Infromation Technology Quiz: Database Management

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| By Sumit Srivastava
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Sumit Srivastava
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Quizzes Created: 2 | Total Attempts: 1,382
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Infromation Technology Quiz: Database Management - Quiz

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Questions and Answers
  • 1. 

    Which of the following gives a logical structure of the database graphically?

    • A.

      Entity-relationship diagram

    • B.

      Entity diagram

    • C.

      Both A and B are correct.

    • D.

      Neither A nor B

    Correct Answer
    A. Entity-relationship diagram
    Explanation
    An entity-relationship diagram is a graphical representation that shows the logical structure of a database. It depicts the entities (objects or concepts) in the database, their relationships with each other, and the attributes (properties) of each entity. This diagram helps in understanding the relationships and dependencies between different entities in the database, making it easier to design, analyze, and communicate about the database structure.

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  • 2. 

    Every weak entity set can be converted into a strong entity set by:  

    • A.

      Using generalization

    • B.

      Adding appropriate attributes

    • C.

      Using aggregation

    • D.

      None of the above

    Correct Answer
    B. Adding appropriate attributes
    Explanation
    Adding appropriate attributes to a weak entity set can convert it into a strong entity set. This is because weak entity sets depend on a strong entity set for their existence and cannot be uniquely identified on their own. By adding appropriate attributes, the weak entity set gains the ability to be uniquely identified and becomes a strong entity set. Generalization and aggregation are not the correct methods for converting a weak entity set into a strong entity set.

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  • 3. 

    In a one-to-many relationship, the entity that is on the one side of the relationship is called a(n) ________ entity.

    • A.

      Parent

    • B.

      Child

    • C.

      Subtype

    • D.

      Instance

    Correct Answer
    A. Parent
    Explanation
    In a one-to-many relationship, the entity that is on the one side of the relationship is called a "Parent" entity. This is because the parent entity can have multiple child entities associated with it, but each child entity can only have one parent entity. The parent entity often acts as the main or primary entity in the relationship, while the child entities depend on the parent entity for their existence or attributes.

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  • 4. 

    Weak entity relationship is represented as:

    • A.

      Underline

    • B.

      Double line

    • C.

      Double diamond

    • D.

      Double rectangle

    Correct Answer
    C. Double diamond
    Explanation
    A weak entity relationship is represented by a double diamond. A weak entity is an entity that cannot exist without the existence of another entity, known as the identifying entity. The double diamond symbolizes this dependency, indicating that the weak entity's existence is reliant on the identifying entity. This notation helps to visually distinguish weak entity relationships from regular entity relationships.

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  • 5. 

    The ER model includes additional concepts like:

    • A.

      Specialization

    • B.

      Generalization

    • C.

      Categorization

    • D.

      All of the Above

    Correct Answer
    D. All of the Above
    Explanation
    The ER model includes additional concepts like specialization, generalization, and categorization. Specialization refers to the process of defining subclasses within an entity class, where each subclass inherits the attributes and relationships of the superclass. Generalization, on the other hand, is the process of defining a superclass that encompasses multiple subclasses. Categorization involves grouping similar entities into categories based on common characteristics. Therefore, all of the above concepts are included in the ER model.

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  • 6. 

    E-R modelling techniques is a:

    • A.

      Top-down approach

    • B.

      Bottom-up approach

    • C.

      Left-right approach

    • D.

      None of the above

    Correct Answer
    A. Top-down approach
    Explanation
    The correct answer is top-down approach because E-R modeling techniques typically involve starting with a high-level view of the system and gradually refining it by breaking it down into smaller and more detailed components. This approach allows for a systematic and organized way of designing the system, starting from the overall structure and then moving towards the specific details.

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  • 7. 

    The total participation by entities is represented in E-R diagram as:

    • A.

      Dashed line

    • B.

      Double line

    • C.

      Double rectangle

    • D.

      Circle

    Correct Answer
    B. Double line
    Explanation
    In an E-R diagram, the total participation by entities is represented by a double line. This indicates that every entity in the relationship must participate fully in the relationship. It means that each entity is connected to the relationship through a mandatory participation, and cannot exist without being associated with the relationship. The double line represents a strong relationship between the entities and emphasizes the requirement for total participation.

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  • 8. 

     Which relationship is used to represent a specialization entity?

    • A.

      ISA

    • B.

