## UGC NET CS 2016 Aug- paper-2

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Question 1 |

The Boolean function [~(~p ∧ q) ∧ ~( ~p ∧ ~q)] ∨ (p ∧ r) is equal to the Boolean function:

q | |

p ∧ r | |

p ∨ q | |

p |

Question 1 Explanation:

Question 2 |

Let us assume that you construct ordered tree to represent the compound proposition (~ (p ∧ q)) ↔ (~ p ∨ ~ q) Then, the prefix expression and post-fix expression determined using this ordered tree are given as ____ and _____ respectively.

↔~∧pq∨ ~ ~ pq, pq∧~p~q~∨↔ | |

↔~∧pq∨ ~ p~q, pq∧~p~q~∨↔ | |

↔~∧pq∨ ~ ~ pq, pq∧~p~~q∨↔ | |

↔~∧pq∨ ~ p~ q, pq∧~p~ ~q∨↔ |

Question 2 Explanation:

Step-1: Given compound proposition is

(~(p ∧ q))↔(~ p ∨ ~ q)

It is clearly specifying that

↔ is a root

(~(p ∧ q)) is left subtree

(~p ∨ ~q) is right subtree.

Step-2: Finally the tree is looks like

Step-3: Prefix operation traverse through Root,left and Right ↔~∧pq∨ ~ p~q Step-4: Postfix operation traverse through Left,Right and Root.

pq∧~p~q~∨↔

(~(p ∧ q))↔(~ p ∨ ~ q)

It is clearly specifying that

↔ is a root

(~(p ∧ q)) is left subtree

(~p ∨ ~q) is right subtree.

Step-2: Finally the tree is looks like

Step-3: Prefix operation traverse through Root,left and Right ↔~∧pq∨ ~ p~q Step-4: Postfix operation traverse through Left,Right and Root.

pq∧~p~q~∨↔

Question 3 |

Let A and B be sets in a finite universal set U. Given the following:
|A – B|, |A ⊕ B|, |A| + |B| and |A ∪ B|
Which of the following is in order of increasing size ?

|A – B| ≤ |A ⊕ B| ≤ |A| + |B| ≤ |A ∪ B| | |

|A ⊕ B| ≤ |A – B| ≤ |A ∪ B| ≤ |A| + |B| | |

|A ⊕ B| ≤ |A| + |B| ≤ |A – B| ≤ |A ∪ B| | |

|A – B| ≤ |A ⊕ B| ≤ |A ∪ B| ≤ |A| + |B| |

Question 3 Explanation:

Step-1: Let A and B be sets in a finite universal set U.

Step-2: |A–B| we can also write into |A|-|A ∩ B| and equivalent Venn diagram is

Step-3: |A⊕B| We can also write into |A|+|B|- 2|A∩B| and equivalent Venn diagram is

Step-4: |A| + |B| We can represented into |A| + |B| and equivalent Venn diagram is

Step-5: |A∪B| We can also write into |A|+|B|- |A∩B| and equivalent Venn diagram is

Step-6: |A – B| ≤ |A ⊕ B| ≤ |A ∪ B| ≤ |A| + |B| is correct order.

Step-2: |A–B| we can also write into |A|-|A ∩ B| and equivalent Venn diagram is

Step-3: |A⊕B| We can also write into |A|+|B|- 2|A∩B| and equivalent Venn diagram is

Step-4: |A| + |B| We can represented into |A| + |B| and equivalent Venn diagram is

Step-5: |A∪B| We can also write into |A|+|B|- |A∩B| and equivalent Venn diagram is

Step-6: |A – B| ≤ |A ⊕ B| ≤ |A ∪ B| ≤ |A| + |B| is correct order.

Question 4 |

What is the probability that a randomly selected bit string of length 10 is a palindrome?

1/64 | |

1/32 | |

1/8 | |

1⁄4 |

Question 4 Explanation:

Palindrome is a number that remains the same when its digits are reversed.

Ex: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121...,

Step-1: Here, string length is 10. If we consider first 5 numbers as random choices like 0 or 1.

Remaining 5 numbers are fixed. Total number of possibilities are 2

Step-2: The probability 2

= 1/2

= 1/32

Ex: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121...,

Step-1: Here, string length is 10. If we consider first 5 numbers as random choices like 0 or 1.

Remaining 5 numbers are fixed. Total number of possibilities are 2

^{ 10} . But we are only considering first 5 choices. The probability is 2^{5} .Step-2: The probability 2

^{5} /2^{ 10}= 1/2

^{5}= 1/32

Question 5 |

Given the following graphs:

Which of the following is correct?

Which of the following is correct?

G _{1} contains Euler circuit and G _{2} does not contain Euler circuit. | |

G _{1} does not contain Euler circuit and G _{2} contains Euler circuit. | |

Both G _{1} and G_{ 2} do not contain Euler circuit. | |

Both G _{ 1} and G_{ 2} contain Euler circuit. |

Question 5 Explanation:

Step-1: G1 have odd number of vertices. So, it is not euler circuit.

Step-2: G2 also have odd number of vertices. So, it not euler circuit.

Question 6 |

The octal number 326.4 is equivalent to

(214.2) _{10} and (D6.8))_{ 16} | |

(212.5) _{10} and (D6.8)) _{ 16} | |

(214.5) _{10} and (D6.8)) _{ 16} | |

(214.5) _{10} and (D6.4)) _{ 16} |

Question 6 Explanation:

(326.4)

Step1: First convert given octal no. to binary number because it will be easier to solve this way.

Step 2: Now convert above binary no. into decimal .

(011010110.100)

= ( 1 * 2

= (214.5)

(326.4)

Step 2: Now convert above binary no. into decimal .

