This section contains carefully selected MCQs and Previous Year Questions with explanations to help students understand concepts and prepare effectively for examinations, interviews, and competitive tests.
Q: 1Consider a relation with attributes {A, B, C, D} and the following functional dependencies:
A → B, B → C, and C → D. If A is the primary key, which of the following is true regarding BCNF?
Option C
Given the relation R(A, B, C, D) with functional dependencies:
A → B
B → C
C → D
Primary Key: A
Check for BCNF:
A relation is in Boyce-Codd Normal Form (BCNF) if for every functional dependency X→Y, X is a candidate key or super key.
So, BCNF ensures that each non—key (non—prime) attributes is dependent only on the candidate key.
A→B: A is the primary key (Super Key): Satisfy BCNF condition.
B→C: B is not a Super Key : Violates BCNF condition.
C→D: C is not a Super Key : Violates BCNF condition.
Check for 3NF:
The relation is in 3NF if there is no transitive dependency i.e., no non—prime attribute can determine another non—prime attribute.
A→B: A is the prime attribute : Satisfy 3NF condition.
B→C: Both B and C are non—prime attributes : Violates 3NF condition.
C→D: Both C and D are non—prime attributes : Violates 3NF condition.
Check for 2NF:
The relation is in 2NF if there is no partial dependency i.e., no proper subset of candidate key can determine the non—prime attribute.
The relation is at least in 2NF, if key contains only one attribute (here A is primary key, only one attribute no subset possible), then there is no partial dependency because one attribute has no subset. So, the relation is in 2NF. Among the given options the correct answer is The relation is not in BCNF.
Q: 2Which of the following normal forms is adequate for designing relational database?
Option B
Normalization organizes relational database tables to minimize data redundancy and prevent anomalies during insert, update, or delete operations.
A database is considered adequately designed when it reaches Third Normal Form (3NF) because it:
Q: 3What are the three main steps involved in normalizing a database up to Third Normal Form (3NF)?
Option B
Normalization up to Third Normal Form (3NF) is carried out in three logical steps.
First, the table is converted to First Normal Form (1NF) by removing repeating groups and multivalued attributes so that each field contains atomic values.
Next, the table is converted to Second Normal Form (2NF) by removing partial dependencies, ensuring that every non-key attribute is fully dependent on the entire primary key.
Finally, the table is converted to Third Normal Form (3NF) by removing transitive dependencies, so that no non-prime attribute can determine another non-prime attribute.
Q: 4Functional dependencies are a generalization of:
Option A
A Functional Dependency (FD) describes a relationship where one set of attributes uniquely determines another set of attributes in a relation.
Key Dependency is a special case of functional dependency in which a key attribute determines all other attributes in the table.
Q: 5Consider R = (A, B, C, D, E, P, G) and the following Functional Dependencies in R:
AB → CD
DE → P
C → E
P → C
B → G
Which one of the following is true?
Option D
First find the candidate key using the given functional dependencies.
Compute closure of AB:
AB+ = {A, B}
AB+ = {A, B, C, D} Using AB→CD
AB+ ={A, B, C, D, E} Using C→E
AB+ ={A, B, C, D, E, P} Using DE→P
AB+ ={A, B, C, D, E, G, P} Using B→G
Thus AB is a candidate key.
Prime attributes = A, B
Non-prime attributes = C, D, E, G, P
For Second Normal Form (2NF), non-prime attributes should not depend on a part of a composite key.
But we have B → G. Here B is only a part of the candidate key (AB) and it determines G, which is a non-prime attribute. This is a partial dependency, which violates 2NF. Therefore, the relation R is not in 2NF.
Q: 6Given functional dependencies on U(P, Q, R, S, T):
P → QR
RS → T
Which dependencies cannot be inferred?
Option C
To see which dependencies follow from the given ones, compute attribute closures using the given FDs: P→QR and RS→T. If an attribute set’s closure contains the right-hand side, that dependency is implied.
(a)
P→R
Find P+ :
P+ = {P}
P+={P,Q,R} Using P→QR
R is in P+ ⇒ P→R inferred.
(b)
PS → T
Find PS+ :
(PS)+={P,S}
(PS)+={P,S,Q,R} Using P → QR
Now, (PS)+ has RS so Using RS→T
(PS)+={P,S,Q,R,T}
T is in (PS)+ ⇒ PS → T inferred.
(c)
R→T
Find R+ :
R+ = {R}
No dependency with just R, so cannot go further.
T is not in R+ ⇒ R→T cannot be inferred.
(d)
PS → Q
Find PS+ :
(PS)+ = {P,S)
(PS)+ = {P,S,Q,R) Using P→QR
Now, (PS)+ has RS so Using RS→T
(PS)+={P,S,Q,R,T}
Q is in (PS)+ ⇒ PS → Q inferred.
Q: 7A relation schema R is in First Normal Form (1NF) if the domains of all its attributes are:
Option C
A relation is said to be in First Normal Form (1NF) when each attribute contains only atomic or indivisible values. This means there are no repeating groups or multivalued attributes in the table.
Q: 8Consider the following relation R and the set of all functional dependencies that hold on it :
R(A,B,C,D,E,F)
AB→C
C→ABDE
ADE→F
Which of the following statements is true about R?
Option B
In the context of database design, Normalization is a process used to organize data efficiently within a database, reducing redundancy and improving data integrity.
For the given relation R(A,B,C,D,E,F) with functional dependencies AB→C, C→ABDE, and ADE→F. We first identify the candidate keys. Here, C and AB are candidate keys because each can uniquely identify all attributes.
The prime attributes (part of key) are A, B, and C, while D, E, and F are non-prime.
