## UGC NET CS 2014 Dec-Paper-2

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Question 1 |

Consider a set A = {1, 2, 3, ........, 1000}. How many members of A shall be divisible by 3 or by 5 or by both 3 and 5 ?

533 | |

599 | |

467 | |

66 |

Question 1 Explanation:

Method-1:

Given data,

-- Set A={1, 2, 3, ........, 1000}

-- Set A shall be divisible by 3=?

-- Set A shall be divisible by 5=?

-- Set A shall be divisible by 3 and 5=?

Step-1: To find divisible by 3 numbers are

=⌊1000/3⌋

= 333

Step-2: To find divisible by 5 numbers are

=⌊1000/5⌋

= 200

Step-3: To find divisible by 3 and 5 numbers are

=⌊1000/(3*5)⌋

= 66

These 66 is already part of 333 and 200. So, we have to exclude it from the list.

Total= 333+200-66

= 467

Note: We are using floor because excluding fraction value.

Method-2:

The above problem is in the form of (AUB) = (A)+(B)-(A∩B)

A=1000/3

B=1000/5

(A∩B)=66

(AUB)=467

Note: Getting this idea in exam hall is very difficult. So, better follow method-1 ratherhan method-2

Given data,

-- Set A={1, 2, 3, ........, 1000}

-- Set A shall be divisible by 3=?

-- Set A shall be divisible by 5=?

-- Set A shall be divisible by 3 and 5=?

Step-1: To find divisible by 3 numbers are

=⌊1000/3⌋

= 333

Step-2: To find divisible by 5 numbers are

=⌊1000/5⌋

= 200

Step-3: To find divisible by 3 and 5 numbers are

=⌊1000/(3*5)⌋

= 66

These 66 is already part of 333 and 200. So, we have to exclude it from the list.

Total= 333+200-66

= 467

Note: We are using floor because excluding fraction value.

Method-2:

The above problem is in the form of (AUB) = (A)+(B)-(A∩B)

A=1000/3

B=1000/5

(A∩B)=66

(AUB)=467

Note: Getting this idea in exam hall is very difficult. So, better follow method-1 ratherhan method-2

Question 2 |

A certain tree has two vertices of degree 4, one vertex of degree 3 and one vertex of degree 2. If the other vertices have degree 1, how many vertices are there in the graph ?

5 | |

n-3 | |

20 | |

11 |

Question 2 Explanation:

Method-1:

Here, they are clearly mentioned that “certain tree”

The tree contain maximum n-1 edges. Here, ‘n’ is number of vertices.

→ According to the handshaking lemma “sum of degrees of all vertices=2|e|”.

→ Two vertices of degree 4, we can write into (2*4)

→ One vertex of degree 3, we can write into (1*3)

→ One vertex of degree 2, we can write into (1*2).

→ Other vertices(X) have degree 1, we can write into (X*1)

Step-1: (2*4)+(1*3)+(1*2)+(X*1)

=2(X+4-1) [Note: ‘X’ value is not given ]

X=7

Step-2: To find total number of vertices, we are adding X+4 because they already given 4 vertices.

=X+4

=7+4

=11

Method-2:

Draw according to given constraints but it not suitable for very big trees

B, C, E, F, H, I, J degree = 1

D degree = 2

A degree = 4

K degree = 4

G degree = 3

Here, they are clearly mentioned that “certain tree”

The tree contain maximum n-1 edges. Here, ‘n’ is number of vertices.

→ According to the handshaking lemma “sum of degrees of all vertices=2|e|”.

→ Two vertices of degree 4, we can write into (2*4)

→ One vertex of degree 3, we can write into (1*3)

→ One vertex of degree 2, we can write into (1*2).

→ Other vertices(X) have degree 1, we can write into (X*1)

Step-1: (2*4)+(1*3)+(1*2)+(X*1)

=2(X+4-1) [Note: ‘X’ value is not given ]

X=7

Step-2: To find total number of vertices, we are adding X+4 because they already given 4 vertices.

=X+4

=7+4

=11

Method-2:

Draw according to given constraints but it not suitable for very big trees

B, C, E, F, H, I, J degree = 1

D degree = 2

A degree = 4

K degree = 4

G degree = 3

Question 3 |

Consider the Graph shown below :

This graph is a __________.

This graph is a __________.

Complete Graph | |

Bipartite Graph | |

Hamiltonian Graph | |

All of the above |

Question 3 Explanation:

Option-A: It is not complete graph because it won’t have n(n-1)/2 edges.

Option-B: It is not Bipartite Graph because it takes more than 2 colours. Bipartite Graph have exactly 2 colours.

Option-C: Hamiltonian path is a path in an undirected or directed graph that visits each vertex exactly once.

Hence, Option-C is correct answer.

Option-B: It is not Bipartite Graph because it takes more than 2 colours. Bipartite Graph have exactly 2 colours.

Option-C: Hamiltonian path is a path in an undirected or directed graph that visits each vertex exactly once.

Hence, Option-C is correct answer.

Question 4 |

A computer program selects an integer in the set {k : 1 ≤ k ≤ 10,00,000} at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k=1 appears in the printout at least once ?