(011010110.100)

= ( 1 * 2

= (214.5)

(326.4)

_{8}= ( ?)Step1: First convert given octal no. to binary number because it will be easier to solve this way.

Step 2: Now convert above binary no. into decimal .

(011010110.100)

_{2}= ( 1 * 2

^{7}) + ( 1 × 2^{ 6}) + ( 1 × 2^{ 4}) + ( 1 × 2^{2}) + ( 1 × 2^{ 1}) · [ 1 × ( 1/2) ]= (214.5)

_{10}(326.4)

_{8}= (?)_{ 16}Step 1: Convert given octal no. into binary no.Step 2: Now convert above binary no. into decimal .

(011010110.100)

_{2}= ( 1 * 2

^{7}) + ( 1 × 2^{ 6}) + ( 1 × 2^{ 4}) + ( 1 × 2^{ 2}) + ( 1 × 2^{ 1}) · [ 1 × ( 1/2) ]= (214.5)

_{10}(326.4)

_{ 8}= (?)_{ 16}Question 7 |

Which of the following is the most efficient to perform arithmetic operations on the numbers?

Sign-magnitude | |

1’s complement | |

2’s complement | |

9’s complement |

Question 7 Explanation:

2’s complement has single representation for zero , but Sign-magnitude, 1’s complement and 9’s complement have two representations for 0 (i.e., both positive zero and negative zero).
While doing arithmetic operations like addition or subtraction using 1's complement(or 9's complement), we have to add an extra carry bit, i.e 1 to the result to get the correct answer. 2's complement doesn't require such extra calculation.

Question 8 |

The Karnaugh map for a Boolean function is given as

The simplified Boolean equation for the above Karnaugh Map is

The simplified Boolean equation for the above Karnaugh Map is

AB + CD + AB’ + AD | |

AB + AC + AD + BCD | |

AB + AD + BC + ACD | |

AB + AC + BC + BCD |

Question 8 Explanation:

Question 9 |

Which of the following logic operations is performed by the following given combinational circuit?

EXCLUSIVE-OR | |

EXCLUSIVE-NOR | |

NAND | |

NOR |

Question 9 Explanation:

Question 10 |

Match the following:

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

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

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

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

Question 10 Explanation:

a. Controlled Inverter : A circuit that transmits a binary word or its1’s complement.

b. Full adder : A circuit that can add 3 bits. It adds two data bits(ai,bi) and also carry bit(ci).

c. Half adder : A logic circuit that adds 2 bits(ai,bi).

d. Binary adder : A circuit that can add two binary numbers(A and B). Collection of Full adders form a binary adder.

b. Full adder : A circuit that can add 3 bits. It adds two data bits(ai,bi) and also carry bit(ci).

c. Half adder : A logic circuit that adds 2 bits(ai,bi).

d. Binary adder : A circuit that can add two binary numbers(A and B). Collection of Full adders form a binary adder.

Question 11 |

Given i= 0, j = 1, k = –1, x = 0.5, y = 0.0 What is the output of given ‘C’ expression ?

x * 3 && 3 || j | k

x * 3 && 3 || j | k

-1 | |

0 | |

1 | |

2 |

Question 11 Explanation:

Step-1: Evaluate x * 3 because multiplication has more priority than remaining operators x * 3→ 1.5

Step-2: && is logical AND. Both the statements are TRUE, then it returns 1 otherwise 0.

1.5 && 3 is TRUE. So, it returns 1.

Step-3: j | k is bitwise OR operator. It returns -1.

Step-4: ((x * 3) && 3) || (j | k) islogical OR operator. It returns 1

Note: The precedence is ((x * 3) && 3) || (j | k)

Step-2: && is logical AND. Both the statements are TRUE, then it returns 1 otherwise 0.

1.5 && 3 is TRUE. So, it returns 1.

Step-3: j | k is bitwise OR operator. It returns -1.

Step-4: ((x * 3) && 3) || (j | k) islogical OR operator. It returns 1

Note: The precedence is ((x * 3) && 3) || (j | k)

Question 12 |

The following ‘C’ statement :
int *f [ ] ( );
declares:

A function returning a pointer to an array of integers. | |

Array of functions returning pointers to integers. | |

A function returning an array of pointers to integers. | |

An illegal statement. |

Question 12 Explanation:

int *f [ ] ( ); It declares array of functions returning pointers to integers.

Question 13 |

If a function is friend of a class, which one of the following is wrong?

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

Friend functions are not members of a class, they are associated with it. | |

Friend functions are members of a class. | |

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

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.

TRUE: function can only be declared a friend by a class itself.

TRUE: Friend functions are not members of a class, they are associated with it.

FALSE:Friend functions are members of a class.

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

TRUE: function can only be declared a friend by a class itself.

TRUE: Friend functions are not members of a class, they are associated with it.

FALSE:Friend functions are members of a class.

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

Question 14 |

In C++, polymorphism requires:

Inheritance only | |

Virtual functions only | |

References only | |

Inheritance, Virtual functions and references |

Question 14 Explanation:

→ Polymorphism means "many forms", and it occurs when we have many classes that are related to each other by inheritance.

→ In C++, polymorphism requires inheritance, virtual functions and references

→ In C++, polymorphism requires inheritance, virtual functions and references

Question 15 |

A function template in C++ provides _____ level of generalization.

4 | |

3 | |

2 | |

1 |

Question 15 Explanation:

→ A function template in C++ provides 2 level of generalization.

→ Templates are a feature of the C++ programming language that allows functions and classes to operate with generic types. This allows a function or class to work on many different data types without being rewritten for each one.

→ Templates are a feature of the C++ programming language that allows functions and classes to operate with generic types. This allows a function or class to work on many different data types without being rewritten for each one.

