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24751.

Raw facts and figures are known as ..........

Answer»

Raw facts and figures are known as data

24752.

Pick the odd one out and give reason a) Calculation b) Storage c) Comparison d) Categorization

Answer»

b) Storage 

It is one of the data processing stage the others are various operations in the stage Process.

24753.

Explain some of the main goals of normalization.

Answer»

Normalization should minimize data redundancy. It should also simplify referential integrity constraints. Normalization will also make it easier to insert, update, and delete data. And finally, it provides better design.

24754.

Define the term ‘bit’?

Answer»

A bit stands for Binary digit. That means either 0 or 1.

24755.

Which one of the following CPU resister helds address of next instruction to be executed by the processor? a) Accumulator b) Instruction Register (IR) c) Memory address Register d) Program Counter (PC)

Answer»

d) Program Counter (PC)

24756.

Fill up the missing digits. 1. (4……)8 = (……110)2 2. (…….7……)8 = (100…….110)2Consider the followingDigits in octalCorresponding binary is 3 bits00001001201030114100510161107111

Answer»

1. 4…….100 and 110……….6

So (46)8 = (100 110)2 

2. 100…….4 

7……111 

110………6 

So (476)8 = (100 111 110)2

24757.

Fill up the missing numbers. 1. (A…….)16 = (……..1001)2 2. (…….B…….)16 = (1000………1111)2Consider the following:Digits in Hexa decimalCorresponding binary is 4 bits00000100012001030011401005010160110810009100110(A)101011(B)101112(C)110013(D)110114(E)111015(F)1111

Answer»

1. A……..1010 

1001………9 

So (A9)16 = (1010 1001)2

2. B…….1011 

1000………8 

1111………F 

So (8BF)16 = (1000 1011 1111)2

24758.

Explain how a scenario is used for business rules.

Answer»

A scenario is used to test business rules. It is a short script that describes how a business reacts to certain situations.

24759.

Describe the naming of Relationships?

Answer»

Relationships are verb phrases and represent actions usually in the present tense. The name should exclude vague names such as "has". The name should explain what action is being taken and possibly why it is important.

24760.

Explain minimum and maximum cardinality.

Answer»

Minimum cardinality is the minimum number of instances of an entity that can be associated with each instance of another entity. Maximum cardinality is the maximum number of instances of an entity that can be associated with each instance of another entity.

24761.

List the three types of business rules and define each of them.

Answer»

A derivation is a statement that is derived from other knowledge. A structured assertion is a statement that expresses some aspect of the static structure of an organization. An action assertion is a statement of a constraint on the actions of an organization.

24762.

Why is modeling time-dependent data with a time stamp important?

Answer»

The values of data may change. A time stamp helps to ensure that the previous value of the data stays in the database after it has changed so that you can see the before and after values through time. Without a time stamp, you will most likely lose some of the history.

24763.

Match the following.1. (106)10a. (171.53125)102. (71.24)8b. (6a)163. (AB.88)16c. (20)84. (10)16d. (10000000)25. (128)10e. (10)166. (16)10f. (57.3125)10

Answer»

1 – b, 

2 – f, 

3 – a, 

4 – c, 

5 – d, 

6 – e

24764.

The base of octal number system is ..........

Answer»

The base of octal number system is 8.

24765.

Explain the difference between total specialization and partial specialization.

Answer»

Total specialization exists when every instance of a supertype must also be an instance of a subtype. Partial specialization exists when every instance of a supertype does not have to be an instance of a subtype.

24766.

Explain the difference between the disjoint and overlap rule.

Answer»

The disjoint rule states an entity instance of a supertype can only be a member of one subtype. The overlap rule states an entity instance of a supertype can be a member of multiple subtypes.

24767.

Arithmetic instructions.

Answer»

Arithmetic Instructions: 

The 8086 provides many arithmetic operations: addition, subtraction, negation, multiplication and comparing two values. 

ADD : 

The add instruction adds the contents of the source operand to the destination operand. 

Eg. ADD AX, 0100H 

ADD AX, BX 

ADD AX, [SI] 

ADD AX, [5000H] 

ADD [5000H], 0100H 

ADD 0100H

ADC : Add with Carry 

This instruction performs the same operation as ADD instruction, but adds the carry flag to the result. 

Eg. ADC 0100H 

ADC AX, BX 

ADC AX, [SI] 

ADC AX, [5000] 

ADC [5000], 0100H 

6

SUB : Subtract 

The subtract instruction subtracts the source operand from the destination operand and the result is left in the destination operand. 

