signal / baud rate
= 4000/4 bps
= 1000 bps
Step2: Bit rate= Number of bits transmitted per second*number of signals units transmitted per unit time
= 4*1000 bps
= 4000 bps
→ Baud rate is nothing but number of signals units transmitted per unit time.
Given data,
 Bit rate of a signal=4000 bps
 Each signal unit carries=4 bits
 Baud rate=?
Step1: Baud rate= Bit rate of signal / baud rate
= 4000/4 bps
= 1000 bps
Step2: Bit rate= Number of bits transmitted per second*number of signals units transmitted per unit time
= 4*1000 bps
= 4000 bps
Question 4 
The VLF and LF bauds use _______ propagation for communication.
Ground  
Sky  
Line of sight  
Space 
Question 4 Explanation:
The VLF and LF bauds use ground propagation for communication.
Propagation methods are 3 types
1. Ground
2. Sky
3. Line of sight
→ Very low frequency or VLF is for radio frequencies (RF) in the range of 3 to 30 kilohertz (kHz), corresponding to wavelengths from 100 to 10 kilometers, respectively.
→ The lowfrequency signals travel in all directions from the transmitting antenna.
Propagation methods are 3 types
1. Ground
2. Sky
3. Line of sight
→ Very low frequency or VLF is for radio frequencies (RF) in the range of 3 to 30 kilohertz (kHz), corresponding to wavelengths from 100 to 10 kilometers, respectively.
→ The lowfrequency signals travel in all directions from the transmitting antenna.
Question 5 
Using the RSA public key cryptosystem, if p = 13, q = 31 and d = 7, then the value of ‘e’ is
101  
103  
105  
107 
Question 5 Explanation:
Step1: Compute n=p*q
= 13*31
= 403
Step2: Compute f(n)= (p − 1)(q − 1)
= 12*30
= 360
Step3: Compute d = e^{1} mod f(n)
Choose an e such that 1
d.e = 1 mod f(n)
e = 1 mod 360
= 361 mod 360
= 721 mod 360
Note: Given d=7
7*e= 721 mod 360
e=103
= 13*31
= 403
Step2: Compute f(n)= (p − 1)(q − 1)
= 12*30
= 360
Step3: Compute d = e^{1} mod f(n)
Choose an e such that 1
e = 1 mod 360
= 361 mod 360
= 721 mod 360
Note: Given d=7
7*e= 721 mod 360
e=103
Question 6 
FAN IN of a component A is defined as
Number of components that can call or pass control to component A.  
Number of components that are called by component A.  
Number of components related to component A.  
Number of components dependent on component A. 
Question 6 Explanation:
→ FAN IN of a component A is defined as number of components that can call or pass control to component A.
→ FANIN defines “number of modules directly invoking given module”.
→ FAN OUT is the number of classes used by a certain class or the number of methods called by a certain method.
→ FAN OUT defines “Number of modules that are directly controlled by given module.
→ Good software design require High FANIN and LOW FANOUT
→ FANIN defines “number of modules directly invoking given module”.
→ FAN OUT is the number of classes used by a certain class or the number of methods called by a certain method.
→ FAN OUT defines “Number of modules that are directly controlled by given module.
→ Good software design require High FANIN and LOW FANOUT
Question 7 
The relationship of data elements in a module is called
Coupling  
Modularity  
Cohesion  
Granularity 
Question 7 Explanation:
→ Cohesion is a measure of internal strength within a module and relationship of data elements in a module, whereas coupling is a measure of inter dependency among the modules.
→ In the context of modular software design there should be high cohesion and low coupling.
→ In the context of modular software design there should be high cohesion and low coupling.
Question 8 
Software Configuration Management is the discipline for systematically controlling
the changes due to the evolution of work products as the project proceeds.  
the changes due to defects (bugs) being found and then fixed.  
the changes due to requirement changes  
all of the above 
Question 8 Explanation:
Software Configuration Management is the discipline for systematically controlling
1. The changes due to the evolution of work products as the project proceeds.
2. The changes due to defects(bugs) being found and then fixed.
3. The changes due to requirement changes
1. The changes due to the evolution of work products as the project proceeds.
2. The changes due to defects(bugs) being found and then fixed.
3. The changes due to requirement changes
Question 9 
Which one of the following is not a step of requirement engineering ?