0.5 | |

0.704 | |

0.632121 | |

0.68 |

Question 4 Explanation:

Probability that the value k=1 appears in the printout at least once is

1-p(x=0)

We will apply the formula of binomial distribution

1-(

=1-(1*1*0.367879)

=0.632121

Hence answer is option C

1-p(x=0)

We will apply the formula of binomial distribution

1-(

^{ 1000000}C_{0} (1/1000000)^{0} * (999999/1000000)^{10^6} )=1-(1*1*0.367879)

=0.632121

Hence answer is option C

Question 5 |

If we define the functions f, g and h that map R into R by :
f(x) = x

^{ 4} , g(x) = √ x^{ 2}+ 1 , h(x) = x^{ 2} + 72, then the value of the composite functions ho(gof) and (hog)of are given asx ^{8} – 71 and x ^{8} – 71 | |

x ^{8} – 73 and x^{8} – 73 | |

x ^{8} + 71 and x ^{8} + 71 | |

x ^{8} + 73 and x ^{8} + 73 |

Question 5 Explanation:

Given f(x) = x

for, ho(gof)

gof=g(f(x))

=g(x

=√(x

ho(gof)=h(gof)

=h(√(x

=(√(x

=x

=x

√ x

for, (hog)of,

hog=h(g(x))

=h(√(x

=(√(x

=x

=x

(hog)of=(hog)(f(x))

=(hog)(x

=(x

=x

Hence, option-D is the correct answer

^{4} , g(x) = √ x^{2}+ 1 , h(x) = x 2 + 72for, ho(gof)

gof=g(f(x))

=g(x

^{4} )=√(x

^{8} +1)ho(gof)=h(gof)

=h(√(x

^{8} +1))=(√(x

^{8} +1) 2 +72=x

^{8} +1+72=x

^{8} +73√ x

^{2}+ 1 , h(x) = x^{ 2} + 72for, (hog)of,

hog=h(g(x))

=h(√(x

^{2} +1)=(√(x

^{2} +1) 2 +72=x

^{2} + 1+72=x

^{2} +73(hog)of=(hog)(f(x))

=(hog)(x

^{4} )=(x

^{4} )^{2} +73=x

^{8} +73Hence, option-D is the correct answer

Question 6 |

The BCD adder to add two decimal digits needs minimum of

6 full adders and 2 half adders | |

5 full adders and 3 half adders | |

4 full adders and 3 half adders | |

5 full adders and 2 half adders |

Question 6 Explanation:

→ Each digit is represented by a 4-bit BCD code.

→ To add two 4-bit number, we need 1 Half Adder(to add LSBs) and 3 Full Adders(remaining three bits of both number along with carry bits).

→ To make the resultant Sum as valid BCD sum, we need to add 0110 to the sum.

→ This can be done with 1 Half adder and 2 Full Adder

(Note: LSB bit of 0110 is always zero. So there is no need of ADDER to add LSBs.)

→ Here, Half adder is used to add next significant bits.

Total 5 Full Adders and 2 Half Adders are needed

→ To add two 4-bit number, we need 1 Half Adder(to add LSBs) and 3 Full Adders(remaining three bits of both number along with carry bits).

→ To make the resultant Sum as valid BCD sum, we need to add 0110 to the sum.

→ This can be done with 1 Half adder and 2 Full Adder

(Note: LSB bit of 0110 is always zero. So there is no need of ADDER to add LSBs.)

→ Here, Half adder is used to add next significant bits.

Total 5 Full Adders and 2 Half Adders are needed

Question 7 |

The Excess-3 decimal code is a self-complementing code because

The binary sum of a code and its 9’s complement is equal to 9. | |

It is a weighted code | |

Complement can be generated by inverting each bit pattern | |

The binary sum of a code and its 10’s complement is equal to 9 |

Question 7 Explanation:

The Excess-3 decimal code is a self-complementing code because complement can be generated by inverting each bit pattern.

Question 8 |

How many PUSH and POP operations will be needed to evaluate the following expression by reverse polish notation in a stack machine (A * B) + (C * D/E) ?

4 PUSH and 3 POP instructions | |

5 PUSH and 4 POP instructions | |

6 PUSH and 2 POP instructions | |

5 PUSH and 3 POP instructions |

Question 8 Explanation:

Given infix notation, we have to change infix notation into postfix notation.

after converting postfix notation the notations are ab* cde /* +

Total 5 PUSH and 4 POP operations performed.

after converting postfix notation the notations are ab* cde /* +

Total 5 PUSH and 4 POP operations performed.

Question 9 |

The range of representable normalized numbers in the floating point binary fractional representation in a 32-bit word with 1-bit sign, 8-bit excess 128 biased exponent and 23-bit mantissa is

2 ^{-128} to (1 – 2^{ –23} ) * 2^{ 127} | |

(1 – 2 ^{ –23} ) * 2 ^{ -127} to 2 ^{128} | |

(1 – 2 ^{–23} ) * 2^{ –127} to 2^{ 23>} | |

2 ^{–129} to (1 – 2^{ –23} ) * 2^{ 127} |

Question 9 Explanation:

The range of representable normalized numbers in the floating point binary fractional representation in a 32-bit word with 1-bit sign, 8-bit excess 128 biased exponent and 23-bit mantissa is 2

^{–129}to (1 – 2^{ –23} ) * 2^{ 127}Question 10 |

The size of the ROM required to build an 8-bit adder/subtractor with mode control, carry input, carry output and two’s complement overflow output is given as