      AIS

    • C.

      ONIS

    • D.

      WHOIS

    Correct Answer
    A. ISA
    Explanation
    The relationship used to represent a specialization entity is ISA. This relationship indicates that one entity is a specialized version of another entity. It is commonly used in inheritance hierarchies, where the specialized entities inherit attributes and behaviors from the general entity.

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  • 9. 

    A _____________ constraint requires that an entity belong to no more than one lower-level entity set.

    • A.

      Disjointness

    • B.

      Uniqueness

    • C.

      Special

    • D.

      Relational

    Correct Answer
    A. Disjointness
    Explanation
    A disjointness constraint requires that an entity belong to no more than one lower-level entity set. This means that an entity can only be associated with one specific lower-level entity set and cannot belong to multiple lower-level entity sets simultaneously. This constraint ensures that the relationships between entities are clear and unambiguous, preventing any confusion or duplication of data.

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  • 10. 

    The completeness constraint may be one of the following: Total generalization or specialization, Partial generalization or specialization . Which is the default?

    • A.

      Total

    • B.

      Partial

    • C.

      Should be specified

    • D.

      Cannot be determined

    Correct Answer
    B. Partial
    Explanation
    The default completeness constraint is partial. This means that unless otherwise specified, a generalization or specialization is assumed to be partial. In a partial generalization or specialization, not all entities or attributes are included in the higher-level entity or specialization. This allows for flexibility and allows for entities or attributes to be excluded if they are not relevant to the higher-level concept.

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  • 11. 

    Which normal form is considered adequate for normal relational database design?

    • A.

      2NF

    • B.

      5NF

    • C.

      4NF

    • D.

      3NF

    Correct Answer
    D. 3NF
    Explanation
    The 3NF (Third Normal Form) is considered adequate for normal relational database design. This normal form ensures that there are no transitive dependencies within a table. It eliminates any non-key attributes that depend on other non-key attributes, promoting data integrity and reducing redundancy. By achieving 3NF, a database design is considered to be well-structured and efficient, making it suitable for most applications.

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  • 12. 

    Consider a schema R(A, B, C, D) and functional dependencies A -> B and C -> D. Then the decomposition of R into R1 (A, B) and R2(C, D) is

    • A.

      Dependency preserving and lossless join

    • B.

      Lossless join but not dependency preserving

    • C.

      Dependency preserving but not lossless join

    • D.

      (d) not dependency preserving and not lossless join

    Correct Answer
    C. Dependency preserving but not lossless join
    Explanation
    The decomposition of R into R1(A, B) and R2(C, D) is dependency preserving because the functional dependencies A -> B and C -> D are still preserved in the decomposed relations. However, it is not lossless join because there is no common attribute between R1 and R2 that can be used to join them and obtain the original relation R.

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  • 13. 

    Relation R with an associated set of functional dependencies, F, is decomposed into BCNF. The redundancy (arising out of functional dependencies) in the resulting set of relations is:

    • A.

      Zero

    • B.

      More than zero but less than that of an equivalent 3NF decomposition 

    • C.

      Proportional to the size of F+

    • D.

      Indeterminate

    Correct Answer
    B. More than zero but less than that of an equivalent 3NF decomposition 
    Explanation
    The correct answer is "More than zero but less than that of an equivalent 3NF decomposition". This means that when a relation is decomposed into BCNF, there will still be some redundancy present in the resulting set of relations. However, this redundancy will be less than what would be present in an equivalent 3NF decomposition. BCNF aims to eliminate redundancy by ensuring that every non-trivial functional dependency is represented by a candidate key. However, it does not guarantee complete elimination of redundancy, unlike 3NF decomposition.

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  • 14. 

    A table has fields F1, F2, F3, F4, and F5, with the following functional dependencies: 1) F1->F3 2) F2->F4 3) (F1,F2)->F5 in terms of normalization, this table is in

    • A.

      1NF

    • B.

      2NF

    • C.

      3NF

    • D.

      NONE OF THE ABOVE

    Correct Answer
    A. 1NF
    Explanation
    The given table is in 1NF (First Normal Form) because it satisfies the requirement that each attribute in a table must contain only atomic values. The functional dependencies F1->F3 and F2->F4 indicate that there are no repeating groups or multivalued dependencies in the table. However, the functional dependency (F1,F2)->F5 suggests that there may be a partial dependency, which violates the second normal form (2NF) requirement. Therefore, the table is not in 2NF or 3NF.

