## CIL Part - B

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Question 1 |

Which of the following statements is TRUE?

The cardinality ratio for a binary relationship specifies the maximum number of relationship instances that an entity can participate in. | |

The cardinality ratio for a binary relationship specifies the minimum number of relationship instances that an entity can participate in. | |

The partial participation constraint is also called existence dependency. | |

The cardinality ratio for a relationship specifies the average number of relationship instances that an entity can participate in. |

Question 1 Explanation:

**Cardinality**: cardinality ratio for a binary relationship specifies the maximum number of relationship instances that an entity can participate in.

**Participation**: Participation for a binary relationship specifies the minimum number of relationship instances that an entity can participate in.

Question 2 |

Which of the following represents the function of a Multiplexer?

Y= A+ B | |

Y = A | B | |

Y = A & B | |

Y = S ? A : B |

Question 2 Explanation:

**Multiplexer**: It also known as a

**data selector**, is a device that selects between several input signals and forwards it to a single output line. A multiplexer of 2

^{n }inputs has n select lines, which are used to select which input line to send to the output. Here in fourth option “S” is used as select line to select either “A” or “B”.

A multiplexer can be used to implement if-else statements.

Question 3 |

The prefix equivalent of the following infix expression is:
a / b - c + d * e - a * c

- + - / a b c * d e * a c | |

+ - - / a b c * d e * a c | |

- + / a b c - * d e * a c | |

- - + / a b c * d e * a c |

Question 3 Explanation:

a / b - c + d * e - a * c

* Start scanning the expression from RHS. * Push the operators on the top of the stack by ensuring that only high priority operator can be pushed over a low priority operator.

* If a low priority operator is encountered in expression while scanning it, and if top of stack have high priority operator then pop it and after that push low priority operator on top of stack.

* Print each operand just after scanning it.

* Start scanning the expression from RHS. * Push the operators on the top of the stack by ensuring that only high priority operator can be pushed over a low priority operator.

* If a low priority operator is encountered in expression while scanning it, and if top of stack have high priority operator then pop it and after that push low priority operator on top of stack.

* Print each operand just after scanning it.

Question 4 |

Which of the following file type arguments is used in fopen() llibrary function to open a new file for both reading and writing and destroys the file if a file with the specified file-name already exists in the current directory ?

“w” | |

“w+” | |

“a+” | |

“r+” |

Question 4 Explanation:

W+ : Create a text file for both reading and writing. If the given file exists, its contents are cleared unless it is a logical file.

A+ : Open a text file in append mode for reading or updating at the end of the file. The fopen() function creates the file if it does not exist.

W : Create a text file for writing. If the given file exists, its contents are destroyed unless it is a logical file.

A : Open a text file in append mode for writing at the end of the file. The fopen() function creates the file if it does not exist and is not a logical file.

R+ : Open a text file for both reading and writing. The file must exist.

A+ : Open a text file in append mode for reading or updating at the end of the file. The fopen() function creates the file if it does not exist.

W : Create a text file for writing. If the given file exists, its contents are destroyed unless it is a logical file.

A : Open a text file in append mode for writing at the end of the file. The fopen() function creates the file if it does not exist and is not a logical file.

R+ : Open a text file for both reading and writing. The file must exist.

Question 5 |

Which of the following statements about IPv6 is FALSE?

IPSec support is an integral part of IPv6. | |

IPv6 has no way to distinguish delay-sensitive packets from bulk data transfers. | |

In comparison to IPA. IPv6 has improved header structure with less processing overhead. | |

IPv6 routers no longer have to fragment packets. |

Question 5 Explanation:

TRUE: IPSec support is an integral part of IPv6.

TRUE: IPv6 has no way to distinguish delay-sensitive packets from bulk data transfers.

FALSE: In comparison to IPA. IPv6 has improved header structure with less processing overhead.

An independent physician association (IPA) is a business entity organized and owned by a network of independent physician practices for the purpose of reducing overhead or pursuing business ventures such as contracts with employers, accountable care organizations (ACO) and/or managed care organizations (MCOs). There are substantial opportunities for innovation in delivery system modeling and benefit design in the creation of physician networks. Specifically, creation of practice networks involving patient-centered medical home (PCMH) practices may accelerate important and necessary changes in health care delivery.

TRUE: IPv6 routers no longer have to fragment packets.

TRUE: IPv6 has no way to distinguish delay-sensitive packets from bulk data transfers.

FALSE: In comparison to IPA. IPv6 has improved header structure with less processing overhead.

An independent physician association (IPA) is a business entity organized and owned by a network of independent physician practices for the purpose of reducing overhead or pursuing business ventures such as contracts with employers, accountable care organizations (ACO) and/or managed care organizations (MCOs). There are substantial opportunities for innovation in delivery system modeling and benefit design in the creation of physician networks. Specifically, creation of practice networks involving patient-centered medical home (PCMH) practices may accelerate important and necessary changes in health care delivery.

TRUE: IPv6 routers no longer have to fragment packets.

Question 6 |

Which of the following does not represent the total number of multiplications for multiplying four matrices of orders

20*2,2*30, 30*12 and 12*8?

1232 | |

3680 | |

10320 | |

8850 |

Question 6 Explanation:

Optimal Parenthesization is : (A((BC)D))

Optimal Cost is : 1232

Optimal Cost is : 1232

Question 7 |

To schedules are said to be______. If the order of any two conflicting operations is same in both the schedules.

conflict equivalent | |

schema equivalent | |

result equivalent | |

view equivalent |

Question 7 Explanation:

**View-serializability**of a schedule is defined by equivalence to a serial schedule (no overlapping transactions) with the same transactions, such that respective transactions in the two schedules read and write the same data values ("view" the same data values).

**Conflict-serializability**is defined by equivalence to a serial schedule (no overlapping transactions) with the same transactions, such that both schedules have the same sets of respective chronologically ordered pairs of conflicting operations (same precedence relations of respective conflicting operations).

Question 8 |

Which of the following data structures allows both addition and deletion of items from either end?

Double Ended Queue | |

Queue | |

Priority Queue | |

Stack |

Question 8 Explanation:

**Double Ended Queue**: It is abbreviated to

**deque**and it is an abstract data type that generalizes a queue, for which elements can be added to or removed from either the front (head) or back (tail). It is also often called a

**head-tail linked list**

**Stack**: Stack is an abstract data type that serves as a collection of elements, with two principal operations:

->

**push**, which adds an element to the collection, and

->

**pop**, which removes the most recently added element that was not yet removed.

Both PUSH and POP operations are done from a single end.The order in which elements come off a stack gives rise to its alternative name,

**LIFO (last in, first out)**.

**Queue**: a queue is a collection of entities that are maintained in a sequence and can be modified by the addition of entities at one end of the sequence and removal from the other end of the sequence. It allows addition and deletion of items from fixed different ends.

Question 9 |

Considering the relation schemas R (A, B, C, D) and S (C. D. E. F), what will be the degree of the resultant relation of the following Relational Algebra expression. Where “*” represents the”natural join" operation?

3 | |

4 | |

6 | |

5 |

Question 10 |

Which of the following will be the encoding of string “aabaabaca" using Huffman's coding?

110111011001 | |

110111011111 | |

110111111001 | |

110011011011 |

Question 10 Explanation:

Question 11 |

Which of the following is the time complexity of dynamic programming algorithm to compute the Binomial coefficient

^{n}C_{k}?Θ(nk) | |

Θ(k ^{n}) | |

Θ(n ^{k}) | |

Θ(n+k) |

Question 11 Explanation:

Optimal Substructure: The value of C(n, k) can be recursively calculated using following standard formula for Binomial Coefficients.