2 ^{16} * 8 | |

2 ^{18} * 10 | |

2 ^{16} * 10 | |

2 ^{18} * 8 |

Question 10 Explanation:

The size of the ROM required to build an 8-bit adder/subtractor with mode control, carry input, carry output and two’s complement overflow output is given as 2

^{18} * 10.Question 11 |

What will be the output of the following ‘C’ code ?

main( )

{

int x=128;

printf (“\n%d”, 1 + x++);

}

main( )

{

int x=128;

printf (“\n%d”, 1 + x++);

}

128 | |

129 | |

130 | |

131 |

Question 11 Explanation:

In this program we are using post increment. Post increment will send the value before updating the value.

printf(“\n%d”, 1 + x++); /* x=128 */

Here, 1+128=129

printf(“\n%d”, 1 + x++); /* x=128 */

Here, 1+128=129

Question 12 |

What does the following expression means ?
char *(*(*a[N]) ( )) ( );

a pointer to a function returning array of n pointers to function returning character pointers. | |

a function return array of N pointers to functions returning pointers to characters | |

an array of n pointers to function returning pointers to characters | |

an array of n pointers to function returning pointers to functions returning pointers to characters. |

Question 12 Explanation:

char *(*(*a[N]) ( )) ( ); /* an array of n pointers to function returning pointers to functions returning pointers to characters */

Question 13 |

Which of the following is not a member of class ?

Static function | |

Friend function | |

Const function | |

Virtual function |

Question 13 Explanation:

A friend function of a class is defined outside that class scope but it has the right to access all private and protected members of the class. Even though the prototypes for friend functions appear in the class definition, friends are not member functions.

1. A function can only be declared a friend by a class itself.

2. It can have access to all members of the class, even private ones.

1. A function can only be declared a friend by a class itself.

2. It can have access to all members of the class, even private ones.

Question 14 |

When an array is passed as parameter to a function, which of the following statements is correct ?

The function can change values in the original array. | |

In C, parameters are passed by value, the function cannot change the original value in the array. | |

It results in compilation error when the function tries to access the elements in the array. | |

Results in a runtime error when the function tries to access the elements in the array. |

Question 14 Explanation:

The function can change values in the original array, when an array is passed as parameter to a function because we have to pass it by reference.

Question 15 |

Which of the following differentiates between overloaded functions and overridden functions ?

Overloading is a dynamic or runtime binding and overridden is a static or compile time binding. | |

Overloading is a static or compile time binding and overriding is dynamic or runtime binding. | |

Redefining a function in a friend class is called overloading, while redefining a function in a derived class is called as overridden function. | |

Redefining a function in a derived class is called function overloading, while redefining a function in a friend class is called function overriding. |

Question 15 Explanation:

Question 16 |

Division operation is ideally suited to handle queries of the type :

customers who have no account in any of the branches in Delhi. | |

customers who have an account at all branches in Delhi. | |

customers who have an account in at least one branch in Delhi. | |

customers who have only joint account in any one branch in Delhi |

Question 16 Explanation:

→ The DIVISION operation is defined for convenience for dealing with queries that involve universal quantification or the all condition. For a tuple 't' to appear in the result T of the DIVISION, the values in ‘t’ must appear in R in combination with every tuple in S.

→ Note that in the formulation of the DIVISION operation, the tuples in the denominator relation S restrict the numerator relation R by selecting those tuples in the result that match all values present in the denominator. The DIVISION operation can be expressed as a sequence of π, ×, and – operations as follows:

T1 ← πY(R)

T2 ← πY((S × T1) – R)

T ← T1 – T2

→ Note that in the formulation of the DIVISION operation, the tuples in the denominator relation S restrict the numerator relation R by selecting those tuples in the result that match all values present in the denominator. The DIVISION operation can be expressed as a sequence of π, ×, and – operations as follows:

T1 ← πY(R)

T2 ← πY((S × T1) – R)

T ← T1 – T2

Question 17 |

Which of the following is true ?

I. Implementation of self-join is possible in SQL with table alias.

II. Outer-join operation is basic operation in relational algebra.

III. Natural join and outer join operations are equivalent.

I. Implementation of self-join is possible in SQL with table alias.

II. Outer-join operation is basic operation in relational algebra.

III. Natural join and outer join operations are equivalent.

I and II are correct. | |

II and III are correct. | |

Only III is correct. | |

Only I is correct. |

Question 17 Explanation:

NATURAL JOIN requires that the two join attributes (or each pair of join attributes) have the same name in both relations. If this is not the case, a renaming operation is applied first.

OUTER JOIN: A set of operations, called outer joins, were developed for the case where the user wants to keep all the tuples in R, or all those in S, or all those in both relations in the result of the JOIN, regardless of whether or not they have matching tuples in the other relation.

In SQL, the same name can be used for two (or more) attributes as long as the attributes are in different relations. If this is the case, and a multi table query refers to two (or more) attributes with the same name, we must qualify the attribute name with the relation name to prevent ambiguity. The ambiguity of attribute names also arises in the case of queries that refer to the same relation twice. In this case, we are required to declare alternative relation names, called aliases or tuple variables.

