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CIW 1D0-541 Exam With Confidence Using Practice Dumps

Exam Code:
1D0-541
Exam Name:
CIW v5 Database Design Specialist
Certification:
CIW Web Development Series
Vendor:
CIW
Questions:
124
Last Updated:
Feb 12, 2026
Exam Status:
Stable
CIW 1D0-541

1D0-541: CIW Web Development Series Exam 2025 Study Guide Pdf and Test Engine

Are you worried about passing the CIW 1D0-541 (CIW v5 Database Design Specialist) exam? Download the most recent CIW 1D0-541 braindumps with answers that are 100% real. After downloading the CIW 1D0-541 exam dumps training , you can receive 99 days of free updates, making this website one of the best options to save additional money. In order to help you prepare for the CIW 1D0-541 exam questions and verified answers by IT certified experts, CertsTopics has put together a complete collection of dumps questions and answers. To help you prepare and pass the CIW 1D0-541 exam on your first attempt, we have compiled actual exam questions and their answers. 

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CIW v5 Database Design Specialist Questions and Answers

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Question 1

Consider the Employee relation shown in the exhibit. A database manager wants to set up a view called Emp_Dept that allows users to find employees and their department ID numbers. Which SQL statement will accomplish this?

Options:

A.

CREATE VIEW Emp_Dept AS SELECT

Last_Name, First_Name, Dept_ID FROM

Employee;

B.

UPDATE VIEW Emp_Dept

AS SELECT * FROM

Employee;

C.

UPDATE VIEW Emp_Dept AS SELECT

Last_Name, First_Name, Dept_ID FROM

Employee;

D.

CREATE VIEW Emp_Dept

AS SELECT * FROM

Employee WHERE ID = 0001

AND ID = 0002 AND ID =

0003 AND ID = 0004;

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Question 2

Which subset of Structured Query Language (SQL) is used to create and name database entities?

Options:

A.

Data Query Language

B.

Database Entity Language

C.

Data Definition Language

D.

Data Manipulation Language

Question 3

What is the highest normal form of the relation(s) shown in the exhibit?

Options:

A.

Second normal form

B.

First normal form

C.

Boyce-Codd normal form

D.

Third normal form

E.

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

F.

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

G.

STUDENT( Student_Number: integer NOT NULL Name:

variable length character string length 20 NOT NULL)

Primary Key StudenMMumber 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

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