The University of Texas
CS 327e
Spring 2002
Test 1
NAME:
______________________________________ LAST 4 DIGITS of SSN: __________________
Date: 2002-02-18
Duration: 60 minutes
Total: 100 points, 50 questions, Each Question is weighted
equally, and is worth 2 points. Additionally, there is also a bonus question in
the end.
The correct answers are shown
in BOLD.
1) The DBMS manages the
interaction between the end user and the database.
a) True b)
False
2) There are 3 different
types of anomalies: modification, insertion, and deletion.
a) True b) False
3) The final outcome of a
natural JOIN yields a table that does not include un-matched pairs.
a) True b)
False
4) The proper use of foreign
keys is crucial to exercising data redundancy control.
a) True b) False
5) Cardinality expresses the
specific number of entity occurrences associated with one occurrence of the
related entity.
a) True b)
False
6) The physical model is both
software and hardware independent.
a) True b) False
7) Normalization is a process
for changing attributes to entities.
a) True b) False
8) Because a partial
dependency can exist only if a table's primary key is composed of several
attributes, a table whose primary key consists of only a single attribute must
automatically be in 2NF if it is in 1NF.
a) True b)
False
9) It is possible for a table
in Second Normal Form (2NF) to exhibit transitive dependency, where one or more
attributes may be functionally dependent on nonkey attributes.
a) True b)
False
10) A table is in Third
Normal Form (3NF) if it is in 2NF and it includes no transitive dependencies.
a) True b)
False
11) Normalization should be
part of the database design process.
a) True b)
False
12) Normalization procedures focus on the characteristics of specific entities. It represents a micro view of the entities within the E-R diagram.
a) True b)
False
13) The American National
Standards Institute prescribes a standard SQL - the most recent version is
known as SQL-99.
a) True b)
False
14) Any changes made to the
table contents are not physically saved on disk until you use the SAVE
<table name> command.
a) True b) False
15) You can select partial
table contents by naming the desired fields and by placing restrictions on the
rows to be included in the output.
a) True b)
False
16) SQL allows the use of
logical restrictions on its inquiries such as OR, AND, AND NOT.
a) True b)
False
17) DBMSs are important
because
|
a. |
We need good ways of
managing such data. |
|
b. |
They contain a query
language that makes it possible to produce ad hoc queries. |
|
c. |
They help create an
environment for end users to have access to more data. |
|
d. |
All of the above |
|
e. |
None of the above |
18) A file system is composed
of
|
a. |
Hardware - Software -
Procedure |
|
b. |
People - Hardware - Data |
|
c. |
Procedures - Hardware -
Software - Data |
|
d. |
Procedures - Hardware -
Software - Data - People |
19) A primary key
|
a. |
Consists of only one field. |
|
b. |
Has the same value for all
records. |
|
c. |
Must contain a unique
value for each record within the table. |
|
d. |
Is defined automatically. |
20) Of the database system
environment, which of the following is not considered part of the people
component?
|
a. |
Salesmen |
|
b. |
Administrators |
|
c. |
Designers |
|
d. |
End Users |
21) The most important
advantages of the hierarchical database model are
|
a. |
Conceptual simplicity,
security, integrity, diversity, and security. |
|
b. |
Security, efficiency,
diversity, simplicity, and integrity. |
|
c. |
Integrity, efficiency,
diversity, independence, and simplicity. |
|
d. |
Conceptual simplicity,
security, independence, integrity, and efficiency. |
22) Using network
terminology, a relationship is called a (n)
|
a. |
Member. |
|
b. |
Owner. |
|
c. |
Set. |
|
d. |
Table. |
23) The network database
models have
|
a. |
A navigational system that
yields simple design. |
|
b. |
A simple system that
promotes efficiency. |
|
c. |
An owner/member
relationship that promotes database integrity. |
|
d. |
a and b |
|
e. |
b and c |
24) The entity relationship
model
|
a. |
Has unlimited constraint
representation. |
|
b. |
Has unlimited relationship
representation. |
|
c. |
Has data manipulation
language. |
|
d. |
Has visual
representation that makes it an effective communications tool. |
25) Nulls, if used
improperly, can create problems because they can represent:
|
a. |
An unknown attribute value. |
|
b. |
A known, but missing,
attribute value. |
|
c. |
A "not
applicable" condition. |
|
d. |
All of the above |
|
e. |
None of the above |
26) The entity integrity rule
requires that
|
a. |
All entries are unique. |
|
b. |
A part of the key may be
null. |
|
c. |
Foreign key values do not
reference primary key values. |
|
d. |
Duplicate object values are
allowed. |
27) The referential integrity
rule requires that
|
a. |
Every null foreign key
value must reference an existing primary key value. |
|
b. |
It makes it possible for an
attribute to have a corresponding value. |
|
c. |
Every non-null foreign
key value must reference an existing primary key value. |
|
d. |
It makes it possible to
delete a row in one table whose primary key does not have a matching foreign
key value in another table. |
28) The key's role is based
on a concept known as
|
a. |
Consistency. |
|
b. |
Availability. |
|
c. |
