LPI Linux Essentials (010-160, v1.6) (010-160) 010-160 Questions 301–323 | Page 5/5

301

MCQmedium

In the context of the open source community, what is the significance of the term 'fork'?

A.A type of software license that forbids distribution.

B.A project that diverges from the original source to move in a new direction.

C.A way to kill a process in the Linux terminal.

D.A method for merging two disparate codebases into one.

AnswerB

A fork occurs when developers take the existing source code and continue it independently, creating a divergent project with its own goals and governance. This differs from a simple copy because the new project evolves separately from the original upstream codebase.

Why this answer

In open source, a fork is a copy of a project's source code that diverges from the original to pursue a different direction, often due to disagreements or different goals. The forked project maintains its own development, governance, and release cycle while preserving the original license terms. This is distinct from a branch, which is a temporary divergence within the same repository.

Exam trap

The trap is confusing the open source term 'fork' with the Unix fork() system call or with branching/merging, leading candidates to pick process-related or merge-related answers.

How to eliminate wrong answers

Option A is wrong because a fork is not a license type; licenses govern use and distribution, and forking is a right granted by many open source licenses. Option C is wrong because killing a process in Linux uses signals like SIGTERM or SIGKILL via kill, not the term fork (though fork() is a process-creation syscall, unrelated to open source project forking). Option D is wrong because merging codebases is a different operation (e.g., git merge), and a fork specifically represents divergence, not unification.

Full explanation →

302

Multi-Selectmedium

Which THREE of the following are environment variables commonly seen in a Linux shell?

Select 3 answers

A.USER

B.ROOT

C.HOME

D.FILE

E.PATH

AnswersA, C, E

USER holds the current logged-in account's username and is exported by the login process, making it a standard shell environment variable. It is inherited by child processes, matching the stem's criterion of variables commonly present in a Linux shell.

Why this answer

USER is a standard environment variable that holds the username of the currently logged-in user, so option A is correct. HOME is a standard environment variable that specifies the current user's home directory path, making option C correct. PATH is a standard environment variable that lists the directories the shell searches for executable commands, so option E is correct.

ROOT (B) is not a standard environment variable; it is a common username or directory name, not a shell variable. FILE (D) is not a standard environment variable either; it is only a special variable inside Makefiles, not a general Linux shell environment variable.

Exam trap

The trap is picking plausible-sounding names like ROOT or FILE that resemble real variables but are not part of the standard Linux environment — candidates must recall the actual canonical set.

Full explanation →

303

Multi-Selectmedium

A community college is designing an introductory course on Linux and open source culture. The instructor wants to include activities that reflect the collaborative practices of the open source community. (Choose two.)

Select 2 answers

A.Asking students to fork a project on a public hosting platform and open a pull request with improvements.

B.Requiring students to reverse engineer a proprietary application and publish its source code.

C.Having students purchase a commercial license before reading any project documentation.

D.Having students submit bug reports and patches to a project's public issue tracker and mailing list.

E.Instructing students to keep all modifications private and never share changes with the upstream project.

AnswersA, D

Forking and submitting pull requests is the standard contribution workflow on platforms like GitHub and GitLab. It teaches branching, version control, code review, and community etiquette while producing a visible, reviewable contribution that project maintainers can discuss and merge.

Why this answer

Open source collaboration centers on transparent, public participation. Filing bug reports and patches through public trackers, and forking repositories to submit pull requests, are authentic activities that teach the tools and social norms of community development. Reverse engineering proprietary code, hoarding changes, and requiring paid licenses contradict those practices.

Exam trap

The trap here is assuming that any activity involving software counts as open source collaboration, when the defining factor is public, license-respecting contribution.

Full explanation →

304

Multi-Selectmedium

You are hardening a server and must ensure that ordinary users cannot read the contents of `/etc/shadow`. Which two statements about `/etc/shadow` access are correct? (Choose two.)

Select 2 answers

A.It is readable by all users because the passwd command must consult it during a normal login.

B.It can be read by any user who supplies the correct password through the sudo prompt.

