Quick Answer
The answer is umask 026. This value prevents world-readable log files by subtracting from the default file creation base of 666, yielding permissions of 640 (rw-r-----), which removes read access for "others" while preserving group read access. On the LPIC-2 exam, this question tests your understanding of how umask interacts with file creation masks in security-sensitive directories like /var/log, where world-readable logs pose a data exposure risk. A common trap is confusing umask with chmod—remember that umask specifies what to deny, not what to grant. For system-wide application, the umask is set in a shell initialization file such as /etc/profile or /etc/bash.bashrc. A helpful memory tip: think "026 = Others get Zero read, Group gets 2 (write), Owner gets 6 (read+write)," or simply recall that to block world-read, the last digit of the umask must be at least 2.
This LPIC-2 practice question tests your understanding of system security. Match the stated requirement to the specific cloud service, access model, or configuration option — many options are valid in isolation but not for this scenario. 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 security audit reveals that the /var/log directory contains world-readable log files that may contain sensitive information. The administrator wants to ensure new files created in /var/log are not readable by others, without affecting existing file permissions. Which umask value should be set system-wide?
umask 027
Why wrong: Results in 640 for files but 640 for directories? Actually 666-027=639, which is unusual.
umask 022
Why wrong: Results in 644 (rw-r--r--), still world-readable.
umask 077
Why wrong: Results in 600 (rw-------), too restrictive for logs.
umask 026
Results in default file permissions 640 (rw-r-----).
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
umask 026
Option D (umask 026) is correct because it sets the default permissions for new files to 640 (rw-r-----), which removes read access for 'others' while preserving read access for group members. The umask value is subtracted from the base permissions (666 for files), so a umask of 026 results in 666 - 026 = 640, meeting the requirement that new files in /var/log are not world-readable without altering existing file permissions.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
umask 027
Why it's wrong here
Results in 640 for files but 640 for directories? Actually 666-027=639, which is unusual.
umask 022
Why it's wrong here
Results in 644 (rw-r--r--), still world-readable.
umask 077
Why it's wrong here
Results in 600 (rw-------), too restrictive for logs.
umask 026
Why this is correct
Results in default file permissions 640 (rw-r-----).
Related concept
Read the scenario before looking for a memorised answer.
Common exam traps
The trap here is that candidates often choose umask 027 (option A) thinking it provides better security, but they overlook that it removes group read access, which can break legitimate log-reading processes, while the correct answer 026 preserves group read access and only removes world-read access.
Detailed technical explanation
The umask is applied as a bitwise complement to the base permissions; for files, the base is typically 666 (rw-rw-rw-), and for directories, 777 (rwxrwxrwx). The umask value is subtracted from these bases, but note that the execute bit is not automatically set on files, so a umask of 026 yields 666 - 026 = 640 (rw-r-----), which is exactly rw-r----- for files. In practice, system-wide umask is set in /etc/profile, /etc/bashrc, or via pam_umask.so, and a common mistake is to use 027, which overly restricts group access, or 022, which leaves logs world-readable.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
A network engineer segments a warehouse floor into three subnets: 20 scanners, 5 printers, and 2 management hosts. Picking the wrong mask wastes addresses or leaves too few usable hosts. Exam questions test whether you can apply CIDR notation, calculate block size, and identify the correct usable-host range for a given prefix.
What to study next
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FAQ
System Security — This question tests System Security — Read the scenario before looking for a memorised answer..
The correct answer is: umask 026 — Option D (umask 026) is correct because it sets the default permissions for new files to 640 (rw-r-----), which removes read access for 'others' while preserving read access for group members. The umask value is subtracted from the base permissions (666 for files), so a umask of 026 results in 666 - 026 = 640, meeting the requirement that new files in /var/log are not world-readable without altering existing file permissions.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
Read the scenario before looking for a memorised answer.
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Last reviewed: Jun 11, 2026
This LPIC-2 practice question is part of Courseiva's free LPI 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 LPIC-2 exam.
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