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WGU Data Management – Foundations Exam Sample Questions (Q25-Q30):

NEW QUESTION # 25
What is the second step in the implement relationships stage of database design?

A. Implement subtype entities
B. Implement weak entities
C. Implement one-one relationships
D. Specify cascade

Answer: C

Explanation:
Thesecond step in implementing relationshipsis definingone-to-one (1:1) relationshipsbetween entities.
Example Usage:
* Example of a 1:1 relationship:
sql
CREATE TABLE Employees (
EmpID INT PRIMARY KEY,
Name VARCHAR(50)
);
CREATE TABLE EmployeeDetails (
EmpID INT PRIMARY KEY,
Address VARCHAR(255),
FOREIGN KEY (EmpID) REFERENCES Employees(EmpID)
);
* Here, eachemployee has exactly one detail record, creating a1:1 relationship.
Why Other Options Are Incorrect:
* Option A (Implement weak entities) (Incorrect):Weak entities rely on aforeign keyand are implementedlater.
* Option C (Implement subtype entities) (Incorrect):Subtypes arespecial casesandnot implemented in the second step.
* Option D (Specify cascade) (Incorrect):Cascade rules (ON DELETE, ON UPDATE)are defined duringforeign key implementation, not in the second step.
Thus, the correct answer isImplement one-one relationships, as it is thenext logical stepafter defining entities.

NEW QUESTION # 26
Which product has an open-source license in addition to having a non-relational system?

A. MySQL
B. MongoDB
C. SQL Server
D. Oracle Database

Answer: B

Explanation:
MongoDB is aNoSQL (non-relational) databasethat isopen-sourceand supportsdocument-oriented storage
. It allows for flexible schema design and is optimized for big data applications.
* Option A (Correct):MongoDB is aNoSQL databasewith an open-sourceAGPLlicense, making it both free to use and non-relational.
* Option B (Incorrect):SQL Serveris arelational database(RDBMS) developed by Microsoft and isnot open-source.
* Option C (Incorrect):MySQLisopen-sourcebut is arelational database(RDBMS), not a NoSQL system.
* Option D (Incorrect):Oracle Databaseisrelationalandproprietary(not open-source).

NEW QUESTION # 27
Which relationship or association exists between a supertype and its subtype entities?

A. IsA relationship
B. Weak entity
C. Associative entity
D. Strong entity

Answer: A

Explanation:
Indatabase modeling, the relationship between asupertype and its subtypesis called anIsA relationship.
Example Usage:
* AVehicle supertypemay haveCar and Truck subtypes.
Vehicle
### Car
### Truck
* InER diagrams, this is represented as:
Vehicle (Supertype)
|
### Car (Subtype)
### Truck (Subtype)
* SQL Table Implementation:
sql
CREATE TABLE Vehicle (
VehicleID INT PRIMARY KEY,
Make VARCHAR(50),
Model VARCHAR(50)
);
CREATE TABLE Car (
VehicleID INT PRIMARY KEY,
FOREIGN KEY (VehicleID) REFERENCES Vehicle(VehicleID),
EngineType VARCHAR(50)
);
CREATE TABLE Truck (
VehicleID INT PRIMARY KEY,
FOREIGN KEY (VehicleID) REFERENCES Vehicle(VehicleID),
CargoCapacity INT
);
* This structurepreserves the IsA relationshipbetween Vehicle (supertype) and Car/Truck (subtypes).
Why Other Options Are Incorrect:
* Option A (Strong entity) (Incorrect):Strong entitiesdo not rely on a supertype/subtype hierarchy.
* Option C (Associative entity) (Incorrect):Used toresolve many-to-many relationships, not supertype
/subtype relationships.
* Option D (Weak entity) (Incorrect):Weak entitiesdepend on a strong entity, but supertype-subtype relations useinheritance(not dependency).
Thus, the correct answer isIsA relationship, as it describes theinheritance hierarchybetweensupertypes and subtypes.

NEW QUESTION # 28
Which expression can be used to create a temporary name for a table?

A. UNION
B. ALIAS
C. NEW
D. HAVING

Answer: B

Explanation:
Analiasis used in SQL to give atemporaryname to a table or column within a query. It makes queries more readable and helps in cases where a table needs to be referenced multiple times (e.g., in aself-join).
Example Usage:
sql
SELECT e.Name, d.DepartmentName
FROM Employees AS e
JOIN Departments AS d
ON e.DeptID = d.ID;
* Here,Employees is aliased as eandDepartments as d, making the query shorter and clearer.
Why Other Options Are Incorrect:
* Option A (HAVING) (Incorrect):Used to filter grouped results, not create aliases.
* Option B (NEW) (Incorrect):Not a valid SQL keyword for aliasing.
* Option D (UNION) (Incorrect):Combines result sets but does not rename tables.
Thus, the correct answer isALIAS, which allows fortemporary naming of tables or columns.

NEW QUESTION # 29
Which term refers to a path from a top-level block to a bottom-level block?

A. Crow's foot
B. Fan-out
C. Branch
D. Sparse index

Answer: C

Explanation:
In database indexing, abranchrefers to thepath from the top-level block (root node) to a bottom-level block (leaf node)in aB-Tree or B+ Tree index structure.
Example Usage in Indexing:
* AB-Tree indexorganizes data hierarchically, withbranches leading to different parts of the tree.
* When searching for a record, thequery follows a branchfrom theroot node down to the correct leaf node.
Why Other Options Are Incorrect:
* Option A (Fan-out) (Incorrect):Refers tohow many children a node has, not the path.
* Option B (Crow's foot) (Incorrect):Anotation used in ER diagrams, not indexing.
* Option D (Sparse index) (Incorrect):A type of indexstoring only some entries, not the path itself.
Thus, the correct answer isBranch, as it defines thepath from top to bottom in a database index.

NEW QUESTION # 30
......

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