Free Mock Test : DBMS (Database Management System) : Normalization (Test 2)

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Free Mock Test - DBMS (Database Management System) : Normalization (Test 2)

Normalization remains one of the most important and frequently asked topics in DBMS for competitive examinations like NIC, TRE, Banking, SSC, and other technical recruitment tests. Candidates are advised to have a clear conceptual understanding as well as problem-solving ability in this area.
In the ever-evolving landscape of computer-based competitive examinations such as NIC, TRE, and various Banking exams, Database Management System (DBMS) continues to be a high-weightage subject. Among its core topics, Normalization has emerged as a frequently tested and conceptually important area, especially in technical and IT-related papers.

Total Questions : 30

Timer : sec

Question : Which statement is NOT true about First Normal Form?

It removes repeating groups

It eliminates multi-valued attributes

It removes all redundancy

It ensures atomic attributes

Question : A relation has attributes (Student_ID, Courses) where Courses contains multiple values per student. Which is the correct 1NF conversion?

Keep the same table structure

Replace Courses with Course1, Course2, Course3 columns

Create separate rows for each course per student

Remove the Courses attribute

Question : Which of the following transformations converts a relation into 1NF?

Removing transitive dependencies

Splitting multi-valued attributes into separate rows

Eliminating partial dependencies

Combining related tables

Question : Consider a table storing multiple phone numbers in a single column separated by commas. What violation does this represent?

Partial dependency

Multi-valued attribute violation

Transitive dependency

Lossless join violation

Question : Which of the following best describes a relation in First Normal Form (1NF)?

It contains only prime attributes

It contains no transitive dependencies

Each attribute contains only atomic (indivisible) values

It contains no functional dependencies

Question : A table has a single-attribute primary key. Which of the following is correct?

It cannot be in 2NF

It is automatically in 2NF

It always violates 2NF

It must be decomposed to reach 2NF

Question : Which of the following statements is TRUE regarding 2NF?

It eliminates all anomalies completely

It is applicable only when primary key is composite

It removes transitive dependency

It ensures no redundancy at all

Question : Consider a relation STUDENT_COURSE(StudentID, CourseID, StudentName, CourseName)
Primary Key: (StudentID, CourseID)
Which normalization step converts it to 2NF?

Remove transitive dependencies

Remove repeating groups

Separate attributes depending on part of key into new tables

Merge all attributes into one table

Question : Which of the following conditions must be satisfied for a relation to be in Second Normal Form (2NF)?

It must be in 1NF and have no transitive dependency

It must be in 1NF and eliminate partial dependency

It must not have any functional dependency

It must have a single primary key only

Question : A relation R(A, B, C, D) has a composite primary key (A, B). The following dependencies exist:
A → C
(A, B) → D
Which statement is correct?

Relation is in 2NF

Relation is in 1NF but not in 2NF

Relation is in 3NF

Relation violates 1NF

Question : Which of the following is the main objective of 3NF?

Increase redundancy

Allow multi-valued attributes

Remove transitive dependencies and improve data integrity

Eliminate primary keys

Question : Consider attributes: EMP_ID → EMP_ZIP → EMP_CITY.
Why does this violate 3NF?

EMP_ID is not a candidate key

EMP_CITY is a prime attribute

EMP_CITY is transitively dependent on EMP_ID

EMP_ZIP is independent

Question : Which of the following best describes a transitive dependency?

A primary key depends on another key

A non-prime attribute depends on another non-prime attribute

A prime attribute depends on itself

A candidate key depends on a non-key attribute

Question : In a relation R, a functional dependency X → Y satisfies 3NF if:

X is not a super key

Y is always a non-prime attribute

X is a super key OR Y is a prime attribute

Both X and Y are non-prime attributes

Question : Which of the following conditions must be satisfied for a relation to be in Third Normal Form (3NF)?

It must be in 1NF only

It must be in 2NF and contain partial dependency

It must be in 2NF and have no transitive dependency

It must not contain any candidate key

Question : What is the main purpose of converting a relation into BCNF?

To increase redundancy

To remove anomalies due to functional dependencies

To reduce number of tables

To eliminate foreign keys

Question : A relation has FD: A → B, B → C. If A is the candidate key, what is the highest normal form satisfied?

BCNF

3NF but not BCNF

2NF only

Not in 3NF

Question : Which statement correctly describes BCNF compared to 3NF?

BCNF is weaker than 3NF

BCNF allows partial dependency

Every BCNF relation is in 3NF, but not vice versa

BCNF ignores functional dependencies

Question : Consider relation R(X, Y, Z) with FD = {XY → Z, Z → Y}. Which is correct?

R is in BCNF

R is not in 3NF

R is in 3NF but not in BCNF

R is neither in 3NF nor BCNF

Question : Which condition must hold true for a relation to be in BCNF?

Every non-key attribute is fully dependent on the primary key

For every functional dependency X → Y, X must be a super key

The relation must be in 2NF only

No transitive dependency exists

Question : Which of the following is a direct benefit of converting a relation into 4NF?

Increases redundancy for faster queries

Eliminates unnecessary data duplication

Removes candidate keys

Combines multiple entities into one table

Question : A table is in 3NF but still shows redundancy due to independent attributes. What is the main reason it is not in 4NF?

Presence of partial dependency

Presence of transitive dependency

Presence of multivalued dependency

Absence of primary key

Question : Which of the following best describes a multivalued dependency (MVD)?

One attribute determines exactly one value of another

One attribute determines multiple independent values of another

Two attributes together determine a third attribute

All attributes depend on a composite key

Question : A relation R(A, B, C) has multivalued dependencies:
A →→ B and A →→ C.
What is the correct decomposition to achieve 4NF?

(A, B, C) remains unchanged

(A, B) and (B, C)

(A, B) and (A, C)

(B, C) and (A)

Question : Which condition is necessary and sufficient for a relation to be in Fourth Normal Form (4NF)?

It must be in 3NF and free from functional dependencies

It must be in BCNF and have no non-trivial multivalued dependencies

It must have only one candidate key

It must eliminate transitive dependencies only

Question : A relation R has a join dependency (R1, R2, R3) such that none of R1, R2, R3 is a superkey. What can be concluded?

R is in 5NF

R is not in 4NF

R is not in 5NF

R must be in BCNF

Question : Which of the following is TRUE regarding 5NF and 4NF?

Every 4NF relation is in 5NF

Every 5NF relation is in 4NF

5NF ignores multivalued dependencies

4NF removes join dependencies

Question : Which statement about join dependency (JD) is correct?

It ensures that a table cannot be decomposed further

It guarantees lossless decomposition when relations are joined back

It only applies to binary decompositions

It is weaker than multivalued dependency

Question : What is the primary purpose of 5NF decomposition?

Remove partial dependencies

Remove transitive dependencies

Eliminate redundancy caused by join dependencies

Ensure atomic attributes

Question : Which of the following conditions must be satisfied for a relation to be in Fifth Normal Form (5NF)?

It is in 3NF and has no transitive dependency

It is in BCNF and has no multivalued dependency

It is in 4NF and has no non-trivial join dependency

It has only one candidate key

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