C(n, k) = C(n-1, k-1) + C(n-1, k)

C(n, 0) = C(n, n) = 1

Time Complexity: O(nk)

Space complexity: O(nk)

C(n, k) = C(n-1, k-1) + C(n-1, k)

C(n, 0) = C(n, n) = 1

Time Complexity: O(nk)

Space complexity: O(nk)

Question 12 |

Which of the following constraints enforces that the value of the primary key cannot be Null?

Key Constraint | |

Foreign key constraint | |

Entity integrity constraint | |

Domain constraint |

Question 12 Explanation:

The entity integrity constraint states that primary key value can't be null. This is because the primary key value is used to identify individual rows in relation and if the primary key has a null value, then we can't identify those rows.

Question 13 |

Two concurrent processes P and Q execute the following code.

Process P;

while(True){

W:_____

print(’0’);print(’0’);

X:______

}

Process Q;

while(True){

Y:_____

print(’1’);print(’1’);

Z:______

}

Given S and T are binary semaphore variables, and P() and V() as standard “wait“ and “signal” functions respectively.

What should be the semaphore operations W,X,Y and Z for the output string: 11001100…?

W=P(T),X=V(S),Y=P(S),Z=V(T),S=1,T=0 | |

W=P(T),X=V(S),Y=P(S),Z=V(T),S=T=1 | |

W=P(T),X=V(T),Y=P(S),Z=V(S),S=T=1 | |

W=P(T),X=V(T),Y=P(S),Z=V(S),S=1, T=0 |

Question 13 Explanation:

Process P will be executed first and then Process Q can be executed next.

At the process P: W=P(T)

X=V(S)

At the process Q: Y=P(S)

Z=V(T)

Here, S=0, T=1 then the process P executes first and then Q, and both can run on process alternate way start with P.

At the process P: W=P(T)

X=V(S)

At the process Q: Y=P(S)

Z=V(T)

Here, S=0, T=1 then the process P executes first and then Q, and both can run on process alternate way start with P.

Question 14 |

The output of the following 'C' language code is:

void main(){

int x=1, i , y=2;

for(i=0;i<10;i++)

{

x<<1;

y=x+i;

}

printf("%d.%d”,x.y);

}

10,11 | |

1,1 | |

10,1
| |

1,10 |

Question 14 Explanation:

Question 15 |

Which of the following protocols is built on client-server architecture and uses separate control and data connections between the client and the server?

FTP | |

SMTP | |

POP | |

TELNET |

Question 15 Explanation:

FTP protocols is built on client-server architecture and uses separate control and data connections between the client and the server

Question 16 |

An organization has a Class B network and wishes to form subnets for 60 departments. The subnet mask would be:

255.255.64.0 | |

255.255.0.0 | |

255.255.252.0 | |

255.255.255.0 |

Question 16 Explanation:

Organization have 60 departments, and to assign 60 subnet we need 6 bits for subnet. In Class B network first two octet are reserved for NID, so we take first 6 bit of third octet for subnets and subnet mask would be 255.255.11111100.00000000 = 255.255.252.0

Question 17 |

Which of the following is an NP -complete problem?

Turing's Halting problem | |

CNF-Satisfiability problem | |

Presburger Arithmetic problem | |

Number of Hamiltonian circuits in a complete graph of ‘n’ vertices with n>2 |

Question 17 Explanation:

CNF-Satisfiability problem belongs to NP-Complete.

Question 18 |

Which of the following is/are FALSE?

I) Operator precedence parser works on ambiguous grammar

II) Top-down parser works on left recursive, unambiguous and deterministic grammar

III) LL(I) is a non-recursive descent parser

IV) CLR(I) is the most powerful parser

Only II | |

I, II, III and IV | |

ll and IV | |

I, III and IV |

Question 18 Explanation:

I) TRUE: Operator precedence parser works on ambiguous grammar

II) FALSE: Top-down parser works on left recursive, unambiguous and deterministic grammar

Ill) TRUE: LL(I) is a non-recursive descent parser

IV) TRUE: CLR(I) is the most powerful parser

II) FALSE: Top-down parser works on left recursive, unambiguous and deterministic grammar

Ill) TRUE: LL(I) is a non-recursive descent parser

IV) TRUE: CLR(I) is the most powerful parser

Question 19 |

The data type of the variable "var1 " declared in the following 'C' language statement is:
unsigned var1:

int | |

double | |

char | |

float |

Question 19 Explanation:

By default unsigned variables will consider integer data type.

Question 20 |

Which of the following problem cannot be solved using greedy approach?

0-1 knapsack | |

Job scheduling | |

Minimum spanning tree | |

Huffman code |

Question 20 Explanation:

Greedy Problems:

1. Job scheduling

2. Minimum spanning tree

3. Huffman code

Dynamic Programming:

1. O/1 knapsack

2. Optimal Binary search tree

3. Matrix chain multiplication

1. Job scheduling

2. Minimum spanning tree

3. Huffman code

Dynamic Programming:

1. O/1 knapsack

2. Optimal Binary search tree

3. Matrix chain multiplication

Question 21 |

The decimal number 395, when converted into binary occupies______binary digits, whereas when it is represented using BCD codes using BCD codes, occupies_____ binary digits.

12; 9 | |

7; 12 | |

9; 12 | |

12; 7 |

Question 21 Explanation:

Decimal Number= (395)

Total 9 digits of binary number required to convert decimal number

_{10}=(110001011)_{2 }=(613)_{8}= (18B)_{16}Total 9 digits of binary number required to convert decimal number

Question 22 |

Match the following.

List I List II

I)Attenuation P) Loss of energy

II)Shannon capacity Q) Changes in shape of the signal

Ill)Nyquist bit rate R) Noisy channel

IV)Distortion S) Noiseless channel

I-P, II-Q,III-R,IV-S | |

I-P, II-R,III-S,IV-Q | |

I-S, II-R,III-Q,IV-P | |

I-Q, II-P,III-S,IV-R |

Question 22 Explanation:

Attenuation → Changes in shape of the signal

Shannon capacity → Noisy channel

Nyquist bit rate → Noiseless channel

Distortion → Loss of energy

Shannon capacity → Noisy channel

Nyquist bit rate → Noiseless channel

Distortion → Loss of energy

Question 23 |

Which of the following relation schema is always in BCNF?

R( A,B,C,D) | |

R( A,B,C) | |

R( A,B,C,D,E) | |

R( A,B) |

Question 23 Explanation:

BCNF Properties:

1. BCNF is the advance version of 3NF. It is stricter than 3NF.

2. A table is in BCNF if every functional dependency X → Y, X is the super key of the table.

3. For BCNF, the table should be in 3NF, and for every FD, LHS is super key.

4. Two (or) Binary tuples are always be in BCNF

1. BCNF is the advance version of 3NF. It is stricter than 3NF.

2. A table is in BCNF if every functional dependency X → Y, X is the super key of the table.

3. For BCNF, the table should be in 3NF, and for every FD, LHS is super key.

4. Two (or) Binary tuples are always be in BCNF

Question 24 |

Let k

_{2,2}be a complex bipartite graph given below. Which of the following is the total number of paths of length 3 from vertex 1 to vertex 4?