Question 16 |

DBMS provides the facility of accessing data from a database through

DDL | |

DML | |

DBA | |

Schema |

Question 16 Explanation:

The DBMS provides a set of operations or a language called the data manipulation language (DML) for manipulations include retrieval, insertion, deletion, and modification of the data.

Examples of DML:

1. SELECT – is used to retrieve data from a database.

2. INSERT – is used to insert data into a table.

3. UPDATE – is used to update existing data within a table.

4. DELETE – is used to delete records from a database table

(delete one row at a time and can be roll backed)

Examples of DML:

1. SELECT – is used to retrieve data from a database.

2. INSERT – is used to insert data into a table.

3. UPDATE – is used to update existing data within a table.

4. DELETE – is used to delete records from a database table

(delete one row at a time and can be roll backed)

Question 17 |

Relational database schema normalization is NOT for:

reducing the number of joins required to satisfy a query. | |

eliminating uncontrolled redundancy of data stored in the database. | |

eliminating number of anomalies that could otherwise occur with inserts and deletes. | |

ensuring that functional dependencies are enforced. |

Question 17 Explanation:

→ There are many small Relational database schema in the database system. If we want to execute a query then it may require multiple relation access.

→ So to avoid this one solution is to have only one big relation so that number of relation access can be reduced.

→ This solution leads to redundancy of data stored in database and various inert, delete, update anomalies. So the solution of these problems is Normalisation using functional dependency. So option B,C,D are clearly correct according to above explanation.

→ Option A is not correct because normalisation does not reduce the number of joins required to satisfy a query it only tries to eliminate redundancy and anomalous using functional dependency.

→ So to avoid this one solution is to have only one big relation so that number of relation access can be reduced.

→ This solution leads to redundancy of data stored in database and various inert, delete, update anomalies. So the solution of these problems is Normalisation using functional dependency. So option B,C,D are clearly correct according to above explanation.

→ Option A is not correct because normalisation does not reduce the number of joins required to satisfy a query it only tries to eliminate redundancy and anomalous using functional dependency.

Question 18 |

Consider the following statements regarding relational database model:

(a) NULL values can be used to opt a tuple out of enforcement of a foreign key.

(b) Suppose that table T has only one candidate key. If Q is in 3NF, then it is also in BCNF.

(c) The difference between the project operator (Π) in relational algebra and the SELECT keyword in SQL is that if the resulting table/set has more than one occurrences of the same tuple, then Π will return only one of them, while SQL SELECT will return all.

One can determine that:

(a) NULL values can be used to opt a tuple out of enforcement of a foreign key.

(b) Suppose that table T has only one candidate key. If Q is in 3NF, then it is also in BCNF.

(c) The difference between the project operator (Π) in relational algebra and the SELECT keyword in SQL is that if the resulting table/set has more than one occurrences of the same tuple, then Π will return only one of them, while SQL SELECT will return all.

One can determine that:

(a) and (b) are true. | |

(a) and (b) are true | |

(b) and (c) are true | |

(a), (b) and (c) are true |

Question 18 Explanation:

Option(A) is correct. NULL values can be used to opt a tuple out of enforcement of a foreign key.

Option(B) is correct. Suppose that table T has only one candidate key. If Q is in 3NF, then it is also in BCNF.

Option (C) is correct because the main difference between Project operator and SELECT keyword is that SELECT keyword will return Duplicate values if the exist in the result of a query but Project Operator Do not return Duplicate values if the exist in the result of the query.

Option(B) is correct. Suppose that table T has only one candidate key. If Q is in 3NF, then it is also in BCNF.

Option (C) is correct because the main difference between Project operator and SELECT keyword is that SELECT keyword will return Duplicate values if the exist in the result of a query but Project Operator Do not return Duplicate values if the exist in the result of the query.

Question 19 |

Consider the following Entity-Relationship (E-R) diagram and three possible relationship sets (I, II and III) for this E-R diagram:

If different symbols stand for different values (e.g., t1 is definitely not equal to t2 ) , then which of the above could not be the relationship set for the E-R diagram ?

If different symbols stand for different values (e.g., t1 is definitely not equal to t2 ) , then which of the above could not be the relationship set for the E-R diagram ?

I only | |

I and II only | |

II only | |

I, II and III |

Question 19 Explanation:

In the given E-R diagram there is Many-To-Many relationship between entity P and Q it means single value in P can relate to Multiple values in Q and vice-versa. And there is One-To-One
relationship between S and T it means single value in S can relate to single value in T and vice-versa .

In table1: there is One to many relationship between S and T because "s1" in S attribute related with two values "t1" and "t2" in T attribute. so this table is violating One-to-One relationship of S and T as shown in E-R diagram so it can't be the relationship set of given E-R diagram.

In Table 2: There is one to one relationship between S and T and In P and Q also many-to many relationship constraints are not violated. So it could be the relationship set of given E-R diagram.

In Table 3: "p1" in P attribute is related with two values in Q attribute so many to many relationship constraints are not violate here and S and T attributes are also obeying one-to-one relationship constraints. So it could be the relationship set of given E-R diagram.

In table1: there is One to many relationship between S and T because "s1" in S attribute related with two values "t1" and "t2" in T attribute. so this table is violating One-to-One relationship of S and T as shown in E-R diagram so it can't be the relationship set of given E-R diagram.

In Table 2: There is one to one relationship between S and T and In P and Q also many-to many relationship constraints are not violated. So it could be the relationship set of given E-R diagram.

In Table 3: "p1" in P attribute is related with two values in Q attribute so many to many relationship constraints are not violate here and S and T attributes are also obeying one-to-one relationship constraints. So it could be the relationship set of given E-R diagram.