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  • 15. 

    Consider the following functional dependencies in a database. 1) Date_of_Birth->Age 2) Age->Eligibility 3) Name->Roll_number 4)Roll_number->Name 5) Course_number->Course_name 6)Course_number->Instructor 7) (Roll_number, Course_number)->Grade The relation (Roll_number, Name, Date_of_birth, Age) is

    • A.

      In second normal form but not in third normal form 

    • B.

      In third normal form but not in BCNF 

    • C.

      In BCNF   

    • D.

      In none of the above

    Correct Answer
    D. In none of the above
    Explanation
    The given relation (Roll_number, Name, Date_of_birth, Age) is not in any of the above forms because it violates the third normal form (3NF) and Boyce-Codd Normal Form (BCNF). In 3NF, no non-key attribute (attribute not part of the primary key) should depend on another non-key attribute. Here, Age depends on Date_of_Birth, which is a violation. In BCNF, every determinant (attribute that determines another attribute) should be a candidate key. In this case, Age is not a candidate key, but it determines Eligibility, which violates BCNF. Therefore, the relation is not in 3NF or BCNF.

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  • 16. 

    The relation schema Student_Performance (name, courseNo, rollNo, grade) has the following FDs: 1) name, courseNo->grade 2) rollNo, courseNo->grade 3)name->rollNo 4) rollNo->name The highest normal form of this relation scheme is:

    • A.

      2NF

    • B.

      BCNF

    • C.

      3NF

    • D.

      4NF

    Correct Answer
    C. 3NF
    Explanation
    The highest normal form of this relation schema is 3NF. This is because all the functional dependencies in the schema are already in 3NF. The first two functional dependencies (1) name, courseNo->grade and (2) rollNo, courseNo->grade are both already in 3NF as they have a single attribute on the right-hand side. The third functional dependency (3) name->rollNo and the fourth functional dependency (4) rollNo->name are also in 3NF as they have a transitive dependency. Therefore, the relation schema is already in 3NF and cannot be further decomposed.

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  • 17. 

    The relation EMPDT1 is defined with attributes empcode(unique), name, street, city, state, and pincode. For any pincode, there is only one city and state. Also, for any given street, city and state, there is just one pincode. In normalization terms EMPDT1 is a relation in:

    • A.

      1NF only

    • B.

      3NF and hence also in 2NF and 1NF 

    • C.

      BCNF and hence also in 3NF, 2NF and 1NF

    • D.

      2NF and hence also in 1NF

    Correct Answer
    D. 2NF and hence also in 1NF
    Explanation
    The relation EMPDT1 satisfies the conditions for 2NF because it does not have any partial dependencies. This means that all non-key attributes depend on the entire primary key. It also satisfies the conditions for 1NF because it does not have any repeating groups or composite attributes. Therefore, the relation is in 2NF and hence also in 1NF.

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  • 18. 

    Which one of the following statements if FALSE?

    • A.

      A prime attribute can be transitively dependent on a key in a BCNF relation.

    • B.

      Any relation with two attributes is in BCNF

    • C.

      A relation in which every key has only one attribute is in 2NF

    • D.

      A prime attribute can be transitively dependent on a key in a 3 NF relation.

    Correct Answer
    A. A prime attribute can be transitively dependent on a key in a BCNF relation.
    Explanation
    A prime attribute can be transitively dependent on a key in a BCNF relation. This statement is false because in BCNF (Boyce-Codd Normal Form), all non-key attributes must be functionally dependent on the entire primary key, and there should be no transitive dependencies. Transitive dependency occurs when a non-key attribute is functionally dependent on another non-key attribute, which violates BCNF.

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  • 19. 

    Consider the following relational schemes for a library database: Book (Title, Author, Catalog_no, Publisher, Year, Price) Collection (Title, Author, Catalog_no) With the following functional dependencies: I. Title Author -> Catalog_no II. Catalog_no -> Title Author Publisher Year III. Publisher Title Year -> Price Assume {Author, Title} is the key for both schemes. Which of the following statements is true?

    • A.

      Both Book and Collection are in 3NF only

    • B.

      Book is in 2NF and Collection is in 3NF

    • C.

      Both Book and Collection are in BCNF

    • D.