To check which normal form the relation satisfies, we start with 2NF (Second Normal Form), which requires that all non-key (non-prime) attributes are fully functionally dependent on the entire primary key. Since no partial dependency exists here, the relation is in 2NF.
For 3NF, every functional dependency X→Y must satisfy means X is a superkey or Y is a prime attribute.
However, the dependency ADE→F violates this because ADE is not a superkey, and F is not a prime attribute. This means the relation does not satisfy 3NF.
Finally, since the relation is not in 3NF, it also is not in BCNF (Boyce-Codd Normal Form), which is stricter than 3NF.
Q: 9Given below are statements in context of Armstrong's axioms to find logically implied functional dependencies in DBMS.
I. If B→C holds and A is a set of attributes, then AB → AC holds.
II. If A is a set of attributes and B ⊆ A, then A → B holds.
Choose the correct ordered pair to complete the following sentence.
“Statement I is called __________ rule and Statement II is called __________ rule.”
Option B
Armstrong’s Axioms are a set of inference rules used to derive all functional dependencies (FDs) logically implied by a given set of FDs in a database.
Statement I says that if a functional dependency B → C holds, then adding the same set of attributes A to both sides results in AB → AC. This is known as the Augmentation Rule. Statement II says that if B ⊆ A (B is a subset of A), then A → B always holds, which is called the Reflexivity Rule.
| Axiom / Rule | Description | Example |
|---|---|---|
| Reflexivity | If B ⊆ A, then A → B. A set of attributes functionally determines any of its subsets. | If A = {X, Y} and B = {X}, then {X, Y} → {X} |
| Augmentation | If A → B holds, and C is a set of attributes, then AC → BC also holds. Adding same attributes to both sides preserves the dependency. | If A → B and C = {Z}, then AZ → BZ |
| Transitivity | If A → B and B → C, then A → C. Functional dependencies can be combined through an intermediate attribute. | If A → B and B → C, then A → C |
| Union | If A → B and A → C, then A → BC. Dependencies from the same determinant can be combined. | If A → B and A → C, then A → BC |
| Decomposition | If A → BC, then A → B and A → C. A dependency on multiple attributes can be split into separate dependencies. | If A → BC, then A → B and A → C |
| Pseudo Transitivity | If A → B and BC → D, then AC → D. Combines transitivity with augmentation. | If A → B and BC → D, then AC → D |
Q: 10Consider a schema R(A, B, C, D) and functional dependencies A → B, C → D. Then, the decomposition of R into R1 and R2 where R1(AB) and R2(CD), is –
Option A
The decomposition of schema R(A,B,C,D) with functional dependencies A→B and C→D into R1(AB) and R2(CD), is dependency preserving but not lossless join.
Lossless Join: A decomposition is lossless if you can always recover the original relation by joining the decomposed relations, without losing any data. For a lossless join, the intersection of the schemas must be a key in at least one of the decomposed relations.
Dependency Preserving: A decomposition is dependency preserving if every functional dependency in the original relation is still enforceable in at least one of the decomposed relations without needing to join.
Q: 11Tables in second normal form (2NF):
Option D
Second Normal Form (2NF) eliminates partial dependencies in tables with composite primary keys. A table is in 2NF if it is in 1NF (atomic value) and every non-key attribute depends on the entire primary key, not just part of it.
Note:
Hidden Dependencies, also known as Transitive Dependencies, occur when a non-key attribute depends on another non-key attribute rather than directly on the primary key, and these dependencies are eliminated in Third Normal Form (3NF).
Q: 12A relation in which every non-key attribute is fully functionally dependent on the primary key and which as no transitive dependencies is in—
Option B
A relation is in Third Normal Form (3NF) if:
This means each non-key attribute depends only on the primary key and not on other non-key attributes, which helps eliminate redundancy and update anomalies.
| Normal Form | Description | Key Characteristics |
|---|---|---|
| 1NF (First NF) | Atomicity | Each column contains atomic, indivisible values; no repeating groups. |
| 2NF (Second NF) | Full Functional Dependency | Every non-key attribute is fully dependent on the entire primary key. |
| 3NF (Third NF) | No Transitive Dependency | No non-key attribute depends on another non-key attribute. |
| BCNF (Boyce-Codd NF) | Stricter form of 3NF | Every determinant is a candidate key. It resolves some anomalies left by 3NF. |
| 4NF (Fourth NF) | No Multi-valued Dependency | No table contains two or more independent multi-valued dependencies. |
| 5NF (Fifth NF) | Join Dependency | Every join dependency in the table is a consequence of candidate keys. |
Q: 13A functional dependency of the form X → Y is trivial if
Option B
In DBMS, a Functional Dependency (FD) represents a relationship between two sets of attributes in a relation. It is written as X → Y, meaning that the value of attribute set X uniquely determines the value of attribute set Y.
A Trivial functional dependency is one that always holds true and does not provide any new information about the data. It occurs when the dependent attributes (Y) are already included in the determinant (X). In other words, if Y is a subset of X (Y ⊆ X), the dependency is called trivial.
Q: 14DKNF stands for:
Option D
DKNF (Domain-Key Normal Form), is an advanced level of database normalization. A table is said to be in DKNF when all rules about the data can be enforced using only domain rules and key rules.
Domain rules decide what type of values are allowed in a column, and key rules decide how rows are uniquely identified.
Q: 15BCNF stands for:
Option C
In database normalization, BCNF stands for Boyce–Codd Normal Form. BCNF was proposed by Raymond F. Boyce and Edgar F. Codd.
A relation is said to be in BCNF if for every functional dependency, the determinant is a super key.
Means, a relation is in BCNF if for every functional dependency X→Y, X is a candidate key or super key.
So, BCNF ensures that each non—key (non—prime) attributes is dependent only on the candidate key.
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