The answer is 255.255.255.224, or a /27 subnet mask, because it provides exactly 30 usable host addresses—the minimum that supports 25 devices without waste. This works by borrowing 3 bits from the host portion of a Class C address, leaving 5 bits for hosts, which gives 2^5 = 32 total addresses per subnet; subtracting the network and broadcast addresses leaves 30 usable IPs. On the CompTIA Network+ N10-009 exam, this question tests your ability to match subnet masks to real-world requirements, often appearing in scenario-based questions where you must avoid the trap of choosing a /28 (only 14 usable hosts) or a wasteful /26 (62 usable hosts). A quick memory tip: remember the “minus 2” rule for every subnet—usable hosts are always total addresses minus two—and for 25 hosts, the next power of two above 25 is 32, which corresponds to a /27 mask.
N10-009 Networking Concepts Practice Question
This N10-009 practice question tests your understanding of networking concepts. Match the stated requirement to the specific cloud service, access model, or configuration option — many options are valid in isolation but not for this scenario. A key principle to apply: a /27 subnet mask uses 27 network bits and 5 host bits.. 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.
A network engineer needs to assign IP addresses to a new subnet that will support exactly 25 devices. Which subnet mask would provide the minimum number of usable host addresses while still accommodating the requirement?
Clue words in this question
Noticing these words before you look at the options changes how you read each choice.
Clue: "minimum / minimize"
Why it matters: Asks for the least resource use — fewest addresses, smallest subnet, lowest overhead. Eliminate over-provisioned options even if they would technically work.
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
255.255.255.224 (/27)
Option B (255.255.255.224, /27) provides 32 total addresses per subnet, with 30 usable host addresses (2^(32-27)-2 = 30). This is the smallest subnet that can accommodate exactly 25 devices, as /28 yields only 14 usable hosts and /26 yields 62, which is excessive.
Key principle: A /27 subnet mask uses 27 network bits and 5 host bits.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✗
255.255.255.240 (/28)
Why it's wrong here
/28 provides 14 usable hosts, which is insufficient for 25 devices.
✓
255.255.255.224 (/27)
Why this is correct
/27 provides 28 usable hosts, which meets the requirement with minimal waste.
Clue confirmation
The clue word "minimum / minimize" in the question point toward this answer.
Related concept
A /27 subnet mask uses 27 network bits and 5 host bits.
✗
255.255.255.192 (/26)
Why it's wrong here
/26 provides 62 usable hosts, which is more than needed but not the minimum.
✗
255.255.255.128 (/25)
Why it's wrong here
/25 provides 126 usable hosts, far exceeding the need and not the minimal subnet.
Common exam traps
Common exam trap: answer the scenario, not the keyword
Cisco often tests the common misconception that the number of usable hosts equals 2^(host bits) without subtracting the network and broadcast addresses, leading candidates to incorrectly choose a /28 mask thinking 16 addresses are enough for 25 devices.
Detailed technical explanation
How to think about this question
The subnet mask determines the boundary between the network and host portions of an IPv4 address. For a /27 prefix, the host portion is 5 bits, yielding 2^5 = 32 total addresses, with the first address reserved as the network ID and the last as the broadcast address. In real-world scenarios, using the smallest sufficient subnet conserves IP address space, which is critical in VLSM (Variable Length Subnet Mask) designs within a larger address block.
KKey Concepts to Remember
A /27 subnet mask uses 27 network bits and 5 host bits.
The formula for total addresses in a subnet is 2^n, where n is host bits.
Usable host addresses are calculated as 2^n - 2 (excluding network and broadcast).
A /27 subnet provides 30 total addresses, with 28 usable host addresses.
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
A /27 subnet mask uses 27 network bits and 5 host bits.
Real-world example
How this comes up in practice
A network engineer segments a warehouse floor into three subnets: 20 scanners, 5 printers, and 2 management hosts. Picking the wrong mask wastes addresses or leaves too few usable hosts. Exam questions test whether you can apply CIDR notation, calculate block size, and identify the correct usable-host range for a given prefix.
Related glossary terms
Concepts from this question explained
These glossary pages explain the core terms tested in this N10-009 question in full detail.
Networking Concepts — This question tests Networking Concepts — A /27 subnet mask uses 27 network bits and 5 host bits..
What is the correct answer to this question?
The correct answer is: 255.255.255.224 (/27) — Option B (255.255.255.224, /27) provides 32 total addresses per subnet, with 30 usable host addresses (2^(32-27)-2 = 30). This is the smallest subnet that can accommodate exactly 25 devices, as /28 yields only 14 usable hosts and /26 yields 62, which is excessive.
What should I do if I get this N10-009 question wrong?
Review a /27 subnet mask uses 27 network bits and 5 host bits., then practise related N10-009 questions on the same topic to reinforce the concept.
Are there clue words in this question I should notice?
Yes — watch for: "minimum / minimize". Asks for the least resource use — fewest addresses, smallest subnet, lowest overhead. Eliminate over-provisioned options even if they would technically work.
What is the key concept behind this question?
A /27 subnet mask uses 27 network bits and 5 host bits.
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