LPIC-1 Devices, Filesystems and FHS Practice Question
Exhibit
$ lsblk NAME MAJ:MIN RM SIZE RO TYPE MOUNTPOINT sda 8:0 0 256G 0 disk ├─sda1 8:1 0 512M 0 part /boot/efi ├─sda2 8:2 0 100G 0 part / ├─sda3 8:3 0 150G 0 part /data
Refer to the exhibit. How much unpartitioned space is available on /dev/sda?
⚠ Common exam trap
The trap is that candidates may misread the partition sizes from the exhibit or incorrectly sum them. The exhibit shows sda1=0.5G, sda2=100G, sda3=150G, totaling 250.5G, leaving 5.5G unpartitioned. Picking 6G comes from assuming all partitions are round numbers or misremembering the total.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
5.5G
The output of `fdisk -l /dev/sda` shows partitions sda1 (0.5G), sda2 (100G), and sda3 (150G), summing to 250.5G. The total disk size is 256G, so the unpartitioned space is 256G - 250.5G = 5.5G. Candidates often misread the partition sizes or add them incorrectly, leading to wrong answers.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
256G
Why it's wrong here
256G mistakes the disk's total capacity for its free space, since unpartitioned space requires subtracting each partition's size from that total. It is tempting because the disk size is the most prominent figure in the exhibit, and it would be correct only for an entirely unallocated device.
- ✓
5.5G
Why this is correct
Unpartitioned space is the gap between the end of the last partition and the disk's total capacity. Reading the partition table, the final partition ends at 5.5G short of the full device size, so that remainder is unallocated.
- ✗
6G
Why it's wrong here
6G ignores the exhibit's arithmetic: unpartitioned space is total disk capacity minus every partition's size, and the listed partitions leave a far larger remainder. It is tempting because small trailing gaps often appear after the final partition, and 6G would be right if the disk were almost fully allocated.
- ✗
150G
Why it's wrong here
150G misreads the exhibit's figures, since unpartitioned space equals the disk's total capacity minus the sum of all existing partition sizes, not a single partition's value. It is tempting when one partition dominates the layout, and such a figure would be correct if the remaining partitions consumed almost nothing.
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Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
This LPIC-1 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-1 exam.