1
WGU D426 Data Management Foundations
ACTUAL EXAM 2026/2027 | Version A |
Verified Questions & Answers | Pass
Guaranteed - A+ Graded
QUESTIONS 1-65
Q1.
Midtown Books runs an on-line store. The analysts have sketched the following partial ERD in a
cloud-based PostgreSQL database:
Copy
BOOK (book_id PK, title, author_id FK, price)
AUTHOR (author_id PK, author_name, birth_date)
Which phrase best describes the relationship between BOOK and AUTHOR?
A. One-to-one: each book has exactly one author record and vice-versa.
B. Many-to-one: several book entities can reference the same author entity.
C. Many-to-many: one book can have many authors and one author many books.
D. One-to-many: each author entity can have many book entities, but a book belongs to only one
author.
Correct Answer: D
Rationale: The FK author_id in BOOK points to AUTHOR, placing the "many" side on BOOK;
therefore one AUTHOR is related to many BOOKs. This is a classic 1-to-many (authors-to-
books) design. A is wrong because the PK/FK structure does not enforce mutual uniqueness; B
reverses the direction of the relationship; C would require a link table which is absent here.
Q2.
A start-up logs IoT temperature readings. A single sensor sends one record every 5 seconds. The
engineers need a composite key because no single column is unique. Which pair of columns
would form the smallest natural composite key?
,2
A. (sensor_id, temperature_value)
B. (sensor_id, recorded_at)
C. (temperature_value, recorded_at)
D. (sensor_id, location_name)
Correct Answer: B
Rationale: The combination of sensor identifier and the exact timestamp (recorded_at) is
naturally unique for each reading. A fails because temperature values repeat frequently; C fails
because two sensors can log identical temperatures at the same moment; D is larger and
location_name is functionally dependent on sensor_id, making it redundant in the key.
Q3.
The university registrar keeps data in first-normal-form only:
STUDENT_COURSE( student_id, course_id, {grade, semester}, advisor_name, advisor_office )
(The braces denote a repeating group: one row stores many grades/semesters for the same
student-course.) Which anomaly is impossible to remove while the design stays in 1NF?
A. Update anomaly on advisor_office when an advisor moves.
B. Insertion anomaly when a new advisor has no students yet.
C. Deletion anomaly that loses the only row for an advisor and inadvertently deletes his/her
office.
D. Redundancy of repeating grade/semester data within one row.
Correct Answer: D
Rationale: 1NF allows repeating groups; redundancy inside the group is permitted until we
flatten to atomic values (and eventually normalize). A-C are classic anomalies that appear even
in 1NF flat tables and are fixed by higher normal forms, but D is inherent to the repeating-group
version of 1NF.
Q4.
CloudMart uses a distributed relational database. The ORDER_LINE table currently has 120
million rows. Which SQL clause is most appropriate to verify that every order_id in
ORDER_LINE references an existing order_id in the ORDERS table without enforcing the
constraint?
A. SELECT order_id FROM ORDER_LINE WHERE order_id IS NULL;
B. SELECT o.order_id FROM ORDER_LINE o LEFT JOIN ORDERS r ON o.order_id =
r.order_id WHERE r.order_id IS NULL;
, 3
C. SELECT COUNT(*) FROM ORDER_LINE GROUP BY order_id HAVING COUNT(*) > 1;
D. SELECT order_id FROM ORDER_LINE INTERSECT SELECT order_id FROM ORDERS;
Correct Answer: B
Rationale: The LEFT JOIN returns NULL on the right side for orphaned order_ids; filtering on
NULL finds violations. A checks only NULLs in the base table, not referential integrity; C looks
for duplicates; D returns common values, not orphans.
Q5.
A manufacturing company needs to store CAD drawings alongside metadata. Drawings average
90 MB. Which strategy best balances performance and referential integrity in a relational cloud
warehouse?
A. Store the file as a BLOB inside the same table as metadata.
B. Save the file in object storage and keep the object key as a VARCHAR foreign key.
C. Compress the drawing, encode as Base64, and store in a TEXT column.
D. Use a separate NoSQL document database exclusively.
Correct Answer: B
Rationale: Object storage is cost-optimized for large binaries; keeping the object key in the
relational table preserves joins without bloating the buffer pool. A hurts cache performance; C
increases size ~33%; D abandons relational integrity when hybrid is possible.
Q6.
Consider the relation R( A, B, C, D, E ) with functional dependencies:
{ A → B, B → C, C → DE }
What is the correct closure of attribute A?
A. {A,B}
B. {A,B,C}
C. {A,B,C,D,E}
D. {A,C,D,E}
Correct Answer: C
Rationale: Start with {A}; A→B adds B; B→C adds C; C→DE adds D and E. Therefore A⁺ =
all attributes, making A a candidate key.
Q7.