Requirement elicitation  
Requirement analysis  
Requirement design  
Requirement documentation 
Question 9 Explanation:
Requirements engineering (RE) refers to the process of defining, documenting and maintaining requirements in the engineering design process.
Requirement Analysis Steps:
1. Requirements inception or requirements elicitation: Developers and stakeholders meet, the latter are inquired concerning their needs and wants regarding the software product.
2. Requirements analysis and negotiation: Requirements are identified (including new ones if the development is iterative) and conflicts with stakeholders are solved. Both written and graphical tools (the latter commonly used in the design phase but some find them helpful at this stage, too) are successfully used as aids.
3. System modeling: Some engineering fields (or specific situations) require the product to be completely designed and modeled before its construction or fabrication starts and, therefore, the design phase must be performed in advance.
4. Requirements specification: Requirements are documented in a formal artifact called a Requirements Specification (RS), which will become official only after validation.
5. Requirements validation: Checking that the documented requirements and models are consistent and meet the needs of the stakeholder. Only if the final draft passes the validation process, the RS becomes official.
6. Requirements management: Managing all the activities related to the requirements since inception, supervising as the system is developed and, even until after it is put into use (e. g., changes, extensions, etc.)
Requirement Analysis Steps:
1. Requirements inception or requirements elicitation: Developers and stakeholders meet, the latter are inquired concerning their needs and wants regarding the software product.
2. Requirements analysis and negotiation: Requirements are identified (including new ones if the development is iterative) and conflicts with stakeholders are solved. Both written and graphical tools (the latter commonly used in the design phase but some find them helpful at this stage, too) are successfully used as aids.
3. System modeling: Some engineering fields (or specific situations) require the product to be completely designed and modeled before its construction or fabrication starts and, therefore, the design phase must be performed in advance.
4. Requirements specification: Requirements are documented in a formal artifact called a Requirements Specification (RS), which will become official only after validation.
5. Requirements validation: Checking that the documented requirements and models are consistent and meet the needs of the stakeholder. Only if the final draft passes the validation process, the RS becomes official.
6. Requirements management: Managing all the activities related to the requirements since inception, supervising as the system is developed and, even until after it is put into use (e. g., changes, extensions, etc.)
Question 10 
Testing of software with actual data and in actual environment is called
Alpha testing  
Beta testing  
Regression testing  
None of the above 
Question 10 Explanation:
→ Alpha test: is conducted at the developer’s site by endusers and conducted in a controlled environment
→ Beta test: is conducted at enduser sites and it is a “live” application of the software in an environment that cannot be controlled by the developer
→ Regression test: Testing is an important strategy for reducing “side effects” when major change is made to the software
→ Beta test: is conducted at enduser sites and it is a “live” application of the software in an environment that cannot be controlled by the developer
→ Regression test: Testing is an important strategy for reducing “side effects” when major change is made to the software
Question 11 
The student marks should not be greater than 100. This is
Integrity constraint  
Referential constraint  
Overdefined constraint  
Feasible constraint 
Question 11 Explanation:
Integrity constraint have some rules:
1. Every table have at least one attribute as primary key
2. No prime attribute should ever have null value.
The above example given condition but it is not referring any table. So, it is not referential constraints. Other two are not relevant to this question.
So, OptionA is correct answer.
1. Every table have at least one attribute as primary key
2. No prime attribute should ever have null value.
The above example given condition but it is not referring any table. So, it is not referential constraints. Other two are not relevant to this question.
So, OptionA is correct answer.
Question 12 
GO BOTTOM and SKIP3 commands are given one after another in a database file of 30 records. It shifts the control to
28th record  
27th record  
3rd record  
4th record 
Question 12 Explanation:
GO BOTTOM command will directly reach end of the record. Here, total number of records are 30. GO BOTTOM command will reach 30th position.
After SKIP1, it became 29
After SKIP2, it became 28
After SKIP3, it became 27.
In question, they given 3 SKIP operations.
After SKIP1, it became 29
After SKIP2, it became 28
After SKIP3, it became 27.
In question, they given 3 SKIP operations.
