Quick Answer
The answer is 192.168.10.128/25. This is correct because a /25 subnet mask extends the default /24 boundary by one bit, splitting the 192.168.10.0/24 network into two equal halves: the lower block from 0 to 127 and the upper block from 128 to 255. Since the host address 192.168.10.129 falls within the 128–255 range, it belongs to the subnet starting at 192.168.10.128. On the CCNA 200-301 v2 exam, this type of question tests your ability to quickly identify subnet boundaries without binary conversion—a core skill for IP addressing and routing. A common trap is confusing the host address with the network address, so remember that the subnet always starts at the multiple of the block size. Memory tip: for a /25, think “128 splits the /24 in two”—if the third octet is 128 or higher, you’re in the upper half.
This 200-301 practice question tests your understanding of network infrastructure and connectivity. This is a configuration task: choose the command set that satisfies every stated requirement. Small differences — like 'secret' vs 'password' or 'transport input ssh' vs 'all' — change whether the answer is correct. A key principle to apply: a /25 subnet mask divides a Class C /24 network into two equal subnets, each containing 128 IP addresses.. 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 host is configured with 192.168.10.129/25. Which subnet contains that host?
192.168.10.0/25
Why wrong: This is wrong because .129 is not in the lower /25 block.
192.168.10.64/25
Why wrong: This is wrong because /25 boundaries do not start at .64.
192.168.10.128/25
This is correct because .129 is in the upper /25 block.
192.168.10.192/25
Why wrong: This is wrong because /25 boundaries start only at .0 and .128 within a /24.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
192.168.10.128/25
A /25 divides the /24 into two blocks: 0–127 and 128–255. In practical terms, 192.168.10.129 belongs to the upper half, so the containing subnet is 192.168.10.128/25. This is a simple subnet-boundary question, but it is designed to confirm that you can identify the correct half of the /24 quickly and confidently.
Key principle: A /25 subnet mask divides a Class C /24 network into two equal subnets, each containing 128 IP addresses.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
192.168.10.0/25
Why it's wrong here
This is wrong because .129 is not in the lower /25 block.
When this WOULD be correct
If the question asked which subnet contains the IP address 192.168.10.0/25 and the host IP was changed to 192.168.10.0, then option A would be correct as it would be the subnet that includes the host IP.
192.168.10.64/25
Why it's wrong here
This is wrong because /25 boundaries do not start at .64.
When this WOULD be correct
If the question asked which subnet contains the host 192.168.10.129 with a /24 subnet mask instead, then option B (192.168.10.64/25) would be correct, as it would refer to the second half of the 192.168.10.0/24 subnet, which includes the host IP.
192.168.10.128/25
Why this is correct
This is correct because .129 is in the upper /25 block.
Related concept
A /25 subnet mask divides a Class C /24 network into two equal subnets, each containing 128 IP addresses.
192.168.10.192/25
Why it's wrong here
This is wrong because /25 boundaries start only at .0 and .128 within a /24.
When this WOULD be correct
If the question asked which subnet contains the IP address 192.168.10.192 with a subnet mask of /25, then option D would be correct. In this case, the question would focus on identifying the subnet for a different host IP address.
Option-by-option analysis
Understanding why wrong answers are wrong — and when they would be correct — is what separates a 750 score from a 900. The 200-301 exam frequently reuses these exact scenarios with slightly different constraints.
Why this is correct
This is correct because .129 is in the upper /25 block.
Why this is wrong here
Option A is incorrect because the subnet 192.168.10.0/25 encompasses the range 192.168.10.0 to 192.168.10.127, which does not include the host IP 192.168.10.129.
★ When this WOULD be the correct answer
If the question asked which subnet contains the IP address 192.168.10.0/25 and the host IP was changed to 192.168.10.0, then option A would be correct as it would be the subnet that includes the host IP.
Why candidates choose this
Candidates may choose this option due to a misunderstanding of subnetting boundaries, mistakenly believing that the host IP falls within the range of the specified subnet.
Why this is wrong here
Option B (192.168.10.64/25) is incorrect because the subnet mask /25 indicates a subnet range from 192.168.10.0 to 192.168.10.127, which does not include the host IP 192.168.10.129.