2 | |

3 | |

4 | |

1 |

Question 25 |

Consider the list of numbers 1, 2, 3, ..., 1000 is stored in a[0..999]. What will be the total number of comparisons to search x = 501 using the following binary_search() function?

int binary_search(int a[ ], int n, int x)

{

int low=0, high=n-1;

while(low <= high)

{

int m = (low + high) /2:

if(x > a[m])

low = m+1;

else if(x < a[m])

high= m-1:

else

return m;

}

return -1;

}

2 | |

15 | |

17 | |

1 |

Question 26 |

Match the following .

A-III, B-IV, C-I, D-II | |

A-II, B-I, C-IV, D-III | |

A-III, B-I, C-II, D-IV | |

A-II, B-IV, C-I, D-III |

Question 26 Explanation:

Moves suspended process to secondary storage → Medium Term Scheduler

Loades the process into memory for execution → Long Term Scheduler

Moves one of the process to running time → Short Term Scheduler

Allocate CPU to a process → Dispatcher

Question 27 |

Which of the following is the Postorder traversal of the binary tree whose tree Inorder and Preorder traversals are as follows?

In-order : OLCHJBEKNGMADFI

Preorder: KHLOCBJEAGNMFDI

ADFLIBNCJEMGHOK | |

OCLJEBGDIAFNMHK | |

ADFLIBNMGJCEHOK | |

OCLJEBHNMGDIFAK |

Question 27 Explanation:

Inorder: OLCHJBEKNGMADFI

Preorder: KHLOCBJEAGNMFDI

→ Inorder = Left_child, Root, Right_child

Preorder = Root, Left_child, Right_child

Postorder = Left_child, Right_child, Root

→ Since in Pre-order ‘K’ is at 1st position so ‘K’ is the root of the Binary tree.

→ And in In-order if you will see then OLCHJBE are in its LHS and NGMADFI are in its RHS.

Question 28 |

If p is a two-dimensional array having 10 and 20 columns, then which of the following cannot be used to access the element in row 2 and column 5?

P[2][5] | |

*(*(P+2)+5) | |

*(P[2]+5)
| |

*(P+2+5) |

Question 28 Explanation:

__Option (A)__:

P[2][5] represents the element present in column five of row two. So it can be used to access the required given element in question.

__Option (B)__:

*(*(P+2)+5)

*(P+2) → will lead to enter row two.

(*(P+2)+5) will give address of element present at fifth column of row 2.

→ *(*(P+2)+5) will return element present at fifth column of row 2.

Hence option (B) is Correct.

__Option (C)__: *(P[2]+5)

P[2] → will give starting address of row 2

P[2]+5 → will give address of element present at 5th column of row 2.

*(P[2]+5) → will return the element present at 5th column of row 2.

Hence it is correct option.

__Option (D)__: *(P+2+7)

= (P+7)

*(P+7) will return starting address of 7th row.

→ Hence it is not the correct answer.

Question 29 |

Given total number of instances of a resource to be 18, three processes (Pl, P2, and P3) and the resource requirement and allocation table are given below.

Which of the following orders of process execution forms a safe (deadlock free)

Which of the following orders of process execution forms a safe (deadlock free)

< P1,P2,P3> | |

< P3,P1,P2> | |

< P1,P3,P2> | |

< P2,P1,P3> |

Question 29 Explanation:

Question 30 |

What will be the "First" and "Follow" of E and F for the following grammar?

E->TE’

E’->+TE’/ε

T->FT’

T’->*FT’/ε

F->id/(E)

E->TE’

E’->+TE’/ε

T->FT’

T’->*FT’/ε

F->id/(E)

First(E)={id, (,ε},follow(E)={ε,) }, First(F)={id,),$}, Follow(F)={*,$,(} | |

First(E)={id, ( },follow(E)={$,) }, First(F)={id,(}, Follow(F)={*,$,),+} | |

First(E)={id, (,ε},follow(E)={ε,) }, First(F)={id,)}, Follow(F)={*,$,(,+ } | |

First(E)={id, )},follow(E)={$,) }, First(F)={id,(,$}, Follow(F)={*,$,),+} |

Question 30 Explanation:

First (∊) = { id, ( }

Follow (∊) = { $, ) }

First (F) = { id, ( }

Follow (F) = { *, +, $, ) }

Question 31 |

Which of the following recurrence relation can be solved using Master theorem?

T(n) = 64T(n/8) - n | |

T(n) =T(n/2) + n/(log n) | |

T(n) =2 ^{n}T(n/2) + n | |

T(n) =2T(n/2) + 1 |

Question 31 Explanation:

Masters theorem is in the form of aT(n/b)+n^k log^p n

FALSE: T(n) = 64T(n/8) - n : In masters theorem will not support negative sign recurrence relation (64T(n/8) - n ).

FALSE: T(n) =T(n/2) + n/(log n) it is violated “non polynomial difference”. But we can write it n(log n) into n^-1 logn but it violates non polynomial difference.

FALSE: In masters theorem “a” value must be greater than 1. But here they given exponential (2^n).

TRUE: T(n) =2T(n/2) + 1

a=2, b=2 , k=0 and p=0

a>b^k

=O(n^ log_b ^a n)

= O(n)

FALSE: T(n) = 64T(n/8) - n : In masters theorem will not support negative sign recurrence relation (64T(n/8) - n ).

FALSE: T(n) =T(n/2) + n/(log n) it is violated “non polynomial difference”. But we can write it n(log n) into n^-1 logn but it violates non polynomial difference.

FALSE: In masters theorem “a” value must be greater than 1. But here they given exponential (2^n).

TRUE: T(n) =2T(n/2) + 1

a=2, b=2 , k=0 and p=0

a>b^k

=O(n^ log_b ^a n)

= O(n)

Question 32 |

Which of the following statements is TRUE for the grammar given below?

S->(L)/a

L->L.S/S

The grammar can be parsed by LR(0) parser only | |

The grammar can be parsed by LR(0) and SLR(1) parsers
| |

The grammar can be parsed by LL(1) parser only | |

The grammar can be parsed by LL(1) and LR(0) parsers |

Question 32 Explanation:

→ The given grammar can be parsed by LR(0) grammar because there is no S-R conflict or R-R conflict.
→ Since the grammar can be parsed by LR(0) parser hence we can say SLR(1) parser can also parse it.

Question 33 |

Match the following

I-C,II-A,III-B | |

I-C,II-B,III-A | |

I-A,II-B,III-C | |

I-B,II-C,III-A |

Question 33 Explanation:

Question 34 |

The number of tokens in the following

“C” language statement is:

printf(“The number of tokens are %d”, &tcount);

8 | |

9 | |

10 | |

11 |

Question 34 Explanation:

Question 35 |

Suppose that we have an ordered file with r = 30000 records stored on a disk with block size B = 1024 bytes. If file records are of fixed size and are unspanned with record length R = 100 bytes, the blocking factor for the file and the number of blocks needed for the file are __________ and _________ respectively

110, 3000 | |

10, 3000 | |

110, 30000 | |

10, 300ly. |

Question 35 Explanation:

File size = 30,000 records
Block size = 1024 Bytes
Record size = 100 Bytes
No. of records that can be stored in one Block =1024B/100B
=⎣10.24⎦
=10 records
(Note: We take floor because file records are given as unspanned.)
→ No. of Blocks needed for file =
10 records requires ------- 1 Block
1 record requires ------- 1/10 Block
30,000 records ------- 30000/10 Block = 3000 Blocks
Hence option B is correct.

