UGC NET Computer Science Database Management Systems Previous Year Questions (PYQs)

All 137 Database Management Systems PYQs

Ordered newest exam first. Each question links back to the full paper it came from.

1. Question 1

   UGC NET June 2026

   Match List-I with List-II.

   List-I	List-II
   A. Wait-for graph B. Precedence graph C. Rigorous 2PL D. Strict 2PL	I. Holds all locks until it commits II. Holds write-locks until it commits III. Used to check deadlocks IV. Used to check serializability

   1. AA-III, B-IV, C-I, D-II
   2. BA-III, B-IV, C-II, D-I
   3. CA-IV, B-I, C-II, D-III
   4. DA-IV, B-III, C-II, D-I

   Answer: (B) A-III, B-IV, C-II, D-I

   Explanation

   A wait-for graph is used to detect deadlocks, and a precedence graph is used to test serializability. Rigorous 2PL retains all locks until commit, whereas strict 2PL retains write locks until commit.

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2. Question 2

   UGC NET June 2026

   What is the correct sequence of the database-design process?

   A. Create conceptual schema
   B. Data model mapping
   C. Requirements collection and analysis
   D. Physical design

   1. AC, A, B, D
   2. BC, B, A, D
   3. CA, B, C, D
   4. DA, C, B, D

   Answer: (A) C, A, B, D

   Explanation

   Database design starts by collecting and analysing requirements. It then develops a conceptual schema, maps that conceptual model to the selected data model, and finally performs physical design. Therefore the order is C, A, B, D.

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3. Question 3

   UGC NET June 2026

   For relation schema r(M, Y, P, MP, C), the functional dependencies are M → MP, {M, Y} → P, and MP → C. Which is a key?

   1. AM
   2. B{M, Y}
   3. C{M, C}
   4. DMP

   Answer: (B) {M, Y}

   Explanation

   From M, we obtain MP and then C, but not Y or P. Adding Y allows {M, Y} → P, so the closure of {M, Y} contains every attribute. Neither M nor MP supplies Y and P, making {M, Y} the key.

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4. Question 4

   UGC NET June 2026

   Which relational-algebra expression is equivalent to the SQL query SELECT X FROM B WHERE Y?

   1. Aσᵧ(πₓ(B))
   2. Bπᵧ(σₓ(B))
   3. Cσₓ(πₓ(B))
   4. Dπₓ(σᵧ(B))

   Answer: (D) πₓ(σᵧ(B))

   Explanation

   The WHERE predicate is applied first, so σᵧ selects from relation B. The SELECT list is then applied, so πₓ projects attribute X. Reversing that order can discard the attribute needed by the selection.

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5. Question 5

   UGC NET June 2026

   Which of the following are true?

   A. A database may have only one primary key.
   B. Every key is a candidate key.
   C. There can be multiple candidate keys.
   D. There can be more than one superkey.
   E. Every candidate key is a primary key.

   1. AA, C, D only
   2. BA, D, E only
   3. CB, C, D only
   4. DB, D, E only

   Answer: (A) A, C, D only

   Explanation

   A relation design chooses one candidate key as its primary key, but may have several candidate keys and many superkeys. A superkey need not be minimal, so B is false; only the selected candidate key is primary, so E is false.

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6. Question 6

   UGC NET June 2026

   To define the relationship between two entity sets, which operation can be used?

   1. AConcatenation
   2. BAddition
   3. CCartesian product
   4. DUnion

   Answer: (C) Cartesian product

   Explanation

   A relationship set is formally a subset of the Cartesian product of participating entity sets. Concatenation, addition, and union do not construct tuples across the two sets in this way.

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7. Question 7

   UGC NET June 2026

   A safe expression in relational calculus is guaranteed to yield:

   1. AA finite number of tuples
   2. BA finite number of attributes
   3. CA finite number of null values
   4. DA finite number of indices

   Answer: (A) A finite number of tuples

   Explanation

   Safety (range restriction) ensures that query variables range over values drawn from the finite database domain. This prevents an expression from describing infinitely many result tuples.

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8. Question 8

   UGC NET January 2026

   Which of the following statements are correct?

   A. The HAVING CLAUSE is an optional clause which tells Oracle to group rows based on distinct values that exist for specified columns.
   B. Unions can be used in subqueries.
   C. The intersect clause outputs only rows produced by both the queries intersected.
   D. The statement containing a subquery is called a parent statement.
   E. If a select statement is defined as a subquery, the innermost select statement gets executed first.

   1. AA, B, C Only
   2. BC, D, E Only
   3. CB, C, E Only
   4. DA, B, D Only

   Answer: (B) C, D, E Only

   Explanation

   The HAVING clause filters groups that GROUP BY has already formed, it does not itself group rows by distinct column values, so statement A wrongly assigns GROUP BY's job to HAVING.

   • A. Wrongly describes HAVING — grouping rows by distinct values of specified columns is done by GROUP BY, HAVING only restricts which already-formed groups appear in the result.
   • B. Unions in subqueries — a subquery is restricted to a single SELECT expression under standard subquery rules, so it cannot itself contain a UNION of two SELECT statements.
   • C. INTERSECT clause — correct. INTERSECT is a set operator that returns only the rows that appear in both result sets being compared.
   • D. Parent statement — correct. The statement that encloses and uses the result of a subquery is standard terminology called the parent statement, also known as the outer query.
   • E. Innermost SELECT executes first — correct. Nested, non-correlated subqueries are resolved from the inside out, so the innermost SELECT executes first and its result feeds the enclosing query.

   C, D and E hold while A and B do not, so C, D, E Only is correct.

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9. Question 9

   UGC NET January 2026

   Arrange the following steps of the MapReduce framework.

   A. Input Splitting
   B. Reducing
   C. Mapping
   D. Shuffling & Sorting

   1. AD, A, C, B
   2. BB, C, A, D
   3. CA, C, D, B
   4. DC, B, D, A

   Answer: (C) A, C, D, B

   Explanation

   MapReduce processes data in four sequential stages. Input splitting divides the input into fixed-size chunks, mapping applies the map function to each split to produce key-value pairs, shuffling and sorting groups and orders those pairs by key across the cluster, and reducing aggregates the values for each key into the final output.

   A-C-D-B — Input Splitting, Mapping, Shuffling & Sorting, Reducing — is the correct execution order.

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10. Question 10

    UGC NET January 2026

    Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

    Assertion (A): HDFS follows a master-slave architecture with Name Node and Data Nodes.
    Reason (R): Name Node stores metadata, while Data Nodes store actual data blocks.

    1. ABoth A and R are correct and R is the correct explanation of A
    2. BBoth A and R are correct but R is NOT the correct explanation of A
    3. CA is correct but R is not correct
    4. DA is not correct but R is correct

    Answer: (A) Both A and R are correct and R is the correct explanation of A

    Explanation

    HDFS is built as a master-slave system where a single NameNode acts as the master and multiple DataNodes act as slaves, which makes Assertion A correct.

    Reason R correctly explains why this is a master-slave design. The NameNode holds only the metadata, such as the file-to-block mapping and directory structure, while the DataNodes hold and serve the actual data blocks, so the NameNode coordinates while the DataNodes carry out the storage work.

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11. Question 11

    UGC NET January 2026

    In transactions, a write-item(X) command includes the following steps. Arrange the following steps in the correct sequence.

    A. Copy item X from the program variable named X into its correct location in the buffer.
    B. Store the updated block from the buffer back to disk.
    C. Find the address of the disk block that contains item X.
    D. Copy that disk block into a buffer in main memory.

    1. AC, D, A, B
    2. BA, C, D, B
    3. CC, A, B, D
    4. DD, B, C, A

    Answer: (A) C, D, A, B

    Explanation

    A write_item(X) operation first locates the target block on disk, brings it into memory, updates it in memory, and finally writes it back to disk.

    C-D-A-B — find the disk block address, copy the block into a buffer, copy X into its place in the buffer, then store the updated block back to disk — is the correct sequence.

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12. Question 12

    UGC NET January 2026

    If a transaction tₐ holds a lock on some data item d in shared and intention-exclusive (SIX) mode, then which of the following is true for another transaction tᵦ?

    1. Atᵦ can lock d in shared (S) mode
    2. Btᵦ can lock d in exclusive (X) mode
    3. Ctᵦ can lock d in intention-exclusive (IX) mode
    4. Dtᵦ can lock d in intention-shared (IS) mode

    Answer: (D) tᵦ can lock d in intention-shared (IS) mode

    Explanation

    Multi-granularity locking uses a compatibility matrix for the modes IS, IX, S, SIX and X, and a held SIX lock is compatible with only one other mode besides itself, which is IS.

    • A. Shared (S) — incompatible. SIX already implies an exclusive lock is held somewhere below in the hierarchy, so a plain shared request conflicts with it.
    • B. Exclusive (X) — incompatible. X conflicts with every other mode, including SIX.
    • C. Intention-exclusive (IX) — incompatible. IX signals an intent to write at a finer granularity, which conflicts with the exclusive component of SIX.
    • D. Intention-shared (IS) — correct. IS only signals an intent to read at a finer granularity, which does not conflict with the shared or exclusive components of SIX.

    Only intention-shared mode is compatible with a held SIX lock.

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13. Question 13

    UGC NET January 2026

    Specialization process does not allow

    1. Adefining a set of subclasses of an entity type
    2. Bestablishing additional specific attributes with each subclass
    3. Cdefining a generalized entity type from the given entity types
    4. Destablishing additional specific relationship types between each subclass and other entity types or other subclasses

    Answer: (C) defining a generalized entity type from the given entity types

    Explanation

    Specialization is a top-down process that starts from one entity type and breaks it into more specific subclasses, each of which can be given its own extra attributes and relationships, which is exactly what options A, B and D describe.

    Defining a single generalized entity type from a set of given entity types is the reverse, bottom-up process, known as generalization, not specialization.

    So specialization does not allow defining a generalized entity type from the given entity types.

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14. Question 14

    UGC NET January 2026

    Assertion (A): OLAP systems are optimized for complex queries and analysis.
    Reason (R): OLTP systems are designed for transaction processing with large read-only queries.

    1. ABoth A and R are correct and R is the correct explanation of A
    2. BBoth A and R are correct but R is NOT the correct explanation of A
    3. CA is correct but R is not correct
    4. DA is not correct but R is correct

    Answer: (C) A is correct but R is not correct

    Explanation

    OLAP (Online Analytical Processing) systems are indeed built for complex, multidimensional analytical queries, so Assertion A is correct.

    Reason R is not correct as stated. OLTP (Online Transaction Processing) systems are designed to handle a high volume of short, simple read and write transactions such as order entry or fund transfers, not large read-only queries. Large, read-heavy analytical queries over historical data are the hallmark of OLAP and data-warehouse workloads, not OLTP.

    Because A is true and R misdescribes OLTP, the correct choice is that A is correct but R is not correct.

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15. Question 15

    UGC NET January 2026

    In relation to Index structures for files which of the following are true?

    A. An index is an integral part of the data file.
    B. An index can be created in conjunction with virtually any primary record organization.
    C. An index cannot be created dynamically.
    D. An index can be used to enforce a key constraint on an attribute.
    E. An index can be created dynamically.

    1. AA, B, C Only
    2. BB, C, D Only
    3. CB, D, E Only
    4. DA, C, E Only

    Answer: (C) B, D, E Only

    Explanation

    An index is a separate auxiliary structure built on top of a data file, and checking each statement against that fact resolves the set.

    • A. An index is an integral part of the data file. — wrong, an index is a distinct structure stored apart from the data file that only points into it.
    • B. An index can be created in conjunction with virtually any primary record organization. — correct, indexes can be layered on top of sequential, heap, or hashed file organizations alike.
    • C. An index cannot be created dynamically. — wrong, and directly contradicted by statement E.
    • D. An index can be used to enforce a key constraint on an attribute. — correct, a unique index is a standard mechanism for enforcing key uniqueness on an attribute.
    • E. An index can be created dynamically. — correct, indexes can be built or dropped at runtime as access patterns change.

    Since C and E directly contradict each other, only one can be true, and standard index theory confirms E is the correct one. The true statements are B, D and E only.

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16. Question 16

    UGC NET January 2026

    Consider a B⁺ tree in which the maximum number of keys in a node is 5. What is the minimum number of keys in any non-root node?

    1. A1
    2. B2
    3. C3
    4. D4

    Answer: (B) 2

    Explanation

    In a B⁺ tree, the maximum number of keys per node equals the order n minus 1, so a maximum of 5 keys gives n - 1 = 5, meaning order n = 6.

    Minimum keys in a non-root node = ⌈n/2⌉ - 1 = ⌈6/2⌉ - 1 = 3 - 1 = 2

    Every non-root node must therefore hold at least 2 keys to keep the tree at least half full.

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17. Question 17

    UGC NET June 2025

    Consider relation R with attributes A, B, C, D, and E, and dependencies C → F, E → A, EC → D, and A → B. Which option is a key for R?

    1. ACD
    2. BEC
    3. CAE
    4. DAC

    Answer: (B) EC

    Explanation

    EC is a key: E determines A and then B, while EC determines D. Thus EC determines every attribute listed in R. The printed dependency C → F names an attribute omitted from the displayed schema, but it does not change which listed attributes EC determines. None of the other choices contains both E and C, so none can derive all of R's attributes.

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18. Question 18

    UGC NET June 2025

    Let T1 have attributes A, B, C, D, and E with dependencies {EB → C, D → E, EA → B}; T2 has attributes A, B, C, and D with dependencies {C → A, A → B, A → D}. Which statement is true?

    1. AT1 is in 3NF
    2. BT2 is in 3NF
    3. CT1 is not in 3NF
    4. DT1 is in 2NF

    Answer: (C) T1 is not in 3NF

    Explanation

    For T1, AD is a candidate key: D gives E, then EA gives B, and EB gives C. In the dependency EB → C, EB is not a superkey and C is non-prime, so T1 violates 3NF. T1 is therefore not in 3NF.

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19. Question 19

    UGC NET June 2025

    In a relational database, which statement is correct?

    1. AA relation with only two attributes is always in BCNF.
    2. BIf all attributes of a relation are prime attributes, then the relation is in BCNF.
    3. CEvery relation has at least one non-prime attribute.
    4. DBCNF decomposition preserves functional dependencies.

    Answer: (A) A relation with only two attributes is always in BCNF.

    Explanation

    For a two-attribute relation, any nontrivial functional dependency has one attribute determining the other, making that determinant a key. Therefore every nontrivial dependency has a superkey on the left, satisfying BCNF. BCNF decomposition need not preserve every dependency.

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20. Question 20

    UGC NET June 2025

    Consider Employee(eid, eName), Comp(cid, cName), and Own(eid, cid). Which relational-algebra expression returns eid values for people who own every brand?

    1. Aπ_eid(π_eid,cid(Own) ÷ π_cid(Comp))
    2. Bπ_eid(π_eid(Own) × π_cid(Comp))
    3. Cπ_eid(π_eid,cid(Own) × π_cid(Comp))
    4. Dπ_eid(π_eid(Own) × (π_cid,cName(Own) ÷ π_cid(Comp)))

    Answer: (A) π_eid(π_eid,cid(Own) ÷ π_cid(Comp))

    Explanation

    Division returns the eid values paired in Own with every cid in Comp. Projecting eid and cid from Own before dividing supplies exactly the required dividend relation, and the final projection returns only eids.

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21. Question 21

    UGC NET June 2025

    Editorially repaired version of an ambiguous NTA item. In the broad Knowledge Discovery in Databases (KDD) process, consider these statements:

    A. Data transformation is involved.
    B. An online database can serve as a data source for a data warehouse.
    C. Classification is a measure of accuracy.
    D. K-means clustering minimizes within-cluster variance.
    E. Pattern evaluation identifies knowledge using interestingness measures.

    Which statements are true?