★ When this WOULD be the correct answer
If the question asked which subnet contains the host 192.168.10.129 with a /24 subnet mask instead, then option B (192.168.10.64/25) would be correct, as it would refer to the second half of the 192.168.10.0/24 subnet, which includes the host IP.
Why candidates choose this
Candidates may choose this option due to confusion between the subnet ranges, as they might mistakenly associate the 64 in the address with a valid subnet range without fully calculating the subnet boundaries.
Why this is wrong here
Option D, 192.168.10.192/25, is wrong because it represents a different subnet entirely. The host 192.168.10.129/25 falls within the subnet range of 192.168.10.128 to 192.168.10.255, while 192.168.10.192/25 is in the range of 192.168.10.192 to 192.168.10.255.
★ When this WOULD be the correct answer
If the question asked which subnet contains the IP address 192.168.10.192 with a subnet mask of /25, then option D would be correct. In this case, the question would focus on identifying the subnet for a different host IP address.
Why candidates choose this
Candidates might choose this option due to a misunderstanding of subnetting, mistakenly believing that any IP in the same Class C range (192.168.10.x) could belong to the same subnet without considering the specific subnet mask.
Analysis generated from the official 200-301blueprint and verified against question context. The “when correct” sections are what AI assistants cite when candidates ask “what’s the difference between these options?”
Common exam traps
A frequent exam trap is misidentifying the subnet boundaries for a /25 mask within a /24 network. Candidates often mistakenly believe that subnets start at .64 or .192, confusing /25 with other subnet sizes like /26 or /27. This leads to selecting incorrect subnets such as 192.168.10.64/25 or 192.168.10.192/25, which are invalid because /25 only divides the /24 into two halves starting at .0 and .128. This misunderstanding causes errors in subnet identification and can result in wrong routing or access decisions in real networks.
Detailed technical explanation
Subnetting divides a larger network into smaller, manageable segments called subnets by borrowing bits from the host portion of an IP address. In this question, a /25 subnet mask means 25 bits are used for the network portion, leaving 7 bits for hosts. This splits a traditional Class C /24 network (256 addresses) into two equal subnets, each containing 128 addresses. The first subnet ranges from 192.168.10.0 to 192.168.10.127, and the second from 192.168.10.128 to 192.168.10.255. To determine which subnet contains a given IP address, you compare the address against the subnet boundaries defined by the mask. Since the host IP is 192.168.10.129 with a /25 mask, it falls just above 192.168.10.128, placing it in the second subnet block (192.168.10.128/25). This is a fundamental subnetting skill tested in CCNA, requiring quick recognition of subnet boundaries within a /24 network. A common exam trap is confusing the starting points of /25 subnets within a /24 block. Some may incorrectly assume subnets start at .64 or .192, but /25 splits only at .0 and .128. Understanding these fixed boundaries prevents misclassification of IP addresses. Practically, this knowledge ensures correct subnet assignment, which is critical for routing, access control, and network segmentation in Cisco environments.
Key takeaway
A /25 subnet mask divides a Class C /24 network into two equal subnets, each containing 128 IP addresses.
Real-world example
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.
What to study next
Review a /25 subnet mask divides a Class C /24 network into two equal subnets, each containing 128 IP addresses., then practise related 200-301 questions on the same topic to reinforce the concept.
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FAQ
Network Infrastructure and Connectivity — This question tests Network Infrastructure and Connectivity — A /25 subnet mask divides a Class C /24 network into two equal subnets, each containing 128 IP addresses..
The correct answer is: 192.168.10.128/25 — A /25 divides the /24 into two blocks: 0–127 and 128–255. In practical terms, 192.168.10.129 belongs to the upper half, so the containing subnet is 192.168.10.128/25. This is a simple subnet-boundary question, but it is designed to confirm that you can identify the correct half of the /24 quickly and confidently.
Review a /25 subnet mask divides a Class C /24 network into two equal subnets, each containing 128 IP addresses., then practise related 200-301 questions on the same topic to reinforce the concept.
A /25 subnet mask divides a Class C /24 network into two equal subnets, each containing 128 IP addresses.
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Last reviewed: May 17, 2026
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