Question 36 |

if L ={a,b,c} is a language over the set A={a,b,c). Then L3 is:

{ababc,abcab, abc ^{2}, cabab, cabc, c^{2}ab, c^{3}} | |

{ababab,ababc, abcab, abc ^{2}, cabab, c^{2}ab, c^{3}} | |

{ababab,ababc, abcab, abc ^{2}abab, cabab, cabc, c^{2}ab, c^{3}}
| |

{ababab,ababc, abcab, abc ^{2}, cabab, cabc, c^{2}ab, c^{3}} |

Question 37 |

Consider the following Adjacency matrix corresonding to some weighted Graph ‘G’.

What is the weight of the minimum spanning tree for the graph ‘G’?

11 | |

9 | |

10 | |

8 |

Question 37 Explanation:

Question 38 |

Which of the following is used to specify whether the existence of an entity depends on its being related to another entity via the relationship type?

Entity integrity constraint | |

Foreign key constraint | |

Cardinality ratio | |

Participation constraint |

Question 39 |

Logical data independence is the ability to change________schema without having to change_____ Schema

internal; external | |

conceptual; internal | |

conceptual; external | |

internal; conceptual |

Question 39 Explanation:

The three-schema approach provides for three types of schemas with schema techniques based on formal language descriptions:

→External schema for user views

→Conceptual schema integrates external schemata

→Internal schema that defines physical storage structures

→External schema for user views

→Conceptual schema integrates external schemata

→Internal schema that defines physical storage structures

Question 40 |

Which of the following addressing modes is more appropriate for accessing elements of an array?

Index mode | |

Auto increment mode | |

Displacement mode | |

Register mode |

Question 40 Explanation:

→The address of the operand is obtained by adding to the contents of the general register (called index register) a constant value.

→The number of the index register and the constant value are included in the instruction code.

→Index Mode is used to access an array whose elements are in successive memory locations. The content of the instruction code, represents the starting address of the array and the value of the index register, and the index value of the current element.

→By incrementing or decrementing index register different element of the array can be accessed.

→The number of the index register and the constant value are included in the instruction code.

→Index Mode is used to access an array whose elements are in successive memory locations. The content of the instruction code, represents the starting address of the array and the value of the index register, and the index value of the current element.

→By incrementing or decrementing index register different element of the array can be accessed.

Question 41 |

Which of the following ‘C’ language arithmetic expressions has logical error?

-13 % -5 + 3; | |

4 / (-10 % -2) / 3; | |

3 / (-13 % -5) / 3; | |

-5 % 3 / 13; |

Question 41 Explanation:

→In the option B, -10%-2 gives remainder of “0” and then expression becomes 4/0/3.

→4/0 gives divide by zero error.

→4/0 gives divide by zero error.

Question 42 |

Match the following

I-D, II-A, III-B.IV-C | |

I-C, II-A, III-D.IV-B | |

I-B, II-A, III-D.IV-C | |

I-B, II-D, III-A.IV-C |

Question 42 Explanation:

Question 43 |

Which of the following statements is FALSE?

Johnson counter is a synchronous counter | |

Ripple counter is an asynchronous counter. | |

Asynchronous counters are slower than synchronous counters. | |

A counter may count up or count down, but cannot count both up and down. |

Question 43 Explanation:

→A counter is a device which stores (and sometimes displays) the number of times a particular event or process has occurred, often in relationship to a clock.

→Counters are of two types depending upon clock pulse applied. These counters are: Asynchronous counter and Synchronous counter.

→In Asynchronous Counter { also known as Ripple counter} different flip flops are triggered with different clock, not simultaneously!

→While in Synchronous Counter, all flip flops are triggered with same clock simultaneously ; Synchronous Counter is faster than Asynchronous counter in operation.

→

→Synchronous Counter is also called Serial Counter.

→Synchronous Counter will operate in any desired count sequence.

→In Synchronous Counter designing as well implementation are complex due to increasing the number of states.

→Synchronous Counter examples are: Ring counter , Johnson Counter, etc.

→

→Asynchronous Counter produces decoding error.

→Asynchronous Counter is also called Parallel Counter.

→Asynchronous Counter will operate only in fixed count sequence (UP/DOWN).

→In Asynchronous Counter designing as well as implementation is very easy.

→Asynchronous Counter examples are: Ripple UP counter, Ripple DOWN counter, etc.

→Counters are of two types depending upon clock pulse applied. These counters are: Asynchronous counter and Synchronous counter.

→In Asynchronous Counter { also known as Ripple counter} different flip flops are triggered with different clock, not simultaneously!

→While in Synchronous Counter, all flip flops are triggered with same clock simultaneously ; Synchronous Counter is faster than Asynchronous counter in operation.

→

**Synchronous Counter**:- →Synchronous Counter does not produce any decoding errors.→Synchronous Counter is also called Serial Counter.

→Synchronous Counter will operate in any desired count sequence.

→In Synchronous Counter designing as well implementation are complex due to increasing the number of states.

→Synchronous Counter examples are: Ring counter , Johnson Counter, etc.

→

**Asynchronous Counters**:-→Asynchronous Counter produces decoding error.

→Asynchronous Counter is also called Parallel Counter.

→Asynchronous Counter will operate only in fixed count sequence (UP/DOWN).

→In Asynchronous Counter designing as well as implementation is very easy.

→Asynchronous Counter examples are: Ripple UP counter, Ripple DOWN counter, etc.

Question 44 |

Consider the following relation schema R and S along with their tuple sets.

R(A, B) = {

S(A) = {a1, a2, a3}

What is the value of TR / S. where "/" represents the Relational Algebra “division" operation?

T(B) = {b1, b3} | |

T(B) = {b1, b2, b4} | |

T(B) = {b1, b4} | |

T(B) = {b1, b3, b4} |

Question 44 Explanation:

Question 45 |

Considering the following key using a block of five characters, encryption of the message "NETWORKING" using the Transposition Cipher is:

Plaintext : 5 4 3 2 1

Ciphertext 1 2 3 4 5

GNIKROWTEN | |

OGWNTIEKNR | |

OWTENGNIKR | |

NREKTIWNOG |

Question 45 Explanation:

→A transposition cipher is a method of encryption by which the positions held by units of plaintext (which are commonly characters or groups of characters) are shifted according to a regular system, so that the ciphertext constitutes a permutation of the plaintext. That is, the order of the units is changed (the plaintext is reordered).

→According to given question ,

→Plaintext : 5 4 3 2 1

→Ciphertext 1 2 3 4 5

→Given message "NETWO RKING".

→The message is divided into two equal messages of 5 characters length.

→According to cipher text, The reverse of 5 characters of message is OWTEN GNIKR

→According to given question ,

→Plaintext : 5 4 3 2 1

→Ciphertext 1 2 3 4 5

→Given message "NETWO RKING".

→The message is divided into two equal messages of 5 characters length.

→According to cipher text, The reverse of 5 characters of message is OWTEN GNIKR

Question 46 |

If every production is of the form where or of the form , then the grammer is said to be of:

Type 3
| |

Type 1 | |

Type 0 | |

Type 2 |

Question 47 |

Assume that source S and destination D are connected through five intermediate routers R1, R2,R3,R4 and R5. Determine how many times each packet has to visit the Network layer and the Data Link layer during transmission from S to D?

Network Layer: 7 times, Data Link Layer: 12 times | |

Network Layer: 5 times, Data Link Layer: 12 times | |

Network Layer: 12 times, Data Link Layer: 12 times | |

Network Layer: 7 times, Data Link Layer: 7 times |

Question 47 Explanation:

The packet will visit network layer 7 times(5 times at router and 1 time at source and another time at destination) , once at each node [S, R, R,R,R,R, D] and packet will visit Data Link layer 12 times. One time at S and one time at D, then two times for each intermediate router R[5*2=10] as data link layer is used for link to link communication.