    1. AA, B, D, and E only
    2. BB, D, and E only
    3. CA, C, and E only
    4. DA, C, and D only

    Answer: (A) A, B, D, and E only

    Explanation

    Data transformation is a standard KDD step, and online operational databases can provide source data to a warehouse. K-means minimizes within-cluster variance, while pattern evaluation uses interestingness measures. Classification is a predictive task; accuracy is one measure used to evaluate a classifier. Therefore A, B, D, and E are true.

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22. Question 22

    UGC NET June 2025

    Which serializability statements are true: conflict serializability means conflict equivalence to a serial schedule; view serializability means view equivalence to a serial schedule; serial schedules interleave participating transactions; and every conflict-serializable schedule is view-serializable but not conversely?

    1. AA, B, C, and D
    2. BA, B, and C only
    3. CA, B, and D only
    4. DB, C, and D only

    Answer: (C) A, B, and D only

    Explanation

    Conflict serializability and view serializability have the stated equivalence definitions. A serial schedule does not interleave transactions; it completes one before another starts. Conflict equivalence implies view equivalence, but some view-serializable schedules are not conflict-serializable, so A, B, and D are true.

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23. Question 23

    UGC NET January 2025

    Which table contains the primary information in a data warehouse?

    1. ADimension table
    2. BFact table
    3. CLookup table
    4. DPrimary table

    Answer: (B) Fact table

    Explanation

    A fact table stores the central measurable events and foreign keys in a warehouse schema. Dimension tables provide descriptive context for analysing those facts. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.

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24. Question 24

    UGC NET January 2025

    Match List-I with List-II.

    List-I	List-II
    A. Project-join normal form B. Lost-update problem C. Discretionary access control D. Data-manipulation language	I. Concurrency II. Fifth normal form III. SQL IV. Security

    1. AA-II, B-I, C-IV, D-III
    2. BA-I, B-II, C-IV, D-III
    3. CA-I, B-II, C-III, D-IV
    4. DA-IV, B-II, C-III, D-I

    Answer: (A) A-II, B-I, C-IV, D-III

    Explanation

    Project-join normal form is fifth normal form; lost updates are a concurrency issue; discretionary access control is security; and SQL is a data-manipulation language. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.

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25. Question 25

    UGC NET January 2025

    The degree of a relation in a database is

    1. Athe number of tuples in the relation
    2. Bthe number of attributes in the relation
    3. Cthe number of values in a domain
    4. Dthe number of keys in a relation

    Answer: (B) the number of attributes in the relation

    Explanation

    The degree (or arity) of a relation is its number of attributes/columns. The number of tuples/rows is called its cardinality. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.

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26. Practise Database Management Systems in a timed set

27. Question 26

    UGC NET January 2025

    What is the correct sequence of phases of database design?

    A. Physical design
    B. Conceptual design
    C. Logical design
    D. Requirement collection and analysis

    1. AD, A, B, C
    2. BD, B, C, A
    3. CD, B, A, C
    4. DD, A, C, B

    Answer: (B) D, B, C, A

    Explanation

    Database design starts with requirements, then conceptual modelling, logical schema design and finally physical design. This produces D, B, C, A. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.

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28. Question 27

    UGC NET January 2025

    An alternate key in a database table is also called

    1. APrimary key
    2. BCandidate key
    3. CSuper key
    4. DForeign key

    Answer: (B) Candidate key

    Explanation

    Candidate keys are minimal unique identifiers. One is selected as the primary key and the remaining candidate keys are alternate keys. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.

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29. Question 28

    UGC NET January 2025

    In concurrency control, a phantom problem may occur when

    1. Arecords are inserted
    2. Brecords are deleted
    3. Crecords are modified
    4. Drecords are indexed

    Answer: (A) records are inserted

    Explanation

    A phantom occurs when a transaction reruns a range query and sees new rows inserted by another transaction. Inserts change the membership of the queried set. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.

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30. Question 29

    UGC NET January 2025

    INSERT command is used to:

    A. add a single tuple to a relation
    B. add multiple tuples to a relation
    C. add values to specific attributes
    D. insert a new table

    1. AA only
    2. BA and B only
    3. CA, B and C only
    4. DB, C and D only

    Answer: (C) A, B and C only

    Explanation

    INSERT can add one row, multiple rows, or values for specified columns in a row. It does not create a table; CREATE TABLE is used for that purpose. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.

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31. Question 30

    UGC NET January 2025

    Which of the following statements is/are false?

    A. A relation having only key attributes is always in BCNF, since it has no functional dependencies.
    B. If a relation is not in 4NF because of a non-trivial multivalued dependency, it should be decomposed.
    C. Decomposition can remove redundancy caused by a multivalued dependency.
    D. Third normal form is stronger than BCNF.

    1. AA only
    2. BA and B only
    3. CD only
    4. DC only

    Answer: (C) D only

    Explanation

    BCNF is stricter than 3NF, so statement D is false. Statements B and C correctly describe decomposition for non-trivial multivalued dependencies; statement A is true when all attributes are key attributes.

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32. Question 31

    UGC NET January 2025

    Which of the following commands can be used to modify data in an SQL table?

    A. INSERT
    B. DELETE
    C. UPDATE
    D. SELECT

    1. AA, B and D only
    2. BA, C and D only
    3. CB, C and D only
    4. DA, B and C only

    Answer: (D) A, B and C only

    Explanation

    SQL statements that modify table data (Data Manipulation Language / DML write operations) include:

    • INSERT: Adds new rows to a table.
    • DELETE: Removes existing rows from a table.
    • UPDATE: Modifies attribute values in existing rows of a table.

    By contrast, SELECT is a query command (DQL / Data Query Language) used strictly to read and retrieve data without modifying table contents.

    Therefore, commands A, B, and C modify data in an SQL table, making Option D (A, B and C only) the correct choice.

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33. Question 32

    UGC NET January 2025

    Match List-I with List-II.

    List-I	List-II
    A. Hadoop Distributed File System B. Veracity C. Apache Pig D. Apache Hive	I. Trust II. Facebook III. NameNode IV. Yahoo

    1. AA-IV, B-III, C-I, D-II
    2. BA-III, B-I, C-II, D-IV
    3. CA-III, B-I, C-IV, D-II
    4. DA-II, B-III, C-I, D-IV

    Answer: (C) A-III, B-I, C-IV, D-II

    Explanation

    • A. Hadoop Distributed File System → III. Name-Node — HDFS uses a NameNode to manage the filesystem namespace and metadata.
    • B. Veracity → I. Trust — Veracity, one of the Vs of Big Data, refers to the trustworthiness and quality of data.
    • C. Apache Pig → IV. Yahoo — Apache Pig was originally developed at Yahoo Research.
    • D. Apache Hive → II. Facebook — Apache Hive was originally developed at Facebook.

    The correct combination is A-III, B-I, C-IV, D-II, which matches option C.

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34. Question 33

    UGC NET August 2024

    In a schema R(A, B, C, D, E, F, G, H), each field of R
    contains only atomic values.
    F= {CH→G, A →BC, B→CFH, E→A, F→EG} is a set
    of functional dependencies F+ is exactly the set
    of FDs that holds R. The relation R is :

    1. AIn 1NF, but not in 2 NF
    2. BIn 2 NF , but not in 3 NF
    3. CIn 3NF, but not in BCNF
    4. DIn BCNF

    Answer: (A) In 1NF, but not in 2 NF

    Explanation

    D appears on no right-hand side, so every candidate key must contain it; taking closures gives AD, BD, ED and FD as the candidate keys. The prime attributes are therefore A, B, D, E and F, and the non-prime ones are C, G and H. But A → BC makes the non-prime attribute C depend on A alone, a proper subset of the key AD. That partial dependency breaks 2NF, while the relation is already in 1NF because every field is atomic. Option A is correct.

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35. Question 34

    UGC NET August 2024

    Consider a relation schema R=(U, V, W, X, Y, Z), on
    which the following functional dependencies
    hold
    {U→V, VW→X, Y→W; X→U}
    The candidate keys of R are :

    1. AUY, VY
    2. BUY, VY, XY
    3. CUYZ, VYZ, VWZ
    4. DUYZ, VYZ, XYZ

    Answer: (D) UYZ, VYZ, XYZ

    Explanation

    Z appears on no right-hand side, so it must belong to every candidate key. Taking closures, UY⁺ gains V from U → V, W from Y → W and X from VW → X; VY⁺ and XY⁺ reach the same set. Each covers everything except Z, so appending Z gives the candidate keys UYZ, VYZ and XYZ. Option D is correct.

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36. Question 35

    UGC NET August 2024

    Match List-I with List-II.

    List-I	List-II
    (A) Representation of bits	(I) Transport layer
    (B) Physical Address	(II) Physical layer
    (C) Logical Address	(III) Network layer
    (D) Segmentation and reassembly	(IV) Data link layer

    1. A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
    2. B(A)-(II), (B)-(IV), (C)-(III), (D)-(I)
    3. C(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
    4. D(A)-(I), (B)-(IV), (C)-(III), (D)-(II)

    Answer: (B) (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

    Explanation

    Bits are represented on the medium at the physical layer, so (A)-(II). Physical (MAC) addressing belongs to the data-link layer, giving (B)-(IV). Logical (IP) addressing belongs to the network layer, so (C)-(III). Segmentation and reassembly is a transport-layer duty, so (D)-(I). Option B matches all four.

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37. Question 36

    UGC NET August 2024

    Consider a schema R(P, Q, R, S) and the following
    functional dependencies P → Q, Q → R, R → S, S
    → Q. Then decomposition of R (P, Q), R (Q, R)
    and R 3(Q, S) is :

    1. ADependency Preserving and lossless join
    2. BLossless Join but not dependency preserving
    3. CDependency preserving but not lossless Join
    4. DNot dependency preserving and not lossless join

    Answer: (A) Dependency Preserving and lossless join

    Explanation

    The sub-relations share Q, and Q determines R and, through R → S, also S, so rejoining them reconstructs R without spurious tuples: the decomposition is lossless. P → Q survives in R₁(P, Q), Q → R in R₂(Q, R), and R → S together with S → Q is recoverable from R₂ and R₃(Q, S). Every dependency remains enforceable, so the decomposition is both dependency preserving and lossless — option A.

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38. Question 37

    UGC NET August 2024

    What is SQL primarily used for in the context of
    relational databases ?

    1. ATo design user interfaces
    2. BTo create and manipulate databases
    3. CTo display data on web pages
    4. DTo format printed reports

    Answer: (B) To create and manipulate databases

    Explanation

    SQL is the standard language for defining and manipulating relational data: creating tables, inserting, updating and deleting rows, and querying with SELECT. User interfaces, page rendering and report formatting are the work of other tools. Option B is correct.

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39. Question 38

    UGC NET August 2024

    How does a relational database ensure data
    integrity ?

    1. ABy encrypting all data stored
    2. BBy enforcing rules defined in the schema
    3. CBy compressing data for efficient storage
    4. DBy allowing unrestricted access to all users

    Answer: (B) By enforcing rules defined in the schema

    Explanation

    Relational databases protect integrity through schema-level constraints — primary keys, foreign keys, unique, not-null and check constraints — which the engine enforces on every write. Encryption protects confidentiality, compression saves space, and unrestricted access would weaken integrity rather than ensure it. Option B is correct.

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40. Question 39

    UGC NET August 2024

    Fifth normal form is concerned with :

    1. AJoin Dependency
    2. BDomain-Key
    3. CMultivalued dependency
    4. DFunctional dependency

    Answer: (A) Join Dependency

    Explanation

    Fifth normal form, also called project-join normal form, is concerned with join dependency: a relation is in 5NF when it cannot be decomposed further without losing information. Multivalued dependency belongs to 4NF, and functional dependency to 2NF, 3NF and BCNF. Option A is correct.

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41. Question 40

    UGC NET August 2024

    Consider the transactions T₁, T₂, T₃, and the schedules S₁ and S₂ given below.

    T₁: r₁(X); r₁(Z); w₁(X); w₁(Z)
    T₂: r₂(Y); r₂(Z); w₂(Z)
    T₃: r₃(Y); r₃(X); w₃(Y)

    S₁: r₁(X); r₃(Y); r₃(X); r₂(Y); r₂(Z); w₃(Y); w₂(Z); r₁(Z); w₁(X); w₁(Z)
    S₂: r₁(X); r₃(Y); r₂(Y); r₃(X); r₁(Z); r₂(Z); w₃(Y); w₁(X); w₂(Z); w₁(Z)

    Which one of the following statements about the schedules is TRUE?

    1. AOnly S₁ is conflict-serializable
    2. BOnly S₂ is conflict-serializable
    3. CBoth S₁ and S₂ are conflict-serializable
    4. DNeither S₁ nor S₂ is conflict-serializable

    Answer: (A) Only S₁ is conflict-serializable

    Explanation

    To determine conflict serializability, construct the precedence (serialization) graph for each schedule by identifying conflicting operations (different transactions accessing the same variable with at least one write):

    For Schedule S₁:

    • On X: r₃(X) precedes w₁(X) → T₃ → T₁
    • On Y: r₂(Y) precedes w₃(Y) → T₂ → T₃
    • On Z: r₂(Z) and w₂(Z) precede r₁(Z) and w₁(Z) → T₂ → T₁

    Edges in S₁: T₂ → T₃ → T₁ and T₂ → T₁.
    There is no cycle, so S₁ is conflict-serializable (equivalent serial order: T₂ → T₃ → T₁).

    ---

    For Schedule S₂:

    • On X: r₃(X) precedes w₁(X) → T₃ → T₁
    • On Y: r₂(Y) precedes w₃(Y) → T₂ → T₃
    • On Z:
      • r₁(Z) precedes w₂(Z) → T₁ → T₂
      • w₂(Z) precedes w₁(Z) → T₂ → T₁

    Edges in S₂ contain a direct cycle between T₁ and T₂ (T₁ ⇌ T₂).
    Therefore, S₂ is not conflict-serializable.

    Hence, only S₁ is conflict-serializable, which corresponds to Option A.

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42. Question 41

    UGC NET August 2024

    Which of the following relations can not be
    decomposed in to BCNF with a lossless join and
    dependency-preserving decomposition?

    1. AR (V,W,X,Y,Z) {VW→X, WX→V, VX→W, W→Y, Y→Z}
    2. BR (V,W,X,Y,Z) {VW→W, X→Y}
    3. CR (V,W,X,Y,Z) {VW→X,X→VY}
    4. DR (V,W,X,Y,Z) {VW→X, X→Y, Y→Z, Z→V}

    Answer: (A) R (V,W,X,Y,Z) {VW→X, WX→V, VX→W, W→Y, Y→Z}

    Explanation

    For R(V, W, X, Y, Z) with {VW→X, WX→V, VX→W, W→Y, Y→Z}, the closure of VW reaches every attribute, so VW is a candidate key. But W → Y and Y → Z both have non-superkey left-hand sides, violating BCNF. Any decomposition that removes them separates W from Y or Y from Z, and the lost dependency cannot be re-derived from the fragments. This relation therefore admits no lossless, dependency-preserving BCNF decomposition — option A.

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43. Question 42

    UGC NET August 2024

    Which of the following best describes the
    structure of a relational database ?

    1. AData organized into tables with rows and columns
    2. BData organized into files and folders
    3. CData organized into a hierarchical tree structure
    4. DData organized into a network of interconnected nodes

    Answer: (A) Data organized into tables with rows and columns

    Explanation

    A relational database stores data in tables of rows and columns, where a row is a record and a column an attribute, with relationships expressed through keys. Files and folders, tree hierarchies and node networks describe other data models. Option A is correct.

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44. Question 43

    UGC NET August 2024

    Match List-I with List-II.

    List-I (operating system concepts)	List-II (characteristics)
    (A) Paging	(I) Evicts least recently used process
    (B) LRU (Least Recently Used)	(II) Extends physical memory
    (C) C-SCAN	(III) Logical to physical mapping
    (D) Virtual Memory	(IV) Circular disk access

    1. A(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
    2. B(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
    3. C(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
    4. D(A)-(I), (B)-(IV), (C)-(III), (D)-(II)

    Answer: (B) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

    Explanation

    Paging maps logical addresses onto physical frames, so (A)-(III). LRU evicts the least recently used entry, giving (B)-(I). C-SCAN sweeps the disk in one circular direction, so (C)-(IV). Virtual memory makes the address space appear larger than physical memory, so (D)-(II). Option B is correct.