C.Its contents are encrypted with a system-wide key stored in /etc/crypttab, so permissions are largely irrelevant.

D.Its permissions are typically 640 with owner root and group shadow, so only root and members of the shadow group can read it.

E.It stores password hashes in a format that ordinary users could crack offline if they could read the file.

AnswersD, E

The conventional mode for /etc/shadow is 640 owned by root:shadow, which allows root full access and members of the shadow group read access. Ordinary users fall into the others category, which has no permissions, so they cannot read the hashed passwords. This is the standard hardening expectation on most Linux distributions.

Why this answer

On typical Linux systems /etc/shadow is mode 640 with owner root and group shadow, leaving the others category with no access, which is what keeps ordinary users from reading password hashes. Those hashes are salted but still vulnerable to offline cracking if exposed, so read restriction is the key control. Direct reads happen through privileged programs, not through ordinary user access.

Exam trap

The trap here is assuming the login process requires world-readable access to /etc/shadow, when privileged helper programs perform the lookup on the user's behalf.

Full explanation →

305

MCQmedium

A system administrator wants to check the partition table of a specific disk. Which tool is used for this?

A.fsck

B.df

C.mkfs

D.fdisk

AnswerD

fdisk reads and edits the partition table of a specified disk, displaying partitions, their types and sizes. Invoked with a device path, it directly satisfies the requirement to inspect a specific disk's partition table, unlike tools that only report filesystem usage.

Why this answer

fdisk is the classic interactive tool for viewing and editing a disk's partition table, invoked as 'fdisk -l /dev/sda' to list partitions. It reads the MBR or GPT structures on the device and displays partition start/end sectors, sizes, types, and IDs. This makes it the correct tool for inspecting partition layout.

Exam trap

The trap is confusing filesystem-level tools (fsck, mkfs, df) with disk-level tools; candidates must distinguish operations on filesystems from operations on partition tables.

How to eliminate wrong answers

Option A is wrong because fsck checks and repairs filesystem integrity; it operates on filesystems, not partition tables, and can even damage data if run on a mounted filesystem. Option B is wrong because df reports filesystem disk space usage (mounted filesystems), not the underlying partition table. Option C is wrong because mkfs creates a filesystem on a partition or volume; it writes filesystem metadata, not partition entries.

Full explanation →

306

MCQeasy

You are logged into a terminal and want to identify your current working directory. Which command should you execute?

A.pwd

B.whoami

C.ls

D.cd

AnswerA

`pwd` prints the absolute path of the shell's current working directory, directly satisfying the stem's requirement to identify where you are in the filesystem. Unlike `ls`, which lists directory contents, or `cd`, which changes location, `pwd` reports the present directory without altering state.

Why this answer

The pwd (print working directory) command displays the absolute path of the current working directory. It is the standard Linux command for identifying where you are in the filesystem hierarchy.

Exam trap

The trap here is confusing basic navigation commands; candidates might select ls or cd because they are directory-related, but only pwd specifically prints the current working directory.

How to eliminate wrong answers

Option B is wrong because whoami displays the current username, not the directory. Option C is wrong because ls lists the contents of a directory, not the directory path itself. Option D is wrong because cd changes the working directory rather than displaying it.

Full explanation →

307

Multi-Selecthard

Which THREE of the following items are typically found in the /etc directory?

Select 3 answers

A./etc/tmp

B./etc/fstab

C./etc/passwd

D./etc/hosts

E./etc/bin

AnswersB, C, D

/etc/fstab resides in /etc and defines filesystem mount points and options read at boot, making it a standard configuration file found there. It is a plain-text system configuration file, matching the stem's expectation for /etc contents rather than binaries or runtime data.

Why this answer

/etc/fstab (B) is correct because it is the standard configuration file listing filesystems to be mounted at boot, including mount points, filesystem types, and mount options. /etc/passwd (C) is correct because it is the traditional system file storing user account information such as usernames, UIDs, GIDs, home directories, and login shells. /etc/hosts (D) is correct because it provides static hostname-to-IP-address mappings used by the system resolver before DNS. The other options do not belong: /etc/tmp (A) is not a standard file or directory in /etc, and /etc/bin (E) is not standard since executable binaries reside in /bin, /usr/bin, or /usr/local/bin, not under /etc.

