This EX200 practice question tests your understanding of create and configure file systems. Examine the command output carefully: the correct answer depends on what the output actually shows, not on general recall alone. 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.
Exhibit
Refer to the exhibit.
Output of 'pvscan' and 'vgdisplay':
PV /dev/sda2 VG vg00 lvm2 [19.99 GiB / 0 free]
PV /dev/sdb1 VG vg00 lvm2 [19.99 GiB / 19.99 GiB free]
Total: 2 [39.98 GiB] / in use: 2 [39.98 GiB] / in no VG: 0 [0 ]
--- Volume group ---
VG Name vg00
System ID
Format lvm2
Metadata Areas 2
Metadata Sequence No 6
VG Access read/write
VG Status resizable
MAX LV 0
Cur LV 3
Open LV 3
Max PV 0
Cur PV 2
Act PV 2
VG Size 39.98 GiB
PE Size 4.00 MiB
Total PE 10235
Alloc PE / Size 5120 / 20.00 GiB
Free PE / Size 5115 / 19.98 GiB
VG UUID abcd1234
Based on the exhibit, how much free space is available in the volume group vg00?
Exhibit
Refer to the exhibit.
Output of 'pvscan' and 'vgdisplay':
PV /dev/sda2 VG vg00 lvm2 [19.99 GiB / 0 free]
PV /dev/sdb1 VG vg00 lvm2 [19.99 GiB / 19.99 GiB free]
Total: 2 [39.98 GiB] / in use: 2 [39.98 GiB] / in no VG: 0 [0 ]
--- Volume group ---
VG Name vg00
System ID
Format lvm2
Metadata Areas 2
Metadata Sequence No 6
VG Access read/write
VG Status resizable
MAX LV 0
Cur LV 3
Open LV 3
Max PV 0
Cur PV 2
Act PV 2
VG Size 39.98 GiB
PE Size 4.00 MiB
Total PE 10235
Alloc PE / Size 5120 / 20.00 GiB
Free PE / Size 5115 / 19.98 GiB
VG UUID abcd1234
A
19.98 GiB
Correct: matches Free PE size in vgdisplay.
B
10235 MiB
Why wrong: Incorrect: that is total PE.
C
20.00 GiB
Why wrong: Incorrect: that is allocated size.
D
19.99 GiB
Why wrong: Incorrect: that is individual PV size.
E
5115 MiB
Why wrong: Option E (5115 MiB) mistakenly uses the number of Physical Extents (PEs) as the free space in MiB. The actual free space is 5115 PEs × 4 MiB/PE = 20460 MiB = 19.98 GiB, not 5115 MiB.
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
19.98 GiB
Option A is correct because the VG free space is calculated from the total number of Physical Extents (PEs) and the PE size. The output shows 5115 PEs and a PE size of 4.00 MiB, so total free space = 5115 × 4 MiB = 20460 MiB = 19.98 GiB (since 20460 ÷ 1024 = 19.98046875). No Logical Volumes are allocated, so all PEs are free. Option B (10235 MiB) is exactly half of 20470 MiB, a common miscalculation. Option C (20.00 GiB) is the raw PV size, ignoring the PE overhead. Option D (19.99 GiB) is a rounding error. Option E (5115 MiB) confuses the PE count with the free space in MiB.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✓
19.98 GiB
Why this is correct
Correct: matches Free PE size in vgdisplay.
Related concept
Read the scenario before looking for a memorised answer.
✗
10235 MiB
Why it's wrong here
Incorrect: that is total PE.
✗
20.00 GiB
Why it's wrong here
Incorrect: that is allocated size.
✗
19.99 GiB
Why it's wrong here
Incorrect: that is individual PV size.
✗
5115 MiB
Why it's wrong here
Option E (5115 MiB) mistakenly uses the number of Physical Extents (PEs) as the free space in MiB. The actual free space is 5115 PEs × 4 MiB/PE = 20460 MiB = 19.98 GiB, not 5115 MiB.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Red Hat often tests the distinction between raw PV size and usable VG space, trapping candidates who assume the PV size equals the VG free space without accounting for PE size and rounding.
Detailed technical explanation
How to think about this question
In LVM, the Volume Group's free space is calculated as (Total PE count × PE size) minus any allocated extents. The PE size is set at VG creation (default 4 MiB in many distributions) and cannot be changed without recreating the VG. The raw PV size (e.g., 20.00 GiB) is never fully usable because LVM reserves a small amount of metadata (typically a few MiB) and because the PE size may not divide evenly into the PV size, leading to rounding. In this case, 20.00 GiB = 20480 MiB, but with a 4 MiB PE size, only 5115 PEs fit (5115 × 4 = 20460 MiB), leaving 20 MiB unused (not part of the VG). This is a common real-world scenario when adding disks to a VG: the 'free space' shown by vgdisplay is the sum of unallocated PEs, not the raw disk size.
KKey Concepts to Remember
Read the scenario before looking for a memorised answer.
Find the constraint that changes the correct option.
Eliminate answers that are true in general but not in this case.
TExam Day Tips
→Watch for words such as best, first, most likely and least administrative effort.
→Review why wrong options are wrong, not only why the correct option is correct.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
How this comes up in practice
A practitioner preparing for the EX200 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option. Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.
What to study next
Got this wrong? Here's your next step.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
Create and configure file systems — This question tests Create and configure file systems — Read the scenario before looking for a memorised answer..
What is the correct answer to this question?
The correct answer is: 19.98 GiB — Option A is correct because the VG free space is calculated from the total number of Physical Extents (PEs) and the PE size. The output shows 5115 PEs and a PE size of 4.00 MiB, so total free space = 5115 × 4 MiB = 20460 MiB = 19.98 GiB (since 20460 ÷ 1024 = 19.98046875). No Logical Volumes are allocated, so all PEs are free. Option B (10235 MiB) is exactly half of 20470 MiB, a common miscalculation. Option C (20.00 GiB) is the raw PV size, ignoring the PE overhead. Option D (19.99 GiB) is a rounding error. Option E (5115 MiB) confuses the PE count with the free space in MiB.
What should I do if I get this EX200 question wrong?
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
What is the key concept behind this question?
Read the scenario before looking for a memorised answer.
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