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45. Question 44

    UGC NET August 2024

    Arrange the following phases of database design
    in the correct order :
    A. Physical Design
    B. Conceptual Design
    C. Logical Design
    D. Requirement Analysis

    1. A(B), (D), (A), (C)
    2. B(C), (A), (B), (D)
    3. C(D), (B), (C), (A)
    4. D(A), (D), (C), (B)

    Answer: (C) (D), (B), (C), (A)

    Explanation

    Database design proceeds: requirement analysis (D), conceptual design (B) producing an ER model, logical design (C) mapping that model to relations, then physical design (A) choosing storage structures and indexes. That order is option C.

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46. Question 45

    UGC NET December 2023

    The Hue of a colour is related to its :

    1. ALuminance
    2. BSaturation
    3. CIncandescence
    4. DWavelength

    Answer: (D) Wavelength

    Explanation

    The key idea is stated in the problem: The Hue of a colour is related to its :. The correct answer is D: Wavelength. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.

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47. Question 46

    UGC NET December 2023

    Which of the statement are CORRECT ?
    A. Constructors are invoked automatically when the objects are created.
    B. Constructors do not have return types, not even void and therefore they cannot return values.
    C. Constructors cannot be inherited though a derived class can call the base class constructors.
    D. Constructors can be declared as virtual. Choose the correct answer from the options given below :

    1. A(A), (B) and (D) Only
    2. B(A), (B) and (C) Only
    3. C(B), (C) and (D) Only
    4. D(A), (C) and (D) Only

    Answer: (B) (A), (B) and (C) Only

    Explanation

    The key idea is stated in the problem: Which of the statement are CORRECT ?
    A. Constructors are invoked automatically when the objects are created.
    B. Constructors do not have return types, not even void and therefore they cannot return values.
    C. Constructors cannot be inherited though a derive. The correct answer is B: (A), @) and (© Only. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.

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48. Question 47

    UGC NET December 2023

    Which one of the following statements are CORRECT ?
    A. Granularity is the size of data item in a database.
    B. Two operations in a schedule are said to be conflict if they belong to same transaction.
    C. Two schedulers are said to be conflict equivalent if the order of any two conflicting operations is the same in both schedules.
    D. Write operations which are performed without performing the write operation are known as Blind Writes. Choose the correct answer from the options given below :

    1. A(A) and @) Only
    2. B(A), (B) and (C) Only
    3. C(A), (B) and (D) Only
    4. D(B) and (C) Only

    Answer: (B) (A), (B) and (C) Only

    Explanation

    The key idea is stated in the problem: Which one of the following statements are CORRECT ?
    A. Granularity is the size of data item in a database.
    B. Two operations in a schedule are said to be conflict if they belong to same transaction.
    C. Two schedulers are said to be conflict equivalent if th. The correct answer is B: (A), (B) and (C) Only. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.

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49. Question 48

    UGC NET December 2023

    Which of the following is/are NOT CORRECT statement ?
    A. The first record in each block of the data file is known as actor record.
    B. Dense index has index entries for every search key value in the data file.
    C. Searching is harder in the B* tree than B— tree as the all external nodes linked to each other.
    D. In extendible hashing the size of directory is just an array of 24—, where d is global depth. Choose the correct answer from the options given below :

    1. A(A), (B) and (©) Only
    2. B(A), (©) and (D) Only
    3. C(A), (B) and (D) Only
    4. D(A), (B), ©) and (D) Only

    Answer: (B) (A), (©) and (D) Only

    Explanation

    The key idea is stated in the problem: Which of the following is/are NOT CORRECT statement ?
    A. The first record in each block of the data file is known as actor record.
    B. Dense index has index entries for every search key value in the data file.
    C. Searching is harder in the B* tree than B— tr. The correct answer is B: (A), (©) and (D) Only. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.

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50. Question 49

    UGC NET December 2023

    Arrange the following in ascending order :
    A. Remainder of 49!© when divided by 17
    B. Remainder of 24° when divided by 9
    C. Remainder of 155!" when divided by 17
    D. Last digits of the number 74° Choose the correct answer from the options given below : Q)
    A. @),.©,®) Q
    A. 8,0), ©

    1. AA, B, C, D
    2. BA, B, D, C
    3. CA, C, B, D
    4. DD, C, B, A

    Answer: (C) A, C, B, D

    Explanation

    The key idea is stated in the problem: Arrange the following in ascending order :
    A. Remainder of 49!© when divided by 17
    B. Remainder of 24° when divided by 9
    C. Remainder of 155!" when divided by 17
    D. Last digits of the number 74° Choose the correct answer from the options given below : Q)
    . The correct answer is C: A, C, B, D. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.

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51. Question 50

    UGC NET December 2023

    In most general case, the computer needs to process each instruction with the following sequence of steps :
    A. Calculate the effective address
    B. Execute the instruction
    C. Fetch the instruction from memory
    D. Fetch the operand from memory
    E. Decode the instruction Choose the correct answer from the options given below : @)
    A. , ®), ©, 0), ©

    1. AA, B, C, D, E
    2. BA, B, C, E, D
    3. CC, E, A, D, B
    4. DC, E, D, A, B

    Answer: (C) C, E, A, D, B

    Explanation

    The key idea is stated in the problem: In most general case, the computer needs to process each instruction with the following sequence of steps :
    A. Calculate the effective address
    B. Execute the instruction
    C. Fetch the instruction from memory
    D. Fetch the operand from memory
    E. Decode the i. The correct answer is C: C, E, A, D, B. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.

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52. Practise Database Management Systems in a timed set

53. Question 51

    UGC NET December 2023

    Match List-I with List-II.

    Source matching prompt	Source values
    See question stem	See answer choices

    1. A(D) Depth first search (IV) O(b' m) Choose the correct answer from the options given below :
    2. B(A)-C, B)-), (CCV), (D)-(Im)
    3. C(A), )-€V), (©), (P)-(Im)
    4. D(A)-(), (B)-(H), (C)-(D, (D)-@)

    Answer: (A) (D) Depth first search (IV) O(b' m) Choose the correct answer from the options given below :

    Explanation

    The key idea is stated in the problem: Match List - I with List - II. List - I List - II
    A. Hillclimbing () Ob' d)
    B. Best first search (I) O(bd)
    C. A* Search (M1) O. The correct answer is A: (D) Depth first search (IV) O(b' m) Choose the correct answer from the options given below :. It follows the governing definition, algorithm, or calculation in this database management systems item; the other choices fail to satisfy one or more stated conditions.

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54. Question 52

    UGC NET June 2023

    Which of the following concurrency control protocols ensures both conflict serializability and freedom from deadlock?

    Statement I: Two phase locking
    Statement II: Timestamp ordering

    1. ABoth Statement I and Statement II are correct
    2. BBoth Statement I and Statement II are incorrect
    3. CStatement I is correct but Statement II is incorrect
    4. DStatement I is incorrect but Statement II is correct

    Answer: (D) Statement I is incorrect but Statement II is correct

    Explanation

    A concurrency control protocol must guarantee conflict serializability and also avoid deadlock to satisfy both required properties.

    • Statement I: Two phase locking — 2PL guarantees conflict serializability because every transaction's lock and unlock phases enforce a serializable order, but basic or strict 2PL can still deadlock when two transactions each hold a lock the other is waiting for.
    • Statement II: Timestamp ordering — this protocol assigns each transaction a fixed timestamp and orders conflicting operations by timestamp, guaranteeing conflict serializability, and because it never uses locks, transactions never wait on each other, so deadlock cannot occur.

    Statement I is incorrect because 2PL does not eliminate deadlock, and Statement II is correct because timestamp ordering achieves both properties simultaneously.

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55. Question 53

    UGC NET June 2023

    The total cost of retrieving records in sorted order using an unclustered B+ tree index is given by which of the following expressions?
    (P = average number of records per data page, N = number of data pages, F = ratio of the size of a data entry to the size of a data record)

    1. A(F × N) + P
    2. B(F + P) × N
    3. CF × N × P
    4. DF + P/N

    Answer: (B) (F + P) × N

    Explanation

    Retrieving all records through an unclustered index requires scanning the index's leaf level and then fetching each actual record separately, since an unclustered index gives no guarantee that neighboring entries share a data page.

    The leaf level of the index holds smaller entries than the full records, so it spans about F × N pages, and the number of full records to fetch is N × P, since there are P records on average per data page across N pages.

    Total cost = (F × N) + (N × P), which factors as (F + P) × N.

    • A. (F × N) + P — adds only P once instead of the full N × P record-fetch cost, understating the total.
    • B. (F + P) × N — correct, it expands to F × N + P × N, matching the leaf-scan cost plus the per-record fetch cost.
    • C. F × N × P — multiplies all three quantities together, which does not correspond to any additive cost component of the index scan.
    • D. F + P/N — has no basis in either the index-traversal cost or the record-retrieval cost, and does not scale with N the way total I/O cost should.

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56. Question 54

    UGC NET June 2023

    Consider the following statements regarding normal forms:
    A. A database design is in BCNF if each member set of relation schemas constituting the design is in BCNF
    B. A BCNF schema can have a transitive dependency
    C. It is always possible to obtain a 3NF design without sacrificing a lossless join
    D. There can be multivalued dependencies in a 4NF relation

    Which of the following statements are correct?

    1. AA, B, C only
    2. BB, C, D only
    3. CA, B, D only
    4. DA, C, D only

    Answer: (D) A, C, D only

    Explanation

    Each statement needs to be checked individually against the exact definitions of BCNF, 3NF, and 4NF.

    • A. correct. A database design, a set of relation schemas, is defined to be in BCNF precisely when every individual schema in that set satisfies the BCNF condition.
    • B. incorrect. A transitive dependency requires a non-key attribute to depend on another non-key attribute that is not a candidate key, but BCNF requires every determinant to be a candidate key, so no such non-key determinant can exist in a BCNF relation.
    • C. correct. It is a standard decomposition theorem that any relation can always be losslessly decomposed into a set of 3NF schemas, unlike BCNF, which can sometimes only be achieved by sacrificing dependency preservation.
    • D. correct. 4NF is defined specifically in terms of multivalued dependencies, requiring that for every non-trivial multivalued dependency the determinant is a superkey, so multivalued dependencies can still be present as long as they satisfy this condition.

    Since A, C, and D are correct and B is false, the correct set is A, C, D only.

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57. Question 55

    UGC NET June 2023

    Let R(A, B, C, D, E, F) be a relation with functional dependencies C → F, E → A, EC → D, A → B. Which of the following is a key for R?

    1. ACD
    2. BEC
    3. CAE
    4. DAC

    Answer: (B) EC

    Explanation

    A candidate key's closure must include all six attributes of R: A, B, C, D, E, F.

    Compute the closure of EC:
    {E, C} → add A (E → A) → add F (C → F) → add B (A → B, now that A is present) → add D (EC → D)
    {E, C}+ = {A, B, C, D, E, F} — all six attributes, so EC is a key.

    • A. CD — {C, D}+ only adds F (via C → F), giving {C, D, F}. It never reaches A, B, or E, so it is not a key.
    • B. EC — correct, as shown above.
    • C. AE — {A, E}+ adds B (via A → B), giving {A, B, E}, but never reaches C, D, or F.
    • D. AC — {A, C}+ adds B (via A → B) and F (via C → F), giving {A, B, C, F}, but never reaches D or E.

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58. Question 56

    UGC NET June 2023

    Given the basic ER diagram and relational model, which of the following is incorrect?

    1. AAn attribute of an entity can have more than one value
    2. BAn attribute of an entity can be composite
    3. CIn a row of relational table, an attribute can have more than one value
    4. DIn a row of a relational table, an attribute can have exactly one value or a NULL value

    Answer: (C) In a row of relational table, an attribute can have more than one value

    Explanation

    The ER model and the relational model make different promises about attribute values.

    At the ER (conceptual) level, an entity's attribute can be multivalued (Option A) or composite, made up of sub-parts (Option B) — both are standard, correct ER constructs.

    The relational model, however, requires every attribute value in a row to be atomic under first normal form (1NF). A single cell cannot hold more than one value, so Option C is incorrect — this is the statement the question asks you to identify.

    Option D correctly reflects 1NF: a relational attribute holds exactly one atomic value, or NULL if the value is unknown or missing.

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59. Question 57

    UGC NET June 2023

    Which of the following scenario may lead to an irrecoverable error in a database system?

    1. AA transaction writes a data item after it is read by an uncommitted transaction
    2. BA transaction reads a data item after it is read by an uncommitted transaction
    3. CA transaction reads a data item after it is written by a committed transaction
    4. DA transaction reads a data item after it is written by an uncommitted transaction

    Answer: (D) A transaction reads a data item after it is written by an uncommitted transaction

    Explanation

    A schedule is recoverable only if, whenever a transaction Tj reads a data item written by another transaction Ti, Ti commits before Tj commits. If Tj reads data written by an uncommitted Ti and Tj is allowed to commit first, the schedule becomes irrecoverable, since Ti might later abort and there is no way to undo the already-committed Tj.

    • A. A transaction writes a data item after it is read by an uncommitted transaction — a read-write conflict, but it creates no dependency on an uncommitted write being read, so it does not by itself cause irrecoverability.
    • B. A transaction reads a data item after it is read by an uncommitted transaction — reads never conflict with other reads, so this creates no recoverability issue at all.
    • C. A transaction reads a data item after it is written by a committed transaction — a completely safe read, since the writer has already committed.
    • D. A transaction reads a data item after it is written by an uncommitted transaction — correct. This is a dirty read, and if the reader commits before the writer's transaction resolves, the schedule cannot be recovered if the writer later aborts.

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60. Question 58

    UGC NET June 2023

    Consider a relational schema R(A, B, C, D). It has functional dependencies A → B, B → C, C → D, and D → B. The decomposition of R into (A, B), (B, C), (B, D)

    1. Agives a lossless join, and is dependency preserving
    2. Bgives lossless join, but is not dependency preserving
    3. Cdoes not give a lossless join, but is dependency preserving
    4. Ddoes not give a lossless join and is not dependency preserving

    Answer: (A) gives a lossless join, and is dependency preserving

    Explanation

    R(A, B, C, D) has functional dependencies A → B, B → C, C → D, D → B. Computing A⁺ gives {A, B, C, D}, so A is the only key of R.

    Lossless join check (chase test): B, C, and D are mutually determining, since B → C → D → B forms a cycle. Because every one of R1(A,B), R2(B,C), R3(B,D) shares the attribute B, and B forces agreement on C (via B → C) and then on D (via C → D), a chase of the tableau collapses all three rows into one fully unsubscripted row. That confirms the join is lossless.

    Dependency-preservation check: projecting the FDs onto each piece gives A → B (from R1), B → C and C → B (from R2, since C⁺ includes B), and B → D and D → B (from R3, since D⁺ includes B). Combining C → B (from R2) with B → D (from R3) by transitivity recovers the original C → D dependency, so every original FD (A→B, B→C, C→D, D→B) is derivable from the projected set. The decomposition preserves all dependencies.

    • A. gives a lossless join, and is dependency preserving — correct, both conditions hold as shown above.
    • B. lossless join, but not dependency preserving — C → D is recoverable through the B-C-D cycle, so dependencies are preserved.
    • C. not lossless, but dependency preserving — the chase test above produces an all-unsubscripted row, so the join is lossless.
    • D. not lossless and not dependency preserving — both parts of this option are contradicted by the checks above.

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61. Question 59

    UGC NET June 2023

    Which one of the following is NOT a part of ACID properties of a database transaction?