Exam trap

The trap is that /etc/tmp and /etc/bin sound plausible because /tmp and /bin are real directories — candidates must recognize that /etc contains configuration, not temporary files or binaries.

Full explanation →

308

MCQhard

A developer stores a private SSH key at /home/dev/id_rsa and wants to remove all permissions for group and others while keeping full owner access, using octal notation. Which command achieves exactly this?

A.chmod 600 /home/dev/id_rsa

B.chmod 644 /home/dev/id_rsa

C.chmod 400 /home/dev/id_rsa

D.chmod 700 /home/dev/id_rsa

AnswerA

Octal 600 grants read and write to the owner and nothing to group or others, which is the required state for a private key and matches SSH's own expectations. It removes every group and others permission while preserving the owner's full read/write access, exactly as the developer requested.

Why this answer

Octal mode 600 sets rw- for the owner and --- for both group and others. That isolates the private key to its owner while keeping read and write access intact, satisfying the stated requirement and aligning with the restrictive permissions SSH expects for private key material.

Exam trap

The trap here is choosing 400 for a private key, forgetting that the owner then loses the write access the scenario explicitly requires.

Full explanation →

309

MCQeasy

To see the contents of a directory, which command is standard?

A.ls

B.dir

C.list

D.cat

AnswerA

ls is the POSIX-standard utility for listing directory entries, printing filenames to standard output. It exists on every Linux and Unix system, making it the canonical command for viewing a directory's contents in the shell.

Why this answer

`ls` is the standard POSIX/Linux command for listing directory contents, and it is present on every Linux distribution as part of coreutils. It reads directory entries via the readdir() system call and prints them to stdout, with options like -l, -a, and -h for formatting. It is the canonical answer for 'list directory contents' on Linux.

Exam trap

The trap here is confusing DOS/Windows commands (`dir`) with Linux commands, or assuming `cat` can list directories — LFCS/010-160 frequently tests basic command recognition.

How to eliminate wrong answers

Option B is wrong because `dir` is a DOS/Windows command (also available as a GNU coreutils alias on some Linux systems, but not the standard Linux answer). Option C is wrong because `list` is not a Linux command at all — it does not exist in coreutils or any standard shell. Option D is wrong because `cat` concatenates and prints file contents; it does not list directory entries (running `cat` on a directory returns 'Is a directory').

Full explanation →

310

Multi-Selectmedium

A Linux administrator is reviewing user account files and needs to understand where user information is stored. Which TWO of the following files contain user account details? (Choose two.)

Select 2 answers

A./etc/shadow

B./etc/login.defs

C./etc/group

D./etc/gshadow

E./etc/passwd

AnswersA, E

/etc/shadow stores encrypted user passwords and password aging information. It is readable only by root for security. Along with /etc/passwd, it forms the core of user account data. Therefore, it is a correct choice for user account details, especially sensitive authentication data.

Why this answer

User account details are primarily stored in /etc/passwd and /etc/shadow. The former holds general account information, while the latter holds encrypted passwords and aging data. Other files like /etc/group and /etc/gshadow pertain to groups, and /etc/login.defs provides configuration defaults, not actual account records.

Exam trap

The trap here is confusing group-related files with user account files, or assuming that configuration files like login.defs store user data, when they only define policies.

Full explanation →

311

MCQeasy

What is the primary role of the 'Free Software Foundation' (FSF)?

A.Manufacturing servers and hardware for Linux.

B.Providing commercial support for Linux users globally.

C.Advocating for computer user freedom and maintaining the GNU project.

D.Developing Linux distribution patches for security.

AnswerC

The Free Software Foundation promotes users' freedom to run, study, modify and redistribute software, and serves as the steward of the GNU project. This advocacy and GNU maintenance directly matches the question's ask about the organisation's primary role.

Why this answer

The Free Software Foundation (FSF), founded by Richard Stallman in 1985, exists to promote computer user freedom and to defend the rights of free software users. It is the principal organizational sponsor of the GNU Project, which develops much of the userland software (bash, coreutils, GCC, Emacs) that forms a GNU/Linux system.