      Both Book and Collection are in 2NF only

    Correct Answer
    B. Book is in 2NF and Collection is in 3NF
    Explanation
    The given answer states that Book is in 2NF and Collection is in 3NF. This means that Book satisfies the requirements of 2NF, which includes being in 1NF and having no partial dependencies, while Collection satisfies the requirements of 3NF, which includes being in 2NF and having no transitive dependencies. The functional dependencies given in the question indicate that Book has a partial dependency (Catalog_no depends on Title and Author) and Collection has a transitive dependency (Catalog_no determines Title, Author, Publisher, and Year). Therefore, the given answer correctly identifies the normalization levels for each scheme.

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  • 20. 

    Let R(A,B,C,D,E,P,G) be a relational schema in which the following FDs are known to hold: 1) AB->CD 2) DE->P 3) C->E 4) P->C 5) B->G The relation schema R is

    • A.

      Not in 2NF

    • B.

      In 3NF, but not in BCNF

    • C.

      In 2NF, but not in 3NF      

    • D.

      In BCNF  

    Correct Answer
    A. Not in 2NF
    Explanation
    The given relational schema is not in 2NF because there is a partial dependency between the attributes. The FDs (functional dependencies) AB->CD and DE->P indicate that attributes C and P depend on only a part of the candidate key (AB and DE respectively). Therefore, it violates the second normal form (2NF) which requires that non-prime attributes should not be functionally dependent on only a part of the candidate key.

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  • 21. 

    Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values. F={CH→G, A→BC, B→CFH, E→A, F→EG} is a set of functional dependencies (FDs) so that F + is exactly the set of FDs that hold for R. How many candidate keys does the relation R have?

    • A.

      3

    • B.

      5

    • C.

      4

    • D.

      6

    Correct Answer
    C. 4
    Explanation
    The given functional dependencies are CH→G, A→BC, B→CFH, E→A, F→EG. To determine the candidate keys, we need to find the minimal set of attributes that can uniquely identify all other attributes in the relation. By analyzing the given functional dependencies, we can see that the attributes A, B, C, and E are all candidate keys because they can determine all other attributes in the relation. Therefore, the relation R has 4 candidate keys.

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  • 22. 

    Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values. F={CH→G, A→BC, B→CFH, E→A, F→EG} is a set of functional dependencies (FDs) so that F + is exactly the set of FDs that hold for R. The relation R is

    • A.

      In 2NF, but not in 3NF. 

    • B.

      In 3NF, but not in BCNF. 

    • C.

      In 1NF, but not in 2NF. 

    • D.

      In BCNF.

    Correct Answer
    C. In 1NF, but not in 2NF. 
    Explanation
    The given set of functional dependencies (FDs) F={CH→G, A→BC, B→CFH, E→A, F→EG} implies that there are partial dependencies present in the relation R. This means that there are attributes that depend on only a part of the primary key. Specifically, attribute B depends on attribute A, and attribute F depends on attribute E. Therefore, the relation R is in 1NF because it contains atomic values, but it is not in 2NF because it has partial dependencies.

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  • 23. 

    Every time attribute A appears, it is matched with the same value of attribute B, but not the same value of attribute C. Therefore, it is true that:

    • A.

      A → B.

    • B.

      A → C.

    • C.

      A → (B,C).

    • D.

      (B,C) → A.

    Correct Answer
    A. A → B.
    Explanation
    The given information states that whenever attribute A appears, it is always matched with the same value of attribute B, but not the same value of attribute C. This implies that there is a direct relationship between A and B, where A determines the value of B. However, there is no clear relationship between A and C, as A does not determine the value of C. Therefore, the correct answer is A → B, indicating that attribute A determines the value of attribute B.

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  • 24. 

    A functional dependency is a relationship between or among:

    • A.

      Tables

    • B.

      Rows

    • C.

      Relations

    • D.

      Attributes

    Correct Answer
    D. Attributes
    Explanation
    A functional dependency is a relationship between attributes in a database. It describes how the value of one attribute determines the value of another attribute. In other words, if we know the value of one attribute, we can determine the value of another attribute based on this relationship. This relationship exists between attributes, not tables, rows, or relations. Therefore, the correct answer is "Attributes".

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  • 25. 

    If every functional dependency in set E is also in closure of F then this is classified as:

    • A.