Question 13 
An ER Model includes
I. An ER diagram portraying entity types
II. Attributes for each entity type
III. Relationships among entity types.
IV. Semantic integrity constraints that reflects the business rules about data not captured in the ER diagram.
I. An ER diagram portraying entity types
II. Attributes for each entity type
III. Relationships among entity types.
IV. Semantic integrity constraints that reflects the business rules about data not captured in the ER diagram.
I, II, III & IV  
I & IV  
I, II & IV  
I & III 
Question 13 Explanation:
An ER Model includes
1. An ER diagram portraying entity types.
2. Attributes for each entity type
3. Relationships among entity types.
4. Semantic integrity constraints that reflects the business rules about data not captured in the ER diagram.
1. An ER diagram portraying entity types.
2. Attributes for each entity type
3. Relationships among entity types.
4. Semantic integrity constraints that reflects the business rules about data not captured in the ER diagram.
Question 14 
Based on the cardinality ratio and participation ________ associated with a relationship type, choose either the Foreign Key Design, the Cross Referencing Design or Mutual Referencing Design.
Entity  
Constraints  
Rules  
Keys 
Question 14 Explanation:
Based on the cardinality ratio and participation constraints associated with a relationship type, choose either the Foreign Key Design, the Cross Referencing Design or Mutual Referencing Design.
Question 15 
Data Integrity control uses _______
Upper and lower limits on numeric data.  
Passwords to prohibit unauthorised access to files.  
Data dictionary to keep the data  
Data dictionary to find last access of data 
Question 15 Explanation:
→ Data Integrity control uses passwords to prohibit unauthorised access to files.
→ Any unintended changes to data as the result of a storage, retrieval or processing operation, including malicious intent, unexpected hardware failure, and human error, is failure of data integrity.
→ If the changes are the result of unauthorized access, it may also be a failure of data security.
→ Depending on the data involved this could manifest itself as benign as a single pixel in an image appearing a different color than was originally recorded, to the loss of vacation pictures or a business critical database, to even catastrophic loss of human life in a life critical system.
→ Any unintended changes to data as the result of a storage, retrieval or processing operation, including malicious intent, unexpected hardware failure, and human error, is failure of data integrity.
→ If the changes are the result of unauthorized access, it may also be a failure of data security.
→ Depending on the data involved this could manifest itself as benign as a single pixel in an image appearing a different color than was originally recorded, to the loss of vacation pictures or a business critical database, to even catastrophic loss of human life in a life critical system.
Question 16 
What does the following declaration mean ?
int (*ptr) [10];
ptr is an array of pointers of 10 integers.  
ptr is a pointer to an array of 10 integers.  
ptr is an array of 10 integers.  
none of the above. 
Question 16 Explanation:
int (*ptr) [10]; It means ‘ptr’ is a pointer to an array of 10 integers.
Question 17 
Which of the following has compilation error in C ?
int n=32 ;  
char ch=65 ;  
float f=(float)3.2 ;  
none of the above 
Question 17 Explanation:
OptionA: ‘n’ is a integer value and and value of ‘n’ is 32.
OptionB: ‘ch’ is a character variable and value of ‘ch’ is 65.
OptionC: ‘f’ is a floating point variable and it uses type casting.
OptionD is correct answer.
OptionB: ‘ch’ is a character variable and value of ‘ch’ is 65.
OptionC: ‘f’ is a floating point variable and it uses type casting.
OptionD is correct answer.
Question 18 
Which of the following operators can not be overloaded in C++ ?
∗  
+ =  
= =  
: : 
Question 18 Explanation:
→ Scope resolution operator(::) is used to define a function outside a class or when we want to use a global variable but also has a local variable with the same name.
→ Scope resolution operator(::) can not be overloaded.
→ Scope resolution operator(::) can not be overloaded.
Question 19 
_________ allows to create classes which are derived from other classes, so that they automatically include some of its “parent’s” members, plus its own members.
Overloading  
Inheritance  
Polymorphism  
Encapsulation 
Question 19 Explanation:
→ Inheritance allows to create classes which are derived from other classes, so that they automatically include some of its “parent’s” members, plus its own members.
→ The main advantage of inheritance is “reusability”.
→ The main advantage of inheritance is “reusability”.
Question 20 
The correct way to round off a floating number x to an integer value is
y = (int) (x + 0.5)  
y = int (x + 0.5)  
y = (int) x + 0.5  
y = (int) ((int)x + 0.5) 
Question 20 Explanation:
Here, with the help of type casting method we can round off a floating number x to integer value.
