Major links



Quicklinks


📌 Quick Links
[ DBMS ] [ SQL ] [ DDB ] [ ML ] [ DL ] [ NLP ] [ DSA ] [ PDB ] [ DWDM ] [ Quizzes ]


Wednesday, February 17, 2016

Anna University MCA MC 1654 – DATABASE MANAGEMENT SYSTEMS Question May June 2008

MC 1654 – Database Management Systems - May June 2008 Question Paper / Anna University MCA Exam May June 2008 Question Paper / MC1654 DBMS May June 2008 Question Paper / MC1654 Database Management Systems Previous Year MCA Question Paper / Anna University Affiliated Colleges MCA Question Papers / MC1654 DBMS Regulation 2005 MCA Question Paper / MCA Second Semester Question Paper





Question Paper Code :
M.C.A DEGREE EXAMINATION, MAY/JUNE 2008
MC 1654 – DATABASE MANAGEMENT SYSTEMS
(Regulation 2005)
Time : Three Hours                                                         Maximum : 100 marks
Answer ALL Questions
PART A – (10 X 2 = 20 marks)

1. Write any four major disadvantages of keeping organization information in a file processing system.
2. Define a weak and strong entity set.
3. What is a condition Box in the context of Query By example (QBE)?
4. Write the advantages and Disadvantages of 3 NF.
5. Difference between mirroring and striping.
6. What are Bitmap indices?
7. How can you measure the cost of a query?
8. List some of the useful statistical information about database relations as maintained by DBMS catalog.
9. Write any four properties of a transaction.
10. What is fuzzy check point?

PART B – (5 X 16 = 80 marks)

11. (a) (i) Explain two methods for dealing with the deadlock problem ?
(ii) Write short note on shadow paging and buffer management.
Or
(b) (i) Explain the distinction between the terms Serial Schedules and Serializable Schedules.
(ii) During its execution, a transaction passes through several stages, until it finally commits or aborts. List all possible sequences of stages through which a transaction may pass.
12. (a) Explain various data models with suitable examples for each.
Or
(b) (i) Construct an E-R diagram with all major components for a banking enterprises with entity set customer, branch, loan, payment, account, employee with other own assumption.
(ii) Write short note on history of database system
13. (a) (i) Explain the fourth Normal Form (4NF).
(ii) Write short note on Embedded SQL.
Or
(b) (i) Explain about Functional Dependencies (FD’s).
ii) Write short note on Embedded SQL.
14. (a) (i) Explain the physical characteristics of Disk.
(ii) What is RAID? Explain all the RAID levels.
Or
(b) Explain B+ Tree index files.
15. (a) What do you mean by external sorting? Explain the merge sort technique and discuss its performance.
Or
(b) How can you perform heuristic optimization? Explain the various heuristics in query optimization with examples

*************



Tuesday, February 16, 2016

Super key candidate key primary key in normalization

Keys that are used in normalization process - Define super key - Define candidate key - Define primary key - Define secondary key - Define composite key


Various keys used in the normalization process



It is the combination of two or more attributes that can uniquely identify a record within a table. There may be many super keys in a table.
Candidate KeyClick here to read more….

Candidate key is an attribute or set of attributes that can be used to uniquely identify any records in a table.
Technically, a candidate key is a minimal super key. That is, it is a super key for which no proper subset can be a key. A candidate key is a minimal set of attributes that can be used to identify a tuple (record) uniquely.
We may have many candidate keys in a table.

A primary key is one of the candidate keys of the table that is most appropriate to be the main reference key for the table. As a rule of thumb, we may try to choose the smallest among the candidate keys as the primary key.
Secondary / Alternate Key – Click here to read more….

All the candidate keys that are part of a table other than the primary key are called as secondary keys or alternate keys.
Composite Key – Click here to read more….

It is a candidate key (or primary key) that consists of more than one attribute (at least two attributes) to uniquely identify records of the table. Here, the component attributes are called prime attributes.






Monday, February 15, 2016

Database management systems - Normalization and normal forms quiz

Database management system quiz - DBMS quiz with answers - DBMS multiple choice questions - Database normalization quiz - DBMS normalization process interview questions - Normalization and normal forms quiz in DBMS - Normal forms in Database interview questions


Normalization QUIZ


1. Assume that a relation R has the following properties. What is the normal form of R?
            No multi-valued attributes
            No partial key dependencies
(a) First Normal Form
(b) Second Normal Form
(c) Third Normal Form
(d) Boyce-Codd Normal Form
2. Assume that a relation R has the following properties. What is the normal form of R?
            Has no partial key dependencies
            Has multi-valued attributes
(a) First Normal Form
(b) Second Normal Form
(c) Third Normal Form
(d) None of the above
3. All functional dependencies must _________
(a) have attributes of the same table
(b) have a single attribute on its LHS
(c) have a single attribute on its RHS
(d) All of the above
4. Which one of the following is always a single-valued attribute?
(a) a person’s phone number
(b) a person’s address
(c) a person’s age
(d) a person’s car

Let R (A, B, C, D, E, F) be a relation with the functional dependencies A → B, A → C, CD → E, CD → F, B → E. Use this information to answer questions 5, 6, and 7.

5. Which of the following does not hold on R?
(a) A → E
(b) CD → EF
(c) AD → F
(d) B → CD
6. Which of the following is holding on R?
(a) A → C
(b) B → A
(c) CD → A
(d) C → D
7. Which of the following does not hold on R?
(a) BC → D
(b) AD → B
(c) BC → E
(d) ACD → BEF
8. A functional dependency that exist between two non-key attributes is called _____
(a) Non-transitive dependency
(b) Transitive dependency
(c) Partial transitive dependency
(d) None of the above
9. Assume that a relation R has the following properties. What is the normal form of R?
            Has no multi-valued attributes
            Has no partial key dependencies
            Has attributes with atomic domains
            Has transitive dependencies
(a) 1NF & 2NF
(b) 1NF, 2NF, & 3NF
(c) 1NF & 3NF
(d) BCNF
10. Assume a relation R with the following 3 attributes; one attribute as primary key, one attribute as foreign key and one non-key attribute. Also, none of them is multi-valued attribute. What is the maximum normal form R can be in?
(a) 1NF
(b) 2NF
(c) 3NF
(d) BCNF



Go to QUIZZ in Database / DBMS / Advanced DBMS page



Please visit, subscribe and share 10 Minutes Lectures in Computer Science

Featured Content

Multiple choice questions in Natural Language Processing Home

MCQ in Natural Language Processing, Quiz questions with answers in NLP, Top interview questions in NLP with answers Multiple Choice Que...

All time most popular contents