      FD is covered by E

    • B.

      E is covered by F

    • C.

      F is covered by E

    • D.

      F+ is covered by E

    Correct Answer
    B. E is covered by F
    Explanation
    If every functional dependency in set E is also in the closure of F, it means that the closure of F contains all the functional dependencies in E. This implies that all the functional dependencies in E can be derived from F, indicating that E is covered by F.

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  • 26. 

    Inst_dept (ID, name, salary, dept name, building, budget) is decomposed into instructor (ID, name, dept name, salary) department (dept name, building, budget) This comes under

    • A.

      Lossy-join decomposition

    • B.

      Lossy decomposition

    • C.

      Lossless-join decomposition

    • D.

      Both a and b

    Correct Answer
    D. Both a and b
    Explanation
    This decomposition is both lossy-join and lossless-join. It is lossy-join because the original relation Inst_dept cannot be reconstructed from the two decomposed relations (instructor and department) without losing information. It is also lossless-join because the natural join of the decomposed relations (instructor and department) will result in the original relation Inst_dept. Therefore, the correct answer is both a and b.

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  • 27. 

    Consider a relation R(A,B,C,D,E) with the following functional dependencies: ABC -> DE and D -> AB The number of superkeys of R is:

    • A.

      2

    • B.

      7

    • C.

      10

    • D.

      15

    Correct Answer
    C. 10
    Explanation
    The given relation R has two functional dependencies: ABC -> DE and D -> AB. A superkey is a set of attributes that can uniquely identify each tuple in a relation. To find the number of superkeys, we need to consider all possible combinations of attributes. In this case, we have 5 attributes (A, B, C, D, E). The total number of possible combinations is 2^5 = 32. However, we need to exclude the empty set and any single attribute sets because they cannot uniquely identify each tuple. Therefore, the number of superkeys of R is 32 - 1 - 5 = 26.

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  • 28. 

    In a two-phase locking protocol, a transaction release locks in ......... phase.

    • A.

      Shrinking phase

    • B.

      Growing phase

    • C.

      Running phase

    • D.

      Initial phase

    Correct Answer
    A. Shrinking pHase
    Explanation
    In a two-phase locking protocol, the transaction releases locks in the shrinking phase. This phase occurs after the transaction has completed its execution and is ready to release the locks it acquired during the growing phase. During the shrinking phase, the transaction releases the locks in a reverse order of their acquisition, ensuring that no other transaction can access the released resources. This allows for better concurrency control and prevents conflicts between transactions.

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  • 29. 

    ............ protocol grantees that a set of transactions becomes serializable.

    • A.

      Two phase locking

    • B.

      Two phase commit

    • C.

      Transaction locking

    • D.

      Checkpoints

    Correct Answer
    A. Two pHase locking
    Explanation
    Two phase locking is a protocol that guarantees that a set of transactions becomes serializable. This protocol ensures that each transaction acquires all the necessary locks before it starts executing and releases all the locks only after it completes. The protocol consists of two phases: the growing phase, where transactions acquire locks, and the shrinking phase, where transactions release locks. By following this protocol, conflicts and inconsistencies between transactions are avoided, ensuring that the set of transactions can be executed in a serializable manner.

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  • 30. 

    The situation in which a transaction holds a data item and waits for the release of data item held by some other transaction, which in turn waits for another transaction, is called ...

    • A.

      Serialiable schedule

    • B.

      Process waiting

    • C.

      Concurrency

    • D.

      Deadlock

    Correct Answer
    D. Deadlock
    Explanation
    A deadlock occurs when multiple transactions are waiting for each other to release the data items they hold. In this situation, none of the transactions can proceed, resulting in a deadlock. This can happen when there is a circular dependency among the transactions, causing them to wait indefinitely for each other. Deadlocks can severely impact the performance and efficiency of a system, and proper deadlock detection and prevention mechanisms are necessary to avoid them.

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  • 31. 

    The database system must take special actions to ensure that transactions operate properly without interference from concurrently executing database statements. This property is referred to as:

    • A.

      Atomicity

    • B.

      Durability

    • C.

      Isolation

    • D.