An alias can follow the keyword "AS" , as shown below

SELECT E.Fname, E.Lname, S.Fname, S.Lname

FROM EMPLOYEE AS E, EMPLOYEE AS S

WHERE E.Super_ssn=S.Ssn;

In above query relation names E and S, called aliases or tuple variables, for the EMPLOYEE relation.

From above explanation it is clear that only option (D) is correct.

OUTER JOIN: A set of operations, called outer joins, were developed for the case where the user wants to keep all the tuples in R, or all those in S, or all those in both relations in the result of the JOIN, regardless of whether or not they have matching tuples in the other relation.

In SQL, the same name can be used for two (or more) attributes as long as the attributes are in different relations. If this is the case, and a multi table query refers to two (or more) attributes with the same name, we must qualify the attribute name with the relation name to prevent ambiguity. The ambiguity of attribute names also arises in the case of queries that refer to the same relation twice. In this case, we are required to declare alternative relation names, called aliases or tuple variables.

An alias can follow the keyword "AS" , as shown below

SELECT E.Fname, E.Lname, S.Fname, S.Lname

FROM EMPLOYEE AS E, EMPLOYEE AS S

WHERE E.Super_ssn=S.Ssn;

In above query relation names E and S, called aliases or tuple variables, for the EMPLOYEE relation.

From above explanation it is clear that only option (D) is correct.

Question 18 |

What kind of mechanism is to be taken into account for converting a weak entity set into strong entity set in entity-relationship diagram ?

Generalization | |

Aggregation | |

Specialization | |

Adding suitable attributes |

Question 18 Explanation:

→ Entity types that do not have key attributes of their own are called weak entity types. In contrast,regular entity types that do have a key attribute are called strong entity types. Entities belonging to a weak entity type are identified by being related to specific entities from another entity type in combination with one of their attribute values. We call this other entity type the identifying or owner entity type, and we call the relationship type that relates a weak entity type to its owner the identifying relationship of the weak entity type.

→ A weak entity type always has a total participation constraint (existence dependency) with respect to its identifying relationship because a weak entity can not be identified without an owner entity.

→ A weak entity type always has a total participation constraint (existence dependency) with respect to its identifying relationship because a weak entity can not be identified without an owner entity.

Question 19 |

The best normal form of relation scheme R(A, B, C, D) along with the set of functional dependencies F = {AB → C, AB → D, C → A, D → B} is

Boyce-Codd Normal form | |

Third Normal form | |

Second Normal form | |

First Normal form |

Question 19 Explanation:

AB → C

AB → D

C → A

D → B

A + ={A}

B + ={B}

C + ={C,A}

D + ={D,B}

Since single attributes can't determine all the attributes of relation. Find closure of combination of two attributes to check whey could be the key for the relation or not.

AB + ={A,B,C,D}

AC + ={A,C}

AD + ={A,D,B,C}

BC + ={B,C,A,D}

AB + ={A,B,C,D}

AC + ={A,C}

AD + ={A,D,B,C}

BC + ={B,C,A,D}

BD + ={B,D}

CD + ={C,D,B,A}

So, AB,AD,BC,CD are the candidate keys for the given relation. All attributes are key attributes and since last two functional dependencies are violating BCNF property (that LHS if a functional dependency should be a super key) so the given relation is in 3NF only.

AB → D

C → A

D → B

A + ={A}

B + ={B}

C + ={C,A}

D + ={D,B}

Since single attributes can't determine all the attributes of relation. Find closure of combination of two attributes to check whey could be the key for the relation or not.

AB + ={A,B,C,D}

AC + ={A,C}

AD + ={A,D,B,C}

BC + ={B,C,A,D}

AB + ={A,B,C,D}

AC + ={A,C}

AD + ={A,D,B,C}

BC + ={B,C,A,D}

BD + ={B,D}

CD + ={C,D,B,A}

So, AB,AD,BC,CD are the candidate keys for the given relation. All attributes are key attributes and since last two functional dependencies are violating BCNF property (that LHS if a functional dependency should be a super key) so the given relation is in 3NF only.

Question 20 |

Identify the minimal key for relational scheme R(A, B, C, D, E) with functional dependencies F = {A → B, B → C, AC → D}

A | |

AE | |

BE | |

CE |

Question 20 Explanation:

Given functional dependencies are F={A → B, B → C, AC → D}

Step-1: Attribute A and E are not present in the right hand side of any production. So, every key should include AE to be a candidate key.

Step-2: To check whether AE could be a candidate key or not.

(AE) + = {A,B,C,D,E}

Since AE can uniquely identify each attribute and we can say this is the candidate key for given relation.

Step-1: Attribute A and E are not present in the right hand side of any production. So, every key should include AE to be a candidate key.

Step-2: To check whether AE could be a candidate key or not.

(AE) + = {A,B,C,D,E}

Since AE can uniquely identify each attribute and we can say this is the candidate key for given relation.

Question 21 |

Convert the following infix expression into its equivalent postfix expression (A + B^ D) / (E – F) + G

ABD^ + EF – / G+ | |

ABD + ^EF – / G+ | |

ABD + ^EF / – G+ | |

ABD^ + EF / – G+ |

Question 21 Explanation:

→ According to priority (or) infix expression we can write expression like this ((A + B^D) / (E – F)) + G

→ Actual tree structure is

→ Actual tree structure is

Question 22 |

You have to sort a list L, consisting of a sorted list followed by a few ‘random’ elements. Which of the following sorting method would be most suitable for such a task ?