Exam trap

The trap is confusing the FSF (advocacy and GNU project stewardship) with commercial Linux vendors (Red Hat, Canonical) or with the Linux Foundation, which is a separate entity focused on kernel development and corporate collaboration.

How to eliminate wrong answers

Option A is wrong because the FSF is a nonprofit advocacy and development organization, not a hardware manufacturer — it does not produce servers or Linux hardware. Option B is wrong because commercial Linux support is provided by companies like Red Hat, Canonical, and SUSE, not by the FSF, which is explicitly non-commercial and does not offer enterprise support contracts. Option D is wrong because the FSF does not develop distribution-specific security patches; distributions like Debian and Red Hat handle their own patching, while the FSF focuses on GNU software and license stewardship.

Full explanation →

312

MCQmedium

A junior administrator is reading documentation for a command-line utility and sees that it is licensed under the GPLv3. The administrator wants to modify the code and distribute the modified version to a colleague outside the company. Which statement best describes the obligation imposed by the GPLv3 in this situation?

A.The modified version may be kept proprietary as long as the original author is credited in the documentation.

B.No obligations apply because the colleague is an individual rather than a company or legal entity.

C.The modified version can be relicensed under any OSI-approved license, including a permissive one such as MIT.

D.The modified version must be distributed under the same GPLv3 license, with the complete corresponding source code made available.

AnswerD

The GPLv3 is a copyleft license, so distributing a modified version requires releasing it under the same license and providing the corresponding source code. Because the administrator is distributing (not merely using internally), the reciprocal obligations are triggered. This matches the scenario's condition of sharing the modified utility with a colleague outside the organization.

Why this answer

The GPLv3 uses copyleft to ensure that modified versions distributed to others remain free software. When the administrator shares the modified utility with someone outside the company, that act of distribution triggers the requirement to license the derivative under GPLv3 and provide the corresponding source code. Internal use without distribution would not trigger these obligations, but external sharing does.

Exam trap

The trap here is assuming that crediting the original author or choosing a different OSI-approved license can replace the copyleft obligation to release source under the same license.

Full explanation →

313

MCQmedium

A developer working in /srv/project wants to create a symbolic link named latest that points to the file releases/v2.1/app.tar.gz, keeping the link inside /srv/project. Which command creates this link correctly?

A.cp -s releases/v2.1/app.tar.gz latest

B.ln -s latest releases/v2.1/app.tar.gz

C.ln -s releases/v2.1/app.tar.gz latest

D.ln releases/v2.1/app.tar.gz latest

AnswerC

The -s option tells ln to create a symbolic link whose contents store the path releases/v2.1/app.tar.gz, and the final argument latest is the new link name in the current directory. This matches the developer's request for a symlink inside /srv/project.

Why this answer

Symbolic links are created with ln -s, and the operand order is target first, link name second. Naming the relative path releases/v2.1/app.tar.gz as the target and latest as the new link places the symlink in the current directory and stores the relative path, which is what the developer intended.

Exam trap

The trap here is forgetting that ln without -s makes a hard link, which behaves very differently and cannot span filesystems or target directories.

Full explanation →

314

MCQeasy

Which of the following is considered a 'downstream' consumer of the Linux kernel?

A.An Ubuntu Linux distribution

B.Linus Torvalds

C.The GCC Compiler project

D.The Linux Kernel Mailing List

AnswerA

An Ubuntu Linux distribution consumes the kernel maintained upstream, packaging it with userland tools and delivering it to end users. This downstream relationship satisfies the stem's requirement by identifying a consumer that builds upon released kernel code.

Why this answer

A downstream consumer is an entity that takes the upstream Linux kernel source and builds derivative products from it. Ubuntu takes kernel source from kernel.org (upstream), applies its own patches, and ships it as a distribution — making it a classic downstream consumer.

Exam trap

The trap is misreading 'downstream consumer' as anyone who uses Linux — candidates may pick the mailing list or Torvalds because they associate them with the kernel, but the term specifically means an entity that consumes upstream kernel source to build derivative products.