Eg. SUB AX, 0100H 

SUB AX, BX 

SUB AX, [5000H] 

SUB [5000H], 0100H 

SBB : Subtract with Borrow 

The subtract with borrow instruction subtracts the source operand and the borrow flag (CF) which may reflect the result of the previous calculations, from the destination operand 

Eg. SBB AX, 0100H 

SBB AX, BX 

SBB AX, [5000H] 

SBB [5000H], 0100H 

INC : Increment 

This instruction increases the contents of the specified Register or memory location by 1. Immediate data cannot be operand of this instruction. 

Eg. INC AX 

INC [BX] 

INC [5000H] 

DEC : Decrement 

The decrement instruction subtracts 1 from the contents of the specified register or memory location. 

Eg. DEC AX 

DEC [5000H]

NEG : Negate 

The negate instruction forms 2’s complement of the specified destination in the instruction. The destination can be a register or a memory location. This instruction can be implemented by inverting each bit and adding 1 to it. 

Eg. NEG AL 

AL = 0011 0101 35H Replace number in AL with its 2’s complement

AL = 1100 1011 = CBH 

CMP : Compare 

This instruction compares the source operand, which may be a register or an immediate data or a memory location, with a destination operand that may be a 7 register or a memory location 

Eg. CMP BX, 0100H 

CMP AX, 0100H 

CMP [5000H], 0100H 

CMP BX, [SI] 

CMP BX, CX

MUL :Unsigned Multiplication Byte or Word 

This instruction multiplies an unsigned byte or word by the contents of AL. 

Eg. MUL BH ; (AX) (AL) x (BH) 

MUL CX ; (DX)(AX) (AX) x (CX) 

MUL WORD PTR [SI] ; (DX)(AX) (AX) x ([SI]) 

IMUL :Signed Multiplication 

This instruction multiplies a signed byte in source operand by a signed byte in AL or a signed word in source operand by a signed word in AX. 

Eg. IMUL BH 

IMUL CX 

IMUL [SI] 

CBW : Convert Signed Byte to Word 

This instruction copies the sign of a byte in AL to all the bits in AH. AH is then said to be sign extension of AL. 

Eg. CBW AX= 0000 0000 1001 1000 Convert signed byte in AL signed word in AX. 

Result in AX = 1111 1111 1001 1000 

CWD : Convert Signed Word to Double Word 

This instruction copies the sign of a byte in AL to all the bits in AH. AH is then said to be sign extension of AL. 

Eg. CWD 

Convert signed word in AX to signed double word in DX : AX

 DX= 1111 1111 1111 1111 

Result in AX = 1111 0000 1100 0001 

DIV : Unsigned division This instruction is used to divide an unsigned word by a byte or to divide an unsigned double word by a word. 

Eg. DIV CL ; Word in AX / byte in CL ; 

Quotient in AL, remainder in AH 

DIV CX ; Double word in DX and AX / word ; in CX, and Quotient in AX, ; 

remainder in DX

 8 

AAA : ASCII Adjust After Addition 

The AAA instruction is executed aftr an ADD instruction that adds two ASCII coded operand to give a byte of result in AL. The AAA instruction converts the resulting contents of Al to a unpacked decimal digits. 

Eg. ADD CL, DL ; [CL] = 32H = ASCII for 2 

; [DL] = 35H = ASCII for 5 ; 

Result [CL] = 67H

MOV AL, CL ; Move ASCII result into AL since ; 

AAA adjust only [AL] 

AAA ; [AL]=07, unpacked BCD for 7 

AAS : ASCII Adjust AL after Subtraction 

This instruction corrects the result in AL register after subtracting two unpacked 

ASCII operands. The result is in unpacked decimal format. The procedure is similar to 

AAA instruction except for the subtraction of 06 from AL. 

AAM : ASCII Adjust after Multiplication 

This instruction, after execution, converts the product available In AL into unpacked BCD format. 

Eg. MOV AL, 04 ; AL = 04 

MOV BL ,09 ; BL = 09 

MUL BL ; AX = AL*BL ; AX=24H 

AAM ; AH = 03, AL=06 

AAD : ASCII Adjust before Division 

This instruction converts two unpacked BCD digits in AH and AL to the equivalent binary number in AL. This adjustment must be made before dividing the two unpacked BCD digits in AX by an unpacked BCD byte. In the instruction sequence, this instruction appears Before DIV instruction. 

Eg. AX 05 08 

AAD result in AL 00 3A 58D = 3A H in AL 

The result of AAD execution will give the hexadecimal number 3A in AL and 00 in AH. 