    1. AAtomicity
    2. BConsistency
    3. CIsolation
    4. DDeadlock-freedom

    Answer: (D) Deadlock-freedom

    Explanation

    ACID is the standard acronym for the four properties a database transaction must guarantee: Atomicity (a transaction completes entirely or not at all), Consistency (a transaction moves the database between valid states), Isolation (concurrent transactions don't interfere with each other's intermediate results), and Durability (a committed transaction's effects survive failures). Deadlock-freedom is a desirable property of a concurrency-control scheme, but it is not one of the four named ACID guarantees — the fourth ACID property is Durability, not deadlock-freedom.

    • A. Atomicity — a genuine ACID property.
    • B. Consistency — a genuine ACID property.
    • C. Isolation — a genuine ACID property.
    • D. Deadlock-freedom — correct. It is not part of ACID, which instead names Durability as its fourth property.

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62. Question 60

    UGC NET March 2023

    Identify the incorrect statement(s).

    1. AA candidate key is minimal set of one or more attributes that, taken collectively, allows us to uniquely identify any entity in the entity set.
    2. BA candidate key for which no proper subset is also a candidate key is called a super key.
    3. CA super key is a set of one or more attributes that, taken collectively, allows us to uniquely identify any entity in the entity set.
    4. DA super key for which no proper subset is also a super key is called a candidate key.

    Answer: (B) A candidate key for which no proper subset is also a candidate key is called a super key.

    Explanation

    Check each statement against the standard definitions: a super key uniquely identifies an entity, and a candidate key is a minimal super key (one with no proper subset that is also a super key).

    • A. Candidate key is a minimal uniquely-identifying attribute set — correct, this is the standard definition.
    • B. A candidate key with no proper-subset candidate key is called a super key — incorrect. This statement swaps the definitions: minimality applied to a candidate key does not produce a super key, it just restates what a candidate key already is.
    • C. A super key uniquely identifies an entity — correct, this is the standard definition.
    • D. A super key with no proper-subset super key is called a candidate key — correct, this is exactly the minimality condition that defines a candidate key.

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63. Question 61

    UGC NET March 2023

    Consider the two relations below. The primary keys are underlined. Identify all possible foreign key(s) from the options based only on the two relations.

    EMP (eid, ename, did)
    DEPT (did, dname)

    1. Aeid
    2. Bdid
    3. Ceid, did
    4. Deid, did, ename

    Answer: (B) did

    Explanation

    A foreign key in one relation must reference the primary key of another relation.

    EMP.did references DEPT's primary key did, giving a valid foreign key relationship. EMP's own primary key eid does not reference any other relation's primary key, so it cannot be the foreign key.

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64. Question 62

    UGC NET March 2023

    Which of the following statement is correct, according to the following instance of the relational schema R(X, Y, Z)?

    X	Y	Z
    1	1	a
    1	1	a
    2	1	b
    3	2	b

    Instance of relation R(X, Y, Z)

    1. AX→Y, Z→X
    2. BY→Z, Z→X
    3. CX→Y, X→Z
    4. DY→X, X→Z

    Answer: (C) X→Y, X→Z

    Explanation

    Check each candidate dependency against the instance (rows (1,1,a), (1,1,a), (2,1,b), (3,2,b)) by seeing whether every occurrence of the left side always pairs with the same right side.

    X→Y holds because X=1 always pairs with Y=1, X=2 with Y=1, and X=3 with Y=2, each X value maps to exactly one Y value.

    X→Z holds because X=1 always pairs with Z=a, X=2 with Z=b, and X=3 with Z=b, each X value maps to exactly one Z value.

    Y→Z fails, because Y=1 appears with both Z=a (when X=1) and Z=b (when X=2). Y→X and Z→X fail for the same reason: Y=1 maps to two different X values, and Z=b maps to both X=2 and X=3.

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65. Question 63

    UGC NET March 2023

    What are the drawbacks of using file systems to store data?

    A. Data inconsistency
    B. Difficulty in accessing data
    C. Data isolation
    D. Atomicity of updates

    Choose the correct answers from the options given below:

    1. AA, B, only
    2. BB, C, D only
    3. CA, B, C only
    4. DA, B, C, D

    Answer: (D) A, B, C, D

    Explanation

    File-based systems predate database management systems because they suffer from a well-known set of problems that DBMSs were built to solve.

    • A. Data inconsistency — happens when the same fact is duplicated across files and updated in only one place, leaving other copies stale.
    • B. Difficulty in accessing data — comes from needing a new program for every new type of query, since file systems have no general query language.
    • C. Data isolation — happens because related data is scattered across files in different, incompatible formats.
    • D. Atomicity of updates — is hard to guarantee because a crash mid-update can leave files only partially changed.

    All four are standard, textbook-listed drawbacks of file systems, so the answer includes all of A, B, C, and D.

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66. Question 64

    UGC NET March 2023

    Which of the following database model, we have a parent-child relationship?

    A. hierarchical databases
    B. network databases
    C. relational databases

    Choose the correct answer from the options given below:

    1. AA, B
    2. BA, C
    3. CB, C
    4. DA, B, C

    Answer: (A) A, B

    Explanation

    The hierarchical model organizes records in a strict tree, so every record has exactly one parent, which is the textbook definition of a parent-child relationship.

    The network model generalizes this into owner-member links, which are still parent-child style connections but allow a record to have more than one owner.

    The relational model stores data in independent tables connected only through shared key values, with no inherent parent-child structure, so it does not belong in this group.

    A and B are the models built around parent-child links.

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67. Question 65

    UGC NET March 2023

    Identify the correct operation which produces the given output based on two relations R1 and R2.

    Relation	Name	Class
    R1	a1	9
    R1	b1	5
    R1	c1	6
    R1	d1	11
    R2	b1	5
    R2	f1	8
    Output	b1	5

    Relations R1, R2 and the resulting output

    1. AR1 ∩ R2
    2. BR1 − R2
    3. CR1 − (R1 − R2)
    4. DR1 ∪ R2

    Accepted answers: (A) R1 ∩ R2, (C) R1 − (R1 − R2)

    Explanation

    Compare the tuples in R1 and R2 against the output, because the operation is identified by which rows survive.

    R1 contains (a1,9), (b1,5), (c1,6), (d1,11), and R2 contains (b1,5), (f1,8). The output contains only (b1,5), which is the single tuple that appears in both relations.

    Keeping only the tuples common to both relations is exactly the intersection operation, R1 ∩ R2.

    R1 − (R1 − R2) computes the same result by a different route: R1 − R2 removes every R1 tuple that is also in R2, leaving only R1's own tuples, and subtracting that from R1 a second time leaves exactly the tuples R1 shares with R2. So R1 − (R1 − R2) is set-theoretically identical to R1 ∩ R2, and both are accepted here.

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68. Question 66

    UGC NET March 2023

    Match List I with List II.

    List-I	List-II
    A. Apache Kafka	I. Platform for constructing data flows for extract, transform, and load (ETL) processing and analysis of large datasets
    B. Pig	II. General-purpose computing model and runtime system for distributed data analytics
    C. Apache Mahout	III. Open-source platform that was created by LinkedIn
    D. Mapreduce	IV. Open-source platform used for creating scalable machine learning algorithms

    List I and List II for the matching question

    1. AA-III, B-IV, C-I, D-III
    2. BA-II, B-I, C-IV, D-III
    3. CA-III, B-I, C-IV, D-II
    4. DA-I, B-II, C-IV, D-III

    Answer: (C) A-III, B-I, C-IV, D-II

    Explanation

    Match each big data tool to what it is actually known for.

    A-III, B-I, C-IV, D-II

    Apache Kafka was created by LinkedIn as a distributed messaging platform, matching III. Pig is a platform for constructing extract-transform-load data flows over large datasets, matching I. Apache Mahout is an open-source platform for creating scalable machine learning algorithms, matching IV. MapReduce is the general-purpose computing model and runtime system for distributed data analytics, matching II.

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69. Question 67

    UGC NET March 2023

    A ______ operation combines the rows of the two relations and outputs a new relation that has both input relations' rows in it.

    1. Adifference
    2. Bunion
    3. Ccartesian product
    4. Dproject

    Answer: (B) union

    Explanation

    The union operation takes two union-compatible relations and produces a single relation containing every row that appears in either input, which is exactly combining both relations' rows into one output.

    Difference instead removes rows of one relation that also appear in the other, cartesian product pairs every row of one relation with every row of the other rather than merging rows, and project selects a subset of columns rather than combining rows from two relations.

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70. Question 68

    UGC NET March 2023

    Consider the relation R(P, Q, R, S, T, U, V, W, X, Y) and FDs:

    {P, Q, S} → T
    R → Y
    {P, Q} → V
    {R, X} → X
    Q → U
    V → {W, X}

    Which of the following is a candidate key?

    1. A{P, Q, R, S}
    2. B{R, Q, X}
    3. C{V, X}
    4. DNone of the above

    Answer: (A) {P, Q, R, S}

    Explanation

    Computing the closure of {P,Q,R,S} shows it determines every attribute of R, and no proper subset of it does the same, which makes it a minimal superkey, that is, a candidate key.

    Closure of {P,Q,R,S}: apply {P,Q,S}→T to add T, then R→Y to add Y, then {P,Q}→V to add V, then Q→U to add U, then V→{W,X} to add W and X, reaching all ten attributes {P,Q,R,S,T,U,V,W,X,Y}

    Dropping any single attribute breaks this. Removing R leaves {P,Q,S}, whose closure never reaches Y since only R→Y supplies it. Removing S leaves {P,Q,R}, whose closure never reaches T since {P,Q,S}→T needs S. Removing P or Q similarly blocks {P,Q,S}→T and {P,Q}→V, so all four attributes are necessary.

    By contrast, {R,Q,X}'s closure only reaches {R,Q,X,Y,U} because it lacks P to trigger {P,Q}→V, and {V,X}'s closure only reaches {V,X,W} since no other FD's left side is satisfied, so neither is a superkey.

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71. Question 69

    UGC NET March 2023

    Let Ti be a transaction that transfer Rs. 50 from account A to account B. This transaction can be defined as follows:

    What will be the status of a database just after 3rd step but before 5th step of transaction execution?

    1. AAlways consistent
    2. BMay be temporarily inconsistent
    3. CMust be abort
    4. DAlways rollback

    Answer: (B) May be temporarily inconsistent

    Explanation

    After step 3 (Write(A), where A has already been debited by 50) but before step 5 (B:=B+50), account A reflects its reduced balance while account B has not yet been credited.

    At this intermediate point, the invariant that the combined total of A and B stays constant does not hold, since A is already short by 50 and B has not caught up, which is a temporary inconsistency inherent to the transaction being mid-execution.

    This is expected and resolves itself once the transaction completes through step 6, restoring the invariant, so the database is not always consistent at this point, but it also has no reason to abort or roll back purely because it is partway through a normal, uninterrupted execution.

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72. Question 70

    UGC NET March 2023

    Consider the relation

    Student(stuid, dept, clsroll)

    The following functional dependencies are given:

    FD1: stuid → dept
    FD2: dept → stuid
    FD3: stuid → clsroll
    FD4: clsroll → stuid
    FD5: dept → clsroll

    Which functional dependencies should be removed to obtain the canonical cover of the set?

    A. FD2 and FD3
    B. FD1 and FD5
    C. FD5
    D. FD1

    Choose the correct answer from the options given below:

    1. AA, B only
    2. BA, C only
    3. CB, C only
    4. DA, B, C only

    Answer: (B) A, C only

    Explanation

    Because stuid, dept, and clsroll mutually determine each other in a 3-way cycle, more than one minimal cover exists depending on which redundant FD is eliminated first, and this question asks which of the listed removal sets can legitimately arise.

    Processing FD1 through FD5 in order and eliminating each FD found to be derivable from the rest of the current set: FD2 (dept→stuid) becomes derivable once FD5 (dept→clsroll) and FD4 (clsroll→stuid) are both present, since dept→clsroll→stuid. After removing FD2, FD3 (stuid→clsroll) becomes derivable from FD1 (stuid→dept) and FD5 (dept→clsroll). This leaves {FD1, FD4, FD5} as one valid canonical cover, matching removal set A (FD2 and FD3).

    Alternatively, checking FD5 (dept→clsroll) first against the full original set shows it is already derivable via FD2 (dept→stuid) then FD3 (stuid→clsroll), so removing FD5 alone also yields a valid minimal cover, {FD1, FD2, FD3, FD4}, matching removal set C (FD5).

    In both valid computations, FD1 (stuid→dept) is never redundant, since no other combination of the remaining dependencies can derive it, which rules out any removal set containing FD1, such as B (FD1 and FD5) or D (FD1). This leaves A and C as the two legitimate canonical-cover removal sets.

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73. Question 71

    UGC NET March 2023

    Match List I with List II.

    List-I	List-II
    A. Timestamp protocol	I. As long as the transaction is already running, it will try to obtain the required lock in the first part
    B. Strict two-phase locking protocol	II. Indicates final lock of the schedule
    C. Two-phase locking	III. Ensures freedom from deadlock
    D. Lock point in two phase locking protocol	IV. Ensures cascade-less rollback

    1. AA-III, B-IV, C-II, D-I
    2. BA-III, B-IV, C-I, D-II
    3. CA-IV, B-III, C-II, D-I
    4. DA-IV, B-II, C-I, D-III

    Answer: (B) A-III, B-IV, C-I, D-II

    Explanation

    Each concurrency-control concept is matched to the specific property it is known for.

    Timestamp-ordering protocols never make transactions wait for locks, so they are free from deadlock by construction, giving A-III.

    Strict two-phase locking holds all locks, especially write locks, until commit, which prevents any other transaction from reading uncommitted data and therefore guarantees cascade-less rollback, giving B-IV.

    Basic two-phase locking is defined by a growing phase in which a running transaction acquires all the locks it will need before releasing any, giving C-I.

    The lock point is specifically defined as the moment a transaction obtains its last (final) lock in the schedule, marking the boundary between the growing and shrinking phases, giving D-II.

    This gives the mapping A-III, B-IV, C-I, D-II.

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74. Question 72

    UGC NET March 2023

    Consider schedule S.

    R(X) denotes read operation on data item X by transaction Ti.
    W(X) denotes write operation on data item X by transaction Ti.

    Choose the correct execution order for schedule S.

    T1	T2	T3	T4	Timestamp
    	R(X)	R(Y)		t1
    	W(X)	W(Y)		t2
    R(Y)			R(X)	t3
    W(Y)			W(X)	t4
    		R(X)		t5
    		W(X)		t6

    Read/write schedule S for transactions T1-T4

    1. AT3-T2-T1-T4
    2. BT2-T3-T1-T4
    3. CT2-T4-T3-T1
    4. DT4-T2-T3-T1

    Answer: (C) T2-T4-T3-T1

    Explanation

    Building a precedence graph from every conflicting pair of operations on the same data item, in the order they occur in S, reveals a single valid serial order.

    On data item X: T2 reads and writes X at t1-t2, T4 reads and writes X at t3-t4, and T3 reads and writes X at t5-t6. Since T2's write at t2 comes before T4's read and write at t3-t4, T2 must precede T4. Since T4's operations at t3-t4 come before T3's operations at t5-t6, T4 must precede T3.

    On data item Y: T3 reads and writes Y at t1-t2, and T1 reads and writes Y at t3-t4. Since T3's write at t2 comes before T1's read and write at t3-t4, T3 must precede T1.

    Chaining these constraints together gives T2 before T4, T4 before T3, and T3 before T1, which fixes the unique serial order T2-T4-T3-T1.

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75. Question 73

    UGC NET March 2023

    Given below are the two statements:

    Statement I: Any retrieval request that is specified in the basic relational algebra can also be specified in relational calculus.

    Statement II: Any retrieval request that is specified in the relational calculus can also be specified in basic relational algebra.

    In the light of the above statements, choose the correct answer from the options given below:

    1. ABoth Statement I and Statement II are true.
    2. BBoth Statement I and Statement II are false.
    3. CStatement I is true but Statement II is false.
    4. DStatement I is false but Statement II is true.

    Answer: (A) Both Statement I and Statement II are true.

    Explanation

    Codd's equivalence result establishes that relational algebra and relational calculus have the same expressive power, so any query expressible in one can also be expressed in the other. Statement I holds directly, since relational algebra is by construction always expressible in relational calculus.

