LPIC-2 Block Devices, Filesystems and Advanced Storage Practice Test 4 — 15 Questions

Question 1

Match each security tool to its function.

Accepted Answer

iptables: A firewall tool used to set up, maintain, and inspect the tables of IP packet filter rules.. Correct matches: iptables is a firewall, nmap is a network scanner, tcpdump is a packet analyzer. Common confusions include misattributing file integrity checking to iptables, IP banning to nmap, and port scanning to tcpdump.

Answer

iptables: A tool for monitoring file system integrity.

Answer

nmap: A tool for banning IPs after failed login attempts.

Answer

tcpdump: A tool for scanning open ports.

Question 2

A Linux administrator notices that a newly added disk /dev/sdc is not recognized by the system. Which command should be used to rescan the SCSI bus without rebooting?

Accepted Answer

echo '- - -' > /sys/class/scsi_host/host0/scan. Writing '- - -' to /sys/class/scsi_host/host0/scan triggers a SCSI bus rescan on the specified host adapter (host0) without requiring a reboot. The three dashes represent 'channel, target, LUN' wildcards, instructing the kernel to scan all SCSI IDs on that bus to detect newly added devices like /dev/sdc.

Answer

echo 'scan' > /proc/scsi/scsi

Answer

blockdev --rereadpt /dev/sdc

Answer

service scsi restart

Question 3

A database server experiences performance degradation. Analysis shows high I/O wait on a LVM logical volume that uses striping across two physical volumes. What is the most effective way to improve performance?

Accepted Answer

Add a cache layer using lvmcache on a fast SSD. Adding a cache layer using lvmcache on a fast SSD is the most effective way to reduce high I/O wait because it places frequently accessed (hot) data on the faster SSD while keeping the bulk of data on the slower striped HDDs. This directly addresses the I/O bottleneck without requiring data migration or downtime, and it leverages the existing striped layout for sequential throughput while improving latency for random reads/writes.

Answer

Increase the stripe size and recreate the LV

Answer

Add an additional PV to the VG and extend the LV

Answer

Convert the LV to use linear mapping instead of striping

Question 4

An administrator creates a new XFS filesystem on /dev/vg/data and needs to mount it at /data with noatime and a stripe unit of 1 MiB. Which /etc/fstab entry is correct?

Accepted Answer

/dev/vg/data /data xfs defaults,noatime,sunit=2048 0 0. The `sunit` mount option specifies the stripe unit in 512-byte blocks. A stripe unit of 1 MiB equals 2048 blocks (1 MiB = 1024 KiB = 2048 * 512 bytes). The `noatime` option disables access time updates. This entry correctly mounts the XFS filesystem at /data with the required parameters.

Answer

/dev/vg/data /data xfs defaults,noatime,swidth=2048 0 0

Answer

/dev/vg/data /data xfs defaults,noatime,allocsize=1m 0 0

Answer

/dev/vg/data /data xfs defaults,noatime,sunit=1024 0 0

Question 5

Which two commands can be used to create a swap space on a partition /dev/sdb2? (Choose two.)

Accepted Answer

mkswap /dev/sdb2. Option A, mkswap /dev/sdb2, is correct because mkswap is the standard util-linux command that initializes a partition as swap space by writing the swap signature and header to the device, which is exactly what is needed to create swap on /dev/sdb2. Option D, mkswap -L swap /dev/sdb2, is also correct because it performs the same swap-space initialization while additionally assigning the label "swap" to the device, and the -L option is a valid mkswap flag that does not prevent the swap area from being created. Option B, swapon /dev/sdb2, is incorrect because swapon only activates an already-initialized swap area; it does not create the swap signature. Option C, dd if=/dev/zero of=/dev/sdb2 bs=1M count=1024, is incorrect because dd merely writes zeros to the partition and does not format it as swap. Option E, mkfs.swap /dev/sdb2, is incorrect because there is no standard mkfs.swap helper; swap is initialized with mkswap, not a mkfs variant.