How to eliminate wrong answers

Option B is wrong because Linus Torvalds is the upstream maintainer who originates kernel releases, not a consumer of them. Option C is wrong because the GCC Compiler project is a separate toolchain project that compiles the kernel but does not consume kernel source as a product — it is a build dependency, not a downstream distribution. Option D is wrong because the Linux Kernel Mailing List is the communication channel where kernel development happens (upstream), not a consumer of kernel code.

Full explanation →

315

MCQhard

You want to search for a file named 'notes.txt' starting from the root directory. Which command is most efficient?

A.grep -r notes.txt /

B.find / -name notes.txt

C.search notes.txt /

D.ls -R / | grep notes.txt

AnswerB

find / -name notes.txt begins traversal at the root directory and recurses through every subdirectory, matching the filename exactly. This guarantees the file is located wherever it resides, satisfying the requirement to search from root.

Why this answer

The `find` command is purpose-built for filesystem traversal and can search by name, size, time, permissions, and more. Running `find / -name notes.txt` starts at the root directory and recursively descends every subdirectory, matching files whose basename is exactly 'notes.txt'. It is the most efficient and correct tool because it operates directly on directory entries via the kernel, without spawning extra processes per file.

Exam trap

The trap is confusing filename search with content search — candidates see 'search for a file' and pick grep because grep is the famous search tool, forgetting that grep searches inside files while find searches for the files themselves.

How to eliminate wrong answers

Option A is wrong because `grep -r notes.txt /` searches file *contents* for the string 'notes.txt', not filenames — it would return files containing that text, and it would also try to read binary files and /proc entries, producing errors. Option C is wrong because `search` is not a Linux command at all; it is a Windows/DOS concept. Option D is wrong because `ls -R /` recursively lists directory contents and pipes to grep, which is far less efficient (spawns ls, buffers entire output, then filters) and can miss files due to output truncation or permission errors.

Full explanation →

316

MCQeasy

What is the purpose of the /proc directory in Linux?

A.To store configuration files for boot.

B.To store user passwords.

C.To provide information about kernel and processes.

D.To store temporary system files.

AnswerC

/proc is a virtual pseudo-filesystem exposing kernel and process state as readable files, satisfying the need for runtime system information. Each running process appears as a numbered subdirectory, and files such as /proc/cpuinfo or /proc/meminfo report kernel data without occupying disk space.

Why this answer

The /proc directory is a virtual filesystem (procfs) that provides a real-time interface to kernel data structures and running process information. It does not exist on disk — the kernel generates its contents dynamically, exposing details like CPU info (/proc/cpuinfo), memory usage (/proc/meminfo), and per-process data under /proc/[PID]/. This makes it the correct answer for providing kernel and process information.

Exam trap

The trap is confusing /proc with /etc or /tmp — candidates who memorize 'config files go in /etc' may incorrectly map boot configuration to /proc, missing that /proc is a virtual, kernel-generated filesystem, not a storage location.

How to eliminate wrong answers

Option A is wrong because boot configuration files live in /boot (kernel images, GRUB config) and /etc (systemd units, fstab), not /proc. Option B is wrong because user password hashes are stored in /etc/shadow (and historically /etc/passwd), which are regular on-disk files with strict permissions. Option D is wrong because temporary system files are stored in /tmp and /var/tmp, which are writable directories on disk, whereas /proc is a virtual in-memory filesystem.

Full explanation →

317

Multi-Selectmedium

Which TWO commands are commonly used to check network connectivity to a host?

Select 2 answers

A.arp

B.fdisk

C.ping

D.traceroute

E.ssh

AnswersC, D

Ping uses ICMP echo request and reply packets to verify reachability and round-trip latency to a specific host, directly satisfying the connectivity-check requirement. It confirms whether the target responds at the network layer, independent of higher-level services, making it a standard first-line diagnostic tool.