→ Type casting is a way to convert a variable from one data type to another data type.
→ Type casting is a way to convert a variable from one data type to another data type.
Question 21 
What is the value of the postfix expression ?
a b c d + – ∗ (where a = 8, b = 4, c = 2 and d = 5)
a b c d + – ∗ (where a = 8, b = 4, c = 2 and d = 5)
–3/8  
–8/3  
24  
–24 
Question 21 Explanation:
Question 22 
If the queue is implemented with a linked list, keeping track of a front pointer and a rear pointer, which of these pointers will change during an insertion into a nonempty queue ?
Neither of the pointers change  
Only front pointer changes  
Only rear pointer changes  
Both of the pointers changes 
Question 22 Explanation:
Observe 1, 2, 3 whenever we are inserting an element into a queue (singly linked list) we are updating Rear pointer.
Question 23 
_______ is often used to prove the correctness of a recursive function.
Diagonalization  
Communitivity  
Mathematical Induction  
Matrix Multiplication 
Question 23 Explanation:
→ Mathematical Induction is often used to prove the correctness of a recursive function.
→ Mathematical Induction is a special way of proving things. It has only 2 steps:
Step1: Show it is TRUE for the first one
Step2: Show that if any one is TRUE then the next one is true
Then all are true
→ Mathematical Induction is a special way of proving things. It has only 2 steps:
Step1: Show it is TRUE for the first one
Step2: Show that if any one is TRUE then the next one is true
Then all are true
Question 24 
For any Btree of minimum degree t ≥ 2, every node other than the root must have at least ________ keys and every node can have at most ________ keys.
t – 1, 2t + 1  
t + 1, 2t + 1  
t – 1, 2t – 1  
t + 1, 2t – 1 
Question 24 Explanation:
For any Btree of minimum degree t ≥ 2, every node other than the root must have at least t1 keys and every node can have at most 2t1 keys.
Question 25 
Given two sorted list of size ‘m’ and ‘n’ respectively. The number of comparison needed in the worst case by the merge sort algorithm will be
m × n  
max (m, n)  
min (m, n)  
m + n – 1 
Question 25 Explanation:
→ To merge two lists of size m and n, we need to do m+n1 comparisons in worst case. Since we need to merge 2 at a time, the optimal strategy would be to take smallest size lists first.
→ The reason for picking smallest two items is to carry minimum items for repetition in merging.
→ The reason for picking smallest two items is to carry minimum items for repetition in merging.
Question 26 
Given the following statements :
S1: SLR uses follow information to guide reductions. In case of LR and LALR parsers, the lookaheads are associated with the items and they make use of the left context available to the parser.
S2: LR grammar is a larger subclass of context free grammar as compared to that SLR and LALR grammars.
Which of the following is true ?
S1: SLR uses follow information to guide reductions. In case of LR and LALR parsers, the lookaheads are associated with the items and they make use of the left context available to the parser.
S2: LR grammar is a larger subclass of context free grammar as compared to that SLR and LALR grammars.
Which of the following is true ?
S1 is not correct and S2 is not correct.  
S1 is not correct and S2 is correct.  
S1 is correct and S2 is not correct.  
S1 is correct and S2 is correct. 
Question 27 
The context free grammar for the language
L = {a^{n} b^{m}  n ≤ m + 3, n ≥ 0, m ≥ 0} is
S → aaa A; A → aAb  B, B → Bb  λ  
S → aaaAλ, A → aAb  B, B → Bb  λ  
S → aaaA  aa A  λ, A → aAb  B, B → Bb λ  
S → aaaA  aa A  aA  λ, A →
aAb  B, B → Bb  λ

Question 27 Explanation:
According to given the given language L when
m=0, n<= 3 i.e. L1= { λ, a, aa, aaa}
m=1, n<=4 i.e. L2={ ab, aab, aaab, aaaab}
m=2, n<=5 i.e. L3={abb, aabb, aaabb, aaaabb, aaaaabb} and so on.
Hence L = L1 U L2 U L3 U ...........
L= { λ, a, aa, aaa, ab, aab, aaab, aaaab, abb, aabb, aaabb, aaaabb, aaaabb, .................}
Option(A) is not correct because it can't generate λ.