Answer

swapon /dev/sdb2

Answer

dd if=/dev/zero of=/dev/sdb2 bs=1M count=1024

Answer

mkfs.swap /dev/sdb2

Question 6

Which three conditions must be met for a filesystem to be mounted by UUID in /etc/fstab? (Choose three.)

Accepted Answer

The UUID must be present in /dev/disk/by-uuid/. Option C is correct because /etc/fstab entries using UUID= rely on the udev-managed symlinks in /dev/disk/by-uuid/, which map each filesystem's UUID to its actual block device node; without that symlink the mount cannot resolve the device. Option D is correct because a UUID only exists if the filesystem was created by a tool that writes one into the superblock (for example mkfs.ext4, mkfs.xfs, or mkswap); filesystems lacking UUID support cannot be referenced by UUID at all. Option E is correct because the UUID string in the fstab entry must exactly match the UUID stored on the filesystem, otherwise mount fails with an error such as 'can't find UUID='. Option A is not required: blkid merely reads and displays existing UUIDs and does not create them, so running it is irrelevant to mounting. Option B is not required: UUIDs are resolved by udev and mount from the device symlinks, not from kernel boot parameters such as root=UUID=, which only affect the root filesystem.

Answer

The blkid command must have been run at least once

Answer

The UUID must be listed in the kernel's boot parameters

Question 7

Refer to the exhibit. The /data filesystem is nearly full. The administrator has added a new disk /dev/sdc with 10GB capacity. The VG currently has no free space. Which steps will increase the available space for /data using LVM and ext4?

Accepted Answer

pvcreate /dev/sdc; vgextend vg /dev/sdc; lvextend -l +100%FREE /dev/vg/data; resize2fs /dev/vg/data. It follows the proper LVM workflow: create a physical volume (pvcreate), add it to the volume group (vgextend), extend the logical volume using all free space in the VG (lvextend -l +100%FREE), and then resize the ext4 filesystem (resize2fs). This ensures the new 10GB disk is fully utilized for /data.

Answer

vgextend vg /dev/sdc; lvextend -L +10G /dev/vg/data; resize2fs /dev/vg/data

Answer

pvcreate /dev/sdc; lvextend -L +10G /dev/vg/data; resize2fs /dev/vg/data

Answer

pvcreate /dev/sdc; vgextend vg /dev/sdc; lvextend -l +50%FREE /dev/vg/data; resize2fs /dev/vg/data

Question 8

A system administrator notices that a new iSCSI target is not appearing as a block device after running iscsiadm. The target is known to be online and reachable. What is the most likely cause?

Accepted Answer

The iSCSI service (iscsid) is not running.. The iSCSI initiator requires the `iscsid` daemon to be running in order to process discovery and login requests from `iscsiadm`. Even if the target is online and reachable, without the `iscsid` service, the kernel's iSCSI layer cannot establish a session or create the corresponding block device (e.g., /dev/sdX). This is the most likely cause because the administrator has confirmed network connectivity and target availability, leaving the service state as the primary suspect.

Answer

The iSCSI initiator name is not set.

Answer

The target IQN is not specified correctly.

Answer

The network interface is down.

Question 9

An administrator needs to configure iSCSI multipathing. After setting up two paths, which command should be used to verify the multipath topology?

Accepted Answer

multipath -l. The `multipath -l` command displays the current multipath topology, showing the aggregated device mapper (dm) devices, their underlying paths, and their status (active/passive or active/active). This is the standard tool for verifying multipath configuration after setting up iSCSI paths, as it reads the multipathd state and presents a human-readable map of the multipath devices.

Answer

lsscsi

Answer

dmsetup ls

Answer

iscsiadm -m session -P 3

Question 10

A filesystem created with mkfs.ext4 on /dev/sda1 has a reserved block count of 5%. How can the administrator reduce it to 1% without reformatting?

Accepted Answer

tune2fs -m 1 /dev/sda1. The `-m` option of tune2fs sets the reserved block percentage for the filesystem. Since the question asks to reduce the reserved block count from 5% to 1% without reformatting, `tune2fs -m 1 /dev/sda1` correctly adjusts the reserved blocks percentage to 1% on the existing ext4 filesystem.