Why this answer

Option C, ping, is correct because it uses ICMP Echo Request and Echo Reply messages to verify basic IP-level reachability and round-trip time to a target host, making it the most common connectivity test. Option D, traceroute, is correct because it sends packets with incrementally increasing TTL values to map each router hop along the path to the destination, which helps diagnose where connectivity breaks down. Option A, arp, only resolves or displays IPv4-to-MAC address mappings in the local ARP cache and does not test end-to-end reachability.

Option B, fdisk, is a disk partitioning utility unrelated to network diagnostics. Option E, ssh, establishes an encrypted remote login session over TCP port 22 but is not a general-purpose connectivity-checking command.

Exam trap

The trap here is confusing tools that display network configuration or establish connections (like arp or ssh) with tools that actively test reachability and path (ping and traceroute).

Full explanation →

318

Multi-Selectmedium

Which TWO of the following options can be used with 'rm' to ensure safe operation?

Select 2 answers

A.-I

B.-v

C.-r

D.-f

E.-i

AnswersA, E

The -I flag prompts once before removing more than three files or recursively deleting, offering a single confirmation rather than per-file prompts. This satisfies the stem's safety requirement by catching broad or recursive deletions before any file is unlinked.

Why this answer

Option A (-I) is correct because it makes rm prompt once before removing more than three files (or recursively), providing a safety net against accidental mass deletion while avoiding the constant prompting of -i. Option E (-i) is correct because it forces rm to prompt for confirmation before every single removal, which is the classic way to ensure safe, deliberate operation. The unmarked options do not belong: -v only prints what is being removed without any confirmation, -r enables recursive deletion of directories (increasing risk rather than safety), and -f forces removal without prompting and ignores nonexistent files, directly defeating safe operation.

Exam trap

The trap is confusing '-i' and '-I' with '-f' and '-r' — candidates often think '-f' is safe because it 'forces' something, but it actually disables prompts, making it dangerous.

Full explanation →

319

MCQeasy

What is the primary difference between 'man' pages and 'info' pages?

A.man is graphical, info is text-based

B.man pages are obsolete

C.man is for system files, info is for users

D.info contains more detailed, hyperlinked documentation

AnswerD

info pages are rendered by the GNU Texinfo system, producing structured documents with hyperlinks, menus and cross-references between nodes. man pages are single flat troff documents without navigable links, so info offers richer, interconnected browsing of detailed GNU documentation.

Why this answer

Info pages are GNU's hypertext documentation system, providing more detailed, structured, and hyperlinked content than traditional man pages. They are navigated with the 'info' command using nodes, menus, and cross-references, making them better suited for complex GNU tools like coreutils and bash. Man pages, by contrast, are single-page, linear reference documents.

Exam trap

The trap is confusing the delivery format (text vs graphical) or audience (system vs user) with the actual distinction, which is the hyperlinked, node-based structure and greater depth of info pages.

How to eliminate wrong answers

Option A is wrong because both man and info are text-based terminal tools — neither is graphical (though both can be viewed in graphical front-ends like yelp or konqueror). Option B is wrong because man pages are not obsolete; they remain the primary reference format on virtually all Unix-like systems and are actively maintained. Option C is wrong because man pages cover both system files (section 5) and user commands (section 1), and info pages likewise cover both user and system topics — the distinction is not about audience.

Full explanation →

320

MCQeasy

A small business owner wants to implement a server that allows for long-term support and stability for critical financial applications. Which distribution strategy should they prioritize?

A.Choosing a distribution that offers an LTS version with a long support lifecycle.

B.Using a desktop-focused distribution for ease of installation.

C.Compiling the kernel from source to match the specific hardware.

D.Selecting a rolling release distribution for frequent kernel updates.

AnswerA

An LTS release freezes package versions and backports only security fixes, so the financial applications run against a stable, unchanging ABI for years. This directly satisfies the stem's demand for long-term support and stability, unlike rolling distributions whose continuous updates risk breaking critical software.

Why this answer

For long-term support and stability of critical financial applications, an LTS (Long Term Support) distribution is ideal because it provides a stable, well-tested platform with security updates for an extended period (often 5-10 years). This minimizes disruptive changes and ensures reliability. Rolling releases and desktop-focused distributions prioritize frequent updates or ease of use, not long-term stability.