      Consistency

    Correct Answer
    C. Isolation
    Explanation
    Isolation refers to the property of a database system that ensures that concurrent transactions do not interfere with each other. It ensures that each transaction is executed in isolation, as if it were the only transaction running. This is achieved by using various concurrency control techniques such as locking and multiversion concurrency control. By maintaining isolation, the database system can prevent issues such as dirty reads, non-repeatable reads, and phantom reads, which can occur when multiple transactions access and modify the same data simultaneously.

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  • 32. 

    Which of the following is a procedure for acquiring the necessary locks for a transaction where all necessary locks are acquired before any are released?

    • A.

      Record controller

    • B.

      Exclusive lock

    • C.

      Authorization rule

    • D.

      Two phase lock

    Correct Answer
    D. Two pHase lock
    Explanation
    The two-phase lock is a procedure for acquiring the necessary locks for a transaction where all necessary locks are acquired before any are released. This means that in the first phase of the locking process, all required locks are acquired, and in the second phase, all locks are released. This ensures that the transaction maintains consistency and avoids any conflicts or inconsistencies that may arise if locks are released before all necessary locks are acquired.

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  • 33. 

    When transaction Ti requests a data item currently held by Tj , Ti is allowed to wait only if it has a timestamp smaller than that of Tj (that is, Ti is older than Tj ). Otherwise, Ti is rolled back (dies). This is:

    • A.

      Wait-die

    • B.

      Wait-wound

    • C.

      Wound-wait

    • D.

      Wait

    Correct Answer
    A. Wait-die
    Explanation
    Wait-die is a technique used in concurrency control to prevent deadlock in a transaction system. It allows a transaction with a smaller timestamp to wait for a data item held by a transaction with a larger timestamp. If a transaction with a smaller timestamp requests a data item held by a transaction with a larger timestamp, it is allowed to wait. However, if a transaction with a larger timestamp requests a data item held by a transaction with a smaller timestamp, the smaller timestamp transaction is rolled back (dies). This ensures that older transactions are given priority and prevents deadlock from occurring.

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  • 34. 

    Immediate database modification technique uses:

    • A.

      Both undo and redo

    • B.

      Undo but no redo

    • C.

      Redo but no undo

    • D.

      Neither undo nor redo

    Correct Answer
    A. Both undo and redo
    Explanation
    Immediate database modification technique uses both undo and redo. Undo allows for the reversal of changes made to the database, ensuring that any incorrect or unwanted modifications can be reverted. Redo, on the other hand, enables the reapplication of changes that were previously undone, ensuring that all modifications are properly recorded and applied to the database. By utilizing both undo and redo, the immediate database modification technique ensures data integrity and consistency.

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  • 35. 

    Consider the following transactions with data items P and Q initialized to zero: T1: read (P) ; read (Q) ; if P = 0 then Q : = Q + 1 ; write (Q) ; T2: read (Q) ; read (P) ; if Q = 0 then P : = P + 1 ; write (P) ; Any non-serial interleaving of T1 and T2 for concurrent execution leads to...

    • A.

      A Serializable Schedule

    • B.

      A Schedule that is not conflict Serializable

    • C.

      A Conflict Serializable Schedule

    • D.

      A Schedule for which a precedence graph cannot be drawn

    Correct Answer
    B. A Schedule that is not conflict Serializable
    Explanation
    A non-serial interleaving of T1 and T2 for concurrent execution leads to a schedule that is not conflict Serializable. Conflict Serializable means that the final result of the schedule is the same as if the transactions were executed serially in some order. In this case, if T1 and T2 are executed concurrently, there is a possibility of conflicts between their read and write operations on the shared data items P and Q. Therefore, the resulting schedule may not be conflict Serializable.

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  • 36. 

    Consider the transactions T1, T2, and 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?

    • A.

      Only S1 is conflict Serializable

    • B.

      Only S2 is conflict Serializable

    • C.

      Both S1 and S2 is conflict Serializable

    • D.

      Neither S1 nor S2 is conflict Serializable

    Correct Answer
    A. Only S1 is conflict Serializable
    Explanation
    The given schedules S1 and S2 are both conflict serializable. Conflict serializability means that the final result of the schedules is the same as if the transactions were executed one after the other in some order. In both S1 and S2, the conflicting operations (read-write or write-read) are executed in the same order as they appear in the original transactions T1, T2, and T3. Therefore, both schedules are conflict serializable.

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  • 37. 

    Consider the following transaction involving two bank accounts x and y. read(x); x := x – 50; write(x); read(y); y := y +50; write(y) The constraint that the sum of the accounts x and y should remain constant is that of:

    • A.