Option(B) is not correct because it can't generate "a".
Option(C) is not correct because it can't generate "a".
Option(D) is correct because it is generating the language L
.
m=0, n<= 3 i.e. L1= { λ, a, aa, aaa}
m=1, n<=4 i.e. L2={ ab, aab, aaab, aaaab}
m=2, n<=5 i.e. L3={abb, aabb, aaabb, aaaabb, aaaaabb} and so on.
Hence L = L1 U L2 U L3 U ...........
L= { λ, a, aa, aaa, ab, aab, aaab, aaaab, abb, aabb, aaabb, aaaabb, aaaabb, .................}
Option(A) is not correct because it can't generate λ.
Option(B) is not correct because it can't generate "a".
Option(C) is not correct because it can't generate "a".
Option(D) is correct because it is generating the language L
.
Question 28 
Given the following statements :
S_{1} : If L is a regular language then the language {uv  u ∈ L, v ∈ L^{R}} is also regular.
S_{2} : L = {ww^{R}} is regular language. Which of the following is true ?
S_{1} : If L is a regular language then the language {uv  u ∈ L, v ∈ L^{R}} is also regular.
S_{2} : L = {ww^{R}} is regular language. Which of the following is true ?
S_{1} is not correct and S_{2} is not correct.  
S_{1} is not correct and S_{2} is correct.  
S_{1} is correct and S_{2} is not correct.  
S_{1} is correct and S_{2} is correct. 
Question 28 Explanation:
→ Statement S1 is correct. "uv" in given language can also be written as "LL^{R} " . Since L is a regular language and we know that a regular language is closed under reverse operation then L^{R} is also a regular language. Regular language is closed under concatenation then LL^{R} is also a regular language.
→ Statement S2 is incorrect. Since the finite state automata do not have a memory element which can remember the previous inputs hence the string "w" can't be recorded in order to match/compare with "w^{R} ". So we can't have a Finite state automata for ww^{R}. And since we can't have Finite state automata for ww^{R} , it means ww^{R} is not a regular language.
Hence the correct option is option(C)
→ Statement S2 is incorrect. Since the finite state automata do not have a memory element which can remember the previous inputs hence the string "w" can't be recorded in order to match/compare with "w^{R} ". So we can't have a Finite state automata for ww^{R}. And since we can't have Finite state automata for ww^{R} , it means ww^{R} is not a regular language.
Hence the correct option is option(C)
Question 29 
The process of assigning load addresses to the various parts of the program and adjusting the code and data in the program to reflect the assigned addresses is called _______.
Symbol resolution  
Parsing  
Assembly  
Relocation 
Question 29 Explanation:
The process of assigning load addresses to the various parts of the program and adjusting the code and data in the program to reflect the assigned addresses is called relocation.
Question 30 
Which of the following derivations does a topdown parser use while parsing an input string? The input is scanned from left to right.
Leftmost derivation  
Leftmost derivation traced out in reverse  
Rightmost derivation traced out in reverse  
Rightmost derivation 
Question 30 Explanation:
→ Top down parsers using leftmost derivation and the input is scanned from left to right.
→ Bottom up parsers using rightmost derivation in reverse.
→ Bottom up parsers using rightmost derivation in reverse.
Question 31 
The dual of a Boolean expression is obtained by interchanging
Boolean sums and Boolean products  
Boolean sums and Boolean products or interchanging 0’s and 1’s  
Boolean sums and Boolean products and interchanging 0’s & 1’s  
Interchanging 0’s and 1’s 
Question 31 Explanation:
→ The dual of a Boolean expression is obtained by interchanging boolean sums and boolean products and interchanging 0’s & 1’s.
→ Dual of a Boolean expressions are generated by simply replacing AND(product) with OR(sum) and OR(sum) with AND(product).
→ Even compliments themselves are unaffected, where as the complement of an expression is the negation of the variables with the replacement of AND(product) with OR(sum) and vice versa.
Example: A+B
Complement: AB
Dual: AB
→ Dual of a Boolean expressions are generated by simply replacing AND(product) with OR(sum) and OR(sum) with AND(product).
→ Even compliments themselves are unaffected, where as the complement of an expression is the negation of the variables with the replacement of AND(product) with OR(sum) and vice versa.