Question 48 |

Which of the following statements about the “DELETE” command is FALSE?

It removes tuples from a relation (table). | |

It removes tuples as well as the relation (table). | |

A missing WHERE clause specifies that all tuples in the relation are to be deleted. | |

Depending on the number of tuples selected by the condition in the WHERE clause, zero, one or several tuples can be deleted by a single DELETE command. |

Question 48 Explanation:

→The DELETE statement is used to delete existing records in a table.
→DELETE Syntax
DELETE FROM table_name WHERE condition;

Question 49 |

Given the Burst Time (BT) of 4 processes P1, P2, P3 and P4 as BT (t1,t2,t3,t4)=(4,8,6,7), smoothening factor()=0.5, and T1=10, what will be the burst time of process P5 for shortest job first scheduling, using the technique of exponential averaging ?

6.875 unit time | |

6.25 unit time | |

5.785 unit time | |

5.5 unit time |

Question 49 Explanation:

Question 50 |

In order to sort list of numbers using radix sort algorithm, we need to get the individual digits of each number ‘n’ of the list. If n is a positive decimal integer, then ith digit , from right , of the number n is:

Question 51 |

Which of the following statements about “total specialization” constraint is TRUE?

Every entity in the subclass must be a member of the superclass. | |

Every entity in the superclass must be a member of at least one subclass in the specialization. | |

At least one entity in the superclass must be a member of at least one subclass in the specialization. | |

Every entity in the superclass must be a member of all subclass in the specialization. |

Question 51 Explanation:

→The total specialization rule demands that every entity in the superclass belong to some subclass.
→The partial specialization rule allows an entity to not belong to any of the subclasses

Question 52 |

In a B -Tree of order m(m>1). every non -leaf node (except root node) has the number of children between:

m/2 and m | |

|m/2| and m | |

[m/2] and m | |

[m/2] and |m/2| |

Question 53 |

Which of the following statements about Secure Shell Protocol (SSH) is FALSE?

SSH cannot be used for file transfer and e-mail tasks. | |

SSH provides a secure client/server communication. | |

SSH runs on top of the TCP/IP layer. | |

SSH is a network protocol. |

Question 53 Explanation:

→Secure Shell (SSH) is a cryptographic network protocol for operating network services securely over an unsecured network.
→ Typical applications include remote command-line, login, and remote command execution, but any network service can be secured with SSH.

Question 54 |

Which of the following statements is INCORRECT with respect to pointers declared in the following ‘C’ language code?

Void main()

Int a[10], *p, *q;

p=&a[5];

q=&a[7];

}

Q-p | |

P+1 | |

Q-3 | |

P+q |

Question 54 Explanation:

→Pointers arithmetic operation is not possible.
→Here p and q are two pointers , which consists of two addresses , adding addresses is invalid.

Question 55 |

Considering the process details given in the following table, what will be the completion order of the processes under the policies FCFS and RR (with CPU quantum of 2 time units)?

FCFS: P1, P2, P3 and RR: P1, P2, P3 | |

FCFS: P3, P2, P1 and RR: P3, P1, P2 | |

FCFS: P3, P2, P1 and RR: P3, P2, P1 | |

FCFS: P1, P3, P2 and RR: P1, P2, P3 |

Question 55 Explanation:

Question 56 |

Which of the following represents the language over the set A={a,b} consisting of all words beginning with one or more a’s and followed by the same number of b’s?

L= {a,ab,ab ^{2},...} | |

L= {a ^{m}b^{n}:m,m>0,n>0} | |

L= {a ^{m}b^{m}:m>0} | |

L= {b ^{m}ab^{n}:m0,n0} |

Question 56 Explanation:

→In the option C, a

→Example :

m=1 ⇒ ab

m=2⇒ aabb and so on.

^{m}b^{m}:m>0, For a given “m” value always it generates one or more “a” s followed by Same number of “b” s.→Example :

m=1 ⇒ ab

m=2⇒ aabb and so on.

Question 57 |

Which of the following is the every -case time complexity of dynamic programming algorithm for “traversing salesperson” problem to traverse 'n’ cities?

Θ(log n2 ^{n}) | |

Θ(n2 ^{n}) | |

Θ(2 ^{n}) | |

Θ(n ^{2}2^{n}) |

Question 57 Explanation:

→ Time complexity of Brute Force technique is O(n!)

→ Time complexity of Dynamic Programming Technique is O(n

→ Space complexity of Dynamic Programming Technique is O(n * 2

→ Time complexity of Dynamic Programming Technique is O(n

^{2}* 2^{n})→ Space complexity of Dynamic Programming Technique is O(n * 2

^{n})Question 58 |

Which of the following statements related to Cache memory organization is FALSE?

In "write through" approach. main memory content is always invalid. | |

In "write back" approach, updates are made only in the cache and it minimizes memory writes. | |

Least Recently Used (LRU) replacement algorithm can be used in associative and set associative mappings.
| |

For direct mapping, no replacement algorithm is needed. |

Question 58 Explanation:

→Write through is a storage method in which data is written into the cache and the corresponding main memory location at the same time.

→The cached data allows for fast retrieval on demand, while the same data in main memory ensures that nothing will get lost if a crash, power failure, or other system disruption occurs.

→The cached data allows for fast retrieval on demand, while the same data in main memory ensures that nothing will get lost if a crash, power failure, or other system disruption occurs.

Question 59 |

If D={R1,R2} is a decomposition of R, and F is the set of functional dependencies on R, then which of the following ensures that the decomposition D is lossless (nonadditive)?

((R1∩ R2) →(R1-R2)) ∈ F ^{+} | |

((R1 ∩R2) → (R1 - R2)) ∉ F ^{+} | |

((R1 ∪ R2) →(R1 - R2)) ∈ F ^{+} | |

((R1∩R2) →R1 ) ∈ F ^{+} |

Question 59 Explanation:

Question 60 |

The 9’s compliment of the decimal number 139452 is:

971658 | |

971659 | |

860547 | |

860548 |

Question 60 Explanation:

→To obtain the 9’s complement of any number we have to subtract the number with (10n – 1) where n = number of digits in the number, or in a simpler manner we have to divide each digit of the given decimal number with 9

→Subtract each and every digit of given number by digit “9”.

999999

139452

________

860547

→Subtract each and every digit of given number by digit “9”.

999999

139452

________

860547

Question 61 |

A flip-flop has a 20-nano second delay from the time its CP input goes from 1 to 0 to the time uses these flip-flops?

20 ns | |

320 ns | |

36 ns | |

16 ns |

Question 61 Explanation:

Delay for one flip-flop = 20ns
Since in question they are asking about the number of flip-flops(i.e.n>1).
So the delay for n- flip flops will be a multiple of 20 and “n” will be greater than 1.
If you will see the given options then option (B) is correct because 320 is a multiple of 20(i.e. 20* 16 = 320) and n =16(i.e. n>1)

Question 62 |

Let L be the language on A={a,b,c} which consists of all words of the form w=a

^{r}b^{s}c^{t}where r,s,t>0. Which of the following is the valid regular expression ‘r’ such that L=L(r)?r= a*b*c | |

r= aa*bb*c | |

r= aa*b*cc | |

r= aa*bb*cc* |

Question 62 Explanation:

→From the given question, w=a

→Substitute in r=s=t=1 in w then L= abc

→So option D gives the regular expression abc

^{r}b^{s}c^{t}where r,s,t>0 which means that the minimum value is r=s=t=1,→Substitute in r=s=t=1 in w then L= abc

→So option D gives the regular expression abc

Question 63 |

Match the following.

