- A
result = [] for x in range(20): if x % 2 == 0: result.append(x**2)
This loop explicitly does the same filtering and squaring.
- B
result = map(lambda x: x**2, filter(lambda x: x % 2 == 0, range(20)))
Why wrong: While functional, it returns a map object, not a list; but list(map(...)) would be equivalent. However, the question asks for equivalent code, and this answer is not a complete list conversion. The loop is the most direct equivalent.
- C
result = [] for x in range(20): result.append(x**2)
Why wrong: This appends squares of all numbers, not just even ones.
- D
result = [x**2 for x in range(20) if x % 2 != 0]
Why wrong: This selects odd numbers, the opposite condition.
Question 257 of 1,000
200-901 Software Development and Design Practice Question
This 200-901 practice question tests your understanding of software development and design. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
A Python script uses a list comprehension: [x**2 for x in range(20) if x % 2 == 0]. Which of the following is equivalent?
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
✓
result = [] for x in range(20): if x % 2 == 0: result.append(x**2)
Option A is correct because it directly translates the list comprehension into an equivalent for-loop with a conditional append. The comprehension `[x**2 for x in range(20) if x % 2 == 0]` iterates over numbers 0–19, filters for even numbers (x % 2 == 0), squares each, and collects the results in a list. Option A's explicit loop and conditional produce the exact same sequence of appended values.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
result = [] for x in range(20): if x % 2 == 0: result.append(x**2)
Why this is correct
This loop explicitly does the same filtering and squaring.
Related concept
Read the scenario before looking for a memorised answer.
- ✗
result = map(lambda x: x**2, filter(lambda x: x % 2 == 0, range(20)))
Why it's wrong here
While functional, it returns a map object, not a list; but list(map(...)) would be equivalent. However, the question asks for equivalent code, and this answer is not a complete list conversion. The loop is the most direct equivalent.
- ✗
result = [] for x in range(20): result.append(x**2)
Why it's wrong here
This appends squares of all numbers, not just even ones.
- ✗
result = [x**2 for x in range(20) if x % 2 != 0]
Why it's wrong here
This selects odd numbers, the opposite condition.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Cisco often tests the distinction between list comprehensions with and without a filtering condition, and the trap here is that candidates may overlook the `if x % 2 == 0` filter and choose Option C, which omits the condition entirely.
Detailed technical explanation
How to think about this question
List comprehensions in Python are syntactic sugar that compile to bytecode similar to an explicit for-loop with an if clause, but they execute faster due to internal optimizations in CPython's `LIST_APPEND` opcode. The `filter` and `map` approach in Option B is also equivalent but returns an iterator in Python 3 unless wrapped in `list()`, which the original comprehension does automatically. Understanding this equivalence is crucial for debugging and performance tuning in automation scripts for Cisco platforms like NX-OS or IOS-XE.
KKey Concepts to Remember
- Read the scenario before looking for a memorised answer.
- Find the constraint that changes the correct option.
- Eliminate answers that are true in general but not in this case.
TExam Day Tips
- Watch for words such as best, first, most likely and least administrative effort.
- Review why wrong options are wrong, not only why the correct option is correct.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
How this comes up in practice
A practitioner preparing for the 200-901 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option. Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.
What to study next
Got this wrong? Here's your next step.
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FAQ
Questions learners often ask
What does this 200-901 question test?
Software Development and Design — This question tests Software Development and Design — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: result = [] for x in range(20): if x % 2 == 0: result.append(x**2) — Option A is correct because it directly translates the list comprehension into an equivalent for-loop with a conditional append. The comprehension `[x**2 for x in range(20) if x % 2 == 0]` iterates over numbers 0–19, filters for even numbers (x % 2 == 0), squares each, and collects the results in a list. Option A's explicit loop and conditional produce the exact same sequence of appended values.
What should I do if I get this 200-901 question wrong?
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
What is the key concept behind this question?
Read the scenario before looking for a memorised answer.
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Last reviewed: Jul 4, 2026
This 200-901 practice question is part of Courseiva's free Cisco 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 200-901 exam.
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