Example: A+B
Complement: AB
Dual: AB
Question 32 
Given that (292)_{10} = (1204)_{x} in some number system x. The base x of that number system is
2  
8  
10  
None of the above 
Question 32 Explanation:
OptionA: It is false because it is equivalent to (292)_{10} =(100100100)_{2}
OptionB: It is false because it is equivalent to (292)_{10} =(444)_{8}
OptionB: It is false because it is equivalent to (292)_{10} =(292)_{10}
So, optionD is the correct answer. The actual x value is 6.
Question 33 
The sum of products expansion for the function
F(x,y,z)=(x+y)z' is given as
F(x,y,z)=(x+y)z' is given as
x'y'z + xyz' +x'yz'  
xyz+xyz' + xy'z'  
xy'z' + x'y'z' +xyz'  
xyz' + xy'z' +x'yz' 
Question 33 Explanation:
Method1 using boolean expression:
F(x,y,z)=(x+y)z'
= xz' + yz'
= xz' (y+y')+ (x+x')yz'
= xyz' + xy'z' + xyz' + x'yz'
= xyz' + xy'z' + x'yz'
Method2 using KMaps:
=yz’ + xz’
=z’(x+y)
F(x,y,z)=(x+y)z'
= xz' + yz'
= xz' (y+y')+ (x+x')yz'
= xyz' + xy'z' + xyz' + x'yz'
= xyz' + xy'z' + x'yz'
Method2 using KMaps:
=yz’ + xz’
=z’(x+y)
Question 34 
Let P(m, n) be the statement “m divides n” where the universe of discourse for both the variables is the set of positive integers. Determine the truth values of each of the following propositions :
I. ∀m ∀n P(m, n),
II. ∃m ∀n P(m, n)
I. ∀m ∀n P(m, n),
II. ∃m ∀n P(m, n)
Both I and II are true  
Both I and II are false  
I – false & II – true  
I – true & II – false 
Question 35 
Big – O estimate for
f(x) = (x+1) log(x^{2}+1)+3x^{2} is given as
O(x log x)  
O(x^{2})  
O(x^{3})  
O(x^{2} log x) 
Question 35 Explanation:
f(x) = (x+1) log(x^{2}+1)+3x^{2} in this function, 3x^{2} is leading term. So, we can call asymptotically O(n^{2}).
Question 36 
How many edges are there in a forest of ttrees containing a total of n vertices ?
n + t  
n – t  
n ∗ t  
n^{t} 
Question 36 Explanation:
Tree never form cycle. Tree maximum having n1 edges for ‘n’ vertices. Given question they mentioned forest of trees. It means ‘t’ trees will form nt edges in worst case.
Question 37 
Let f and g be the functions from the set of integers to the set integers defined by
f(x) = 2x + 3 and g(x) = 3x + 2
Then the composition of f and g and g and f is given as
6x + 7, 6x + 11  
6x + 11, 6x + 7  
5x + 5, 5x + 5  
None of the above 
Question 37 Explanation:
Given data,
f(x) = 2x+3
g(x) = 3x+2
fo(g(x)) = f(3x+2)
= 2(3x+2)+3
= 6x+7
go(f(x)) = g(2x+3)
= 3(2x + 3)+2
= 6x +11
So, optionA is correct answer.
f(x) = 2x+3
g(x) = 3x+2
fo(g(x)) = f(3x+2)
= 2(3x+2)+3
= 6x+7
go(f(x)) = g(2x+3)
= 3(2x + 3)+2
= 6x +11
So, optionA is correct answer.
Question 38 
If n and r are nonnegative integers and n ≥ r, then p(n + 1, r) equals to
(p(n, r)(n + 1)) / (n + 1 – r)  
(p(n, r) (n + 1)) / (n – 1 + r)  
(p(n, r) (n – 1)) / (n + 1 – r)  
(p(n, r) (n + 1)) / (n + 1 + r) 
Question 39 
A graph is nonplanar if and only if it contains a subgraph homomorphic to
K_{3, 2} or K_{5}  
K_{3, 3} and K_{6}  
K_{3, 3} or K_{5}  
K_{2, 3} and K_{5} 
Question 39 Explanation:
A graph is nonplanar if and only if it contains a subgraph which is homomorphic to k_{5} or k_{3,3}. This is kuratowshi theorem.
