LFCS User and Group Management Practice Test 5 — 15 Questions

Question 1

Which command will display all groups a specific user belongs to, including both primary and supplementary groups?

Accepted Answer

groups username. The 'groups' command is the standard utility to list all group memberships for a given user, showing both the primary group (from /etc/passwd) and any supplementary groups (from /etc/group). It queries the system's group database directly, making it the correct and simplest choice for this task.

Answer

cat /etc/passwd | grep username

Answer

id -g username

Answer

grep username /etc/group

Question 2

An administrator needs to create a user 'john' with a home directory in /data/home/john, a UID of 1500, and membership in the group 'developers' as a secondary group. The group 'developers' already exists. Which single command accomplishes this?

Accepted Answer

useradd -u 1500 -d /data/home/john -G developers john. Ly uses useradd with -u to set UID 1500, -d to specify home directory /data/home/john, and -G to add john to the existing supplementary group 'developers'. This is a single command that accomplishes all requirements. Option A uses adduser which is distribution-specific and may not support --uid or --group flags. Option B uses two separate commands (useradd then usermod), not a single command. Option C uses lowercase -g which sets the primary group, not secondary membership, so it would make 'developers' the primary group instead of adding it as a secondary group.

Answer

adduser --uid 1500 --home /data/home/john --group developers john

Answer

useradd john; usermod -u 1500 -d /data/home/john -G developers john

Answer

useradd -u 1500 -d /data/home/john -g developers john

Question 3

A user 'alice' has a umask of 027 in her .bashrc, but the system administrator wants to enforce a umask of 007 for all users in the 'staff' group. Where should the administrator place the umask command to ensure it cannot be overridden by users?

Accepted Answer

In /etc/pam.d/common-session with pam_umask.so. Using pam_umask.so in /etc/pam.d/common-session sets the umask at the PAM (Pluggable Authentication Modules) level, which is applied before any user shell configuration files are read. This makes it impossible for users to override the umask in their personal dot files like .bashrc. Options A (/etc/bash.bashrc), B (/etc/skel/.bashrc), and C (/etc/profile) are all shell configuration files that are processed before or after user files, but users can still override the umask by modifying their own .bashrc or .profile after these files are sourced. Therefore, only a PAM-based solution ensures enforcement that cannot be bypassed by regular users.

Answer

/etc/bash.bashrc

Answer

/etc/skel/.bashrc

Answer

/etc/profile

Question 4

Which TWO commands can be used to display the group membership of a user? (Choose two.)

Accepted Answer

id -Gn username. Options A and D are correct. The `id -Gn` command displays all group names (primary and supplementary) for a user. The `groups` command also lists all group memberships for a user. Option B shows the user's entry in /etc/passwd, which includes primary group ID but not supplementary groups. Option C shows only the primary group ID (numeric). Option E lists groups where the user is explicitly listed in /etc/group, but it misses the primary group if not listed there.

Answer

cat /etc/passwd | grep username

Answer

id -g username

Answer

grep username /etc/group

Question 5

Which THREE of the following are valid methods to temporarily switch to a different user account without logging out entirely? (Choose three.)

Accepted Answer

sudo -u username -s. Options A, B, and E are correct. `sudo -u username -s` launches a shell as the specified user. `su - username` starts a login shell as that user. `runuser -l username -c 'bash'` runs a command as another user (common in scripts). Option C (`newgrp groupname`) changes the group, not the user. Option D (`login username`) requires a full login, which logs out the current session.

Answer

newgrp groupname

Answer

login username

Question 6

Which TWO commands are used to modify user account attributes such as password age, expiration, or lock status? (Choose two.)

Accepted Answer

chage. Options A and D are correct. chage is used for password aging (e.g., -M for max days, -E for expiry). usermod can lock/unlock accounts (-L/-U) and set account expiration (-e). Option B (useradd) creates new users, not modify existing. Option C (chsh) changes the user's login shell. Option E (passwd) changes passwords and can set password aging with -x, but does not manage account lock status or account expiration date.

Answer

useradd

Answer

chsh

Answer

passwd

Question 7

Refer to the exhibit. An administrator adds user 'frank' to the group 'projectx' by editing /etc/group directly and changing the line to 'projectx:x:500:carol,dave,frank'. After saving, the administrator runs 'groups frank' and sees only 'frank' in the output. Why does frank not appear in the group 'projectx'?