I-Q, II-S, III-R, IV-P
| |

I-S, II-Q, III-P, IV-R | |

I-S, II-Q, III-R, IV-P | |

I-Q, II-S, III-P, IV-R |

Question 64 |

The listing of nodes after applying the preorder traversal over the following binary tree is:

A,B,D,H,I,E,J,K,C,F,G,L | |

H,D,I,B,J,E,K,A,L,F,C,G | |

H,I,D,J,K,E,B,L,F,G,C,A | |

A,B,D,H,I,E,J,K,C,F,L,G |

Question 64 Explanation:

In the Pre-order traversal , the nodes are traversed in the order “root-left-right”.

So option-D is correct.

So option-D is correct.

Question 65 |

Consider the following relation schema R along with the tuples.

Employee(name, salary) ={< e1, 10000>,< e2, 5000>,< e3, 2500>, < e4, 7500>,< e5, 8900>,< e6, 9800>}

What is the output of following SQL query?

SELECT name , MAX( salary) FROM Employee WHERE salary<( SELECT MAX (salary) FROM Employee;

< e3, 2500> | |

< e5, 8900> | |

< e6, 9800> | |

< e1, 10000> |

Question 65 Explanation:

The above query is to find the employee whose salary is the second highest

Question 66 |

Maximum number of edges in a simple graph with ‘n’ vertices and ‘k’ components is:

[(n-2)(n-1)/2]+1 | |

[(n-k)(n-k+1)]/2 | |

n ^{2} | |

[(n)(n-1)]/2 |

Question 66 Explanation:

A simple graph with n vertices and k components can have at most have (n-k)(n-k+1)/2 edges.

Question 67 |

The number of elements in a one-dimensional array with lowest and highest index values as -1024 and 1024 is:

2048
| |

0 | |

1024 | |

2049 |

Question 67 Explanation:

→The number of elements in [-n:n] is n+n+1=2n+1

→Example:[-2:2] consists of -2,-1,0,1,2 indexes.

→Example:[-2:2] consists of -2,-1,0,1,2 indexes.

Question 68 |

Total number of spanning tree of a complete graph of 4 vertices k

_{4}is:15 | |

16 | |

3 | |

17 |

Question 68 Explanation:

The number of spanning trees of Kn is n

^{n-2}= n^{4-2}=n^{2}=4^{2}=16Question 69 |

Which of the following relationships holds for non-random-access memory, where T

_{n}= average time to read or write n bits, T_{a}= average access time, n=number of bits and r=transfer rate in bits per second?T _{a}=T_{n}+ n/r | |

T _{n}=T_{a}+ n/r | |

T _{a}=T_{n}+ r/n | |

T _{n}=T_{a}+ r/n |

Question 69 Explanation:

Question 70 |

Match the following.

I-B,II-C,III-D,IV-A | |

I-C,II-B,III-D,IV-A | |

I-A,II-D,III-B,IV-C | |

I-B,II-A,III-D,IV-C |

Question 70 Explanation:

The standard RAID levels comprise a basic set of RAID (redundant array of independent disks) configurations that employ the techniques of striping, mirroring, or parity to create large reliable data stores from multiple general-purpose computer hard disk drives (HDDs). The most common types are RAID 0 (striping), RAID 1 (mirroring) and its variants, RAID 5 (distributed parity), and RAID 6 (dual parity).

RAID levels and their associated data formats are standardized by the Storage Networking Industry Association (SNIA) in the Common RAID Disk Drive Format (DDF) standard

RAID levels and their associated data formats are standardized by the Storage Networking Industry Association (SNIA) in the Common RAID Disk Drive Format (DDF) standard

Question 71 |

Consider an information exchange scenario where Anthony is the sender and Bond is the intended recipient of the data.

Match the following appropriately.

I-C, II-B, III-D, IV-A | |

I-C, II-D, III-A, IV-B | |

I-C, II-D, III-B, IV-A | |

I-A, II-D, III-B, IV-C |

Question 71 Explanation:

Message Authentication: Bond needs to be sure of anthony’s identity and that an imposter has not sent the message.

Message confidentiality: The transmitted message must make sense to only bond and to all others it must be garbage.

Message Integrity: The message must arrive at the bond’s side exactly as it was sent. Message Non Repudiation: Anthony must not be able to deny sending a message that he or she in fact, did send

Options C is correct.

Message confidentiality: The transmitted message must make sense to only bond and to all others it must be garbage.

Message Integrity: The message must arrive at the bond’s side exactly as it was sent. Message Non Repudiation: Anthony must not be able to deny sending a message that he or she in fact, did send

Options C is correct.

Question 72 |

If the address of the operand is embedded in the instruction code itself, then the addressing mode is termed as:

Register mode | |

Direct mode | |

Displacement mode | |

Immediate mode |

Question 72 Explanation:

In direct mode, the effective address of operand is present in the instruction itself.

Single memory reference to access data.

No additional calculations to find the effective address of the operand.

Single memory reference to access data.

No additional calculations to find the effective address of the operand.

Question 73 |

Consider the process details given in the following table.

A new pre-emptive scheduling algorithm is proposed, i.e., the Longest Remaining Time Next (LRTN), wherein ties are broken by giving priority to the process with the highest priority

.

Calculate the following.

i. Turnaround time for Process P2

ii. Response time for Process P3

(i)20, (ii)6 | |

(i)19, (ii)3 | |

(i)19, (ii)8 | |

(i)20, (ii)7 |

Question 73 Explanation:

Question 74 |

Under normal circumstances, the cardinality ratio of the binary relationship “Write” relating “Author” and “Book” entities is:

1 : N | |

N : 1 | |

1 : 1 | |

M : N |

Question 74 Explanation:

The relationship WRITES shows that one author can write many books and one book can be written by many authors. So the cardinality ratio is M:N

Question 75 |

If T is a binary tree with N nodes, then the number of levels is at least:

|log _{2} (N+1) | |

N-1 | |

N | |

[log _{2} (N+1)] |

Question 75 Explanation:

Question 76 |

Which of the following declares ‘pf’ as a pointer to a function, which returns an integer quantity and requires two integer arguments ?

int *pf(int, int); | |

int (*pf)(int, int); | |

(int *) pf(int, int); | |

int ( int *pf(int, int)); |

Question 76 Explanation:

int (*pf)(int, int) , Here ‘pf’ declares as a pointer to a function, which returns an integer quantity and requires two integer arguments

Question 77 |

The output of the following ‘c’ language code is :

Void main(){

Char arr[10];

arr=”world”;

printf(“%s”, arr);

world | |

World5 | |

World10 | |

L-value required error |

Question 77 Explanation:

→Here “arr” is starting address of the array.

→We can’t store array values by using the syntax , arr=”world”.

→The above syntax will give error.

→We can’t store array values by using the syntax , arr=”world”.

→The above syntax will give error.

Question 78 |

Interrupt generated due to which of the following operations does not belong to program-related interrupt category?

Division by zero | |

Attempt to execute an illegal machine instruction | |

Memory parity error | |

Reference outside a user’s allowed memory space |

Question 78 Explanation:

Memory parity errors can be caused by failing hardware or outside elements disrupting how computer memory functions.

Fixing parity errors involves removing the outside cause or failing hardware.

A memory parity error means that one or more stored data values carries a different value when it's recalled than when it was stored.

According to Cisco, parity errors are a type of data corruption.

Parity errors offset the charge value and can bring back invalid or incorrect commands for the computer.

Fixing parity errors involves removing the outside cause or failing hardware.