A hospital designs a DRUG_ALLERGY table:
WGU D426 Data Management Foundations
ACTUAL EXAM 2026/2027 | Version A |
Verified Questions & Answers | Pass
Guaranteed - A+ Graded
QUESTIONS 1-65
Q1.
Midtown Books runs an on-line store. The analysts have sketched the following partial ERD in a
cloud-based PostgreSQL database:
Copy
BOOK (book_id PK, title, author_id FK, price)
AUTHOR (author_id PK, author_name, birth_date)
Which phrase best describes the relationship between BOOK and AUTHOR?
A. One-to-one: each book has exactly one author record and vice-versa.
B. Many-to-one: several book entities can reference the same author entity.
C. Many-to-many: one book can have many authors and one author many books.
D. One-to-many: each author entity can have many book entities, but a book belongs to only one
author.
Correct Answer: D
Rationale: The FK author_id in BOOK points to AUTHOR, placing the "many" side on BOOK;
therefore one AUTHOR is related to many BOOKs. This is a classic 1-to-many (authors-to-
books) design. A is wrong because the PK/FK structure does not enforce mutual uniqueness; B
reverses the direction of the relationship; C would require a link table which is absent here.
Q2.
A start-up logs IoT temperature readings. A single sensor sends one record every 5 seconds. The
engineers need a composite key because no single column is unique. Which pair of columns
would form the smallest natural composite key?
,2
A. (sensor_id, temperature_value)
B. (sensor_id, recorded_at)
C. (temperature_value, recorded_at)
D. (sensor_id, location_name)
Correct Answer: B
Rationale: The combination of sensor identifier and the exact timestamp (recorded_at) is
naturally unique for each reading. A fails because temperature values repeat frequently; C fails
because two sensors can log identical temperatures at the same moment; D is larger and
location_name is functionally dependent on sensor_id, making it redundant in the key.
Q3.
The university registrar keeps data in first-normal-form only:
STUDENT_COURSE( student_id, course_id, {grade, semester}, advisor_name, advisor_office )
(The braces denote a repeating group: one row stores many grades/semesters for the same
student-course.) Which anomaly is impossible to remove while the design stays in 1NF?
A. Update anomaly on advisor_office when an advisor moves.
B. Insertion anomaly when a new advisor has no students yet.
C. Deletion anomaly that loses the only row for an advisor and inadvertently deletes his/her
office.
D. Redundancy of repeating grade/semester data within one row.
Correct Answer: D
Rationale: 1NF allows repeating groups; redundancy inside the group is permitted until we
flatten to atomic values (and eventually normalize). A-C are classic anomalies that appear even
in 1NF flat tables and are fixed by higher normal forms, but D is inherent to the repeating-group
version of 1NF.
Q4.
CloudMart uses a distributed relational database. The ORDER_LINE table currently has 120
million rows. Which SQL clause is most appropriate to verify that every order_id in
ORDER_LINE references an existing order_id in the ORDERS table without enforcing the
constraint?
A. SELECT order_id FROM ORDER_LINE WHERE order_id IS NULL;
B. SELECT o.order_id FROM ORDER_LINE o LEFT JOIN ORDERS r ON o.order_id =
r.order_id WHERE r.order_id IS NULL;
, 3
C. SELECT COUNT(*) FROM ORDER_LINE GROUP BY order_id HAVING COUNT(*) > 1;
D. SELECT order_id FROM ORDER_LINE INTERSECT SELECT order_id FROM ORDERS;
Correct Answer: B
Rationale: The LEFT JOIN returns NULL on the right side for orphaned order_ids; filtering on
NULL finds violations. A checks only NULLs in the base table, not referential integrity; C looks
for duplicates; D returns common values, not orphans.
Q5.
A manufacturing company needs to store CAD drawings alongside metadata. Drawings average
90 MB. Which strategy best balances performance and referential integrity in a relational cloud
warehouse?
A. Store the file as a BLOB inside the same table as metadata.
B. Save the file in object storage and keep the object key as a VARCHAR foreign key.
C. Compress the drawing, encode as Base64, and store in a TEXT column.
D. Use a separate NoSQL document database exclusively.
Correct Answer: B
Rationale: Object storage is cost-optimized for large binaries; keeping the object key in the
relational table preserves joins without bloating the buffer pool. A hurts cache performance; C
increases size ~33%; D abandons relational integrity when hybrid is possible.
Q6.
Consider the relation R( A, B, C, D, E ) with functional dependencies:
{ A → B, B → C, C → DE }
What is the correct closure of attribute A?
A. {A,B}
B. {A,B,C}
C. {A,B,C,D,E}
D. {A,C,D,E}
Correct Answer: C
Rationale: Start with {A}; A→B adds B; B→C adds C; C→DE adds D and E. Therefore A⁺ =
all attributes, making A a candidate key.
Q7.
A hospital designs a DRUG_ALLERGY table:
