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This section includes 30 Mcqs, each offering curated multiple-choice questions to sharpen your Database knowledge and support exam preparation. Choose a topic below to get started.
1. |
Distributing the columns of a table into several separate physical records is known as horizontal partitioning. |
A. | 1 |
B. | |
Answer» C. | |
2. |
The three newest database architectures in use today are relational, multidimensional and hierarchical. |
A. | 1 |
B. | |
Answer» C. | |
3. |
Data-volume and frequency-of-use statistics are not critical inputs to the physical database design process. |
A. | 1 |
B. | |
Answer» C. | |
4. |
Some advantages of partitioning include: efficiency, security, and load balancing. |
A. | 1 |
B. | |
Answer» B. | |
5. |
A pointer is a detailed coding scheme recognized by system software for representing organizational data. |
A. | 1 |
B. | |
Answer» C. | |
6. |
It is usually not very important to design the physical database to minimize the time required by users to interact with the information systems. |
A. | 1 |
B. | |
Answer» C. | |
7. |
In general, larger block sizes are used for online transaction processing applications and smaller block sizes are used for databases with a decision support or data warehousing system. |
A. | 1 |
B. | |
Answer» C. | |
8. |
A field represents each component of a composite attribute. |
A. | 1 |
B. | |
Answer» B. | |
9. |
A multidimensional database model is used most often in which of the following models? |
A. | Data warehouse |
B. | Relational |
C. | Hierarchical |
D. | Network |
Answer» B. Relational | |
10. |
What is the best data type definition for Oracle when a field is alphanumeric and has a fixed length? |
A. | VARCHAR2 |
B. | CHAR |
C. | LONG |
D. | NUMBER |
Answer» C. LONG | |
11. |
Hashing algorithm converts a primary key value into a record address. |
A. | 1 |
B. | |
Answer» B. | |
12. |
The primary goal of physical database design is data processing efficiency. |
A. | 1 |
B. | |
Answer» B. | |
13. |
The fastest read/write time and most efficient data storage of any disk array type is: |
A. | RAID-0. |
B. | RAID-1. |
C. | RAID-2. |
D. | RAID-3. |
Answer» B. RAID-1. | |
14. |
A bitmap index is an index on columns from two or more tables that come from the same domain of values. |
A. | 1 |
B. | |
Answer» C. | |
15. |
Denormalization and clustering can work well to minimize data access time for small records. |
A. | 1 |
B. | |
Answer» B. | |
16. |
An extent is a contiguous section of disk storage space. |
A. | 1 |
B. | |
C. | 1 |
D. | |
Answer» B. | |
17. |
Which of the following improves a query's processing time? |
A. | Write complex queries. |
B. | Combine a table with itself. |
C. | Query one query within another. |
D. | Use compatible data types. |
Answer» E. | |
18. |
Sequential retrieval on a primary key for sequential file storage has which of the following features? |
A. | Very fast |
B. | Moderately fast |
C. | Slow |
D. | Impractical |
Answer» B. Moderately fast | |
19. |
Which of the following are integrity controls that a DBMS may support? |
A. | Assume a default value in a field unless a user enters a value for that field. |
B. | Limit the set of permissible values that a field may assume. |
C. | Limit the use of null values in some fields. |
D. | All of the above. |
Answer» E. | |
20. |
When storage space is scarce and physical records cannot span pages, creating multiple physical records from one logical relation will minimize wasted storage space. |
A. | 1 |
B. | |
C. | 1 |
D. | |
Answer» B. | |
21. |
Which of the following is not a factor to consider when switching from small to large block size? |
A. | The length of all of the fields in a table row. |
B. | The number of columns |
C. | Block contention |
D. | Random row access speed |
Answer» C. Block contention | |
22. |
A rule of thumb for choosing indexes for a relational database includes which of the following? |
A. | Indexes are more useful on smaller tables. |
B. | Indexes are more useful for columns that do not appear frequently in the WHERE clause in queries. |
C. | Do not specify a unique index for the primary key of each table. |
D. | Be careful indexing attributes that have null values. |
Answer» E. | |
23. |
Two basic constructs to link one piece of data with another piece of data: sequential storage and pointers. |
A. | 1 |
B. | |
C. | 1 |
D. | |
Answer» B. | |
24. |
If a denormalization situation exists with a many-to-many or associative binary relationship, which of the following is true? |
A. | All fields are stored in one relation. |
B. | All fields are stored in two relations. |
C. | All fields are stored in three relations. |
D. | All fields are stored in four relations. |
Answer» C. All fields are stored in three relations. | |
25. |
The blocking factor is: |
A. | a group of fields stored in adjacent memory. |
B. | the number of physical records per page. |
C. | attributes grouped together by the same primary key. |
D. | attributes grouped together by the same secondary key. |
Answer» C. attributes grouped together by the same primary key. | |
26. |
Selecting a data type involves which of the following? |
A. | Maximize storage space |
B. | Represent most values |
C. | Improve data integrity |
D. | All of the above. |
Answer» D. All of the above. | |
27. |
What is the best data type definition for Oracle when a field is alphanumeric and has a length that can vary? |
A. | VARCHAR2 |
B. | CHAR |
C. | LONG |
D. | NUMBER |
Answer» B. CHAR | |
28. |
Which of the following is an advantage of partitioning? |
A. | Complexity |
B. | Inconsistent access speed |
C. | Extra space |
D. | Security |
Answer» E. | |
29. |
A secondary key is which of the following? |
A. | Nonunique key |
B. | Primary key |
C. | Useful for denormalization decisions |
D. | Determines the tablespace required |
Answer» B. Primary key | |
30. |
If a denormalization situation exists with a one-to-one binary relationship, which of the following is true? |
A. | All fields are stored in one relation. |
B. | All fields are stored in two relations. |
C. | All fields are stored in three relations. |
D. | All fields are stored in four relations. |
Answer» B. All fields are stored in two relations. | |