      Atomicity

    • B.

      Consistency

    • C.

      Durability

    • D.

      Isolation

    Correct Answer
    B. Consistency
    Explanation
    The constraint that the sum of the accounts x and y should remain constant is that of consistency. Consistency ensures that the data remains in a valid state before and after the execution of a transaction. In this case, the sum of the accounts x and y should remain the same before and after the transaction. If the transaction deducts 50 from x, it should add 50 to y to maintain the consistency of the total sum of the accounts.

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  • 38. 

    Which level of RAID refers to disk mirroring with block striping?

    • A.

      RAID level 1

    • B.

      RAID level 2

    • C.

      RAID level 3

    • D.

      RAID level 4

    Correct Answer
    A. RAID level 1
    Explanation
    RAID level 1 refers to disk mirroring with block striping. In this level of RAID, data is duplicated on two separate disks, providing redundancy and fault tolerance. The data is also striped across the disks, allowing for improved read and write performance.

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  • 39. 

    A unit of storage that can store one or more records in a hash file organization is denoted as:

    • A.

      Buckets

    • B.

      Disk Pages

    • C.

      Blocks

    • D.

      Nodes

    Correct Answer
    A. Buckets
    Explanation
    A unit of storage that can store one or more records in a hash file organization is denoted as "buckets". In a hash file organization, records are distributed across different buckets based on a hashing function. Each bucket can hold multiple records, and the hashing function determines which bucket a record should be placed in. This allows for efficient retrieval of records based on their keys, as the hashing function helps narrow down the search to a specific bucket. Therefore, "buckets" is the correct answer for this question.

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  • 40. 

    The file organization which allows us to read records that would satisfy the join condition by using one block read is:

    • A.

      Heap file organization

    • B.

      Sequential file organization

    • C.

      Clustering file organization

    • D.

      Hash file organization

    Correct Answer
    C. Clustering file organization
    Explanation
    Clustering file organization allows us to read records that would satisfy the join condition by using one block read. In clustering file organization, the records that are related to each other based on a common attribute value are stored physically close to each other on disk. This reduces the number of disk I/O operations required to retrieve records that satisfy a join condition, as the related records can be read in a single block read. This organization is particularly useful for improving the performance of join operations in database systems.

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  • 41. 

    The highest level in the hierarchy of data organization is called:

    • A.

      Databank

    • B.

      Database

    • C.

      Datafile

    • D.

      Datarecord

    Correct Answer
    B. Database
    Explanation
    The highest level in the hierarchy of data organization is called a database. A database is a structured collection of data that is organized and stored in a way that allows for efficient retrieval, manipulation, and management. It typically consists of multiple interconnected data files that are organized and structured using a database management system (DBMS). A database provides a centralized and integrated view of data, allowing users to efficiently store, retrieve, and analyze large amounts of information.

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  • 42. 

    Which of the following hardware component is the most important to the operation of database management system?

    • A.

      High Resolution Video Display

    • B.

      Printer

    • C.

      High Speed Large Capacity Disk

    • D.

      Mouse

    Correct Answer
    C. High Speed Large Capacity Disk
    Explanation
    The most important hardware component for the operation of a database management system is a high-speed large capacity disk. This is because a database management system requires storage space to store and retrieve large amounts of data efficiently. A high-speed large capacity disk allows for faster data access and retrieval, which is crucial for the smooth functioning of a database management system.

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  • 43. 

    What operator tests column for the absence of data?

    • A.

      EXISTS operator

    • B.

      NOT operator

    • C.

      IS NULL operator

    • D.

      None of these

    Correct Answer
    C. IS NULL operator
    Explanation
    The IS NULL operator is used to test if a column contains no data. It returns true if the column has a null value and false otherwise. This operator is commonly used in SQL queries to filter out rows that have missing or unknown values in a specific column.

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  • 44. 

    In SQL, which command(s) is(are) used to change a table's storage characteristics?

    • A.

      ALTER TABLE

    • B.

      MODIFY TABLE

    • C.

      CHANGE TABLE

    • D.

      All of the above

    Correct Answer
    A. ALTER TABLE
    Explanation
    The correct answer is ALTER TABLE. This command is used in SQL to make changes to the structure of a table, including changing its storage characteristics. The other options, MODIFY TABLE and CHANGE TABLE, are not valid SQL commands. Therefore, the correct command to use in this case is ALTER TABLE.

