DBMS MCQ Quiz in मल्याळम - Objective Question with Answer for DBMS - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 9, 2025

നേടുക DBMS ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക DBMS MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest DBMS MCQ Objective Questions

Top DBMS MCQ Objective Questions

DBMS Question 1:

Which of following is NOT an aggregate function in SQL?

  1. Min()
  2. RTrim()
  3. Sum()
  4. Max()

Answer (Detailed Solution Below)

Option 2 : RTrim()

DBMS Question 1 Detailed Solution

Concept:

In SQL, the function - avg, min, max, sum, count is called as aggregate function.

RTrim() is NOT an aggregate function in SQL.

Explanation:

Aggregate functions return a single value after calculating from a group of values.

avg() function returns the average values.

→ max() returns the maximum value.

→ sum() returns the sum of values.

Example:

SELECT MIN(salary)

FROM employees

GROUP BY Department;

DBMS Question 2:

A prime attribute of a relation schema R is an attribute that appears

  1. in all candidate keys of R
  2. in some candidate key of R
  3. in a foreign key of R
  4. only in the primary key of R

Answer (Detailed Solution Below)

Option 2 : in some candidate key of R

DBMS Question 2 Detailed Solution

  • Attributes of the relation which exist in at least one of the possible candidate keys, are called prime or key attributes
  • Candidate key is a minimal super key and a Super key is a set of attributes that uniquely identify a tuple in a relation.
  • Therefore, a prime attribute of a relation scheme R is an attribute that appears in some candidate key of R


Important Point:

  • Primary key is selected from a set of candidate keys of a relation.

Mistake Points 
Option 1) is not correct. Because a prime attribute is an attribute that can appear in any candidate key. The attribute need not appear in all candidate keys of R. Example:- if A is a prime attribute then for a relationship it is not required for A to be present in all candidate keys of relation R.

DBMS Question 3:

DBMS is comprised of tables that are made up of rows called ______ and columns called ______.

  1. Fields, records
  2. Records, fields
  3. Address, fields
  4. Ranges, sheet

Answer (Detailed Solution Below)

Option 2 : Records, fields

DBMS Question 3 Detailed Solution

The correct answer is Records, fields

Key Points

  • In the Relational database model, a table is a collection of data elements organized in terms of rows and columns. A table is also considered as a convenient representation of relations. But a table can have a duplicate row of data while a true relation cannot have duplicate data.

Additional Information 

  • Fields and records are two basic components of a database, which is an organized collection of information, or data. The term "fields" refers to columns or vertical categories of data; the term "records" refers to rows or horizontal groupings of unique field data. Examples of both database components are seen in older, printed databases and also in modern, computer-based databases.
  • A single entry in a table is called a Tuple or Record or Row. A tuple in a table represents a set of related data.
  • A table consists of several records(row), each record can be broken down into several smaller parts of data known as Attributes.

DBMS Question 4:

A ______ is a diagram that shows the hierarchical relationship between objects.

  1. Bar
  2. Pie 
  3. Dendrogram
  4. Factor analysis

Answer (Detailed Solution Below)

Option 3 : Dendrogram

DBMS Question 4 Detailed Solution

A Dendrogram is a diagram that shows the hierarchical relationship between objects.
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Dendrogram tool:

  • Dendrogram for clustering retail items is a tool used in customer relationship management (CRM) for implementing homogeneous schemes for all the items in one cluster.
  • In agglomerative hierarchical clustering, dendrograms are developed based on the concept of `distance' between the entities or, groups of entities.
  • These entities may be the customers, retail items, business units, etc. as per the business problem. 
    F1 Pratiksha Shetty Anil 12-05.21 D13
  • In the above example, we can see that E and F are most similar, as the height of the link that joins them together is the smallest.
  • The next two most similar objects are A and B.

 

The dendrogram tool is also used in CRM in conjunction with Cluster Analysis as represented below:

  • Here, observations are allocated to clusters by drawing a horizontal line through the dendrogram.
  • Observations that are joined together below the line are in clusters.
  • In the example above, we have two clusters.
  • One cluster combines A and B, and a second cluster combines C, D, E, and F.

DBMS Question 5:

Which one of the following is used to represent the supporting many-one relationships of a weak entity set in an entity-relationship diagram?

  1. Diamonds with double/bold border
  2. Rectangles with double/bold border
  3. Ovals with double/bold border
  4. Ovals that contain underlined identifiers

Answer (Detailed Solution Below)

Option 1 : Diamonds with double/bold border

DBMS Question 5 Detailed Solution

Concept: 

An entity set which has determined an attribute or set of attributes to be a primary key is called a strong entity set.

If an entity set does not have enough attributes to form a primary key, it is called a weak entity set.

Explanation:

In E-R diagram, the many-to-one relationship with a weak entity set is represented by diamond with double borders.

F1 Raju.S 16-04-2020 Savita D 1

Important Point:

The relationship with weak entity set would be represented by diamonds with double/bold borders.

The weak entity set itself would be represented by rectangle with double/bold border.

DBMS Question 6:

Consider the following student relation.

Student relation has Name, Subject1_Marks, and Subject2_Marks attributes.

Name Subject1_Marks Subject2_Marks
A 10 5
B Null 10
C 20 15
D Null 20
E 30 Null
F Null 25

Query1:

Select count (*)

From student;

Query2:

Select count (10)

From student;

Query1 prints the Q1 and Query2 prints Q2. Which of the following condition is true?