Determination. |
|
d. |
Uniqueness. |
29) A primary key
|
a. |
Must be defined in every
table. |
|
b. |
Is always the first field
in each table. |
|
c. |
Must be numeric. |
|
d. |
Must be unique. |
30) A Relational operator
that yields all possible pairs of rows from two tables is known as a
|
a. |
Union. |
|
b. |
Intersect. |
|
c. |
Difference. |
|
d. |
Product. |
|
e. |
Project. |
31) A Relational operator
that yields all rows in one table that are not found in the other table is a
|
a. |
Union. |
|
b. |
Intersect. |
|
c. |
Difference. |
|
d. |
Product. |
|
e. |
Project. |
32) In the context of a
database table, the statement "A determines B" indicates that
|
a. |
Knowing the value of
attribute A you can not look up the value of attribute B. |
|
b. |
You do not need to know the
value of attribute A in order to look up the value of attribute B. |
|
c. |
Knowing the value of
attribute B you can look up the value of attribute A. |
|
d. |
Knowing the value of
attribute A you can look up the value of attribute B. |
33) In sophisticated
application development software, nulls can create problems when using
function(s) such as
|
a. |
Count |
|
b. |
Average |
|
c. |
Sum |
|
d. |
a & c |
|
e. |
a & b & c |
34) When designing a new
database, it is a good idea to
|
a. |
Avoid data redundancy. |
|
b. |
Include redundant fields. |
|
c. |
Include a common field in
all tables. |
|
d. |
Use composite keys. |
35) A super key that does not
contain a subset of attributes that is itself a super key is called a
|
a. |
Candidate key. |
|
b. |
Primary key. |
|
c. |
Super key. |
|
d. |
Secondary key. |
|
e. |
Foreign key. |
36) A Relational operator
that combines all rows from two tables is considered to be a
|
a. |
Union. |
|
b. |
Intersect. |
|
c. |
Difference. |
|
d. |
Product. |
|
e. |
Project. |
37) Database designers employ
data models as communications tools to facilitate the interaction among
|
a. |
The designers. |
|
b. |
The applications
programmers. |
|
c. |
The users. |
|
d. |
All of the above. |
38) The conflicts between
design efficiency, information requirements, and processing speed are often
resolved through
|
a. |
Conversion from 1NF to 2NF. |
|
b. |
Conversion from 2NF to 3NF. |
|
c. |
Compromises that include
denormalization. |
|
d. |
Conversion from 3NF to 4NF. |
39) Domains can be used when:
|
a. |
The set of possible values
is relatively small. |
|
b. |
The attribute
characteristic is not sufficiently important to justify the creation of a new
entity. |
|
c. |
The objective is to
standardize the data characteristic for an attribute appearing in multiple
tables |
|
d. |
The attribute
characteristic is not sufficiently important to justify the creation of a new
one. |
|
e. |
all of the above. |
40) The SQL command that lets
you save your work to disk, is
|
a. |
Insert. |
|
b. |
Select. |
|
c. |
Commit. |
|
d. |
Update. |
|
e. |
Rollback. |
41) The SQL command that lets
you list the table contents is
|
a. |
Insert. |
|
b. |
Select. |
|
c. |
Commit. |
|
d. |
Update. |
|
e. |
Rollback. |
42) The SQL command that
enables you to make changes in the data is
|
a. |
Insert. |
|
b. |
Select. |
|
c. |
Commit. |
|
d. |
Update. |
|
e. |
Rollback. |
43) To list all the contents
of a PRODUCT table you would use
|
a. |
LIST * FROM PRODUCT; |
|
b. |
SELECT * FROM PRODUCT; |
|
c. |
DISPLAY * FROM PRODUCT; |
|
d. |
SELECT ALL FROM PRODUCT; |
44) To remove the value
‘2238/QPD’ from the product table you must use the following command
|
a. |
DELETE FROM PRODUCT WHERE P_CODE = '2238/QPD'; |
|
b. |
REMOVE FROM PRODUCT WHERE P_CODE = '2238/QPD'; |
|
c. |
ERASE FROM PRODUCT WHERE P_CODE = '2238/QPD'; |
|
d. |
ROLLBACK FROM PRODUCT WHERE P_CODE = '2238/QPD'; |
45) To select partial table
contents you must use the command
|
a. |
SELECT <column(s)> FROM <Table
name> WHERE <item>; |
|
b. |
LIST <column(s)> FROM <Table name> WHERE <Conditions>; |
|
c. |
SELECT <column(s)> FROM <Table name> WHERE <Conditions>; |
|
d. |
LIST<column(s)> FROM <Table name> WHERE <Item>; |
46) To create a query using a
mathematical symbol to find all codes but code (21344) from the product table
you must write it as
|
a. |
SELECT P_DESCRIPT,
P_INDATE, P_PRICE, V_CODE FROM PRODUCT WHERE V_CODE <> 21344; |
|
b. |
SELECT P_DESCRIPT,
P_INDATE, P_PRICE, V_CODE FROM PRODUCT WHERE V_CODE <= 21344; |
|
c. |
SELECT P_DESCRIPT,
P_INDATE, P_PRICE, V_CODE FROM PRODUCT WHERE V_CODE = 21344; |
|
d. |
SELECT P_DESCRIPT,
P_INDATE, P_PRICE, V_CODE FROM PRODUCT WHERE V_CODE => 21344; |
47) ANSI-standard SQL allows
the use of special operator in conjunction with WHERE clause. Such special operator used to define a range
limit is
|
a. |
Between. |
|
b. |
Null. |
|
c. |
Like. |
|
d. |
In. |
|
e. |
Exists. |
48) The basic SQL aggregate
function that gives the number or rows containing not null values for the given
column is
|
a. |
COUNT |
|
b. |
MIN |
|
c. |
MAX |
|
d. |
SUM |
|
e. |
AVG |
49) The basic SQL aggregate
function that gives the total of all values for a selected attribute in a given
column is
|
a. |
COUNT |
|
b. |
MIN |
|
c. |
MAX |
|
d. |
SUM |
|
e. |
AVG |
50) A HAVING clause can be
used to replace a WHERE clause.
a) TRUE b) FALSE
51) Bonus Question: (2
points)
IBM bought another database
company in 2001. The DBMS that they got as part of the acquisition is
a) DB2 b) UDB c) SQL Anywhere d) Informix e) Teradata