Where 3A is the hexadecimal Equivalent of 58 (decimal). 

DAA : Decimal Adjust Accumulator 

This instruction is used to convert the result of the addition of two packed BCD numbers to a valid BCD number. The result has to be only in AL. 

Eg. AL = 53 CL = 29 

ADD AL, CL ; AL (AL) + (CL) ; 

AL 53 + 29 ; AL 7C DAA ; 

AL 7C + 06 (as C>9) ; 

AL 82 

DAS : Decimal Adjust after Subtraction 

This instruction converts the result of the subtraction of two packed BCD numbers to a valid BCD number. The subtraction has to be in AL only. 

Eg. AL = 75, BH = 46 

SUB AL, BH ; AL 2 F = (AL) - (BH) ; 

AF = 1 

DAS ; AL 2 9 (as F>9, F - 6 = 9)

24768.

The following statement Demorgan’s theorem of Boolean algebra. Identify and state ‘Break the line, change the sign’.

Answer»

Demorgan’s theorems, 

Demorgan’s first theorem, 

\(\overline{x+y}\) = \(\bar x .\bar y\) 

Demorgan’s second theorem, 

\(\overline{x-y} = \bar x + \bar y\) 

24769.

Explain the difference between an ERD and EER.

Answer»

An EER includes everything in an ERD and an EER allows for more complex relationships than an ERD. An EER allows for object-oriented data modeling and include supertypes and subtypes entities and inheritance.

24770.

List some of the chrematistics of good data definitions.

Answer»

Definitions are gathered from the same sources and should be accompanied diagrams. A definition will include special conditions, examples, how the data is created, whether the data can change, who owns the data, whether the data is optional, and whether the data can be broken into something more atomic.

24771.

What is networking Topology? Write the various types of networking topology. Distinguish between Star, Ring and Tree Topology.

Answer»

Networking Topology: It is a layout of interconnection of nodes and their workflow in a network.

Types of Networking Topology:

a) Linear 

b) Star 

c) Ring 

d) Tree 

e) Mesh 

f) Hybrid

Difference between Star, Ring and Tree Network Topology (Three difference)

1) Star: Computers are connected to a central hub through a dedicated link or a common path.

Ring: Every nodegets connected to its neighbouring nodes on both the sides.

Tree: It comprised multiple star topologies as a linear or bus network. The control hub of each star topology is directly connected to the central backbone of bus and every individual hub works as a root of tree for all computers.

2) Star: Error detection is very easy.

Ring: Error detection is highly difficult.

Tree: Error detection is very easy.

3) Star: An error in any node donot cause disruption of the network

Ring: An error in any node can cause disruption of the whole network.

Tree: Failure of backbone can cause total disruption of network

24772.

Briefly explain an ERD.

Answer»

An ERD is a detailed logical representation of the data for an organization. The ERD includes entities, attributes, relationships, and cardinalities. An ERD is the mechanism where an entity-relationship model is displayed.

24773.

Following are the numbers in various number systems. Two of the numbers are same. Identify them:1. (310)8 2. (1010010)2 3. (C8)16 4. (201)10OR Consider the following Boolean expression: (B’ + A)’ = B . A’ Identify the law behind the above expression and prove it using algebriac

Answer»

1. (310)8 = 3 * 82 + 1 * 81 + 0 * 80 

= 192 + 8 + 0 

= (200)10

2. (1010010)2 = 1 × 26 + 0 × 25 + 1 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 0 × 20 

= 64 + 0 + 16 + 0 + 0 + 2 + 0 

= (82)10 

3. (C8)16 = C × 16 + 8 × 160

= 12 × 16 + 8 × 1 

= 192 + 8 

= (200)10 

Here (a) (310)8 and (C8)16 are same

OR 

This is De Morgan’s law (B’ + A’) = (B’)’ . A’ 

= B . A’

Hence it is proved

24774.

The number of numerals or symbols used in a number system is its ...........

Answer»

The number of numerals or symbols used in a number system is its  Base

24775.

What is the difference between program module and an object?

Answer»

A program module is a self-contained independent program segment only it does not provide security to data whereas an object is a collection of data members and member functions that operate on data and data is provided with security.

24776.

How 0 and null value are different

Answer»

Zero is a number value. It is a definite with precise mathematical properties. (You can do arithmetic on it)

NULL means the absence of any value. You can't do anything with it except test for it.

24777.

Raw facts and figures are known as .......

Answer»

Raw facts and figures are known as data

24778.