A memory parity error means that one or more stored data values carries a different value when it's recalled than when it was stored.

According to Cisco, parity errors are a type of data corruption.

Parity errors offset the charge value and can bring back invalid or incorrect commands for the computer.

Question 79 |

The counter implemented by the following circuit diagram where inputs to the NAND gate are the outputs of the B and C flip-flops, is:

MOD-7 Counter | |

MOD-6 Counter | |

MOD-8 Counter | |

MOD-4 Counter |

Question 79 Explanation:

→ In question, CLR is given as Input.

→ CLR is asynchronous input pin.

→ If CLR=1, all flip-flop outputs are reset to ‘0’.

→ When counter will start counting, it will count from 0 to 5 and at 6th clock, flip-flop output are cleared.

→ Hence given counter is MOD6 counter.

→ CLR is asynchronous input pin.

→ If CLR=1, all flip-flop outputs are reset to ‘0’.

→ When counter will start counting, it will count from 0 to 5 and at 6th clock, flip-flop output are cleared.

→ Hence given counter is MOD6 counter.

Question 80 |

What will be the minimum Hamming distance for the following coding scheme?

1 | |

2 | |

3 | |

4 |

Question 80 Explanation:

The minimum Hamming distance is the smallest Hamming distance between all possible pairs.

The Hamming distance can easily be found if we apply the XOR operation on the two words and count the number of 1s in the result. Note that the Hamming distance is a value greater than zero

The minimum hamming distance is between the words 10 and 11

10101 exor 11100 = 01001 which is “2”.

The Hamming distance can easily be found if we apply the XOR operation on the two words and count the number of 1s in the result. Note that the Hamming distance is a value greater than zero

The minimum hamming distance is between the words 10 and 11

10101 exor 11100 = 01001 which is “2”.

Question 81 |

In clustering index, the number of index entries is same as the number of:

Attributes in data file | |

Records in data file | |

Blocks in data file | |

Distinct index field values |

Question 81 Explanation:

An Index is a key built from one or more columns in the database that speeds up fetching rows from the table or view.

Cluster index is a type of index which sorts the data rows in the table on their key values. In the Database, there is only one clustered index per table.

A clustered index defines the order in which data is stored in the table which can be sorted in only one way. So, there can be an only a single clustered index for every table.

In an RDBMS, usually, the primary key allows you to create a clustered index based on that specific column.

Cluster index is a type of index which sorts the data rows in the table on their key values. In the Database, there is only one clustered index per table.

A clustered index defines the order in which data is stored in the table which can be sorted in only one way. So, there can be an only a single clustered index for every table.

In an RDBMS, usually, the primary key allows you to create a clustered index based on that specific column.

Question 82 |

Which of the following operators can be used if a portion of a given bit patterns needs to be copied to a new word, while the remainder of the new word is filled with 0’s?

Logical AND | |

Bitwise AND | |

Bitwise OR | |

Bitwise XOR |

Question 82 Explanation:

Bitwise AND is a binary operator that operates on two equal-length bit patterns. If both bits in the compared position of the bit patterns are 1, the bit in the resulting bit pattern is 1, otherwise 0.

Bitwise OR is also a binary operator that operates on two equal-length bit patterns, similar to bitwise AND. If both bits in the compared position of the bit patterns are 0, the bit in the resulting bit pattern is 0, otherwise 1.

Bitwise XOR also takes two equal-length bit patterns. If both bits in the compared position of the bit patterns are 0 or 1, the bit in the resulting bit pattern is 0, otherwise 1.

Bitwise OR is also a binary operator that operates on two equal-length bit patterns, similar to bitwise AND. If both bits in the compared position of the bit patterns are 0, the bit in the resulting bit pattern is 0, otherwise 1.

Bitwise XOR also takes two equal-length bit patterns. If both bits in the compared position of the bit patterns are 0 or 1, the bit in the resulting bit pattern is 0, otherwise 1.

Question 83 |

Let each process P

_{i}, i= 1 to 7 executes the following code. Repeat P(mutex); CS V(mutex); Forever The process P_{8}executes the following code: Repeat V(mutex); CS V(mutex); Forever What is the maximum number of processes that can be present in the critical section at any point of time? Given that the initial value of binary semaphore variable “mutex” is 1.7 | |

8 | |

1 | |

10 |

Question 84 |

Which of the following is the best-case time complexity of Floyd’s algorithm for finding shortest paths in a graph with ‘n’ vertices?

Θ(n ^{2}) | |

Θ(1) | |

Θ(n) | |

Θ(n ^{3}) |

Question 84 Explanation:

Floyd–Warshall's Algorithm is used to find the shortest paths between all pairs of vertices in a graph, where each edge in the graph has a weight which is positive or negative. The biggest advantage of using this algorithm is that all the shortest distances between any 2 vertices could be calculated in O(V

^{3}), where V is the number of vertices in a graph.Question 85 |

State TRUE or FALSE for the following.

(i) A sine wave with a phase of 180 starts at time 0 with a zero amplitude. The amplitude is decreasing

(ii) If a signal changes instantaneously, its frequency is zero.

(iii) In bipolar encoding, we use three voltage levels: positive, negative and zero.

(iv) Infrared signals can be used for short can be used for short-range communication in a closed area using ground propagation.

(i) TRUE, (ii) FALSE, (iii) TRUE, (iv) FALSE | |

(i) TRUE, (ii) TRUE, (iii) TRUE, (iv) FALSE | |

(i) FALSE, (ii) TRUE, (iii) FALSE, (iv) FALSE | |

(i) TRUE, (ii) TRUE, (iii) FALSE, (iv) FALSE |

Question 86 |

Consider the characters and their frequency counts given in the following table.

Using the Huffman coding technique, which of the following is the valid code for character ‘c’?

Using the Huffman coding technique, which of the following is the valid code for character ‘c’?

11111 | |

1110 | |

11110 | |

110 |

Question 86 Explanation:

Question 87 |

A block of addresses is granted to a small organization, If one of the addresses is 210.32.64.79/26, then what will be the values of the following?

(i) First address

(ii) Last address

(iii) Total number of addresses

(i) 210.32.64.64, (ii) 210.32.64.127, (iii)64
| |

(i) 210.32.64.64, (ii) 210.32.64.255, (iii)32
| |

(i) 210.32.64.79, (ii) 210.32.64.255, (iii)64 | |

(i) 210.32.64.64, (ii) 210.32.64.79, (iii)128 |

Question 87 Explanation:

→Given address is CIDR network address 210.32.64.79/26.

→Classless addressing treats the IP address as a 32 bit stream of ones and zeroes, where the boundary between network and host portions can fall anywhere between bit 0 and bit 31.Classless addressing system is also known as CIDR(Classless Inter-Domain Routing).

→The first address in the block can be found by setting the rightmost 32 − n bits to 0s.

→The last address in the block can be found by setting the rightmost 32 − n bits to 1s.

→The number of addresses in the block can be found by using the formula 2

→Here “n” is 26.

→Classless addressing treats the IP address as a 32 bit stream of ones and zeroes, where the boundary between network and host portions can fall anywhere between bit 0 and bit 31.Classless addressing system is also known as CIDR(Classless Inter-Domain Routing).

→The first address in the block can be found by setting the rightmost 32 − n bits to 0s.

→The last address in the block can be found by setting the rightmost 32 − n bits to 1s.

→The number of addresses in the block can be found by using the formula 2

^{32−n}.→Here “n” is 26.

Question 88 |

A_____takes a directed ______ graph G and produce a linear ordering of all its vertices such that for every directed edge in G, the vertex v comes before the vertex w in the ordering.

