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FC0-U71 Practice Question: A program needs to display a message based on a…
A program needs to display a message based on a user's age. If age is 18 or over, it displays 'Adult'; otherwise, it displays 'Minor'. Which control structure should be used?
⚠ Common exam trap
CompTIA often tests the distinction between selection (if-else) and iteration (loops), so the trap here is that candidates might confuse the 'condition' in a while loop with the conditional logic needed for a binary decision, leading them to incorrectly choose a loop structure.
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
✓
If-else statement
The if-else statement is the correct control structure because it evaluates a single condition (age >= 18) and executes one block of code if true ('Adult') and another block if false ('Minor'). This is the standard branching mechanism in most programming languages for binary decisions based on a comparison.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
If-else statement
Why this is correct
If-else evaluates a condition and executes one of two blocks.
- ✗
For loop
Why it's wrong here
For loop is for repeating a block a fixed number of times, not for a simple condition.
- ✗
Switch-case statement
Why it's wrong here
Switch-case is used when there are many discrete values, not a range condition like age >= 18.
- ✗
While loop
Why it's wrong here
While loop repeats until a condition is false; it's not suitable for a single decision.
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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.