    Statement II holds for safe relational calculus expressions, which is the form relational calculus is restricted to in standard database theory precisely so that every calculus query corresponds to an equivalent algebra expression, and an unrestricted (unsafe) calculus expression is not considered a valid query in the first place. Under this standard restriction, both statements are true, which is why relational algebra and relational calculus are described as equivalent in expressive power. This equivalence claim depends on the safety restriction being assumed rather than stated explicitly here, so it is worth confirming which convention the source course uses before treating this as settled.

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76. Question 74

    UGC NET March 2023

    Use the following schema of the academic institution relational database:

    student (rollno, name, degree, year, sex, deptno, advisor)
    department (deptid, name, hod, phone)
    professor (empid, name, sex, startyear, deptno, phone)
    course (courseid, cname, credits, deptno)
    enrollment (rollno, courseid, sem, year, grade)
    teaching (empid, courseid, sem, year, classroom)
    prerequisite (precourseid, courseid)

    deptno is a foreign key in the student, professor and course relations referring to deptid of department. advisor is a foreign key in student referring to empid of professor. hod is a foreign key in department referring to empid of professor. rollno is a foreign key in enrollment referring to rollno of student. courseid is a foreign key in enrollment and teaching referring to courseid of course. empid is a foreign key of teaching referring to empid of professor. precourseid and courseid are foreign keys in prerequisite referring to courseid of course.

    Which of the following queries would retrieve the roll number and names of students who have enrolled for all the prerequisite courses of the course number 324?

    1. ASelect s.rollno, s.name From student s, enrollment e Where e.rollno = s.rollno and e.courseid = ANY (select precourseid from prerequisite p where p.courseid = "324")
    2. BSelect s.rollno, s.name From student s, enrollment e Where e.rollno = s.rollno and e.courseid = ALL (select precourseid from prerequisite p where p.courseid = "324")
    3. CSelect s.rollno, s.name From student s Where NOT EXISTS (select * from prerequisite p where p.courseid = "324" and NOT EXISTS (select * from enrollment e where e.courseid = p.precourseid and e.rollno = s.rollno))
    4. DSelect s.rollno, s.name From student s Where EXISTS (select * from prerequisite p where p.courseid = "324") and EXISTS (select * from enrollment e where e.courseid = p.precourseid and e.rollno = s.rollno)

    Answer: (C) Select s.rollno, s.name From student s Where NOT EXISTS (select * from prerequisite p where p.courseid = "324" and NOT EXISTS (select * from enrollment e where e.courseid = p.precourseid and e.rollno = s.rollno))

    Explanation

    Finding students enrolled in every prerequisite of course 324 is a universal ('for all') condition, which in SQL is standardly expressed with the double-negation, NOT EXISTS wrapped around another NOT EXISTS, division pattern rather than with ANY or ALL.

    • A. uses = ANY — this only requires a match with at least one prerequisite course, which finds students who took at least one prerequisite rather than every one of them.
    • B. uses = ALL — a single enrollment row's courseid cannot simultaneously equal every value in a multi-row prerequisite list, so this comparison does not correctly express enrollment in every prerequisite.
    • C. correct. The inner NOT EXISTS finds any prerequisite course of 324 that the student has not enrolled in, and the outer NOT EXISTS then requires that no such missing prerequisite exists, which is exactly the standard relational-division idiom for "enrolled in all prerequisites."
    • D. uses EXISTS and EXISTS — this only requires that course 324 has some prerequisite and that the student has enrolled in some course matching an uncorrelated reference, which does not enforce enrollment in every prerequisite either.

    Since C correctly implements the universal quantification needed here, it is the right query.

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77. Question 75

    UGC NET March 2023

    Use the following schema of the academic institution relational database:

    student (rollno, name, degree, year, sex, deptno, advisor)
    department (deptid, name, hod, phone)
    professor (empid, name, sex, startyear, deptno, phone)
    course (courseid, cname, credits, deptno)
    enrollment (rollno, courseid, sem, year, grade)
    teaching (empid, courseid, sem, year, classroom)
    prerequisite (precourseid, courseid)

    advisor is a foreign key in student referring to empid of professor.

    Given query: select s.rollno from student s where s.sex = ALL (select f.sex from professor f where f.empid = s.advisor)

    Consider the following statements about the given query, which retrieves the set of all students whose gender is the same as the gender of their advisor:

    S1. It has a correlated subquery
    S2. It has an uncorrelated subquery
    S3. If we replace ALL by ANY in the above query, its result will be different
    S4. If we replace ALL by ANY in the above query, its result will be same.

    Identify all the statements that are TRUE:

    1. AS1 and S3 are true
    2. BS1 and S4 are true
    3. CS2 and S3 are true
    4. DS2 and S4 are true

    Answer: (B) S1 and S4 are true

    Explanation

    The subquery references s.advisor, a column from the outer query's table s, so it must be re-evaluated for every outer row, making it a correlated subquery. S1 is true and S2, which claims the opposite, is false.

    Since empid is the key of professor, the subquery select f.sex from professor f where f.empid=s.advisor returns at most one row for any given student. When a subquery returns a single value, comparing with = ALL and = ANY reduce to the exact same equality test against that one value, so swapping ALL for ANY here does not change the result. S4 is true and S3, which claims the result would differ, is false.

    Since S1 and S4 are the true statements, the correct option is S1 and S4 are true.

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78. Practise Database Management Systems in a timed set

79. Question 76

    UGC NET March 2023

    Use the following schema of the academic institution relational database:

    department (deptid, name, hod, phone)
    course (courseid, cname, credits, deptno)

    deptno is a foreign key in course referring to deptid of department.

    Which of the following queries would compute, for each department, the total credits of all the courses offered by the department?

    1. Aselect deptid, name, sum(credits) as totalcredits from department, course where deptid=deptno group by deptid, name;
    2. Bselect deptid, name, count(credits) as totalcredits from department, course where deptid=deptno group by deptid, name;
    3. Cselect deptid, name, sum(credits) as totalcredits from department, course where deptid=deptno
    4. Dselect deptid, name, sum(credits) as totalcredits from department, course group by deptid, name having deptid=deptno;

    Answer: (A) select deptid, name, sum(credits) as totalcredits from department, course where deptid=deptno group by deptid, name;

    Explanation

    Computing a per-department total requires first joining department to course on the matching key, then grouping by department, and summing credits within each group.

    • A. correct. It joins department and course on deptid=deptno in the WHERE clause, then groups by deptid and name, and sums credits within each group, which is exactly the standard join-then-aggregate pattern.
    • B. uses count(credits) — this counts how many courses each department offers rather than adding up their credit values, so it does not compute total credits.
    • C. has no GROUP BY at all — mixing an aggregate function like sum with non-aggregated columns such as deptid and name, with no grouping clause, does not produce one correctly aggregated row per department.
    • D. moves the join condition into HAVING and drops the WHERE join — without first joining department to course, grouping department and course together directly cross-joins every department with every course before any filtering, which does not correctly restrict courses to their own department.

    Since only A performs the join before grouping and aggregating correctly, it is the right query.

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80. Question 77

    UGC NET March 2023

    Use the following schema of the academic institution relational database:

    department (deptid, name, hod, phone)
    course (courseid, cname, credits, deptno)

    deptno is a foreign key in course referring to deptid of department.

    Which of the following queries would retrieve the dept id and name of all the departments that are such that the total of the credits of all the offered courses by the department is strictly greater than 40?

    1. Aselect deptid, name, sum(credits) as totalcredits from department, course where totalcredits > 40 group by deptid, name having deptid = deptno;
    2. Bselect deptid, name, sum(credits) as totalcredits from department, course where deptid = deptno and totalcredits > 40 group by deptid, name;
    3. Cselect deptid, name, sum(credits) as totalcredits from department, course group by deptid, name having deptid = deptno and totalcredits > 40;
    4. Dselect deptid, name, sum(credits) as totalcredits from department, course where deptid = deptno group by deptid, name having totalcredits > 40;

    Answer: (D) select deptid, name, sum(credits) as totalcredits from department, course where deptid = deptno group by deptid, name having totalcredits > 40;

    Explanation

    A computed aggregate alias like totalcredits does not exist yet when the WHERE clause is evaluated, since WHERE filters rows before grouping and aggregation happen, so it can only be referenced afterward, in HAVING.

    • A and B reference totalcredits inside WHERE — at that point in query evaluation, sum(credits) has not been computed yet, so this is not a valid way to filter on the aggregated total.
    • C moves the join condition deptid=deptno into HAVING and has no WHERE at all — without joining department to course first, the query cross-joins every department with every course, so the grouped sums are not correctly restricted to each department's own courses.
    • D. correct. It performs the join in WHERE first (deptid=deptno), then groups by department, and only then filters the aggregated total in HAVING (totalcredits > 40), which is exactly where a computed aggregate can legally be filtered.

    Since only D places the join before grouping and the aggregate filter in HAVING, it is the right query.

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81. Question 78

    UGC NET March 2023

    Use the following schema of the academic institution relational database:

    student (rollno, name, degree, year, sex, deptno, advisor)
    course (courseid, cname, credits, deptno)
    enrollment (rollno, courseid, sem, year, grade)

    Given query:

    select s.rollno, s.name
    from student s
    where EXISTS (select c.courseid from course c where c.deptno=5 and
    EXISTS (select e.* from enrollment e
    where e.courseid = c.courseid and
    e.rollno = s.rollno));

    Which one of the following statements is correct with respect to the given query?

    1. AIt retrieves the roll number and names of students who have not enrolled for all courses offered by department number 5.
    2. BIt retrieves the roll number and names of students who have enrolled for all courses offered by department number 5.
    3. CIt retrieves the roll number and names of students who have enrolled for at least one course offered by department number 5.
    4. DIt retrieves the roll number and names of students who are such that there exists at least one course offered by department number 5 which is not enrolled by the student.

    Answer: (C) It retrieves the roll number and names of students who have enrolled for at least one course offered by department number 5.

    Explanation

    The outer EXISTS asks whether there is at least one course c in department 5 for which the inner EXISTS also holds, and the inner EXISTS asks whether the student has an enrollment matching that specific course.

    A single EXISTS is an existential ('at least one') quantifier, not a universal ('for all') one, so this query only requires finding one department-5 course the student has enrolled in, not every one of them.

    This matches retrieving students who have enrolled for at least one course offered by department number 5, not the stronger "enrolled in all" or "not enrolled in some" readings described in the other options.

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82. Question 79

    UGC NET October 2022

    Which mechanism in XML allows organizations to specify globally unique names as element tags in documents?

    1. Aroot
    2. Bheader
    3. Cschema
    4. Dnamespace

    Answer: (D) namespace

    Explanation

    An XML namespace binds a URI to a prefix, and that URI guarantees the element and attribute names using that prefix are globally unique even when documents from different organizations reuse the same tag names, avoiding naming collisions when documents are combined.

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83. Question 80

    UGC NET October 2022

    Let R (ABCDEFGH) be a relation schema and F be the set of dependencies F = {A → B, ABCD → E, EF → G, EF → H and ACDF → EG}. The minimal cover of a set of functional dependencies is

    1. AA → B, ACD → E, EF → G, and EF → H
    2. BA → B, ACD → E, EF → G, EF → H and ACDF → G
    3. CA → B, ACD → E, EF → G, EF → H and ACDF → E
    4. DA → B, ABCD → E, EF → H and EF → G

    Answer: (A) A → B, ACD → E, EF → G, and EF → H

    Explanation

    Build the minimal cover step by step.

    First split ACDF → EG into ACDF → E and ACDF → G, since every right-hand side must be a single attribute.

    Next remove extraneous left-hand-side attributes. In ABCD → E, computing the closure of {A,C,D} already yields B (through A → B) and then E (through the reduced ABCD → E), so B is extraneous and this becomes ACD → E.

    Now check ACDF → E: since ACD → E already holds, ACDF → E follows automatically by augmentation, so it is redundant and can be dropped. The same closure argument, extended with EF → G, shows ACDF → G is also implied once ACD → E and EF → G are present, so it drops too.

    What remains, A → B, ACD → E, EF → G, EF → H, has no further redundant or extraneous parts, since removing any one of these four breaks the ability to derive E, G, or H.

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84. Question 81

    UGC NET October 2022

    A trigger is

    1. AA statement that enables to start DBMS.
    2. BA statement that is executed by the user when debugging an application program.
    3. CA condition the system tests for the validity of the database user.
    4. DA statement that is executed automatically by the system as a side effect of modification to the database.

    Answer: (D) A statement that is executed automatically by the system as a side effect of modification to the database.

    Explanation

    A trigger is a stored procedure that the database management system fires automatically whenever a specified modification, an insert, update, or delete, occurs on a table, without any explicit user invocation. It has nothing to do with starting the DBMS, manual debugging, or user-login validation.

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85. Question 82

    UGC NET October 2022

    In reference to Big data, consider the following database:
    A. Memcached
    B. Couch DB
    C. Infinite graph
    Choose the most appropriate answer from the options given below:

    1. A(A) and (B) only
    2. B(B) and (C) only
    3. C(C) and (A) only
    4. D(A), (B) and (C)

    Answer: (D) (A), (B) and (C)

    Explanation

    Big data NoSQL systems are commonly grouped by data model, and all three named systems are standard examples across those categories. Memcached is a key-value store, CouchDB is a document-oriented database, and InfiniteGraph is a graph database, so all three belong on a list of Big Data database technologies.

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86. Question 83

    UGC NET October 2022

    Match List I with List II regarding normal forms.

    List-I	List-II
    A. BCNF	I. Removes multivalued dependency
    B. 3NF	II. Not always dependency preserving
    C. 2NF	III. Removes transitive dependency
    D. 4NF	IV. Removes partial functional dependency

    1. A(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
    2. B(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
    3. C(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
    4. D(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

    Answer: (C) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

    Explanation

    BCNF decomposition is well known for not always preserving every original functional dependency.
    3NF eliminates transitive dependencies on the key.
    2NF eliminates partial functional dependencies on part of a composite key.
    4NF eliminates multivalued dependencies.

    Matching gives A-II, B-III, C-IV, D-I.

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87. Question 84

    UGC NET October 2022

    Consider the following statements:
    Statement I: Conservative 2 PL is a deadlock-free protocol.
    Statement II: Thomas's write rule enforces conflict serializability.
    Statement III: Timestamp ordering protocol ensures serializability based on the order of transaction timestamps.
    Which of the following is correct?

    1. AStatement I, Statement II true and Statement III false
    2. BStatement I, Statement III true and Statement II false
    3. CStatement I, Statement II false and Statement III true
    4. DStatement I, Statement II and Statement III true

    Answer: (B) Statement I, Statement III true and Statement II false

    Explanation

    Statement I is true. Conservative (static) two-phase locking requires a transaction to acquire all its locks before starting execution, which removes the wait-and-hold pattern that causes deadlock.

    Statement II is false. Thomas's write rule deliberately relaxes conflict serializability by allowing certain outdated writes to be ignored (the thin thomas write rule) rather than treated as a conflict, so it guarantees view serializability, not conflict serializability.

    Statement III is true. Basic timestamp ordering enforces an execution order equivalent to the order in which transactions received their timestamps, which is exactly its serializability guarantee.

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88. Question 85

    UGC NET October 2022

    Consider the following statements:
    Statement I: Composite attributes cannot be divided into smaller subparts.
    Statement II: Complex attribute is formed by nesting composite attributes and multi-valued attributes in an arbitrary way.
    Statement III: A derived attribute is an attribute whose values are computed from other attribute.
    Which of the following is correct?

    1. AStatement I, Statement II and Statement III are true
    2. BStatement I true and Statement II, Statement III false
    3. CStatement I, Statement II true and Statement III false
    4. DStatement I false and Statement II, Statement III true

    Answer: (D) Statement I false and Statement II, Statement III true

    Explanation

    Statement I is false. It describes a simple (atomic) attribute, not a composite one. A composite attribute is defined by the fact that it can be divided into smaller subparts, such as splitting Name into First Name and Last Name.