Bubble sort | |

Selection sort | |

Quick sort | |

Insertion sort |

Question 22 Explanation:

Here, they given sorted list followed by a few ‘random’ elements.

So, we can eliminate option-B and Option-C because it gives worst complexity(O(n 2 )) when array is already sorted.

Bubble sort and insertion sort will give best case complexity(O(n)) but here they given small constraint is “there are few random elements”. So, insertion sort is more appropriate answer.

So, we can eliminate option-B and Option-C because it gives worst complexity(O(n 2 )) when array is already sorted.

Bubble sort and insertion sort will give best case complexity(O(n)) but here they given small constraint is “there are few random elements”. So, insertion sort is more appropriate answer.

Question 23 |

The directory can be viewed as ________ that translates filenames into their directory entries.

Symbol table | |

Partition | |

Swap space | |

Cache |

Question 23 Explanation:

The directory can be viewed as symbol table that translates filenames into their directory entries.

Symbol table can be organized in many ways using some operations:

1.Search for a file or multiple files

2.Create a file

3.Delete a file

4.List a directory

5.Rename a file

6.Traverse the file system

Symbol table can be organized in many ways using some operations:

1.Search for a file or multiple files

2.Create a file

3.Delete a file

4.List a directory

5.Rename a file

6.Traverse the file system

Question 24 |

Consider an array A[20, 10], assume 4 words per memory cell and the base address of array A is 100. What is the address of A[11, 5] ? Assume row major storage.

560 | |

565 | |

570 | |

575 |

Question 24 Explanation:

Here, Row major order generally follows offset formula to find address of the location.

Offset= (Row number * Total number of columns) + Column Number

Step-1: Add the Base address to calculate the effective address. Given matrix is A[20][10], Size of memory word is 4 bytes, address of the location to find A[11][5], Base address is 100.

Step-2: Offset = (11*10) + 5

= 115 (multiply with 4 words)

= 115*4

= 460

Step 3: Initial base address is 100

=Offset+base address

= 460 +100

= 560.

Offset= (Row number * Total number of columns) + Column Number

Step-1: Add the Base address to calculate the effective address. Given matrix is A[20][10], Size of memory word is 4 bytes, address of the location to find A[11][5], Base address is 100.

Step-2: Offset = (11*10) + 5

= 115 (multiply with 4 words)

= 115*4

= 460

Step 3: Initial base address is 100

=Offset+base address

= 460 +100

= 560.

Question 25 |

A full binary tree with n leaves contains

n nodes | |

log _{2} n nodes | |

2n –1 nodes | |

2 ^{ n} nodes |

Question 25 Explanation:

A Binary Tree is full if every node has 0 or 2 children. So, in such case, the binary tree with n leaves contains a total of 2*n-1 nodes.

There are 4 leaf nodes in the above tree.

So, total number of nodes are 2*4-1

=8-1

=7

There are 4 leaf nodes in the above tree.

So, total number of nodes are 2*4-1

=8-1

=7

Question 26 |

The period of a signal is 10 ms. What is its frequency in Hertz ?

10 | |

100 | |

1000 | |

10000 |

Question 26 Explanation:

First we change 10 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10

10 ms=10*10

=10

Frequency(f)= 1/T

= 1/10

=100 Hz

^{−3} kHz).10 ms=10*10

^{-3} s=10

^{ -2} sFrequency(f)= 1/T

= 1/10

^{-2} Hz=100 Hz

Question 27 |

In a classful addressing, first four bits in Class-A IP address is

1010 | |

1100 | |

1011 | |

1110 | |

None of the above |

Question 27 Explanation:

The class-A address starts with 0 to 127 but we never used 0 and 127 numbers. The numbers in binary format is 00000001 to 01111110. So, no option is starts with 0.

Class-A is 0

Class-B is 10

Class-C is 110

Class-D is 1110

Class-E is 1111

Class-A is 0

Class-B is 10

Class-C is 110

Class-D is 1110

Class-E is 1111

Question 28 |

Which of the following algorithms is not a broadcast routing algorithm ?

Flooding | |

Multi Destination routing | |

Reverse path forwarding | |

All of the above |

Question 28 Explanation:

Flooding: Sending packets to all neighbors in the network.

Multi Destination routing: It is also sending packets to multi destinations in the network.

Reverse path forwarding (RPF): Only sends packet forward if it is received from the next node in the shortest path back to the sender.

Multi Destination routing: It is also sending packets to multi destinations in the network.

Reverse path forwarding (RPF): Only sends packet forward if it is received from the next node in the shortest path back to the sender.

Question 29 |

An analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many data elements are carried by each signal element ?

0.336 bits/baud | |

3 bits/baud | |

120,00,000 bits/baud | |

None of the above |

Question 29 Explanation:

Given data,

-- Baud rate = 2000 baud

-- Bit rate = 6000 bps

-- data elements are carried by each signal element=?

Step-1: To find data elements are carried by each signal element using bits per baud.

Bits per baud = bit rate / baud rate

= 6000 / 2000

= 3 bits/baud

-- Baud rate = 2000 baud

-- Bit rate = 6000 bps

-- data elements are carried by each signal element=?

