￥ 295.00 1D0-541 Exam

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Exam : CIW 1D0-541
Title : CIW v5 Database Design Specialist


1. Consider the Information Engineering diagram shown in the exhibit. Which DBDL definition best describes this diagram?
A. BUILDING(Building_ID, Bldg_Name, Location, Room_Count)
Primary Key Building_ID
RESIDENT(R_ID, Room_Num, Res_Name, Building_ID)
Primary Key R_ID
B. BUILDING(Building_ID, Bldg_Name, Location, Room_Count)
Primary Key BUILDING
RESIDENT(R_ID, Room_Num, Res_Name, Building_ID)
Primary Key RESIDENT
C. BUILDING(Building_ID, Bldg_Name, Location, Room_Count)
Primary Key BUILDING
Foreign Key BUILDING(Building_ID) references RESIDENT(Building_ID)
RESIDENT(R_ID, Room_Num, Res_Name, Building_ID)
Primary Key RESIDENT
D. BUILDING(Building_ID, Bldg_Name, Location, Room_Count)
Primary Key Building_ID
RESIDENT(R_ID, Room_Num, Res_Name, Building_ID)
Primary Key R_ID
Foreign Key Building_ID references BUILDING(Building_ID)
Answer: D

2. What is the highest normal form of the relation(s) shown in the exhibit?
A. Second normal form
B. First normal form
C. Boyce-Codd normal form
D. Third normal form
Answer: A

3. Which three pieces of information did E.F. Codd describe as necessary to retrieve a data value from a relational database?
A. Attribute, domain, and tuple
B. Entity, relation name, and domain
C. Table name, primary key, and entity
D. Attribute, relation name, and primary key
Answer: D

4. Several SQL operations are performed by User 1 to access the Fee information for Bowling in the Act_Fee relation (shown in the exhibit). The first access returns a fee of 50. An unrelated SQL operation by another user updates the Bowling fee to 60. The second access by User 1 returns a fee of 60. What problem has occurred?
A. Rollback
B. Deadlock
C. Dirty read
D. No problem has occurred.
Answer: D

5. Consider the entity-relation (ER) diagram shown in the exhibit. When the logical database design phase is completed, which of the following is a valid DBDL description of the base relations for the ER diagram?
A. STUDENT(
Student_Number: integer NOT NULL
Name: variable length character string length 20 NOT NULL)
Primary Key Student_Number
CLASS(
Class_Num: integer NOT NULL
Class_Name: integer NOT NULL)
Primary Key Class_Num
B. STUDENT(
Student_Number: integer NOT NULL
Name: variable length character string length 20 NOT NULL)
Primary Key Student_Number
CLASS(
Class_Num: integer NOT NULL
Class_Name: integer NOT NULL)
Primary Key Class_Num
Foreign Key Class_Num References STUDENT
C. STUDENT(
Student_Number: integer NOT NULL
Name: variable length character string length 20 NOT NULL)
Primary Key Student_Number
STU_CLASS(
Student_Number: integer NOT NULL
Class_Num: integer NOT NULL)
Primary Key Student_Number
CLASS(
Class_Num: integer NOT NULL
Class_Name: integer NOT NULL)
Primary Key Class_Num
D. STUDENT(
Student_Number: integer NOT NULL
Name: variable length character string length 20 NOT NULL)
Primary Key Student_Number
STU_CLASS(
Student_Number: integer NOT NULL
Class_Num: integer NOT NULL)
Primary Key Student_Number, Class_Num
CLASS(
Class_Num: integer NOT NULL
Class_Name: integer NOT NULL)
Primary Key Class_Num
Answer: D