  1. Q1
  2. Q1>Q2
  3. Q2>Q1
  4. Q1=Q2

Answer (Detailed Solution Below)

Option 4 : Q1=Q2

DBMS Question 6 Detailed Solution

The correct answer is option 4.

Concept:

Count():

It takes one set of values and gives one output. The COUNT() function returns the number of records returned by a select query.

Note: If you specify a column name instead of (*), NULL values will not be counted.

Syntax:

COUNT(expression)

Explanation:

The given table is,

Consider the following student relation.

Student relation has Name, Subject1, and Subject2 attributes.

Name Subject1_Marks Subject2_Marks
A 10 5
B Null 10
C 20 15
D Null 20
E 30 Null
F Null 25

Query1:

Select count (*)

From student;

Explanation:

The COUNT(*) function counts the total number of rows in the student relation, regardless of whether they contain NULL values. Here no complete row is null. So it prints the 6 as output.

Query2:

Select count (10)

From student;

Explanation:

Note:

Count(constant) prints the number of rows in a relation.

The query2 has count(10) prints the number of rows that are not null. Here no complete row is null. So it prints the 6 as output.

Hence the correct answer is Q1=Q2.

DBMS Question 7:

In CRM, the full form of RDBMS is ________.

  1. Repairable Database Management System
  2. Relational Database Management System
  3. Relational Database Memory System
  4. Recurring Database Memory System

Answer (Detailed Solution Below)

Option 2 : Relational Database Management System

DBMS Question 7 Detailed Solution

The correct answer is Rational Database Management System

Key Points

  • Customer relationship management (CRM) is a process in which a business or other organization administers its interactions with customers, typically using data analysis to study large amounts of information.
  • CRM systems compile data from a range of different communication channels, including a company's website, telephone, email, live chat, marketing materials and more recently, social media.
  • They allow businesses to learn more about their target audiences and how to best cater for their needs, thus retaining customers and driving sales growth.
  • CRM may be used with past, present or potential customers.

  • A relational database management system (RDBMS) in CRM is a type of database that stores and provides access to data points that are related to one another.
  • Relational databases are based on the relational model, an intuitive, straightforward way of representing data in tables.
  • In a relational database, each row in the table is a record with a unique ID called the key.
  • The columns of the table hold attributes of the data, and each record usually has a value for each attribute, making it easy to establish the relationships among data points.

DBMS Question 8:

In context of database, let T1 and T2 are two concurrent transactions. Consider the following sequence of operations on data X :

T1 : R(X)   T1: W(X)   T2 : R(X) T2 : W(X)

This is a __________ problem.

  1. Dirty Read
  2. Lost Update
  3. Incorrect Summary
  4. Unrepeatable Read

Answer (Detailed Solution Below)

Option 1 : Dirty Read

DBMS Question 8 Detailed Solution

Dirty Reads

  • When a transaction is allowed to read a row that has been modified by another transaction which is not committed yet that time Dirty Reads occurred.
  • It is mainly occurred because of multiple transactions at a time which is not committed.
  • The sequence of operations on data X is a  dirty read problem.

Lost update

  • It occurs when two different transactions are trying to update the same column on the same row within a database at the same time.
  • In other words, The update done to a data item by a transaction is lost as it is overwritten by the update done by another transaction.

Unrepeatable Read Problem
The unrepeatable problem occurs when two or more read operations of the same transaction read different values of the same variable.

Incorrect Summary Problem

where one transaction is applying the aggregate function on some records while another transaction is updating these records. The aggregate function may calculate some values before the values have been updated and others after they are updated.

Hence, the correct answer is Option1.

DBMS Question 9:

_______ symbol is used to see every column of a table.

  1. /
  2. _ _
  3. *
  4. !

Answer (Detailed Solution Below)

Option 3 : *

DBMS Question 9 Detailed Solution

The * symbol is used to see every column of a table.

 For Example, Let us consider a table Table1

ID NAME AGE ADDRESS SALARY
1 Akash 24 Bijnor 12000
2 sema 23 Delhi 15000
3 diya 33 Banglore 33000
4 Badal 29 Odisha 40000

If you want to fetch only some specific columns from the table, then we can use this query,

Select ID,NAME,AGE from Table1;    / Syntax is Select (column_name1, coloumn_name2.....coloum_name) from table_name;

ID NAME AGE
1 Akash 24
2 sema 23
3 diya 33
4 Badal 29

If you want to fetch all the fields of the Table1 table, then you should use the following query 

Select * from Table1;    Syntax is Select * from table_name;

ID NAME AGE ADDRESS SALARY
1 Akash 24 Bijnor 12000
2 sema 23 Delhi 15000
3 diya 33 Banglore 33000
4 Badal 29 Odisha 40000

Therefore Option 3 is correct

DBMS Question 10:

If in a relation R(A, B, C, D, E) CD and E are the keys than the number of super key possible is _____.

  1. 8
  2. 16
  3. 20
  4. 24

Answer (Detailed Solution Below)

Option 3 : 20

DBMS Question 10 Detailed Solution

Concepts:

Any superset of a key (primary key or candidate key) is a superkey.

Data:

n → total number of attributes = 5

n(s) → total number of super keys

n(CD) → number of super keys when CD as the key

n(E) → number of super keys when E as the key

Formula:

n(S) = n(CD ∪ E) = n(CD) + n(E) – n(CDE)

Calculation:

\({\rm{n}}\left( {\rm{S}} \right) = {2^{5 - 2\;}} + {2^{5 - 1\;}} - {2^{5 - 3}}\;\)

\({\rm{n}}\left( {\rm{S}} \right) = 8 + 16 - 4 = 20\)

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