The correct answer is the script snippet that uses `id 'jdoe' &>/dev/null` to check if the user exists, then `useradd -m 'jdoe' && echo 'Created' || echo 'Failed'` to create the user with a home directory and verify success. This works because the `id` command returns a zero exit status if the user exists, and redirecting its output to `/dev/null` suppresses any messages, making the check clean and silent. The `&&` operator ensures the creation message only runs if `useradd` succeeds, while `||` catches a failure, directly addressing the requirement to verify successful creation. On the Red Hat Certified System Administrator EX200 exam, this tests your understanding of exit statuses, conditional operators, and user management with `useradd -m`—a common task in RHEL system administration. A frequent trap is forgetting the `-m` flag, which omits the home directory, or using `grep /etc/passwd` instead of `id`, which is less efficient. Memory tip: think "ID then ADD" — check identity first, then add with `-m` for home.
EX200 Create simple shell scripts Practice Question
This EX200 practice question tests your understanding of create simple shell scripts. 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 system administrator needs to create a shell script that checks if the user 'jdoe' exists in the system and, if not, creates the user with a home directory. The script should also verify that the creation was successful. Which of the following script snippets correctly implements this logic?
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
if id 'jdoe' &>/dev/null; then echo 'Exists'; else useradd -m 'jdoe' && echo 'Created' || echo 'Failed'; fi
Option B is correct because it uses `id` to check for the user's existence (redirecting output to /dev/null to suppress messages), then uses `useradd -m` to create the user with a home directory. The `&&` and `||` operators ensure that success or failure of the creation is explicitly reported, fulfilling the requirement to verify successful creation.
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.
✗
if grep -q '^jdoe:' /etc/passwd; then echo 'Exists'; else useradd 'jdoe' && echo 'Created'; fi
Why it's wrong here
useradd without -m does not create home directory.
✓
if id 'jdoe' &>/dev/null; then echo 'Exists'; else useradd -m 'jdoe' && echo 'Created' || echo 'Failed'; fi
Why this is correct
Correctly checks existence, creates with home dir, and verifies.
Related concept
Read the scenario before looking for a memorised answer.
✗
if ! id 'jdoe' &>/dev/null; then useradd -m 'jdoe'; else echo 'Exists'; fi
Why it's wrong here
Logic inverted: if user does NOT exist, create; else echo. But no verification of success.
[ -z ] checks if the output is empty, but the command substitution may not work as expected; also no error handling.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Red Hat often tests the misconception that grepping /etc/passwd is sufficient for user existence checks, but the trap here is that modern systems may use remote authentication sources, so `id` is the correct command to query all NSS sources.
Trap categories for this question
Command / output trap
[ -z ] checks if the output is empty, but the command substitution may not work as expected; also no error handling.
Detailed technical explanation
How to think about this question
The `id` command queries the Name Service Switch (NSS) configured in /etc/nsswitch.conf, making it the correct way to check for user existence across all backends (local files, LDAP, SSSD, etc.). The `useradd -m` flag ensures the home directory is created per the /etc/default/useradd skeleton, and the `&&`/`||` chain leverages exit codes (0 for success, non-zero for failure) to provide deterministic feedback.
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 EX200 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.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
Create simple shell scripts — This question tests Create simple shell scripts — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: if id 'jdoe' &>/dev/null; then echo 'Exists'; else useradd -m 'jdoe' && echo 'Created' || echo 'Failed'; fi — Option B is correct because it uses `id` to check for the user's existence (redirecting output to /dev/null to suppress messages), then uses `useradd -m` to create the user with a home directory. The `&&` and `||` operators ensure that success or failure of the creation is explicitly reported, fulfilling the requirement to verify successful creation.
What should I do if I get this EX200 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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Question Discussion
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