Accepted Answer

The user 'frank' is still logged into the same session; he must log out and log back in for the new group to be recognized.. The `groups` command shows the current effective group membership of the user's shell session, which is cached at login time. When `/etc/group` is edited directly, the change is written to disk immediately, but the running shell for user 'frank' still holds the old group membership list from the login process. The new group 'projectx' will only appear after 'frank' logs out and logs back in, or starts a new login shell (e.g., `su - frank`).

Answer

Editing /etc/group directly is not a valid method; 'usermod -aG' must be used instead.

Answer

The group 'projectx' has a GID conflict with another group.

Answer

The 'groups' command reads only from /etc/group and the change should appear immediately.

Question 8

Refer to the exhibit. The shadow entry for user 'carol' shows 18000 in the third field. What does the value 18000 represent?

Accepted Answer

The number of days since the password was last changed.. The third field in /etc/shadow is the date of last password change, expressed as the number of days since January 1, 1970 (epoch). A value of 18000 corresponds to approximately April 2025. Option A is incorrect because the account expiration days are stored in the eighth field. Option B is incorrect because the minimum days between password changes is the fourth field. Option D is incorrect because the warning days before expiration is the seventh field.

Answer

The number of days until the account expires.

Answer

The minimum number of days required between password changes.

Answer

The number of days before password expiration that the user is warned.

Question 9

Refer to the exhibit. What is the primary group ID of user 'charlie'?

Accepted Answer

1005 (the same as UID). The primary group ID for user 'charlie' is 1005. By default, Linux creates a private user group for each new user with a GID equal to the user's UID. Both option C, which adds explanatory context, and option D, which provides the numeric value alone, are factually correct. Therefore, both are acceptable answers.

Answer

1000

Answer

Charlie Brown

Question 10

After attempting to log in as user 'alice', the system rejects the password. The admin checks /etc/passwd and sees 'alice:x:1001:1001::/home/alice:/bin/bash'. The /etc/shadow shows 'alice:!!:18200:0:99999:7:::', indicating the account is locked. Which command will unlock the account?

Accepted Answer

usermod -U alice. The `usermod -U alice` command unlocks the user account by removing the `!` from the password hash field in `/etc/shadow`. The `!!` in the shadow file indicates the account is locked (password disabled), and `-U` explicitly unlocks it, restoring the password hash to an active state.

Answer

passwd -S alice

Answer

passwd -u alice

Answer

chage -d 0 alice

Question 11

Which command adds an existing user to a supplementary group without removing the user from other groups?

Accepted Answer

usermod -aG groupname username. The correct command is `usermod -aG groupname username`. The `-a` option (append) combined with `-G` (supplementary groups) adds the user to the specified group without affecting existing supplementary group memberships. Option A (`groupmod -a`) is incorrect because `groupmod` modifies group attributes, not user membership. Option B (`-A`) is not a valid `usermod` option. Option D (`-g`) changes the user's primary group, not supplementary groups.

Answer

groupmod -a username groupname

Answer

usermod -A groupname username

Answer

usermod -g groupname username

Question 12

A user must change their password at next login per security policy. The admin wants to expire the password immediately. Which command accomplishes this?

Accepted Answer

chage -d 0 username. Chage -d 0 sets the last password change date to the epoch, forcing a password change on next login.

Answer

passwd -f username

Answer

usermod -p '' username

Answer

chage -M 90 username

Question 13

After deleting user 'alice', the system administrator wants to also remove the home directory and mail spool. Which command should be used?

Accepted Answer

userdel -r alice. The correct option is B, userdel -r alice. The -r flag removes the user's home directory and mail spool, along with the user account. This is the standard flag for removing user data along with the account.

Answer

userdel -Z alice

Answer

userdel -f alice

Answer

userdel --remove alice

Question 14

Which file stores the encrypted password (or password hash) for user accounts?

Accepted Answer

/etc/shadow. The /etc/shadow file contains the encrypted password and password aging information. The /etc/passwd file historically contained passwords but now uses 'x' placeholder.

Answer

/etc/group

Answer

/etc/passwd

Answer

/etc/gshadow

Question 15

An admin needs to change the primary group of user 'alice' from 'grp1' to 'grp2', but alice should still be a member of 'grp1' as a supplementary group. Which command accomplishes this?

Accepted Answer

usermod -g grp2 -aG grp1 alice. The usermod -g grp2 changes the primary group to grp2, and -aG grp1 adds grp1 as a supplementary group without removing other supplementary groups, satisfying the requirement that alice remains a member of grp1.

Answer

usermod -s /bin/bash alice

Answer

usermod -g grp2 alice

Answer

usermod -G grp2 alice