Step-1: To find data elements are carried by each signal element using bits per baud.

Bits per baud = bit rate / baud rate

= 6000 / 2000

= 3 bits/baud

Question 30 |

How many distinct stages are there in DES algorithm, which is parameterized by a 56-bit key ?

16 | |

17 | |

18 | |

19 |

Question 30 Explanation:

Total 19 distinct stages are there in DES algorithm, which is parameterized by a 56-bit key(and +8 parity bits). The block size is 64 bits and uses 16 rounds.

Question 31 |

Shift-Reduce parsers perform the following :

Shift step that advances in the input stream by K(K > 1) symbols and Reduce step that applies a completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol. | |

Shift step that advances in the input stream by one symbol and Reduce step that applies a completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol. | |

Shift step that advances in the input stream by K(K = 2) symbols and Reduce step that applies a completed grammar rule to form a single tree. | |

Shift step that does not advance in the input stream and Reduce step that applies a completed grammar rule to form a single tree. |

Question 31 Explanation:

Shift-Reduce parsers perform shift step that advances in the input stream by one symbol and Reduce step that applies a completed grammar rule to some recent parse trees, joining them together as one tree with a new root symbol.

Question 32 |

Which of the following is true ?

Canonical LR parser is LR (1) parser with single look ahead terminal | |

All LR(K) parsers with K > 1 can be transformed into LR(1) parsers. | |

Both (A) and (B) | |

None of the above |

Question 32 Explanation:

TRUE: Canonical LR parser is LR (1) parser with single look ahead terminal

TRUE: All LR(K) parsers with K > 1 can be transformed into LR(1) parsers.

TRUE: All LR(K) parsers with K > 1 can be transformed into LR(1) parsers.

Question 33 |

In a two-pass assembler, symbol table is

Generated in first pass | |

Generated in second pass | |

Not generated at all | |

Generated and used only in second pass |

Question 33 Explanation:

→ In a two-pass assembler, symbol table is generated in first pass.

→ The first pass of the assembler reads and processes the assembly program one line at a time. In processing a single line of the assembly program the assembler can make addition(s) to the symbol table, add a (possibly partial) SML instruction to the Simpletron's memory.

→ The purpose of the second pass is to complete the partial instructions written in the first pass.

→ The first pass of the assembler reads and processes the assembly program one line at a time. In processing a single line of the assembly program the assembler can make addition(s) to the symbol table, add a (possibly partial) SML instruction to the Simpletron's memory.

→ The purpose of the second pass is to complete the partial instructions written in the first pass.

Question 34 |

Debugger is a program that

allows to examine and modify the contents of registers | |

does not allow execution of a segment of program | |

allows to set breakpoints, execute a segment of program and display contents of register | |

All of the above |

Question 34 Explanation:

Debugger is a program that allows to set breakpoints, execute a segment of program and display contents of register.

Question 35 |

The following Context-Free Grammar (CFG) :

S → aB | bA

A → a | as | bAA

B → b | bs | aBB

will generate

S → aB | bA

A → a | as | bAA

B → b | bs | aBB

will generate

odd numbers of a’s and odd numbers of b’s | |

even numbers of a’s and even numbers of b’s | |

equal numbers of a’s and b’s | |

different numbers of a’s and b’s |

Question 35 Explanation:

Option A is false: as it generate string "abab" which has even number of a's and b's.

Option B is false: as it generate string "ab" which has odd number of a's and b's.

Option D is false: as it always generate one "a" along with one "b". So it will always generate equal number of a's and b's and will never generate different number of a's and b's.

Hence option C is correct.

Option B is false: as it generate string "ab" which has odd number of a's and b's.

Option D is false: as it always generate one "a" along with one "b". So it will always generate equal number of a's and b's and will never generate different number of a's and b's.

Hence option C is correct.

Question 36 |

Consider the following justifications for commonly using the two-level CPU scheduling :

I. It is used when memory is too small to hold all the ready processes.

II. Because its performance is same as that of the FIFO.

III. Because it facilitates putting some set of processes into memory and a choice is made from that.

IV. Because it does not allow to adjust the set of in-core processes.

I. It is used when memory is too small to hold all the ready processes.

II. Because its performance is same as that of the FIFO.

III. Because it facilitates putting some set of processes into memory and a choice is made from that.

IV. Because it does not allow to adjust the set of in-core processes.

I, III and IV | |

I and II | |

III and IV | |

I and III |

Question 36 Explanation:

Two-level CPU scheduling are using for giving situations:

1. When memory is too small to hold all the ready processes.

2. It facilitates putting some set of processes into memory and a choice is made from that

1. When memory is too small to hold all the ready processes.

2. It facilitates putting some set of processes into memory and a choice is made from that

Question 37 |

A specific editor has 200 K of program text, 15 K of initial stack, 50 K of initialized data, and 70 K of bootstrap code. If five editors are started simultaneously, how much physical memory is needed if shared text is used ?

1135 K | |

335 K | |

1065 K | |

320 K |

Question 37 Explanation:

Given data,

-- Program text=200 K

-- Initial stack=15 K

-- Initialized data=50 K

-- Bootstrap code=70 K

-- Physical memory needed=?

Step-1: Here, given constraint that, all five editors are started simultaneously.