Answer

tune2fs -r 1 /dev/sda1

Answer

dumpe2fs -r 1 /dev/sda1

Answer

resize2fs -r 1 /dev/sda1

Question 11

Which file contains the filesystem table that is read during system boot to mount filesystems?

Accepted Answer

/etc/fstab. /etc/fstab is the filesystem table file that is read by the system during boot (via mount -a or systemd-fstab-generator) to automatically mount filesystems. It contains entries specifying devices, mount points, filesystem types, mount options, dump and pass fields, and is the standard configuration file for static filesystem mounting.

Answer

/etc/mtab

Answer

/proc/mounts

Answer

/etc/mounts

Question 12

Which THREE of the following are correct statements about LVM thin provisioning? (Select THREE)

Accepted Answer

Snapshots of thin volumes use the same pool.. Option B is correct because thin provisioning lets you create thin volumes whose combined virtual size exceeds the actual free space in the thin pool, i.e., it overcommits physical storage, with space allocated on demand as data is written. Option D is correct because a thin volume's virtual size can exceed the pool's physical size, since blocks are allocated lazily from the pool only when written. Option A is correct because snapshots of thin volumes are themselves thin volumes that draw their changed blocks from the same thin pool as their origin, sharing that pool's free space. Option C is wrong because shrinking a thin pool below the space already consumed by its thin volumes and snapshots risks data loss and is not a safe operation. Option E is wrong because the thin pool must exist first; thin volumes are created within an existing thin pool, not before it.

Answer

A thin pool can be resized smaller without data loss.

Answer

Thin volumes must be created before the thin pool.

Question 13

Which THREE of the following are correct about RAID 1 in mdadm? (Select THREE)

Accepted Answer

It can have a hot spare.. Option B is correct because mdadm RAID 1 arrays support hot spares: a spare device can be added with `mdadm --add` and will automatically replace a failed member, and it can be designated as a spare during creation with `--spare-devices`. Option C is correct because RAID 1 is mirroring, which by definition needs at least two block devices to hold identical copies of the data, so `mdadm --create --level=1` requires a minimum of two member disks. Option D is correct because RAID 1's defining characteristic is mirroring, i.e., every write is duplicated to all member disks so each holds a complete copy of the data. Option A is not correct because striping (distributing data across disks without duplication) is RAID 0, not RAID 1. Option E is not correct because the write-intent bitmap is an optional mdadm feature used to speed resynchronization after a crash or disk replacement, not an inherent property of RAID 1.

Answer

It provides striping.

Answer

It uses a bitmap for resync.

Question 14

An administrator runs the command shown. However, when a user tries to execute a binary from /mnt/tmp, it fails with 'Permission denied'. What is the most likely cause?

Accepted Answer

The noexec mount option prevents execution.. The `noexec` mount option explicitly prevents execution of any binaries on the filesystem, regardless of their permission bits. When a filesystem is mounted with `noexec`, the kernel will refuse to execute any file on that mount point, returning 'Permission denied' even if the file has execute permissions set. This is the most likely cause because the administrator's command likely mounted `/mnt/tmp` with `noexec` (e.g., via `mount -o noexec` or a default mount option).

Answer

The binary is not executable.

Answer

The tmpfs filesystem does not support execution.

Answer

The nosuid option prevents setuid binaries.

Question 15

An administrator observes that only one path is shown as active for mpathb. What is the most likely cause?

Accepted Answer

The blacklist rule excludes the other path.. When only one path is active for a multipath device (mpathb), the most likely cause is that the other path is being excluded by a blacklist rule in the multipath configuration. The blacklist can filter out devices based on WWID, device name, or vendor/product, preventing them from being aggregated into the multipath map. This results in the path being visible to the system but not used by multipathd.

Answer

The find_multipaths setting is incorrect.

Answer

The SCSI devices are not properly connected.

Answer

The multipathd service is not running.