Question 20 |

Consider a database table R with attributes A and B. Which of the following SQL queries is illegal ?

SELECT A FROM R; | |

SELECT A, COUNT(*) FROM R; | |

SELECT A, COUNT(*) FROM R GROUP BY A; | |

SELECT A, B, COUNT(*) FROM R GROUP BY A, B; |

Question 20 Explanation:

The aggregate functions can't be used without Group By clause.

Common aggregate functions include : COUNT,AVG,MAX,MIN,SUM

Common aggregate functions include : COUNT,AVG,MAX,MIN,SUM

Question 21 |

Consider an implementation of unsorted singly linked list. Suppose it has its representation with a head and a tail pointer (i.e. pointers to the first and last nodes of the linked list). Given the representation, which of the following operation can not be implemented in O(1) time ?

Insertion at the front of the linked list. | |

Insertion at the end of the linked list. | |

Deletion of the front node of the linked list. | |

Deletion of the last node of the linked list. |

Question 21 Explanation:

→ Linked list best case complexity is O(1) Possibilities are

1. Insertion at the front of the linked list

2. Insertion at the end of the linked list

3. Deletion of the front node of the linked list

→ Linked list worst case complexity is O(n)

Possibilities are

1. Deletion of the last node of the linked list.

Note: We have to traverse entire linked list to get corresponding location. So, it traversing ‘n’ elements, it takes O(n) time worst case situation.

1. Insertion at the front of the linked list

2. Insertion at the end of the linked list

3. Deletion of the front node of the linked list

→ Linked list worst case complexity is O(n)

Possibilities are

1. Deletion of the last node of the linked list.

Note: We have to traverse entire linked list to get corresponding location. So, it traversing ‘n’ elements, it takes O(n) time worst case situation.

Question 22 |

Consider an undirected graph G where self-loops are not allowed. The vertex set of G is {(i, j) | 1 ≤ i ≤ 12, 1 ≤ j ≤ 12}. There is an edge between (a,b) and (c, d) if |a – c| ≤ 1 or | b–d| ≤ 1. The number of edges in this graph is:

726 | |

796 | |

506 | |

616 |

Question 22 Explanation:

The total number of vertices in the graph is 12*12=144. The vertices are allowed to connect in both horizontal and vertical directions which are separated by at most 1 distance.

If we observe the graph, it looks like a 12 by 12 grid. Each corner vertex has a degree of 3 and we have 4 corner vertices. 40 external vertices of degree 5 and remaining 100 vertices of degree 8. From Handshaking theorem, sum of the degrees of the vertices is equal to the 2*number of edges in the graph.

⇒ (4*3) + (40*5) + (100*8) = 2*E

⇒ 1012=2*E

⇒ E=506

If we observe the graph, it looks like a 12 by 12 grid. Each corner vertex has a degree of 3 and we have 4 corner vertices. 40 external vertices of degree 5 and remaining 100 vertices of degree 8. From Handshaking theorem, sum of the degrees of the vertices is equal to the 2*number of edges in the graph.

⇒ (4*3) + (40*5) + (100*8) = 2*E

⇒ 1012=2*E

⇒ E=506

Question 23 |

The runtime for traversing all the nodes of a binary search tree with n nodes and printing them in an order is:

O(lg n) | |

O(n lg n) | |

O(n) | |

O(n ^{2} ) |

Question 23 Explanation:

__Possibility-1:__

→ Traversing all the nodes of a binary search tree with n nodes and printing them in order using inorder traversal.

→ Inorder traversal will print all elements in sorted order. It takes worst case time complexity is O(n).

__Possibility-2:__

Without sorted elements chance to form either left skewed or right skewed tree. It takes worst case time complexity is O(n).

Question 24 |

Consider the following statements:

S1: A queue can be implemented using two stacks.

S2: A stack can be implemented using two queues.

Which of the following is correct?

S1: A queue can be implemented using two stacks.

S2: A stack can be implemented using two queues.

Which of the following is correct?

S 1 is correct and S 2 is not correct. | |

S 1 is not correct and S 2 is correct. | |

Both S 1 and S 2 are correct. | |

Both S 1 and S 2 are incorrect. |

Question 24 Explanation:

→ Implementing queue by using two stack:

(1) When calling the enqueue method, simply push the elements into the stack 1.

(2) If the dequeue method is called, push all the elements from stack 1 into stack 2, which reverses the order of the elements. Now pop from stack 2.

→ A stack can be implemented using two queues

(1) When calling the enqueue method, simply push the elements into the stack 1.

(2) If the dequeue method is called, push all the elements from stack 1 into stack 2, which reverses the order of the elements. Now pop from stack 2.

→ A stack can be implemented using two queues

Question 25 |

Given the following prefix expression:

* + 3 + 3 ↑ 3 + 3 3 3

What is the value of the prefix expression ?

* + 3 + 3 ↑ 3 + 3 3 3

What is the value of the prefix expression ?

2178 | |

2199 | |

2205 | |

2232 |

Question 25 Explanation:

Insertion start from last element because in prefix, first element is root node.

Note: ↑ is cap symbol.

Note: ↑ is cap symbol.

Question 26 |

Which of the following statements is not true with respect to microwaves?

Electromagnetic waves with frequencies from 300 GHz to 400 Thz. | |

Propagation is line-of-sight. | |

Very high-frequency waves cannot penetrate walls. | |

Use of certain portions of the band requires permission from authorities. |

Question 26 Explanation:

→ Microwaves are a form of electromagnetic radiation with wavelengths ranging from about one meter to one millimeter; with frequencies between 300 MHz (1 m) and 300 GHz (1 mm).