Breadth first search; acyclic | |

Topological sort; acyclic | |

Breadth first search; cyclic | |

Topological sort; cyclic |

Question 88 Explanation:

Topological sorting of vertices of a Directed Acyclic Graph is an ordering of the vertices
V1,v2,...vn in such a way, that if there is an edge directed towards vertex vj from vertex
vi , then vi comes before vj

Question 89 |

Match the following.

I-S, II-R, III-P, IV-Q | |

I-S, II-Q, III-P, IV-R | |

I-S, II-P, III-Q, IV-R | |

I-P, II-Q, III-R, IV-S |

Question 89 Explanation:

→Array,Stack and Queue are examples of linear data structure .

→Tree and Graph are examples of non-linear data structure.

→A binary heap is a complete binary tree; that is, all levels of the tree, except possibly the last one (deepest) are fully filled, and, if the last level of the tree is not complete, the nodes of that level are filled from left to right.

→Tree and Graph are examples of non-linear data structure.

→A binary heap is a complete binary tree; that is, all levels of the tree, except possibly the last one (deepest) are fully filled, and, if the last level of the tree is not complete, the nodes of that level are filled from left to right.

Question 90 |

Which of the following statements is TRUE for the function prototype declaration given below?

Int *(*P)(char *Q[]);

P is a function that accepts an argument which is a character array and returns a pointer to an integer quantity. | |

P is a function that accepts an argument which is a pointer to a character array and returns a pointer to an integer quantity. | |

P is a pointer to a function that accepts an argument which is an array of character pointers, and returns a pointer to an integer quantity. | |

P is a pointer to function that accepts an argument which is a character array and returns a pointer to an integer quantity . |

Question 90 Explanation:

P is a pointer to a function that accepts an argument which is an array of character pointers, and returns a pointer to an integer quantity.

Question 91 |

Consider the following segment of codes related to process creation. How many times the message “ child process created” will be printed?

#include

Void main(){

fork();fork();fork();

printf(“child process created”);

}

9 | |

7 | |

8 | |

3 |

Question 91 Explanation:

System call fork() is used to create processes. It takes no arguments and returns a process ID. The purpose of fork() is to create a new process, which becomes the child process of the caller. After a new child process is created, both processes will execute the next instruction following the fork() system call.

Number process created are 2

Number process created are 2

^{number of fork() calls}= 2^{3}=8Question 92 |

What is the output when the following segment of ‘c’ code is executed?

Void main (){

Float a = 123.456

printf(“%7.2f, %7.3f, %12e, a, a, a);

}

123.450, 123.4560, 1.234560e+02 | |

123.46, 123.456, 1.234560e+02 | |

123.456000, 123.456, 0.1234560e+03 | |

123.45, 123.4560, 1.234560e+02 |

Question 92 Explanation:

→The conversion specifier %7.2f tells printf to display the value as a 7-character wide field with 2 digits following the decimal point.

→The %e format uses scientific notation, i.e. one digit before the decimal separator and an exponent for scaling

→The %e format uses scientific notation, i.e. one digit before the decimal separator and an exponent for scaling

Question 93 |

Which of the following is a seven-bit code?

Biquinary code | |

BCD code | |

2421 code | |

Excess-3 code |

Question 93 Explanation:

Bi-quinary coded decimal is a numeral encoding scheme used in many abacuses and in some early computers, including the Colossus.

The term bi-quinary indicates that the code comprises both a two-state (bi) and a five-state (quinary) component.

The term bi-quinary indicates that the code comprises both a two-state (bi) and a five-state (quinary) component.

Question 94 |

Which of the following attributes can be considered as composite, single-valued and key attribute?

Age | |

Date of birth | |

Gender | |

Enrollment number |

Question 94 Explanation:

**Types of Attributes**

Simple attribute − Simple attributes are atomic values, which cannot be divided further. For example, a student's phone number is an atomic value of 10 digits.

Composite attribute − Composite attributes are made of more than one simple attribute. For example, a student's complete name may have first_name and last_name.

Derived attribute − Derived attributes are the attributes that do not exist in the physical database, but their values are derived from other attributes present in the database. For example, average_salary in a department should not be saved directly in the database, instead it can be derived. For another example, age can be derived from data_of_birth.

Single-value attribute − Single-value attributes contain single value. For example − Social_Security_Number.

Multi-value attribute − Multi-value attributes may contain more than one values. For example, a person can have more than one phone number, email_address, etc.

Key is an attribute or collection of attributes that uniquely identifies an entity among entity set.

Question 95 |

Which of the following statements about the following binary tree is FALSE

It is a binary serach tree | |

It is a complete binary tree | |

Nodes ‘J’ and ‘K’ are siblings | |

Node ‘B’ is the ancestor of node ‘J’ |

Question 95 Explanation:

A binary search tree is a rooted binary tree, whose internal nodes each store a key (and optionally, an associated value) and each have two distinguished sub-trees, commonly denoted left and right.

The tree additionally satisfies the binary search property, which states that the key in each node must be greater than or equal to any key stored in the left sub-tree, and less than or equal to any key stored in the right sub-tree.

Only option A is false

The tree additionally satisfies the binary search property, which states that the key in each node must be greater than or equal to any key stored in the left sub-tree, and less than or equal to any key stored in the right sub-tree.

Only option A is false

Question 96 |

The default storage class for functions in ‘C’ language is:

Static | |

Register | |

Extern | |

Auto |

Question 96 Explanation:

The default storage class for variable is auto and function is extern.

We can use functions in multiple files with the help of header files.

We can use functions in multiple files with the help of header files.

Question 97 |

If T is a binary tree with number of levels as L, then the number of leaf nodes in the binary tree is at most :

2 ^{L+1} | |

2L | |

2 ^{L } | |

2 ^{L-1} |

Question 97 Explanation:

Let T be a binary tree with L levels. Then the number of leaves is at most 2

^{L}-1.Question 98 |

If a connected graph G does not contain any vertex whose removal disconnects the rest of the graph, then G is called:

Biconnected graph | |

Separable graph | |

Forest | |

Diagraph |

Question 98 Explanation:

A biconnected graph is a connected and "nonseparable" graph, meaning that if any one vertex were to be removed, the graph will remain connected. Therefore a biconnected graph has no articulation vertices.

Question 99 |

Consider a cache memory organization with m lines in which the cache is divided into v sets, each of which consists of k lines. The set associative mapping technique reduces to direct mapping when:

v=m and k=m | |

v=1 and k=m | |

v=m and k=1 | |

v=1 and k=1 |

Question 99 Explanation:

Question 100 |

Consider a memory unit of size 96k x 16, where first component represents the total number of words and that the second component represents the number of bits per word. What will be the number if address lines and input-output data lines?

16 address lines, 7 data lines | |

7 address lines, 16 data lines | |

17 address lines, 16 data lines | |

16 address lines, 17 data lines |

Question 100 Explanation:

Given memory unit is 96K x 16

96 is greater than 64 and less than 128 ,So we need to consider 128.

96= 2x2x2x2x2x3 which is equivalent to 2 ^ 7

96K which is 2^7 x 2^10 =2^17

So address lines are 17

Word size is 16

So data lines are 2 ^16.

So option C is correct .

96 is greater than 64 and less than 128 ,So we need to consider 128.

96= 2x2x2x2x2x3 which is equivalent to 2 ^ 7

96K which is 2^7 x 2^10 =2^17

So address lines are 17

Word size is 16

So data lines are 2 ^16.

So option C is correct .

There are 100 questions to complete.