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  • 45. 

    Which of the following statement on the view concept in SQL is invalid?

    • A.

      All views are not updatable

    • B.

      The views may be referenced in an SQL statement whenever tables are referenced

    • C.

      The views are instantiated at the time they are referenced and not when they are defined

    • D.

      The definition of a view should not have GROUP BY clause in it

    Correct Answer
    D. The definition of a view should not have GROUP BY clause in it
    Explanation
    The given statement on the view concept in SQL that is invalid is "the definition of a view should not have GROUP BY clause in it." This is incorrect because views can have a GROUP BY clause in their definition. The GROUP BY clause is used to group rows based on a specified column or columns, and it can be used in the definition of a view to create a summarized or aggregated view of the data.

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  • 46. 

    Find the temperature in increasing order of all cities:

    • A.

      SELECT city FROM weather ORDER BY temperature;

    • B.

      SELECT city, temperature FROM weather;

    • C.

      SELECT city, temperature FROM weather ORDER BY temperature;

    • D.

      SELECT city, temperature FROM weather ORDER BY city;

    Correct Answer
    C. SELECT city, temperature FROM weather ORDER BY temperature;
    Explanation
    This query selects both the city and temperature columns from the weather table and orders the result set by the temperature column in ascending order. This means that the cities will be listed in increasing order based on their respective temperatures.

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  • 47. 

    Find the names of these cities with temperature and condition whose condition is neither sunny nor cloudy:

    • A.

      SELECT city, temperature, condition FROM weather WHERE condition NOT IN ('sunny', 'cloudy');

    • B.

      SELECT city, temperature, condition FROM weather WHERE condition NOT BETWEEN ('sunny', 'cloudy');

    • C.

      SELECT city, temperature, condition FROM weather WHERE condition IN ('sunny', 'cloudy');

    • D.

      SELECT city, temperature, condition FROM weather WHERE condition BETWEEN ('sunny', 'cloudy');

    Correct Answer
    A. SELECT city, temperature, condition FROM weather WHERE condition NOT IN ('sunny', 'cloudy');
    Explanation
    The correct answer is "SELECT city, temperature, condition FROM weather WHERE condition NOT IN ('sunny', 'cloudy')". This query will retrieve the names of cities along with their temperature and condition where the condition is neither sunny nor cloudy. It uses the NOT IN operator to exclude the conditions 'sunny' and 'cloudy' from the result set.

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  • 48. 

    Which of the following is not true of the traditional approach to information processing:

    • A.

      There is common sharing of data among the various applications

    • B.

      It is file oriented

    • C.

      Programs are dependent on the file

    • D.

      It is inflexible

    Correct Answer
    A. There is common sharing of data among the various applications
    Explanation
    The traditional approach to information processing does not involve common sharing of data among the various applications. In this approach, each application has its own separate data files and does not share data with other applications. This lack of data sharing can lead to inefficiencies and duplication of data, as each application must maintain its own copy of the data it needs.

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  • 49. 

    The information about data in a database is called _______.

    • A.

      Metadata

    • B.

      Hyperdata

    • C.

      Teradata

    • D.

      None of these

    Correct Answer
    A. Metadata
    Explanation
    Metadata refers to the information about the data in a database. It includes details such as the structure, format, and characteristics of the data. Metadata provides valuable insights into the data, allowing users to understand and effectively utilize the information stored in the database. It plays a crucial role in data management and helps in organizing, accessing, and analyzing the data efficiently. Hyperdata and Teradata are not relevant to the question, and therefore, the correct answer is metadata.

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  • 50. 

    A data dictionary is a special file that contains?

    • A.

      The names of all fields in all files

    • B.

      The data types of all fields in all files

    • C.

      The widths of all fields in all files

    • D.

      All of the mentioned

    Correct Answer
    D. All of the mentioned
    Explanation
    A data dictionary is a special file that contains all the mentioned information. It includes the names of all fields in all files, the data types of all fields in all files, and the widths of all fields in all files. This comprehensive file serves as a reference guide for understanding the structure and organization of the data within a database or system.

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Quiz Review Timeline +

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Apr 27, 2017
    Quiz Created by
    Sumit Srivastava

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