hardMultiple ChoiceObjective-mapped
FC0-U71 Practice Question: Refer to the exhibit
Network Topology
Refer to the exhibit. A database administrator runs the following query:
SELECT c.Name, SUM(o.Quantity) AS TotalItems FROM Customers c LEFT JOIN Orders o ON c.ID = o.CustomerID GROUP BY c.Name;
What is the result for 'Alice'?
⚠ Common exam trap
The trap here is that candidates may mistakenly pick the quantity of a single order (2 or 5) instead of computing the SUM, or incorrectly assume that a LEFT JOIN would produce NULL for Alice when she actually has matching orders.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
7
The LEFT JOIN ensures that Alice's record from the Customers table is retained even if there are no matching rows in Orders. The SUM(o.Quantity) aggregates the Quantity values from the joined Orders rows for Alice. Since Alice has two orders with quantities 2 and 5, the sum is 7, making option D correct.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
2
Why it's wrong here
This is the quantity of only the Laptop order.
- ✗
NULL
Why it's wrong here
NULL would appear only if Alice had no orders, but she has two.
- ✗
5
Why it's wrong here
This is the quantity of only the Mouse order.
- ✓
7
Why this is correct
SUM of both orders (2+5=7).
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JA
Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
This FC0-U71 practice question is part of Courseiva's free CompTIA certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the FC0-U71 exam.