→ The electromagnetic spectrum covers electromagnetic waves with frequencies ranging from below one hertz to above 10 25 hertz, corresponding to wavelengths from thousands of kilometers down to a fraction of the size of an atomic nucleus.

→ The electromagnetic spectrum covers electromagnetic waves with frequencies ranging from below one hertz to above 10 25 hertz, corresponding to wavelengths from thousands of kilometers down to a fraction of the size of an atomic nucleus.

Question 27 |

In a fast Ethernet cabling, 100 Base-TX uses ____ cable and maximum segment size is_____.

twisted pair, 100 metres | |

twisted pair, 200 metres | |

fibre optics, 1000 metres | |

fibre optics, 2000 metres |

Question 27 Explanation:

→ The "100" in the media type designation refers to the transmission speed of 100 Mbit/s, while the "BASE" refers to baseband signalling.

→ The letter following the dash ("T" or "F") refers to the physical medium that carries the signal (twisted pair or fiber, respectively), while the last character ("X", "4", etc.) refers to the line code method used.

→ Fast Ethernet is sometimes referred to as 100BASE-X, where "X" is a placeholder for the FX and TX variants.

→ The letter following the dash ("T" or "F") refers to the physical medium that carries the signal (twisted pair or fiber, respectively), while the last character ("X", "4", etc.) refers to the line code method used.

→ Fast Ethernet is sometimes referred to as 100BASE-X, where "X" is a placeholder for the FX and TX variants.

Question 28 |

A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ?

1 Mbps | |

2 Mbps | |

10 Mbps | |

12 Mbps |

Question 28 Explanation:

Given data,

Bandwidth= 10 Mbps

Frames per minute=12000

Each frame per bits=10000

Throughput=?

Step-1: Here, they given each frame per minute. So, convert into seconds is

(12000*10000) / 60

Step-2: 120000000 / 60

= 2000000

It is nothing but 2Mbps

Bandwidth= 10 Mbps

Frames per minute=12000

Each frame per bits=10000

Throughput=?

Step-1: Here, they given each frame per minute. So, convert into seconds is

(12000*10000) / 60

Step-2: 120000000 / 60

= 2000000

It is nothing but 2Mbps

Question 29 |

Match the following

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

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

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

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

Question 29 Explanation:

Session layer → Manage dialogue control

Application layer → Virtual terminal software

Presentation layer → Semantics of the information transmitted

Transport layer → Flow control

Application layer → Virtual terminal software

Presentation layer → Semantics of the information transmitted

Transport layer → Flow control

Question 30 |

Which of the following protocols is used by email server to maintain a central repository that can be accessed from any machine ?

POP3 | |

IMAP | |

SMTP | |

DMSP |

Question 30 Explanation:

→ IMAP protocol is used by email server to maintain a central repository that can be accessed from any machine.

→ Internet Message Access Protocol (IMAP) is an Internet standard protocol used by email clients to retrieve email messages from a mail server over a TCP/IP connection.

→ IMAP was designed with the goal of permitting complete management of an email box by multiple email clients, therefore clients generally leave messages on the server until the user explicitly deletes them.

→ An IMAP server typically listens on port number 143. IMAP over SSL (IMAPS) is assigned the port number 993.

→ POP3 is designed to delete mail on the server as soon as the user has downloaded it. However, some implementations allow users or an administrator to specify that mail be saved for some period of time. POP can be thought of as a "store-and-forward" service.

→ SMTP is used for connecting to outbound servers to send email while POP3 and IMAP are used to connect to incoming servers to retrieve messages.

→ DMSP is Distributed Mail System for Personal Computers.

→ Internet Message Access Protocol (IMAP) is an Internet standard protocol used by email clients to retrieve email messages from a mail server over a TCP/IP connection.

→ IMAP was designed with the goal of permitting complete management of an email box by multiple email clients, therefore clients generally leave messages on the server until the user explicitly deletes them.

→ An IMAP server typically listens on port number 143. IMAP over SSL (IMAPS) is assigned the port number 993.

→ POP3 is designed to delete mail on the server as soon as the user has downloaded it. However, some implementations allow users or an administrator to specify that mail be saved for some period of time. POP can be thought of as a "store-and-forward" service.

→ SMTP is used for connecting to outbound servers to send email while POP3 and IMAP are used to connect to incoming servers to retrieve messages.

→ DMSP is Distributed Mail System for Personal Computers.

Question 31 |

The number of strings of length 4 that are generated by the regular expression (0

^{ +} 1^{ +} | 2^{+} 3^{+} )*, where | is an alternation character and {+, *} are quantification characters, is:08 | |

09 | |

10 | |

12 |

Question 31 Explanation:

Consider a=0

Hence (0

Now since the min string in 0

Similarly the min string in 2

We require 4 length strings.

using "b" one time only → 2

Using two time, i.e., aa, ab, ba, bb

aa→ 0101

ab→ 0123

ba→ 2301

bb→ 2323

'Hence total 10 strings {0001, 0011, 0111, 2223, 2233, 2333, 0101, 0123, 2301, 2323}

^{+}1^{+}and b=2^{+}3^{+}Hence (0

^{+}1^{+}| 2^{+}3^{+})* = (0^{+}1^{+}+ 2^{+}3^{+})* = (a+b)*Now since the min string in 0

^{+}1^{+}is "01" hence "a" has min length two.Similarly the min string in 2

^{+}3^{+}is "23" , hence "b"has min length two.We require 4 length strings.

^{+}We can generate 4 length strings using "a" only, or using "b" only or using both.^{+}Using "a" one time only → 0^{+}1^{+}only: {0001, 0011, 0111}using "b" one time only → 2

^{+}3^{+}only : {2223, 2233, 2333}Using two time, i.e., aa, ab, ba, bb

aa→ 0101

ab→ 0123

ba→ 2301

bb→ 2323

'Hence total 10 strings {0001, 0011, 0111, 2223, 2233, 2333, 0101, 0123, 2301, 2323}

Question 32 |

The content of the accumulator after the execution of the following 8085 assembly language program, is

MVI A, 35H

MOV B, A

STC

CMC

RAR

XRA B

MVI A, 35H

MOV B, A

STC

CMC

RAR

XRA B

00H | |

35H | |

EFH | |

2FH |

Question 32 Explanation:

MVI A, 35H ← Accumulator A = 35H = 0011 0101

MOV B, A ← B=A;

B = 35H= 0011 0101

STC ← Set the carry flag; Carry C=1

CMC ← Complement the carry flag; Carry C=0

RAR ← Rotate the Accumulator(A) right through Carry C

A C = 0011 0101 0 ← before Rotation

A C = 0001 1010 1 ← After Rotation

XRA B ← A= A ⊕ B

A 0001 1010

B 0011 0101

---------------

A ⊕ B = 0010 1111

--------------

= 2 F

MOV B, A ← B=A;

B = 35H= 0011 0101

STC ← Set the carry flag; Carry C=1

CMC ← Complement the carry flag; Carry C=0

RAR ← Rotate the Accumulator(A) right through Carry C

A C = 0011 0101 0 ← before Rotation

A C = 0001 1010 1 ← After Rotation

XRA B ← A= A ⊕ B

A 0001 1010

B 0011 0101

---------------

A ⊕ B = 0010 1111

--------------

= 2 F

Question 33 |

In compiler optimization, operator strength reduction uses mathematical identities to replace slow math operations with faster operations. Which of the following code replacements is an illustration of operator strength reduction ?

Replace P + P by 2 * P or Replace 3 + 4 by 7. | |

Replace P * 32 by P << 5 | |

Replace P * 0 by 0 | |

Replace (P << 4) – P by P * 15 |

Question 33 Explanation:

Option (A) is not correct because multiplication operation can be performed faster using Left-Shift(<<) operator instead of "+" operator and "3+4=7" is the example of Folding machine independent optimization method instead of operator strength reduction .

Option(B) is correct because here to speedup the multiplication operation, * operator is replaced by Left-Shift operator.

Option(C) is not correct because the method used for compiler optimization is not correct. For given statements to perform compiler optimization, we simply eliminate such statements from the code and this method of elimination is called as Algebraic Simplification.

Option(D) is also not correct because bitwise operator is more faster than multiplication.

Option(B) is correct because here to speedup the multiplication operation, * operator is replaced by Left-Shift operator.

Option(C) is not correct because the method used for compiler optimization is not correct. For given statements to perform compiler optimization, we simply eliminate such statements from the code and this method of elimination is called as Algebraic Simplification.

Option(D) is also not correct because bitwise operator is more faster than multiplication.

Question 34 |

Which of the following are the principles tasks of the linker?

I. Resolve external references among separately compiled program units.

II. Translate assembly language to machine code.

III. Relocate code and data relative to the beginning of the program.

IV. Enforce access-control restrictions on system libraries

I. Resolve external references among separately compiled program units.

II. Translate assembly language to machine code.

III. Relocate code and data relative to the beginning of the program.

IV. Enforce access-control restrictions on system libraries

I and II | |

I and III | |

II and III | |

I and IV |

Question 34 Explanation:

Linker:

A linker or link editor is a computer utility program that takes one or more object files generated by a compiler and combines them into a single executable file, library file, or another 'object' file.

Principles:

1. Resolve external references among separately compiled program units

2. Relocate code and data relative to the beginning of the program.

→ Assembler, Translate assembly language to machine code.

A linker or link editor is a computer utility program that takes one or more object files generated by a compiler and combines them into a single executable file, library file, or another 'object' file.

Principles:

1. Resolve external references among separately compiled program units

2. Relocate code and data relative to the beginning of the program.

→ Assembler, Translate assembly language to machine code.

Question 35 |

Which of the following is FALSE?

The grammar S→aS|aSbS|∈, where S is the only non-terminal symbol, and ∈ is the null string, is ambiguous. | |

An unambiguous grammar has same leftmost and rightmost derivation. | |

An ambiguous grammar can never be LR(k) for any k. | |

Recursive descent parser is a top-down parser. |

Question 35 Explanation:

Option-A: TRUE: S is non terminal symbol, ∈ is the null string, is ambiguous.

Option-B: FALSE: An unambiguous grammar has same leftmost and rightmost derivation.

Option-C: TRUE: An ambiguous grammar can never be LR (k) for any k, because LR(k) algorithm aren’t designed to handle ambiguous grammars. It would get stuck into undecidability problem, if employed upon an ambiguous grammar, no matter how large the constant k is.

Option-D: TRUE: Recursive descent parser is a top-down parser

Option-B: FALSE: An unambiguous grammar has same leftmost and rightmost derivation.

Option-C: TRUE: An ambiguous grammar can never be LR (k) for any k, because LR(k) algorithm aren’t designed to handle ambiguous grammars. It would get stuck into undecidability problem, if employed upon an ambiguous grammar, no matter how large the constant k is.

Option-D: TRUE: Recursive descent parser is a top-down parser

Question 36 |

Consider a system with seven processes A through G and six resources R through W.
Resource ownership is as follows :
process A holds R and wants T
process B holds nothing but wants T
process C holds nothing but wants S
process D holds U and wants S & T
process E holds T and wants V
process F holds W and wants S
process G holds V and wants U
Is the system deadlocked ?
If yes, ______ processes are deadlocked.

No | |

Yes, A, B, C | |

Yes, D, E, G | |

Yes, A, B, F |

Question 36 Explanation:

Visual inspection shows that D, E, and G are deadlocked.