    Statement II is true. Nesting composite and multivalued attributes together in any combination is exactly how a complex attribute is built.

    Statement III is true. A derived attribute, such as Age computed from Date of Birth, has its value calculated from another stored attribute rather than stored directly.

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89. Question 86

    UGC NET October 2022

    Which of the following relational algebra query computes the Sid's of sailors with age over 20 who have not reserved a red boat?

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. Aπ sid (σ age > 20 Sailors) − π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors)
    2. Bπ sid ((σ color ≠ red ∧ age > 20 (Boats ⋈ Sailors ⋈ Reserves)))
    3. Cπ sid (σ age > 20 Sailors) − π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors)
    4. Dπ sid (σ age > 20 Sailors) ∧ π sid ((σ color ≠ red Boats) ⋈ Reserves ⋈ Sailors)

    Accepted answers: (A) π sid (σ age > 20 Sailors) − π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors), (C) π sid (σ age > 20 Sailors) − π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors)

    Explanation

    The correct strategy is set difference: start from every sailor with age over 20, then subtract out every sailor who has reserved at least one red boat, regardless of age.

    π sid (σ age > 20 Sailors) gives the sids of sailors older than 20.
    π sid ((σ color = red Boats) ⋈ Reserves ⋈ Sailors) gives the sids of sailors who reserved a red boat.

    Subtracting the second set from the first gives exactly the sailors over 20 who have not reserved a red boat.

    Options 1 and 3 are printed as byte-identical text in the source PDF itself, confirmed directly against the page image, and both show this correct expression. The color-comparison approaches in options 2 and 4 are unsound, since a sailor who has reserved both a red boat and a non-red boat would still wrongly satisfy a per-row color ≠ red condition on some other reservation.

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90. Question 87

    UGC NET October 2022

    Which of the following relational algebra query/queries computes/compute the names of sailors who have reserved a red boat?
    Q1: π sname ((σ color = red Boats) ⋈ Reserves ⋈ Sailors)
    Q2: π sname (π sid ((π bid σ color = red Boats) ⋈ Reserves) ⋈ Sailors)
    Q3: π sname ((σ color = red Reserves) ⋈ Boats ⋈ Sailors)

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. ABoth Q1 and Q2
    2. BBoth Q2 and Q3
    3. COnly Q1
    4. DOnly Q2

    Answer: (A) Both Q1 and Q2

    Explanation

    Q1 filters Boats to red boats, joins with Reserves and Sailors, and projects the name, a direct and correct approach.

    Q2 first finds the bids of red boats, joins those bids with Reserves to get the sids who reserved one, then joins with Sailors for the name, an equivalent, correct two-step version of the same logic.

    Q3 applies the condition color = red directly to Reserves, but Reserves has no color attribute at all, that attribute only exists on Boats, so this expression is malformed and cannot be correct.

    Only Q1 and Q2 correctly compute the required names.

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91. Question 88

    UGC NET October 2022

    Which of the following relational algebra query/queries computes/compute the name of sailors who have reserved boat 103?
    Q1: π sname ((σ bid = 103 Boats) ⋈ Sailors)
    Q2: π sname (σ bid = 103 (Reserves ⋈ Sailors))
    Q3: π sname ((σ bid = 103 Reserves) ⋈ Sailors)

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. ABoth Q1 and Q3
    2. BBoth Q2 and Q3
    3. COnly Q3
    4. DOnly Q2

    Answer: (B) Both Q2 and Q3

    Explanation

    Boats and Sailors share no join attribute, so Q1's direct join of a filtered Boats with Sailors is not a meaningful relational algebra expression and cannot produce the right names.

    Q2 joins Reserves with Sailors on sid first (bringing bid along from Reserves), then filters the joined result to bid = 103, which correctly isolates sailors who reserved that boat.

    Q3 filters Reserves to bid = 103 first, then joins the smaller filtered result with Sailors on sid, reaching the same correct answer more efficiently.

    Only Q2 and Q3 give the correct names.

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92. Question 89

    UGC NET October 2022

    Which of the following relational algebra query computes the names of sailor who have reserved all boats?

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. Aρ (Tempsids, (π bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors)
    2. Bρ (Tempsids, (π sid,bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors)
    3. Cρ (Tempsids, (π sid Sailors) / π bid Boats) π sname ((Tempsids) ⋈ Sailors)
    4. Dρ (Tempsids, (π sid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Boats)

    Answer: (B) ρ (Tempsids, (π sid,bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors)

    Explanation

    Relational division needs a dividend that pairs the sid and bid attributes together, so the dividend must come from Reserves and include both columns.

    • A. ρ (Tempsids, (π bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors) — wrong. The dividend keeps only the bid column, so dividing it by Boats bid column cannot produce any sid values.
    • B. ρ (Tempsids, (π sid,bid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Sailors) — correct. Dividing the sid-bid pairs from Reserves by every bid in Boats yields exactly the sids that appear with all boats, and joining with Sailors retrieves their names.
    • C. ρ (Tempsids, (π sid Sailors) / π bid Boats) π sname ((Tempsids) ⋈ Sailors) — wrong. The dividend is drawn from Sailors instead of Reserves, so it divides the wrong relation and cannot capture which boats a sailor reserved.
    • D. ρ (Tempsids, (π sid Reserves) / π bid Boats) π sname ((Tempsids) ⋈ Boats) — wrong. The dividend keeps only sid with no bid to divide against, and the final join is with Boats instead of Sailors, so sname could not even be retrieved.

    Only option B pairs sid with bid in the dividend and joins the result with Sailors, so it correctly computes sailors who reserved all boats.

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93. Question 90

    UGC NET October 2022

    Which of the following relational algebra query computes the names of sailors who have reserved a red and a green boat?

    Consider the relational schema of Sailors S, Reserves R, and Boats B given below.

    Table 1: Sailors S

    Sid	Sname	Rating	Age
    22	Dustin	7	45.0
    29	Brutus	1	33.0
    31	Lubber	8	55.5
    32	Andy	8	25.5
    58	Rusty	10	35.0
    64	Horatio	7	35.0
    71	Zorba	10	16.0
    74	Horatio	9	35.0
    85	Art	3	25.5
    95	Bob	3	63.5

    Table 2: Reserves R

    Sid	Bid	Day
    22	101	10/10/98
    22	102	10/10/98
    22	103	10/8/98
    22	104	10/7/98
    31	102	11/10/98
    31	103	11/6/98
    31	104	11/12/98
    64	101	9/5/98
    64	102	9/8/98
    74	103	9/8/98

    Table 3: Boats B

    Bid	Bname	Color
    101	Interlake	blue
    102	Interlake	red
    103	Clipper	green
    104	Marine	red

    1. Aρ (Tempred, π sid ((σ color = 'red' Boats) ⋈ Reserves)), ρ (Tempgreen, π sid ((σ color = 'green' Boats) ⋈ Reserves)), π sname (Tempred ∩ Tempgreen) ⋈ Sailors
    2. Bρ (Tempboats 2, (σ color = 'red' Boats) ∩ (σ color = 'green' Boats)) π sname (Tempboats 2 ⋈ Reserves)
    3. Cπ sname ((σ color = 'red' Boats) ∩ (σ color = 'green' Boats) (Tempboats 2 ⋈ Sailors ⋈ Reserves))
    4. Dρ (Tempboats 2, (σ color = 'red' ∧ color = 'green' Boats)) π sname (Tempboats 2 ⋈ Sailors)

    Answer: (A) ρ (Tempred, π sid ((σ color = 'red' Boats) ⋈ Reserves)), ρ (Tempgreen, π sid ((σ color = 'green' Boats) ⋈ Reserves)), π sname (Tempred ∩ Tempgreen) ⋈ Sailors

    Explanation

    The correct approach separately computes which sailors reserved a red boat and which reserved a green boat, and only then intersects the two sets of sailor ids.

    Tempred holds the sids of sailors who reserved a red boat, and Tempgreen holds the sids of sailors who reserved a green boat. Intersecting them gives exactly the sailors who did both, and joining with Sailors retrieves their names.

    The other options intersect the Boats relation directly on color, but no single boat row can be both red and green at once, so intersecting (σ color = 'red' Boats) with (σ color = 'green' Boats), or selecting color = 'red' AND color = 'green' on Boats, always produces an empty relation. Those approaches can never return a sailor.

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94. Question 91

    UGC NET November 2021

    The order of a leaf node in a B+ tree is the maximum number of (value, data-record-pointer) pairs it can hold. Given block size 1 KB (1024 bytes), data-record pointer 7 bytes, value field 9 bytes, and block pointer 6 bytes, what is the order of the leaf node?

    1. A63
    2. B64
    3. C67
    4. D68

    Answer: (A) 63

    Explanation

    A B+ tree leaf holds m (value + record-pointer) pairs plus one block pointer to the next leaf: m x (9 + 7) + 6 <= 1024, so 16m <= 1018, m <= 63.6. The order is 63.

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95. Question 92

    UGC NET November 2021

    Which DBMS component ensures that concurrent execution of multiple operations on the database results in a consistent database state?

    1. ALogs
    2. BBuffer manager
    3. CFile manager
    4. DTransaction processing system

    Answer: (D) Transaction processing system

    Explanation

    The transaction processing (transaction management and concurrency-control) subsystem enforces the ACID properties, scheduling concurrent transactions so the result is equivalent to some serial order. Logs support recovery, the buffer manager moves pages, and the file manager handles disk space.

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96. Question 93

    UGC NET November 2021

    Statement I: A relational database schema represents the logical design of the database.
    Statement II: The current snapshot of a relation only provides the degree of the relation.

    In light of the above statements, choose the correct answer from the options given below:

    1. AStatement I is TRUE but Statement II is FALSE
    2. BStatement I is FALSE but Statement II is TRUE
    3. CBoth Statement I and Statement II are FALSE
    4. DBoth Statement I and Statement II are TRUE

    Answer: (A) Statement I is TRUE but Statement II is FALSE

    Explanation

    The schema (tables, attributes, constraints) is the logical design (Statement I true). A snapshot/instance of a relation shows both its degree (number of attributes) and its cardinality (number of tuples), not the degree alone, so Statement II is false.

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97. Question 94

    UGC NET November 2021

    A fixed-length record file is ordered on the key field and occupies B disk blocks to store R records. What is the average number of block accesses to find a record using binary search?

    1. AB / 2
    2. BB + R
    3. CB
    4. Dlog2(B)

    Answer: (D) log2(B)

    Explanation

    Binary search over B ordered blocks halves the search range each step, so it needs at most ceil(log2 B) block accesses; the average access cost is therefore O(log2 B).

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98. Question 95

    UGC NET November 2021

    Given STUDENT(Rollno, Name, courseno) and COURSE(courseno, coursename, capacity), with coursename unique. Which of the queries below find the total number of students enrolled in each course together with its coursename?

    -- Query A
    SELECT coursename, count(*) 'total' FROM STUDENT NATURAL JOIN COURSE GROUP BY coursename;
    
    -- Query B
    SELECT C.coursename, count(*) 'total' FROM STUDENT S, COURSE C WHERE S.courseno = C.courseno GROUP BY coursename;
    
    -- Query C
    SELECT coursename, count(*) 'total' FROM COURSE C WHERE courseno IN (SELECT courseno FROM STUDENT);
    

    Choose the correct answer from the options given below:

    1. AA and B only
    2. BC only
    3. CA only
    4. DB only

    Answer: (A) A and B only

    Explanation

    A and B both join STUDENT with COURSE on courseno and group by coursename, so count(*) gives the enrolment per course. C never joins with STUDENT - it only filters courses that have at least one student and counts COURSE rows, so it cannot report the number of students per course.

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99. Question 96

    UGC NET November 2021

    STUDENT and COURSE each have 3 attributes; COURSE has 3 records and STUDENT has 5 records. The relational-algebra query R = STUDENT x COURSE is executed. Match List-I with List-II for R.

    List-I	List-II
    A. Degree of table R	I. 15
    B. Cardinality of table R	II. NIL
    C. Foreign key of relation STUDENT	III. 6
    D. Foreign key of relation COURSE	IV. Courseno

    1. AA-III, B-I, C-IV, D-II
    2. BA-I, B-III, C-IV, D-II
    3. CA-I, B-IV, C-III, D-II
    4. DA-I, B-III, C-II, D-IV

    Answer: (A) A-III, B-I, C-IV, D-II

    Explanation

    The Cartesian product has degree 3 + 3 = 6 (III) and cardinality 3 x 5 = 15 (I). In the schema STUDENT.courseno refers to COURSE, so its foreign key is Courseno (IV); COURSE references no other table, so it has no foreign key - NIL (II).

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100. Question 97

     UGC NET November 2021

     A B-tree indexes a database file. Search-key field = 10 bytes, block size = 1024 bytes, record pointer = 9 bytes, block pointer = 8 bytes. Let K be the order of an internal node and L the order of a leaf node. Then (K, L) = ______.

     1. A(57, 53)
     2. B(50, 52)
     3. C(60, 64)
     4. D(34, 31)

     Answer: (A) (57, 53)

     Explanation

     Internal node: K block pointers and K-1 keys fit in a block: 8K + 10(K - 1) <= 1024, so 18K <= 1034, K = 57. Leaf node: L (key + record-pointer) entries plus one block pointer: 19L + 8 <= 1024, so 19L <= 1016, L = 53.

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101. Question 98

     UGC NET November 2021

     Given R(x, y, z, w) with functional dependencies F = { x -> y, z -> w } and all attributes atomic.

     A. R is in First Normal Form.
     B. R is in Second Normal Form.
     C. The primary key of R is xz.

     Choose the correct answer from the options given below:

     1. AC only
     2. BB and C only
     3. CA and C only
     4. DB only

     Answer: (C) A and C only

     Explanation

     Since the closure of xz is xyzw and neither x nor z alone is a superkey, the primary key is xz (C true). Atomic attributes make R 1NF (A true). But x -> y and z -> w are partial dependencies of non-prime attributes on part of the key, violating 2NF (B false).

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102. Question 99

     UGC NET November 2021

     A part is either manufactured in-house, purchased from an external supplier, or both. A batchnumber is kept for manufactured parts and a supplier name for purchased parts. Which constraints must be modelled for this class/subclass structure?

     1. ATotal specialization and overlapping constraints
     2. BDisjoint constraint only
     3. CPartial participation
     4. DPartial participation and disjoint constraints

     Answer: (A) Total specialization and overlapping constraints

     Explanation

     Every part is manufactured or purchased (or both), so every superclass entity belongs to at least one subclass - total specialization. A part can be both manufactured and purchased, so the subclasses can overlap - the overlapping constraint (not disjoint).

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103. Question 100

     UGC NET November 2021

     A transaction in the ACTIVE state during concurrent execution transitions next to which state?

     A. FAILED
     B. TERMINATED
     C. PARTIALLY COMMITTED
     D. COMMITTED
     E. ACTIVE

     Choose the correct answer from the options given below:

     1. AA only
     2. BEither A or C only
     3. CC only
     4. DD only

     Answer: (B) Either A or C only

     Explanation

     From ACTIVE a transaction moves either to PARTIALLY COMMITTED (after its final statement executes) or to FAILED (if an error aborts it). It cannot jump straight to COMMITTED or TERMINATED.

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104. Practise Database Management Systems in a timed set

105. Question 101

     UGC NET November 2021

     Which language is used to create a database schema?

     1. ADML
     2. BDDL
     3. CHTML
     4. DXML

     Answer: (B) DDL

     Explanation

     Data Definition Language (CREATE, ALTER, DROP) defines and modifies schema objects such as tables, indexes and views. DML (SELECT, INSERT, UPDATE, DELETE) manipulates the data within that schema.

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106. Question 102

     UGC NET November 2020

     For S = (U, V, W, X, Y, Z), the functional dependencies are {U → V, VW → X, Y → W, X → U}. Which listed sets are candidate keys?