So, all editors to perform all above operations. It need physical memory is

= Program text + Initial stack + Initialized data + Bootstrap code

= 200 K + 15 K + 50 K + 70 K

= 335 K

-- Program text=200 K

-- Initial stack=15 K

-- Initialized data=50 K

-- Bootstrap code=70 K

-- Physical memory needed=?

Step-1: Here, given constraint that, all five editors are started simultaneously.

So, all editors to perform all above operations. It need physical memory is

= Program text + Initial stack + Initialized data + Bootstrap code

= 200 K + 15 K + 50 K + 70 K

= 335 K

Question 38 |

Which of the following conditions does not hold good for a solution to a critical section problem ?

No assumptions may be made about speeds or the number of CPUs. | |

No two processes may be simultaneously inside their critical sections. | |

Processes running outside its critical section may block other processes. | |

Processes do not wait forever to enter its critical section |

Question 38 Explanation:

As per the given options, Option-C is the most appropriate one because all situations is somehow related to critical section problem. But Option-C is not relevant.

Question 39 |

For the implementation of a paging scheme, suppose the average process size be ‘x’ bytes, the page size be ‘y’ bytes, and each page entry requires ‘z’ bytes. The optimum page size that minimizes the total overhead due to the page table and the internal fragmentation loss is given by

x/2 | |

xz/2 | |

√2xz | |

√ xz/ 2 |

Question 39 Explanation:

Since the average number of pages required per process will be x/y and the amount of space required by the page table will be (x/y)*z. The amount of space lost due to internal fragmentation is y/2. So total space wastage is

Loss(L)=(x/y)*e + y/2

To find the value of ‘y’ that yields the minimal values, take rst derivative with respect to ‘y’ and set the resulting equation to zero. (dL/dy) =0

y=√(2xz)

Loss(L)=(x/y)*e + y/2

To find the value of ‘y’ that yields the minimal values, take rst derivative with respect to ‘y’ and set the resulting equation to zero. (dL/dy) =0

y=√(2xz)

Question 40 |

In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m.sec. if an empty frame is available or if the replaced page is not modified, and it takes 20 m.secs., if the replaced page is modified. What is the average access time to service a page fault assuming that the page to be replaced is modified 70% of the time ?

11.6 m.sec. | |

16.4 m.sec. | |

28 m.sec. | |

14 m.sec. |

Question 40 Explanation:

Given data,

-- Time to serve empty page fault = 8 m.sec

-- Time to serve modified page fault= 20 m.sec

-- Percentage of page modification= 70%

= 70/100

= 0.7

-- Page to be replaced is not modified= 1-0.7

= 0.3

-- Average access time to service a page fault=?

Step-1: Average access time = (Not modified*Time to serve page fault) + (Modified time * percentage of page modification)

= (0.3*8) + (0.7*20)

= 2.4 + 14

= 16.4 m.sec

-- Time to serve empty page fault = 8 m.sec

-- Time to serve modified page fault= 20 m.sec

-- Percentage of page modification= 70%

= 70/100

= 0.7

-- Page to be replaced is not modified= 1-0.7

= 0.3

-- Average access time to service a page fault=?

Step-1: Average access time = (Not modified*Time to serve page fault) + (Modified time * percentage of page modification)

= (0.3*8) + (0.7*20)

= 2.4 + 14

= 16.4 m.sec

Question 41 |

__________ are applied throughout the software process.

Framework activities | |

Umbrella activities | |

Planning activities | |

Construction activities |

Question 41 Explanation:

→ The umbrella activities occur throughout the software process they are applied evenly across the process, the analysis encompasses a set of work tasks (eg. requirement gathering,
elaboration, negotiation specification and validation).

→ These activities include Software project tracking and control, Risk management, Software quality assurance, and formal technical reviews, measurement, Software configuration management, reusability management and work product preparation and production.

→ These activities include Software project tracking and control, Risk management, Software quality assurance, and formal technical reviews, measurement, Software configuration management, reusability management and work product preparation and production.

Question 42 |

Requirement Development, Organizational Process Focus, Organizational Training, Risk Management and Integrated Supplier Management are process areas required to achieve maturity level

Performed | |

Managed | |

Defined | |

Optimized |

Question 42 Explanation:

Managed maturity level focuses on software metrics. Two metrics are

1. Product Metric: It measures the characteristics of product being developed. Product metrics are size, reliability, understandability etc..,

2. Process Metrics: It reflect the effectiveness of process. Process metrics are average defect correction on time, number of failures detected during testing per LOC etc.,

This level focuses on improving product and process quality. The Key process areas are quantitative process metric and software quality management.

1. Product Metric: It measures the characteristics of product being developed. Product metrics are size, reliability, understandability etc..,

2. Process Metrics: It reflect the effectiveness of process. Process metrics are average defect correction on time, number of failures detected during testing per LOC etc.,

This level focuses on improving product and process quality. The Key process areas are quantitative process metric and software quality management.

Question 43 |

The software _________ of a program or a computing system is the structure or structures of the system, which comprise software components, the externally visible properties of those components, and the relationships among them.

Design | |

Architecture | |

Process | |

Requirement |

Question 43 Explanation:

The software Architecture of a program or a computing system is the structure or structures of the system, which comprise software components, the externally visible properties of those components, and the relationships among them.