Question 37 |

Suppose that the virtual Address space has eight pages and physical memory with four page frames. If LRU page replacement algorithm is used, _____ number of page faults occur with the reference string.
0 2 1 3 5 4 6 3 7 4 7 3 3 5 5 3 1 1 1 7 2 3 4 1

11 | |

12 | |

10 | |

9 | |

None of the above |

Question 37 Explanation:

Total number of page frames are 13.

None: Actually they given options are wrong. Excluded for evaluation.

Question 38 |

Consider a system having ‘m’ resources of the same type. These resources are shared by three processes P

_{1} , P_{ 2}and P_{3} which have peak demands of 2, 5 and 7 resources respectively. For what value of ‘m’ deadlock will not occur ?70 | |

14 | |

13 | |

7 | |

None of the above |

Question 38 Explanation:

A requires 3, B-4, C-7;

→ If A have 2, B have 3, C have 6 then deadlock will occur i.e., 2+3+6=11.

→ If we have one extra resource then deadlock will not occur i.e., 11+1=12.

→ If we have equal (or) more than 12 resources then deadlock will never occur.

Note: Actual answer given option-B but it satisfies option-A,B,C.

→ If A have 2, B have 3, C have 6 then deadlock will occur i.e., 2+3+6=11.

→ If we have one extra resource then deadlock will not occur i.e., 11+1=12.

→ If we have equal (or) more than 12 resources then deadlock will never occur.

Note: Actual answer given option-B but it satisfies option-A,B,C.

Question 39 |

Five jobs A, B, C, D and E are waiting in Ready Queue. Their expected runtimes are 9, 6, 3,5 and x respectively. All jobs entered in Ready queue at time zero. They must run in _____order to minimize average response time if 3 < x < 5.

B, A, D, E, C | |

C, E, D, B, A | |

E, D, C, B, A | |

C, B, A, E, D |

Question 39 Explanation:

Shortest job first is the way to minimize average response time.

0 < X ≤ 3: X, 3, 5, 6, 9

3 < X ≤ 5: 3, X, 5, 6, 9

5 < X ≤ 6: 3, 5, X, 6, 9

6 < X ≤ 9: 3, 5, 6, X, 9

X > 9: 3, 5, 6, 9, X

C, E, D, B, A

So, option B is correct.

0 < X ≤ 3: X, 3, 5, 6, 9

3 < X ≤ 5: 3, X, 5, 6, 9

5 < X ≤ 6: 3, 5, X, 6, 9

6 < X ≤ 9: 3, 5, 6, X, 9

X > 9: 3, 5, 6, 9, X

C, E, D, B, A

So, option B is correct.

Question 40 |

Consider three CPU intensive processes P1 , P2 , P3 which require 20, 10 and 30 units of time, arrive at times 1, 3 and 7 respectively. Suppose operating system is implementing Shortest Remaining Time first (preemptive scheduling) algorithm, then _____ context switches are required (suppose context switch at the beginning of Ready queue and at the end of Ready queue are not counted).

3 | |

2 | |

4 | |

5 |

Question 40 Explanation:

P

P

P

_{1} → P_{ 2} is one switchP

_{2} → P_{ 1} is 2nd switchP

_{3} → P_{ 3}is 3rd switchQuestion 41 |

Which of the following is used to determine the specificity of requirements?
[ Where n

_{1} is the number of requirements for which all reviewers have identical interpretations, n_{2} is number of requirements in a specification.]n _{1} / n_{ 2} | |

n _{ 2} / n _{1} | |

n _{1} + n _{ 2} | |

n _{1} - n _{ 2} |

Question 41 Explanation:

T he specificity of requirements determines n

n

n

_{1}/ n_{ 2}n

_{1} is the number of requirements for which all reviewers have identical interpretationsn

_{2} is number of requirements in a specificationQuestion 42 |

The major shortcoming of waterfall model is

The difficulty in accommodating changes after requirement analysis. | |

The difficult in accommodating changes after feasibility analysis. | |

The system testing | |

The maintenance of system. |

Question 42 Explanation:

Advantages:

1. Waterfall model is simple and easy to understand.

2. It works as reference model for others.

Disadvantage:

1. Real projects cannot be sequential

2. Initially all requirements should known i.e requirements are frozen. It may be suited for static projects.

3. Error omission too costly

4. Maintenance is too costly(We have to redesign the whole project from scratch)

5. Customer must have patience

6. It is based on Big-Bang approach

1. Waterfall model is simple and easy to understand.

2. It works as reference model for others.

Disadvantage:

1. Real projects cannot be sequential

2. Initially all requirements should known i.e requirements are frozen. It may be suited for static projects.

3. Error omission too costly

4. Maintenance is too costly(We have to redesign the whole project from scratch)

5. Customer must have patience

6. It is based on Big-Bang approach

Question 43 |

The quick design of a software that is visible to end users leads to _____.

iterative model | |

prototype model | |

spiral model | |

waterfall model |

Question 43 Explanation:

The quick design of a software that is visible to end users leads to prototype model.

Advantages:

1. We can develop software where requirements are unclear

2. Customer satisfaction

Disadvantage:

1. Who pay cost of prototype

2. Required the design expertise

3. Throwaway approach (or) evolutionary approach ( In this approach we develop the software by enhancing protype)

Advantages:

1. We can develop software where requirements are unclear

2. Customer satisfaction

Disadvantage:

1. Who pay cost of prototype

2. Required the design expertise

3. Throwaway approach (or) evolutionary approach ( In this approach we develop the software by enhancing protype)

Question 44 |

For a program of k variables, boundary value analysis yields ______ test cases.