     1. AUY, VY
     2. BUY, VY, XY
     3. CUYZ, VYZ, VWZ
     4. DUYZ, VYZ, XYZ

     Answer: (D) UYZ, VYZ, XYZ

     Explanation

     Z never appears on the right-hand side, so every key must contain Z. UYZ gives V from U and W from Y, then X from VW; VYZ gives W from Y, then X and U; XYZ gives U from X and then V. Each is minimal, yielding option D.

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107. Question 103

     UGC NET November 2020

     Two operations from a schedule conflict when
     A. at least one operation is a write,
     B. both access the same data item, and
     C. they belong to different transactions.

     1. A(A) and (B) only
     2. B(A), (B), and (C) only
     3. C(A), (C), and (D) only
     4. D(C) and (D) only

     Answer: (B) (A), (B), and (C) only

     Explanation

     Conflicting operations are operations of different transactions on the same item where at least one writes. Two reads do not conflict, and operations on different data items do not conflict.

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108. Question 104

     UGC NET November 2020

     Match List I with List IT.

     List I (E-R symbols) List II (Description)
     A. () Key Attribute Type

     1. — 33 (I) Weak Entity Type

     << > (II) Total Participation of Entity in a
     C. relation

     ) <> E, (IV) Multivalue Attribute type

     List I	List II
     Items are listed in the question stem.	Match each item to its stated description or complexity.

     1. AOCR review required
     2. BA-IV, B-I, C-II, D-III
     3. CA-II, B-I, C-IV, D-III
     4. DA-III, B-IV, C-I, D-II

     Answer: (C) A-II, B-I, C-IV, D-III

     Explanation

     The standard E–R notation distinguishes key, weak entity, total participation, and multivalued attributes by their respectively marked symbols; option C gives that correspondence.

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109. Question 105

     UGC NET November 2020

     Using the university table structures, which query returns distinct employee IDs of employees working on projects 20, 30, or 40?

     1. ASELECT EMPID FROM PROJECTWORK WHERE PROJNO = (20,30,40);
     2. BSELECT EMPID FROM PROJECTWORK WHERE PROJNO IN (20,30,40);
     3. CSELECT DISTINCT EMPID FROM PROJECTWORK WHERE PROJNO IN (20,30,40);
     4. DSELECT DISTINCT EMPID FROM PROJECTWORK WHERE PROJNO = 20,30,40;

     Answer: (C) SELECT DISTINCT EMPID FROM PROJECTWORK WHERE PROJNO IN (20,30,40);

     Explanation

     IN tests membership in a set of values, and DISTINCT removes duplicate employee IDs when an employee works on more than one listed project. Equality cannot be used with a comma-separated list.

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110. Question 106

     UGC NET November 2020

     Which statement correctly computes SUM, MAX, MIN, and AVG salaries for the English department?

     Statement I joins EMPLOYEE and DEPARTMENT on DEPTNO = DID and filters DNAME = 'ENGLISH'.
     Statement II filters DNAME = 'ENGLISH' without joining the two tables.

     1. ABoth statements are true.
     2. BBoth statements are false.
     3. CStatement I is true but Statement II is false.
     4. DStatement I is false but Statement II is true.

     Answer: (C) Statement I is true but Statement II is false.

     Explanation

     Statement I supplies the join predicate that relates an employee to a department before filtering by department name. Statement II has no join condition, so it forms an unrelated Cartesian product and is not a correct department-salary query.

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111. Question 107

     UGC NET November 2020

     Which query returns employee ID and name for employees whose salary is greater than the salary of every employee in department 20, ordered by employee ID?
     A. uses SALARY > (subquery) with ORDER BY.
     B. uses SALARY > (subquery) without ORDER BY.
     C. uses SALARY > ALL (subquery) with ORDER BY.

     1. A(A) and (B) only
     2. B(A) and (C) only
     3. C(B) only
     4. D(C) only

     Answer: (D) (C) only

     Explanation

     A scalar comparison with a subquery returning several salaries is not the required “greater than every” test and can be invalid. > ALL (...) expresses the requirement exactly, and only C also provides the requested ordering.

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112. Question 108

     UGC NET November 2020

     Which SQL statement(s) can drop the SALARY column from EMPLOYEE?
     A. ALTER TABLE EMPLOYEE DROP SALARY CASCADE;
     B. ALTER TABLE EMPLOYEE DROP SALARY RESTRICT;
     C. ALTER EMPLOYEE DROP SALARY;

     1. A(A) and (B) only
     2. B(A) and (C) only
     3. C(B) and (C) only
     4. D(A) only

     Answer: (A) (A) and (B) only

     Explanation

     The supported form is ALTER TABLE ... DROP .... CASCADE permits dependent objects to be removed; the other choices either use a restrictive dependency rule or omit the required TABLE keyword in the form expected here.

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113. Question 109

     UGC NET November 2020

     Using the university table structures, the university wants to give every employee in the SCIENCE department a 20% salary rise. Which query is correct?
     A. UPDATE EMPLOYEE SET SALARY = SALARY * 1.20 WHERE DEPTNO IN (SELECT DID FROM DEPARTMENT WHERE DNAME = 'SCIENCE');
     B. UPDATE TABLE EMPLOYEE SET SALARY = SALARY * 1.20 WHERE DNAME = 'SCIENCE';
     C. ALTER TABLE EMPLOYEE SET SALARY = SALARY * 1.20 WHERE DEPTNO IN (SELECT DNAME FROM DEPARTMENT WHERE DNAME = 'SCIENCE');

     1. A(A) and (B) only
     2. B(A) only
     3. C(B) and (C) only
     4. D(C) only

     Answer: (B) (A) only

     Explanation

     A is valid: it updates EMPLOYEE rows whose department number belongs to the SCIENCE department. B uses invalid UPDATE syntax and a column absent from EMPLOYEE; C uses ALTER TABLE rather than UPDATE and compares DEPTNO to a department name.

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114. Question 110

     UGC NET December 2019

     Given two tables EMPLOYEE (EID, ENAME, DEPTNO) and DEPARTMENT (DEPTNO, DEPTNAME), find the most appropriate statement of the query:

     Select count (*) Total
     from EMPLOYEE
     where DEPTNO IN (D1, D2)
     group by DEPTNO
     having count (*) > 5
     

     1. ATotal number of employees in each department D1 and D2
     2. BTotal number of employees of department D1 and D2 if their total is > 5
     3. CDisplay total number of employees in both departments D1 and D2
     4. DThe output of the query must have atleast two rows

     Answer: (A) Total number of employees in each department D1 and D2

     Explanation

     Breaking down the SQL clauses:

     1. where DEPTNO IN (D1, D2) restricts the rows to employees belonging to either department D1 or D2.
     2. group by DEPTNO partitions the filtered rows by department, calculating the employee count separately for each department.
     3. having count (*) > 5 filters the grouped partitions, returning only departments with more than 5 employees.

     • A — correct. Reflects the per-department aggregation of employees for departments D1 and D2.
     • B and C — incorrect because they describe a single combined total across both departments rather than separate counts grouped by department.
     • D — incorrect because the query returns zero, one, or two rows depending on which departments exceed 5 employees.

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115. Question 111

     UGC NET December 2019

     Find the minimum number of tables required for converting the entity relationship diagram into a relational database.

     

     Entity-Relationship diagram with entities R1 and R2 connected by relationship R12

     1. A2
     2. B4
     3. C3
     4. D5

     Answer: (C) 3

     Explanation

     Converting the entity-relationship components into relational tables:

     1. Entity R1 contains attributes A, B, and C, where B is a multivalued attribute (represented by a double oval). Relational design rules require a separate table for multivalued attribute B along with the primary key of R1.
     2. Relationship R12 is a 1:M relationship from R1 to R2 (cardinality 1 on R1 side and M on R2 side). A 1:M relationship can be merged directly into the M-side entity table (R2) by including the primary key of R1 as a foreign key.
     3. Base table for entity R1 with its atomic attributes A and C.

     This yields exactly 3 tables:

     • Table for R1 (A, C)
     • Table for multivalued attribute B (Key(R1), B)
     • Table for R2 merged with relationship R12 (X, Y, FK_R1)

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116. Question 112

     UGC NET December 2019

     Given two tables R1(x, y) and R2(y, z) with 50 and 30 number of tuples respectively. Find maximum number of tuples in the output of natural join between tables R1 and R2 i.e. R1 * R2? (* - Natural Join)

     1. A30
     2. B20
     3. C50
     4. D1500

     Answer: (D) 1500

     Explanation

     In natural join R1(x, y) ⋈ R2(y, z), rows join on the common attribute y:

     • The maximum output size occurs when all 50 tuples in R1 share the identical value for attribute y, and all 30 tuples in R2 also share that exact same value.
     • Under this condition, every tuple of R1 matches with every tuple of R2, behaving as a full Cartesian product on the non-joining attributes.
     • Maximum tuples = 50 × 30 = 1500.

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117. Question 113

     UGC NET December 2019

     Which of the component module of DBMS does rearrangement and possible ordering of operations, eliminate redundancy in query and use efficient algorithms and indexes during the execution of a query?

     1. Aquery compiler
     2. Bquery optimizer
     3. CStored data manager
     4. DDatabase processor

     Answer: (B) query optimizer

     Explanation

     The query optimizer in a relational database management system analyzes query trees, pushes relational algebra operations like selections and projections down the tree, eliminates redundant operations, and chooses the most efficient access paths and join algorithms using catalog statistics.

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118. Question 114

     UGC NET December 2019

     Two concurrent executing transactions T₁ and T₂ are allowed to update same stock item say ‘A’ in an uncontrolled manner. In such scenario, following problems may occur:

     A. Dirty read problem
     B. Lost update problem
     C. Transaction failure
     D. Inconsistent database state

     Which of the following option is correct if database system has no concurrency module and allows concurrent execution of above two transactions?

     1. AA, B and C only
     2. BC and D only
     3. CA and B only
     4. DA, B and D only

     Answer: (D) A, B and D only

     Explanation

     Concurrency anomalies arising from unconstrained interleaved transaction execution include:

     • A. Dirty read problem — reading uncommitted modifications made by another concurrent transaction that subsequently aborts.
     • B. Lost update problem — two transactions simultaneously read and overwrite the same data item, causing the first update to be obliterated.
     • D. Inconsistent database state — partial or unisolated updates violate relational integrity and consistency constraints.

     Transaction failure (C) is a system or logical crash event rather than a concurrency anomaly caused by lack of lock isolation. Thus, A, B, and D only can occur.

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119. Question 115

     UGC NET December 2019

     An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functional dependency set F is given as:

     F = {A → BC, D → E, BC → D, A → D }

     Consider a universal relation R(A, B, C, D, E ) with functional dependency set F. Also all attributes are simple and take atomic values only.

     Minimal cover F' of functional dependency set F is:

     1. AF' = {A → B, A → C, BC → D, D → E}
     2. BF' = {A → BC, B → D, D → E}
     3. CF' = {A → B, A → C, A → D, D → E}
     4. DF' = {A → B, A → C, B → D, C → D, D → E}

     Answer: (A) F' = {A → B, A → C, BC → D, D → E}

     Explanation

     To derive the minimal canonical cover F':

     1. Decompose right-hand side attributes into singletons:

     F₁ = {A → B, A → C, D → E, BC → D, A → D }

     1. Check for redundant functional dependencies:

     Testing A → D using F₁ \ {A → D }:

     A⁺ = {A, B, C, D, E}

     Because attribute D is in A⁺, the dependency A → D is redundant and removed.

     1. Check for extraneous attributes on the left-hand side of BC → D:

     • Testing if B is extraneous: C⁺ = {C}, which does not contain D.
     • Testing if C is extraneous: B⁺ = {B}, which does not contain D.

     Neither B nor C is extraneous. Hence, the resulting minimal cover is:

     F' = {A → B, A → C, BC → D, D → E}

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120. Question 116

     UGC NET December 2019

     An organization needs to maintain database having five attributes A, B, C, D, E with functional dependency set F = {A → BC, D → E, BC → D, A → D }.

     Assume that given table R is decomposed into two tables:
     R₁(A, B, C ) with functional dependency set F₁ = {A → B, A → C}, and
     R₂(A, D, E ) with functional dependency set F₂ = {A → D, D → E}.

     Which of the following option is true with respect to the given decomposition?

     1. ADependency preservation property is followed
     2. BR₁ and R₂ are both in 2 NF
     3. CR₁ is in 2 NF and R₂ is in 3 NF
     4. DR₁ is in 3 NF and R₂ is in 2 NF

     Answer: (D) R₁ is in 3 NF and R₂ is in 2 NF

     Explanation

     Evaluating the normal forms for both decomposed tables:

     In table R₁(A, B, C ), candidate key is A because A⁺ = {A, B, C}. Every determinant is a candidate key, placing R₁ in 3NF and BCNF.

     In table R₂(A, D, E ), candidate key is A because A⁺ = {A, D, E}. Since the candidate key is a single attribute, no partial dependency can exist, ensuring R₂ satisfies 2NF.

     However, in R₂, the dependency D → E has non-superkey determinant D, and E is not a prime attribute. This constitutes a transitive dependency from A through D to E, violating 3NF.

     Furthermore, the dependency BC → D from F is not preserved across R₁ and R₂. Therefore, R₁ is in 3 NF and R₂ is in 2 NF.

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121. Question 117

     UGC NET December 2019

     An organization needs to maintain database having five attributes A, B, C, D, E with functional dependency set F = {A → BC, D → E, BC → D, A → D }.

     Identify primary key of table R with functional dependency set F.

     1. ABC
     2. BAD
     3. CA
     4. DAB

     Answer: (C) A

     Explanation

     Calculating the attribute closure of A:

     A⁺ = {A}

     Applying A → BC yields {A, B, C}.
     Applying BC → D yields {A, B, C, D}.
     Applying D → E yields {A, B, C, D, E}.

     Because A⁺ contains all attributes of R, and A cannot be reduced further, A is a candidate key. Notice that A never appears on the right-hand side of any functional dependency in F, so every candidate key must include A. Any strict superset such as AD or AB is a superkey but not a minimal key. Thus, A is the primary key.

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122. Question 118

     UGC NET December 2019

     An organization needs to maintain database having five attributes A, B, C, D, E with functional dependency set F = {A → BC, D → E, BC → D, A → D }. Consider relation R(A, B, C, D, E ) where all attributes are simple and atomic.

     Identify the normal form to which relation R belongs:

     1. A1 NF
     2. B2 NF
     3. C3 NF
     4. DBCNF

     Answer: (B) 2 NF

     Explanation

     The candidate key of relation R is A, making A the only prime attribute and {B, C, D, E} non-prime attributes.

     Because the candidate key consists of a single attribute, no proper subset of the key exists to cause partial functional dependencies. Relation R is therefore in 2NF.

     To test for 3NF, consider BC → D and D → E. The determinant D is not a superkey, and E is not a prime attribute. Similarly, BC is not a superkey, and D is not prime. These transitive dependencies violate the 3NF condition. Therefore, relation R is in 2 NF.

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123. Question 119

     UGC NET December 2019

     An organization needs to maintain database having five attributes A, B, C, D, E with functional dependency set F = {A → BC, D → E, BC → D, A → D }.

     Identify the redundant functional dependency in F.

     1. ABC → D
     2. BD → E
     3. CA → D
     4. DA → BC

     Answer: (C) A → D

     Explanation

     A functional dependency X → Y is redundant if it can be derived from the remaining functional dependencies in the set.

     Consider A → D. Removing it leaves F' = {A → BC, D → E, BC → D}.

     Computing the closure of A under F':

     A⁺ = {A}

     Applying A → BC gives {A, B, C}.
     Applying BC → D gives {A, B, C, D}.

     Because D belongs to A⁺, the dependency A → D holds transitively through A → BC and BC → D. None of the other three dependencies can be deduced without their explicit inclusion. Therefore, A → D is the redundant functional dependency.

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124. Question 120

     UGC NET June 2019

     In relational database management, which of the following is/are property/properties of candidate key?

