solution - DBMS (Database Management System) : Normalization (Test 2)

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

Solution :
839420260427082651 Correct Answer : It removes all redundancy

Description -

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

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

Solution :
201920260427082614 Correct Answer : Create separate rows for each course per student

Description -

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

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

Solution :
930520260427082532 Correct Answer : Splitting multi-valued attributes into separate rows

Description -

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

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

Solution :
930820260427082455 Correct Answer : Multi-valued attribute violation

Description -

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

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

Solution :
820520260427082413 Correct Answer : Each attribute contains only atomic (indivisible) values

Description -

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

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

Solution :
424520260427082233 Correct Answer : It is automatically in 2NF

Description -

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

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

Solution :
18920260427082145 Correct Answer : It is applicable only when primary key is composite

Description -

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

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

Solution :
608920260427082110 Correct Answer : Separate attributes depending on part of key into new tables

Description -

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

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

Solution :
137320260427081952 Correct Answer : It must be in 1NF and eliminate partial dependency

Description -

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

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?

Solution :
682020260427081903 Correct Answer : Relation is in 1NF but not in 2NF

Description -

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

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

Solution :
143120260427081521 Correct Answer : Remove transitive dependencies and improve data integrity

Description -

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

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

Solution :
814320260427081428 Correct Answer : EMP_CITY is transitively dependent on EMP_ID

Description -

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

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

Solution :
122820260427081333 Correct Answer : A non-prime attribute depends on another non-prime attribute

Description -

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

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

Solution :
161620260427081259 Correct Answer : X is a super key OR Y is a prime attribute

Description -

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

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

Solution :
791620260427081220 Correct Answer : It must be in 2NF and have no transitive dependency

Description -

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

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

Solution :
489220260427081106 Correct Answer : To remove anomalies due to functional dependencies

Description -

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

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

Solution :
134820260427081030 Correct Answer : 3NF but not BCNF

Description -

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

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

Solution :
46020260427080958 Correct Answer : Every BCNF relation is in 3NF, but not vice versa

Description -

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

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

Solution :
238420260427080921 Correct Answer : R is in 3NF but not in BCNF

Description -

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

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

Solution :
396220260427080833 Correct Answer : For every functional dependency X → Y, X must be a super key

Description -

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

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

Solution :
599820260427080726 Correct Answer : Eliminates unnecessary data duplication

Description -

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

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

Solution :
977920260427080648 Correct Answer : Presence of multivalued dependency

Description -

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

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

Solution :
339020260427080604 Correct Answer : One attribute determines multiple independent values of another

Description -

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

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

Solution :
472220260427080528 Correct Answer : (A, B) and (A, C)

Description -

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

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

Solution :
106620260427080421 Correct Answer : It must be in BCNF and have no non-trivial multivalued dependencies

Description -

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

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?

Solution :
349520260427080300 Correct Answer : R is not in 5NF

Description -

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

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

Solution :
526120260427080214 Correct Answer : Every 5NF relation is in 4NF

Description -

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

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

Solution :
741520260427080132 Correct Answer : It guarantees lossless decomposition when relations are joined back

Description -

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

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

Solution :
178620260427080054 Correct Answer : Eliminate redundancy caused by join dependencies

Description -

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

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

Solution :
964620260427080014 Correct Answer : It is in 4NF and has no non-trivial join dependency

Description -

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