4k – 1 | |

4k | |

4k + 1 | |

2 ^{k} – 1 |

Question 44 Explanation:

For a program of k variables, boundary value analysis yields 4k + 1 test cases.

Question 45 |

The extent to which a software performs its intended functions without failures, is termed as__

Robustness | |

Correctness | |

Reliability | |

Accuracy |

Question 45 Explanation:

→ The extent to which a software performs its intended functions without failures, is termed as Reliability.

→ Software reliability is defined as the ability of a system or component to perform its required functions under stated conditions for a specified period of time. It is the probability of a failure free operation of a program for a specified time in a specified environment.

→ Robustness is the ability of a computer system to cope with errors during execution and cope with erroneous input.

→ Correctness of an algorithm is asserted when it is said that the algorithm is correct with respect to a specification.

→ Accuracy of a measurement system is the degree of closeness of measurements of a quantity to that quantity's true value.

→ Software reliability is defined as the ability of a system or component to perform its required functions under stated conditions for a specified period of time. It is the probability of a failure free operation of a program for a specified time in a specified environment.

→ Robustness is the ability of a computer system to cope with errors during execution and cope with erroneous input.

→ Correctness of an algorithm is asserted when it is said that the algorithm is correct with respect to a specification.

→ Accuracy of a measurement system is the degree of closeness of measurements of a quantity to that quantity's true value.

Question 46 |

An attacker sits between the sender and receiver and captures the information and retransmits to the receiver after some time without altering the information. This attack is called as _____.

Denial of service attack | |

Masquerade attack | |

Simple attack | |

Complex attack |

Question 46 Explanation:

→ An attacker sits between the sender and receiver and captures the information and retransmits to the receiver after some time without altering the information. This attack is called as Denial of service attack(DoS).

→ A denial-of-service attack (DoS attack) is a cyber-attack in which the perpetrator seeks to make a machine or network resource unavailable to its intended users by temporarily or indefinitely disrupting services of a host connected to the Internet.

→ A denial-of-service attack (DoS attack) is a cyber-attack in which the perpetrator seeks to make a machine or network resource unavailable to its intended users by temporarily or indefinitely disrupting services of a host connected to the Internet.

Question 47 |

_______ is subject oriented, integrated, time variant, nonvolatile collection of data in support of management decisions.

Data mining | |

Web mining | |

Data warehouse | |

Database Management System |

Question 47 Explanation:

→ Data warehouse is subject oriented, integrated, time variant, nonvolatile collection of data in support of management decisions.

→ A data warehouse (DW or DWH), also known as an enterprise data warehouse (EDW), is a system used for reporting and data analysis, and is considered a core component of business intelligence.

→ DWs are central repositories of integrated data from one or more disparate sources. They store current and historical data in one single place that are used for creating analytical reports for workers throughout the enterprise.

→ A data warehouse (DW or DWH), also known as an enterprise data warehouse (EDW), is a system used for reporting and data analysis, and is considered a core component of business intelligence.

→ DWs are central repositories of integrated data from one or more disparate sources. They store current and historical data in one single place that are used for creating analytical reports for workers throughout the enterprise.

Question 48 |

In Data mining, classification rules are extracted from _______.

Data | |

Information | |

Decision Tree | |

Database |

Question 48 Explanation:

→ Classification rules are extracted from decision tree. The paths from root to leaf represent classification rules.
→ A decision tree is a decision support tool that uses a tree-like model of decisions and their possible consequences, including chance event outcomes, resource costs, and utility.

Question 49 |

Discovery of cross sales opportunities is called as _____.

Association | |

Visualization | |

Correlation | |

Segmentation |

Question 49 Explanation:

→ Association rule learning is a rule-based machine learning method for discovering interesting relations between variables in large databases. It is intended to identify strong rules discovered
in databases using some measures of interestingness.

→ This rule-based approach also generates new rules as it analyzes more data. The ultimate goal, assuming a large enough dataset, is to help a machine mimic the human brain’s feature extraction and abstract association capabilities from new uncategorized data.

→ This rule-based approach also generates new rules as it analyzes more data. The ultimate goal, assuming a large enough dataset, is to help a machine mimic the human brain’s feature extraction and abstract association capabilities from new uncategorized data.

Question 50 |

In Data mining, ______ is a method of incremental conceptual clustering.

STRING | |

COBWEB | |

CORBA | |

OLAD |

Question 50 Explanation:

→ COBWEB is an incremental system for hierarchical conceptual clustering. COBWEB incrementally organizes observations into a classification tree. Each node in a classification tree represents a class (concept) and is labeled by a probabilistic concept that summarizes the attribute-value distributions of objects classified under the node. This classification tree can be used to predict missing attributes or the class of a new object. There are four basic operations COBWEB employs in building the classification tree.

1. Merging Two Nodes

2. Splitting a node

3. Inserting a new node

4. Passing an object down the hierarchy

→ Common Object Request Broker Architecture (CORBA) is an architecture and specification for creating, distributing, and managing distributed program objects in a network. It allows programs at different locations and developed by different vendors to communicate in a network through an "interface broker." CORBA was developed by a consortium of vendors through the Object Management Group (OMG), which currently includes over 500 member companies.

→ STRING and OLAD is irrelevant options.

1. Merging Two Nodes

2. Splitting a node

3. Inserting a new node

4. Passing an object down the hierarchy

→ Common Object Request Broker Architecture (CORBA) is an architecture and specification for creating, distributing, and managing distributed program objects in a network. It allows programs at different locations and developed by different vendors to communicate in a network through an "interface broker." CORBA was developed by a consortium of vendors through the Object Management Group (OMG), which currently includes over 500 member companies.

→ STRING and OLAD is irrelevant options.

There are 50 questions to complete.