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Technical Interview Questions and Answers :: Syntel Inc.

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Tot. Mock Test: 4


Total Qs: 161+

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    71 / 161

    What are the advantages and disadvantages of Normalization?
    Answer:

    There are several advantages of normalization as under:
    * Faster sorting and index creation.
    * A larger number of clustered indexes.
    * Narrower and more compact indexes.
    * Fewer indexes per table, which improves the performance of INSERT, UPDATE, and DELETE statements
    * Fewer null values and less opportunity for inconsistency, which increase database compactness.

    Beside the above benefits there are few disadvantages as well:
    * Increased amount of Normalization increases the amount of complexity of joins between tables and that hinders the performance.

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    72 / 161

    What is Lossless join property?
    Answer:

    It guarantees that the spurious tuple generation does not occur with respect to relation schemas after decomposition.

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    73 / 161

    What is 1 NF (Normal Form)?
    Answer:

    The domain of an attribute must include only atomic (simple, indivisible) values.

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    74 / 161

    What is Fully Functional dependency?
    Answer:

    It is based on the concept of full functional dependency. A functional dependency X -> Y is full functional dependency if removal of any attribute A from X means that the dependency does not hold any more.

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    Answer:

    A relation schema R is in 2NF if it is in 1NF and every non-prime attribute A in R is fully functionally dependent on primary key.

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    Answer:

    A relation schema R is in 3NF if it is in 2NF and for every FD X -> A either of the following is true
    > X is a Super-key of R.
    > A is a prime attribute of R.
    In other words, if every non prime attribute is non-transitively dependent on primary key.

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    77 / 161

    What is BCNF (Boyce-Codd Normal Form)?
    Answer:

    A relation schema R is in BCNF if it is in 3NF and satisfies an additional constraint that for every FD X -> A, X must be a candidate key.

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    Answer:

    A relation schema R is said to be in 4NF if for every Multivalued dependency X ->-> Y that holds over R, one of following is true
    > X is subset or equal to (or) XY = R.
    > X is a super key.

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    Answer:

    A Relation schema R is said to be 5NF if for every join dependency {R1, R2, ..., Rn} that holds R, one the following is true
    > Ri = R for some i.
    > The join dependency is implied by the set of FD, over R in which the left side is key of R.

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    80 / 161

    What is Domain-Key Normal Form?
    Answer:

    A relation is said to be in DKNF if all constraints and dependencies that should hold on the constraint can be enforced by simply enforcing the domain constraint and key constraint on the relation.

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