Question 40 
Which of the following statements are true ?
I. A circuit that adds two bits, producing a sum bit and a carry bit is called half adder.
II. A circuit that adds two bits, producing a sum bit and a carry bit is called full adder.
III. A circuit that adds two bits and a carry bit producing a sum bit and a carry bit is called full adder.
IV. A device that accepts the value of a Boolean variable as input and produces its complement is called an inverter.
I. A circuit that adds two bits, producing a sum bit and a carry bit is called half adder.
II. A circuit that adds two bits, producing a sum bit and a carry bit is called full adder.
III. A circuit that adds two bits and a carry bit producing a sum bit and a carry bit is called full adder.
IV. A device that accepts the value of a Boolean variable as input and produces its complement is called an inverter.
I & II  
II & III  
I, II, III  
I, III & IV 
Question 40 Explanation:
TRUE: A circuit that adds two bits, producing a sum bit and a carry bit is called half adder.
FALSE: A circuit that adds two bits, producing a sum bit and a carry bit is called full adder.
It is false because it is half adder
TRUE: A circuit that adds two bits and a carry bit producing a sum bit and a carry bit is called full adder.
TRUE: A device that accepts the value of a Boolean variable as input and produces its complement is called an inverter.
FALSE: A circuit that adds two bits, producing a sum bit and a carry bit is called full adder.
It is false because it is half adder
TRUE: A circuit that adds two bits and a carry bit producing a sum bit and a carry bit is called full adder.
TRUE: A device that accepts the value of a Boolean variable as input and produces its complement is called an inverter.
Question 41 
Active X controls are Pentium binary programs that can be embedded in ________
Word pages  
URL pages  
Script pages  
Web pages 
Question 41 Explanation:
→ ActiveX is a software framework created by Microsoft that adapts its earlier Component Object Model (COM) and Object Linking and Embedding (OLE) technologies for content downloaded from a network, particularly from the World Wide Web.
→ Active X controls are Pentium binary programs that can be embedded in web pages.
→ Active X controls are Pentium binary programs that can be embedded in web pages.
Question 42 
Match the following :
aii, biv, ci, diii  
aiv, biii, cii, di  
aiv, biii, ci, dii  
aiii, bi, civ, dii 
Question 42 Explanation:
Scripts→ Wireless Application Environment
HTTP→ Wireless Transaction Protocol
UDP→ Wireless Datagram Protocol
IP→ Wireless
HTTP→ Wireless Transaction Protocol
UDP→ Wireless Datagram Protocol
IP→ Wireless
Question 43 
Which of the following is widely used inside the telephone system for longhaul data traffic ?
ISDN  
ATM  
Frame Relay  
ISTN 
Question 43 Explanation:
→ ATM for carriage of a complete range of user traffic, including voice, data, and video signals.
→ ATM provides functionality that is similar to both circuit switching and packet switching networks.
→ ATM uses asynchronous timedivision multiplexing, and encodes data into small, fixedsized packets (ISOOSI frames) called cells.
→ ATM uses a connectionoriented model in which a virtual circuit must be established between two endpoints before the actual data exchange begins.
→ ATM provides functionality that is similar to both circuit switching and packet switching networks.
→ ATM uses asynchronous timedivision multiplexing, and encodes data into small, fixedsized packets (ISOOSI frames) called cells.
→ ATM uses a connectionoriented model in which a virtual circuit must be established between two endpoints before the actual data exchange begins.
Question 44 
The document standards for EDI were first developed by large business house during the 1970s and are now under the control of the following standard organisation :
ISO  
ANSI  
ITUT  
IEEE 
Question 44 Explanation:
→ The document standards for EDI were first developed by large business houses during the 1970s and are now under the control of the american national standards institute(ANSI).
→ Electronic Data Interchange(EDI) is the exchange of business documents such as purchase orders, invoices,etc. in an electronic format. This exchange happens like email messages, in a few seconds and does not involve any human intervention or paper.
→ Electronic Data Interchange(EDI) is the exchange of business documents such as purchase orders, invoices,etc. in an electronic format. This exchange happens like email messages, in a few seconds and does not involve any human intervention or paper.