     P: Uniqueness
     Q: Irreducibility

     1. AP only
     2. BQ only
     3. CBoth P and Q
     4. DNeither P nor Q

     Answer: (C) Both P and Q

     Explanation

     A candidate key is formally defined in relational database theory by two essential conditions:

     • P. Uniqueness — no two distinct tuples in any valid relation state can contain identical values for the candidate key attribute set.
     • Q. Irreducibility (Minimality) — no proper subset of the candidate key possesses the uniqueness property — removing any attribute destroys its uniqueness.

     Both conditions must hold simultaneously.

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125. Question 121

     UGC NET June 2019

     A relation has two attributes Employee_id and Manager_id, where Employee_id is a primary key and manager_id is a foreign key referencing Employee_id with on-delete cascade:

     On deleting the tuple (20, 40), the set of other tuples that must be deleted to maintain the referential integrity of table is

     Employee_id	Manager_id
     20	40
     25	40
     30	35
     35	20
     40	45
     45	25

     1. A(30, 35) only
     2. B(30, 35) and (35, 20) only
     3. C(35, 20) only
     4. D(40, 45) and (25, 40) only

     Answer: (B) (30, 35) and (35, 20) only

     Explanation

     Evaluating the ON DELETE CASCADE constraint step by step:

     1. Deleting tuple (20, 40) removes Employee_id = 20.
     2. Any tuple with Manager_id = 20 must cascade-delete — tuple (35, 20) is therefore deleted.
     3. Deleting (35, 20) in turn removes Employee_id = 35.
     4. Any tuple with Manager_id = 35 must cascade-delete — tuple (30, 35) is therefore deleted.
     5. Employee_id = 30 has no referencing manager rows, terminating the cascade.

     Thus, the additional tuples deleted are (35, 20) and (30, 35).

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126. Question 122

     UGC NET June 2019

     With respect to relational algebra, which of the following operations are included from mathematical set theory?
     (a) Join
     (b) Intersection
     (c) Cartisian product
     (d) Project

     1. A(a) and (d)
     2. B(b) and (c)
     3. C(c) and (d)
     4. D(b) and (d)

     Answer: (B) (b) and (c)

     Explanation

     Relational algebra divides operators into set-theoretic operations and specialized relational operations:

     1. Set-theoretic operations borrowed directly from mathematical set theory include Union, Intersection, Set Difference, and Cartesian Product.
     2. Specialized relational operations include Select, Project, Join, and Division.

     Among the choices, (b) Intersection and (c) Cartesian product come from mathematical set theory.

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127. Question 123

     UGC NET June 2019

     Which of the following key constraints is required for functioning of foreign key in the context of relational databases?

     1. AUnique key
     2. BPrimary key
     3. CCandidate key
     4. DCheck key

     Answer: (B) Primary key

     Explanation

     In relational databases, referential integrity mandates that a foreign key column references a unique identifier in the parent table.

     The primary key serves as the standard required constraint that guarantees both uniqueness and non-nullability, providing an unambiguous reference point that prevents orphaned records across related tables.

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128. Question 124

     UGC NET June 2019

     Hadoop (a big data tool) works with number of related tools. Choose from the following, the common tools included into Hadoop:

     1. AMySQL, Google API and Map reduce
     2. BMap reduce, Scala and Hummer
     3. CMap reduce, H Base and Hive
     4. DMap reduce, Hummer and Heron

     Answer: (C) Map reduce, H Base and Hive

     Explanation

     The Apache Hadoop ecosystem contains complementary modules designed for distributed big data storage and processing:

     • MapReduce — the primary computational engine for parallel batch processing across commodity clusters.
     • HBase — a distributed, scalable NoSQL column-oriented store that sits on top of HDFS.
     • Hive — a data warehouse infrastructure providing SQL-like queries (HiveQL) over distributed files.

     MySQL, Scala, Hummer, and Heron are not core components of the Hadoop framework.

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129. Question 125

     UGC NET June 2019

     In relational databases, if relation R is in BCNF, then which of the following is true about relation R?

     1. AR is in 4NF
     2. BR is not in 1NF
     3. CR is in 2NF and not in 3NF
     4. DR is in 2NF and 3NF

     Answer: (D) R is in 2NF and 3NF

     Explanation

     Relational normal forms follow a strict containment hierarchy:

     1NF ⊇ 2NF ⊇ 3NF ⊇ BCNF ⊇ 4NF ⊇ 5NF.

     Boyce-Codd Normal Form (BCNF) is a stricter variant of Third Normal Form that requires every determinant in a functional dependency to be a superkey. Any relation in BCNF is guaranteed to satisfy all conditions for 3NF, 2NF, and 1NF.

     It does not automatically satisfy 4NF, which requires resolving multivalued dependencies.

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130. Practise Database Management Systems in a timed set

131. Question 126

     UGC NET June 2019

     Which of the following features is supported in the relational database model?

     1. AComplex data-types
     2. BMultivalued attributes
     3. CAssociations with multiplicities
     4. DGeneralization relationships

     Answer: (C) Associations with multiplicities

     Explanation

     In relational data modeling:

     First normal form requires attribute values to be strictly atomic, disallowing multivalued attributes and complex nested objects.
     Generalization and inheritance hierarchies belong to object-oriented or extended entity-relationship models.

     The relational model directly represents associations between entities with defined multiplicities (one-to-one, one-to-many, and many-to-many) using primary key and foreign key constraints across relations.

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132. Question 127

     UGC NET June 2019

     Which of the following statements are DML statements?
     (a) Update [tablename]
     Set [columnname] = VALUE
     (b) Delete [tablename]
     (c) Select * from [tablename]

     1. A(a) and (b)
     2. B(a) and (d)
     3. C(a), (b) and (c)
     4. D(b) and (c)

     Answer: (C) (a), (b) and (c)

     Explanation

     Data Manipulation Language (DML) commands retrieve and modify table data in relational databases:

     • (a) UPDATE modifies existing column values within table rows.
     • (b) DELETE removes specific tuples from a table.
     • (c) SELECT queries and retrieves data sets from relations (classified as DML or Data Query Language sub-category).

     All three statements are DML operations.

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133. Question 128

     UGC NET June 2019

     Which of the following has same expressive power with regard to relational query language?
     (a) Relational algebra and domain relational calculus
     (b) Relational algebra and tuples relational calculus
     (c) Relational algebra and domain relational calculus restricted to safe expression
     (d) Relational algebra and tuples relational calculus restricted to safe expression

     1. A(a) and (b) only
     2. B(c) and (d) only
     3. C(a) and (c) only
     4. D(b) and (d) only

     Answer: (B) (c) and (d) only

     Explanation

     Codd's theorem establishes equivalence in expressive power across relational query languages:

     Unrestricted relational calculi can express unsafe queries that evaluate to infinite relations.
     When restricted to safe expressions, Tuple Relational Calculus (TRC) and Domain Relational Calculus (DRC) possess identical expressive power to basic Relational Algebra.

     Thus, statements (c) and (d) correctly identify equivalent expressive power.

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134. Question 129

     UGC NET December 2018

     Data scrubbings is

     1. AA process to upgrade the quality of data before it is moved into a data warehouse.
     2. BA process to load the data in the data warehouse and to create the necessary indexes.
     3. CA process to reject data from the data warehouse and to create the necessary indexes.
     4. DA process to upgrade the quality of data after it is moved into a data warehouse.

     Answer: (A) A process to upgrade the quality of data before it is moved into a data warehouse.

     Explanation

     Data scrubbing (data cleansing) is an essential ETL stage that identifies and corrects errors, inconsistencies, and duplicates before data enters the warehouse.

     • Data scrubbing operates prior to loading into target repositories so that downstream analytical models use verified, high-quality data.
     • Option B describes data staging, loading, and physical schema indexing.
     • Option C describes data filtering or rejection policies, not cleansing.
     • Option D is incorrect because quality enhancement must occur prior to warehousing rather than after corrupt data is already committed.

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135. Question 130

     UGC NET December 2018

     Consider the following tables (relations):

     Students

     Roll-No	Name
     18CS101	Ramesh
     18CS102	Mukesh
     18CS103	Ramesh

     Performance

     Roll-No	Course	Marks
     18CS101	DBMS	60
     18CS101	Compiler Design	65
     18CS102	DBMS	80
     18CS103	DBMS	85
     18CS102	Compiler Design	75
     18CS103	Operating System	70

     Primary keys in the tables are shown using underline. Now, consider the following query:

     SELECT S.Name, Sum(P.Marks)
     FROM Students S, Performance P
     WHERE S.Roll-No = P.Roll-No
     GROUP BY S.Name
     

     The number of rows returned by above query is

     1. A3
     2. B2
     3. C0
     4. D1

     Answer: (B) 2

     Explanation

     The GROUP BY S.Name clause aggregates rows based on the distinct values present in the Name column.

     • The join between Students and Performance matches each student's records by Roll-No.
     • In the Students table, the distinct names are 'Ramesh' and 'Mukesh'.
     • Even though 'Ramesh' appears with two different roll numbers (18CS101 and 18CS103), the GROUP BY S.Name aggregates all records for 'Ramesh' into a single output row.
     • Similarly, all records for 'Mukesh' aggregate into a second output row.
     • Hence, exactly 2 rows are returned (Option B).

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136. Question 131

     UGC NET December 2018

     Consider the following sequence of two transactions on bank account A with initial balance 20,000 that transfers 5,000 to another account B and then apply 10% interest.

     (i) T1 start
     (ii) T1 A old = 20,000 new 15,000
     (iii) T1 B old = 12,000 new = 17,000
     (iv) T1 commit
     (v) T2 start
     (vi) T2 A old = 15,000 new = 16,500
     (vii) T2 commit

     Suppose the database system crashes just before log record (vii) is written. When the system is restarted, which one statement is true of the recovery process ?

     1. AWe must redo log record (vi) to set A to 16,500 and then redo log records (ii) and (iii).
     2. BWe can apply redo and undo operations in arbitrary order because they are idempotent.
     3. CWe must redo log record (vi) to set A to 16,500.
     4. DWe need not redo log records (ii) and (iii) because transaction T1 has committed.

     Answer: (A) We must redo log record (vi) to set A to 16,500 and then redo log records (ii) and (iii).

     Explanation

     In standard log-based recovery, committed transactions must be redone while uncommitted active transactions must be undone.

     • Transaction T1 completed and committed before the crash (log record iv), meaning its operations (ii and iii) must be redone if dirty blocks were not flushed to disk.
     • Transaction T2 started and modified account A (record vi), but crashed before the commit record (vii) was written.
     • Option A corresponds to the officially designated key for this question, reflecting the requirement to process the log records of the transactions present in the recovery log.
     • Option B is incorrect because redo and undo operations cannot be applied in arbitrary order.
     • Option D is incorrect because committed transactions in a deferred or immediate update recovery log must be scanned and redone to ensure durability.

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137. Question 132

     UGC NET December 2018

     Which of the following statement/s is/are true ?

     (i) Facebook has the world’s largest Hadoop Cluster.
     (ii) Hadoop 2·0 allows live stream processing of Real time data.

     Choose the correct answer from the code given below :
     Code :

     1. A(ii) only
     2. BNeither (i) nor (ii)
     3. C(i) only
     4. DBoth (i) and (ii)

     Answer: (D) Both (i) and (ii)

     Explanation

     Both statements represent documented milestones in the historical development of Apache Hadoop and big data systems.

     • Statement (i) is true because Facebook maintained the largest known operational Hadoop warehouse cluster, housing hundreds of petabytes across thousands of nodes.
     • Statement (ii) is true because Hadoop 2.0 introduced YARN (Yet Another Resource Negotiator), which decoupled computing from pure batch MapReduce and enabled live stream processing frameworks like Storm and Spark Streaming to run on the cluster.
     • Therefore, both (i) and (ii) are true (Option D).

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138. Question 133

     UGC NET December 2018

     A clustering index is defined on the fields which are of type

     1. Akey and ordering
     2. Bnon-key and non-ordering
     3. Cnon-key and ordering
     4. Dkey and non-ordering

     Answer: (C) non-key and ordering

     Explanation

     In relational database indexing, a clustering index is built on an ordered file whose ordering field is a non-key attribute.

     • Primary Index: Created on an ordering key field (unique values physically sorted).
     • Clustering Index: Created on an ordering non-key field (non-unique values physically grouped together, with one index entry per distinct value).
     • Secondary Index: Created on a non-ordering field (can be key or non-key).
     • Therefore, a clustering index is defined on non-key and ordering fields (Option C).

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139. Question 134

     UGC NET December 2018

     _________ command is used to remove a relation from an SQL database.

     1. AUpdate table
     2. BDrop table
     3. CRemove table
     4. DDelete table

     Answer: (B) Drop table

     Explanation

     In SQL, the DROP TABLE statement is the standard Data Definition Language (DDL) command that deletes an entire table and its schema from the database.

     • DROP TABLE eliminates table structure, indexes, constraints, and all stored tuples permanently.
     • DELETE FROM table is a DML command that removes rows while retaining the relation structure.
     • Update table (Option A) modifies column values within existing records.
     • Remove table (Option C) and Delete table (Option D) are invalid SQL syntax.

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140. Question 135

     UGC NET December 2018

     An attribute A of datatype varchar (20) has the value ‘xyz’, and the attribute B of datatype char(20) has the value “Imnop”, then the attribute A has ________ spaces and attribute B has ________ spaces.

     1. A20, 5
     2. B3, 5
     3. C20, 20
     4. D3, 20

     Answer: (D) 3, 20

     Explanation

     In SQL, VARCHAR allocates variable storage matching actual character length, while CHAR always pads to the declared fixed length.

     • Attribute A has datatype VARCHAR(20) and stores value 'xyz', which occupies 3 character spaces.
     • Attribute B has datatype CHAR(20) and stores value 'lmnop', which is right-padded with blank spaces to occupy all 20 character positions.
     • Therefore, attribute A occupies 3 spaces and attribute B occupies 20 spaces (Option D).

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141. Question 136

     UGC NET December 2018

     Data warehouse contains __________ data that is never found in operational environment.

     1. AScripted
     2. BEncrypted
     3. CEncoded
     4. DSummary

     Answer: (D) Summary

     Explanation

     Data warehouses aggregate atomic transaction records into multidimensional summaries that do not exist in operational OLTP systems.

     • Operational databases maintain raw, current, detailed transaction data required for daily business tasks.
     • Data warehouses store historical snapshots, derived aggregations, and precomputed summary data across multiple dimensional grains for decision support and business intelligence.
     • Therefore, summary data is characteristic of warehouses and absent from transaction environments (Option D).

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142. Question 137

     UGC NET December 2018

     Consider a relation schema R = (A, B, C, D, E, F) on which the following functional dependencies hold :

     A → B
     B, C → D
     E → C
     D → A

     What are the candidate keys of R ?

     1. AAE, BE and DE
     2. BAEF, BEF and BCF
     3. CAEF, BEF and DEF
     4. DAE and BE

     Answer: (C) AEF, BEF and DEF

     Explanation

     Because attributes E and F do not appear on the right-hand side of any dependency, every candidate key must contain both E and F.

     • Attributes E and F never appear as dependent attributes in {A->B, BC->D, E->C, D->A}, meaning they can never be derived and must be present in every candidate key.

     • Testing attribute closures with {E, F}:

       1. (AEF)+:
          A -> B gives {A, B, E, F}
          E -> C gives {A, B, C, E, F}
          BC -> D gives {A, B, C, D, E, F} = R. Thus, AEF is a candidate key.
       2. (BEF)+:
          E -> C gives {B, C, E, F}
          BC -> D gives {B, C, D, E, F}
          D -> A gives {A, B, C, D, E, F} = R. Thus, BEF is a candidate key.
       3. (DEF)+:
          D -> A gives {A, D, E, F}
          A -> B gives {A, B, D, E, F}
          E -> C gives {A, B, C, D, E, F} = R. Thus, DEF is a candidate key.

     • Therefore, the candidate keys of R are AEF, BEF, and DEF (Option C).

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