Exam trap

The trap is confusing 'stability' with 'latest features'; candidates may think rolling releases are stable because they are up-to-date, but stability in this context means minimal change and long support.

How to eliminate wrong answers

Option B is wrong because desktop-focused distributions prioritize user experience and frequent application updates, not the stability and long support lifecycle required for critical financial applications. Option C is wrong because compiling the kernel from source is a customization step, not a distribution strategy, and it does not provide long-term support or stability guarantees. Option D is wrong because rolling release distributions provide the latest software continuously, which introduces instability and frequent changes, contrary to the need for long-term stability.

Full explanation →

321

MCQeasy

A user named alice has a primary group alice and is also a member of the group developers. She creates a new file named report.txt in her home directory. The umask is set to 022. What are the permissions of the newly created file?

A.-rw-------

B.-rw-r--r--

C.-rw-rw-r--

D.-rwxr-xr-x

AnswerB

With umask 022, the default creation mode for a regular file (666) has the write bits for group and others removed, resulting in 644, which is -rw-r--r--. The owner has read and write, group has read, and others have read. This is the standard behavior for a newly created file when the umask is 022 and the application does not specify a different mode.

Why this answer

When a regular file is created, the kernel starts with a base mode of 666 (rw-rw-rw-) and applies the umask by clearing the bits specified. Umask 022 clears the write bits for group and others, leaving 644 (rw-r--r--). Thus the new file report.txt will have read and write for the owner, and read-only for group and others.

Exam trap

The trap here is confusing the base mode for files (666) with that for directories (777), leading to incorrect calculation of the resulting permissions.

Full explanation →

322

Multi-Selectmedium

Which THREE of the following are valid directory navigation commands?

Select 3 answers

A.ls

B.mkdir

C.pushd

D.pwd

E.cd

AnswersC, D, E

`pushd` saves the current directory onto a stack before changing to the specified path, letting you return later with `popd`. This stack-based directory switching satisfies the stem's requirement for a valid navigation command, unlike commands that merely list or manipulate files without altering the working directory.

Why this answer

The three valid directory navigation commands are pushd (C), pwd (D), and cd (E). cd (E) changes the current working directory, which is the core navigation command in Linux/Unix shells. pwd (D) prints the absolute pathname of the current working directory, confirming where you are in the directory tree. pushd (C) pushes the current directory onto the directory stack and changes to a new directory, enabling stack-based navigation (paired with popd). The unmarked options do not belong: ls (A) lists directory contents rather than navigating, and mkdir (B) creates a new directory rather than changing location.

Exam trap

The trap is that ls and mkdir are directory-related commands, so candidates lump them in with navigation — the exam expects you to distinguish commands that change your location from those that list or create directories.

Full explanation →

323

MCQeasy

A technician needs to identify the amount of RAM installed in a server without opening the chassis. Which command provides this information from the command line?

A.uptime

B.lsblk

C.free -m

D.df -h

AnswerC

free -m reads /proc/meminfo and reports total, used and available RAM in mebibytes, giving installed memory without physical access. It satisfies the stem's constraint of identifying RAM from the command line without opening the chassis.

Why this answer

`free -m` reads memory statistics from /proc/meminfo and displays total, used, free, shared, buff/cache, and available RAM in mebibytes. It is the standard command-line tool for reporting installed and currently used physical memory without touching the hardware.

Exam trap

The trap is confusing memory-reporting tools with disk-reporting tools — candidates who see 'free' and 'df' both report 'space' may pick `df -h`, forgetting that `df` is filesystem (disk) usage, not RAM.

How to eliminate wrong answers

Option A is wrong because `uptime` reports how long the system has been running, the number of logged-in users, and load averages — it contains no memory information. Option B is wrong because `lsblk` lists block devices (disks and partitions) and their mountpoints, which is storage, not RAM. Option D is wrong because `df -h` reports filesystem disk space usage in human-readable units, which is also storage, not memory.

Full explanation →

LPI Linux Essentials (010-160, v1.6) (010-160) (010-160) — Questions 301–323