Question 44 |

Which one of the following set of attributes should not be encompassed by effective software metrics ?

Simple and computable | |

Consistent and objective | |

Consistent in the use of units and dimensions | |

Programming language dependent |

Question 44 Explanation:

Programming language dependent attributes should not be encompassed by effective software metrics.

Question 45 |

Which one of the following is used to compute cyclomatic complexity ?

The number of regions – 1 | |

E – N + 1, where E is the number of flow graph edges and N is the number of flow graph nodes. | |

P – 1, where P is the number of predicate nodes in the flow graph G. | |

P + 1, where P is the number of predicate nodes in the flow graph G. |

Question 45 Explanation:

Cyclomatic complexity uses 3 formulas

1. Number of regions + 1

2. Predicate + 1

3. Edges-Vertices+2

1. Number of regions + 1

2. Predicate + 1

3. Edges-Vertices+2

Question 46 |

Consider the following statements S1 and S2 :

S1 : A hard handover is one in which the channel in the source cell is retained and used for a while in parallel with the channel in the target cell.

S2 : A soft handover is one in which the channel in the source cell is released and only then the channel in the target cell is engaged.

S1 : A hard handover is one in which the channel in the source cell is retained and used for a while in parallel with the channel in the target cell.

S2 : A soft handover is one in which the channel in the source cell is released and only then the channel in the target cell is engaged.

S1 is true and S2 is not true. | |

S1 is not true and S2 is true. | |

Both S1 and S2 are true. | |

Both S1 and S2 are not true. |

Question 46 Explanation:

→ Hard handover (or) Hard Handoff: Early systems used hard handoff. In a hard handoff, a mobile station only communicates with one base station. When the MS moves from one cell to
another, communication must first be broken with the previous base station before communication can be established with the new one.

→ Soft handover (or) Soft Handoff: New systems use a soft Handoff. In this case, a mobile station can communicate with base stations at the same time. This means that, during handoff, a mobile station may continue with the new base station before breaking off from the old one.

→ Soft handover (or) Soft Handoff: New systems use a soft Handoff. In this case, a mobile station can communicate with base stations at the same time. This means that, during handoff, a mobile station may continue with the new base station before breaking off from the old one.

Question 47 |

Factless fact table in a data warehouse contains

only measures | |

only dimensions | |

keys and measures | |

only surrogate keys |

Question 47 Explanation:

→ Factless fact table in a data warehouse contains only surrogate keys.

→ The factless fact table is a fact table that does not contain any facts. There are two kinds of factless fact tables:

1. Factless fact table describes event or activity.

2. Factless fact table describes a condition, eligibility or coverage.

→ The factless fact table is a fact table that does not contain any facts. There are two kinds of factless fact tables:

1. Factless fact table describes event or activity.

2. Factless fact table describes a condition, eligibility or coverage.

Question 48 |

Which e-business model allows consumers to name their own price for products and services ?

B2 B | |

B2 G | |

C2 C | |

C2 B |

Question 48 Explanation:

Consumer-to-business (C2B) is a business model in which consumers (individuals) create value and businesses consume that value.

Business-to-government (B2G) (or) "public sector marketing": Which encompasses marketing products and services to various government levels through integrated marketing communications techniques such as strategic public relations, branding, marketing communications (marcom), advertising, and web-based communications.

Customer to customer (C2C) markets provide an innovative way to allow customers to interact with each other.

Business-to-business (B2B) is a situation where one business makes a commercial transaction with another.

Business-to-government (B2G) (or) "public sector marketing": Which encompasses marketing products and services to various government levels through integrated marketing communications techniques such as strategic public relations, branding, marketing communications (marcom), advertising, and web-based communications.

Customer to customer (C2C) markets provide an innovative way to allow customers to interact with each other.

Business-to-business (B2B) is a situation where one business makes a commercial transaction with another.

Question 49 |

__________ model is designed to bring prices down by increasing the number of customers who buy a particular product at once.

Economic Order Quantity | |

Inventory | |

Data Mining | |

Demand-Sensitive Pricing |

Question 49 Explanation:

→ Demand-Sensitive Pricing model is designed to bring prices down by increasing the number of customers who buy a particular product at once.

→ Demand sensitivity is referred to the change in demand for a product when its price is changed by a small amount.

Price Elasticity of demand = ((%change in quantity demanded) / (% change in price of a particular product))

→ Demand sensitivity is referred to the change in demand for a product when its price is changed by a small amount.

Price Elasticity of demand = ((%change in quantity demanded) / (% change in price of a particular product))

Question 50 |

Match the following :

a-iii, b-iv, c-i, d-ii | |

a-iii, b-i, c-iv, d-ii | |

a-i, b-ii, c-iii, d-iv | |

a-iv, b-iii, c-ii, d-i |

Question 50 Explanation:

Call control protocol→ Connection management

A-bis → Interface between Base Transceiver Station(BTS) and Base Station Controller(BSC)

BSMAP → Works between Mobile Switching Centre(MSC) and Base Station Subsystem (BSS)

CDMA → Spread spectrum

A-bis → Interface between Base Transceiver Station(BTS) and Base Station Controller(BSC)

BSMAP → Works between Mobile Switching Centre(MSC) and Base Station Subsystem (BSS)

CDMA → Spread spectrum

There are 50 questions to complete.