Processed data is known as .......

Answer»

Processed data is known as Information

24779.

Briefly describe the six database activities that occur during the systems development life cycle.

Answer»

The enterprise modeling that analyzes the current data processing. Conceptual data modeling that identifies entities, relationships, and attributes. The logical database design that identifies data integrity and security requirements. The physical database design and definition that defines the database to a DBMS. The database implementation that installs and converts data from prior systems. Database maintenance that fixes errors in the database and database applications.

24780.

The valid octal constants from the following(i) 0245(ii) 0387(iii) 04.32(iv) –0467(A) (i) and (ii)(B) (iii) and (iv)(C) (ii) and (iii)(D) (i) and (iv) 

Answer»

Correct option -(D) (i) and (iv)

Explanation:-

(i) and (iv) are valid octal constants. 

24781.

Write a C program to convert a binary number to its corresponding octal number

Answer»

A C program to convert a binary number to its binary octal number is as follows:

main()

{

long int bin_no,no;

int oct_no,oct_p,i,rem,pow2,pow8,inter_oct;

clrscr();

printf ("\nEnter the Binary number: -> ");

scanf("%ld",&bin_no);

no=bin_no;

pow8=1;

oct_no=0;

while (no>0)

{

i=0;

inter_oct=0

pow2=1;

while (i<=2)

{

if (no==0)break; 

rem=no%10;

inter_oct+=rem*pow2;

i++;

pow2*=2;

no/=10;

}

oct_no+=inter_oct*pow8;

pow8*=10;

}

printf ("\nOctal Equivalent for %ld = %d",bin_no,oct_no); getch();

}  

24782.

The method of _________ the matter written in Shorthand to the language concerned is known as transcription: (a) Correcting (b) Copying (c) Reading (d) Converting or translating

Answer»

(d) Converting & translating

The method of Converting & translating the matter written in Shorthand to the language concerned is known as transcription.

24783.

What are the symbols used in Binary number system?

Answer»

0 and 1 are the symbols used in Binary number system

24784.

Mention different types of constructors.

Answer»

The three types of constructors are

  1. Default constructor.
  2. Parameterized constructor.
  3. Copy constructor.
24785.

Who can make up a systems or database team? Provide a brief explanation of each.

Answer»

The team includes a system analyst who identifies the need for information services to meet opportunities of the business, database analysts who design the database, users who monitor that the system will meet their needs, programmers who write computer programs, database and data administrators who have responsibility for current and future databases and other technical experts.

24786.

The base of Binary number system is ..........

Answer»

The base of Binary number system is 2

24787.

The base of decimal number system is..........

Answer»

The base of decimal number system is Base.

24788.

Explain the differences of the two principal types of packaged data models.

Answer»

Universal data models are common to many organizations. These models may be useful for similar functions that are used across companies or organizations such as purchasing and accounting. Industry-specific data models are used by specific industries.

24789.

a) inline b) return c)void d) const

Answer»

The 'return statement' is used to return a value in a function.

24790.

In what multilevel and multiple inheritance differ though both contains many base class?

Answer»

The derived class is a power packed class, as it can add additional attributes and methods and thus enhance its functionality.

24791.

Consider the number 23.87. Its LSD is ....

Answer»

Consider the number 23.87. Its LSD is 7

24792.

LSD is ..........

Answer»

Least significant digit

24793.

Explain the systems development life cycle.

Answer»

This is the traditional method to develop, maintain, and replace information systems that includes the project identification and selection, project initiation and planning, analysis, logical design, physical design, implementation, and maintenance steps. The process is a series of steps that are ideally completed in a linear fashion. In reality, the process is not linear and the process requires steps to be revisited and an overlap of steps.

24794.

Find MSD in the decimal number 7854.25

Answer»

Because it has the most weight

24795.

MSD is......

Answer»

Most significant digit

24796.

The base of Binary number system is ........

Answer»

The base of Binary number system is 2

24797.

Briefly explain the five components of an Information Systems Architecture and their relationship to data.

Answer»

Processes that manipulate the data. Networks that transport the data. People who process and use the data. Events and point in time when processes need to be performed with the data. Reasons for events and rules to govern the processing of the data.

24798.

Consider the number 627. Its MSD is ........

Answer»

Consider the number 627. Its MSD is 6

24799.

Consider the number 23.87. Its LSD is .......

Answer»

Consider the number 23.87. Its LSD is 7.

24800.

The base of Hexadecimal number system is ...........

Answer»

The base of Hexadecimal number system is 16