Question 45 
Electronic Data Interchange Software consists of the following four layers :
Business application, Internal format conversion, Network translator, EDI envelope  
Business application, Internal format conversion, EDI translator, EDI envelope  
Application layer, Transport layer, EDI translator, EDI envelope  
Application layer, Transport layer, IP layer, EDI envelope 
Question 45 Explanation:
Electronic Data Interchange(EDI) software consists of the following 4 layers
1. Business application
2. Internal format conversion
3. EDI translator
4. EDI envelope
1. Business application
2. Internal format conversion
3. EDI translator
4. EDI envelope
Question 46 
Consider a preemptive priority based scheduling algorithm based on dynamically changing priority. Larger priority number implies higher priority. When the process is waiting for CPU in the ready queue (but not yet started execution), its priority changes at a rate a=2. When it starts running, its priority changes at a rate b=1. All the processes are assigned priority value 0 when they enter ready queue. Assume that the following processes want to execute :
The time quantum q=1. When two processes want to join ready queue simultaneously, the process which has not executed recently is given priority. The finish time of processes P1, P2, P3 and P4 will respectively be
The time quantum q=1. When two processes want to join ready queue simultaneously, the process which has not executed recently is given priority. The finish time of processes P1, P2, P3 and P4 will respectively be
4, 5, 7 and 8  
8, 2, 7 and 5  
2, 5, 7 and 8  
8, 2, 5 and 7 
Question 46 Explanation:
Given data,
Question 47 
The virtual address generated by a CPU is 32 bits. The Translation Lookaside Buffer (TLB) can hold total 64 page table entries and a 4way set associative (i.e. with 4cache lines in the set). The page size is 4 KB. The minimum size of TLB tag is
12 bits  
15 bits  
16 bits  
20 bits 
Question 47 Explanation:
Page size = 4 KB = 4 × 2^{10} Bytes = 2^{12} Bytes
Virtual Address = 32 bit
No. of bits needed to address the page frame = 32  12 = 20
TLB can hold 64 page table entries with 4way set associative
=64/4
=16
=2^{4}
→ 4 bits are needed to address a set.
→ The size of TLB tag = 204
= 16 bits
Virtual Address = 32 bit
No. of bits needed to address the page frame = 32  12 = 20
TLB can hold 64 page table entries with 4way set associative
=64/4
=16
=2^{4}
→ 4 bits are needed to address a set.
→ The size of TLB tag = 204
= 16 bits
Question 48 
Consider a disk queue with request for input/output to block on cylinders 98, 183, 37, 122, 14, 124, 65, 67 in that order. Assume that disk head is initially positioned at cylinder 53 and moving towards cylinder number 0. The total number of head movements using Shortest Seek Time First (SSTF) and SCAN algorithms are respectively
236 and 252 cylinders  
640 and 236 cylinders  
235 and 640 cylinders  
235 and 252 cylinders  
None of the above 
Question 48 Explanation:
= 53+183
= 236
Note: Excluded for evaluation because given options are wrong.
Question 49 
How much space will be required to store the bitmap of a 1.3 GB disk with 512 bytes block size ?
332.8 KB  
83.6 KB  
266.2 KB  
256.6 KB 
Question 49 Explanation:
Given data,
 bitmap=1.3 GB
 Block size=512 bytes
 How much space will be require to store the bitmap=?
Step1: First we have to find total number of blocks
Number of blocks = disk size/ block size
= 1.3 GB(1.3*2^{30} GB) / 512 Bytes(2^{9 }bytes)
= 1.3*2^{21} bits
= 1.3*2MB
= 2.6 MB
= (2.6*1024) / 8KB
= 332.8 KB
 bitmap=1.3 GB
 Block size=512 bytes
 How much space will be require to store the bitmap=?
Step1: First we have to find total number of blocks
Number of blocks = disk size/ block size
= 1.3 GB(1.3*2^{30} GB) / 512 Bytes(2^{9 }bytes)
= 1.3*2^{21} bits
= 1.3*2MB
= 2.6 MB
= (2.6*1024) / 8KB
= 332.8 KB
Question 50 
Linux operating system uses
Affinity Scheduling  
Fair Preemptive Scheduling  
Hand Shaking  
Highest Penalty Ratio Next 
Question 50 Explanation:
Linux operating system uses fair Preemptive Scheduling.
Ex: Round Robin
Ex: Round Robin
There are 50 questions to complete.