Question 1,839 of 1,389
CCNA Network Infrastructure and Connectivity Practice Question
A host is configured with 192.168.10.129/25. Which subnet contains that host?
⚠ Common exam trap
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.
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
✓
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.
Answer analysis
Option-by-option breakdown
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
The .0/25 subnet spans addresses .0 through .127, with network address .0 and broadcast .127. Since .129 is 2 addresses above that upper boundary, it cannot belong to the lower half. The /25 split of the /24 assigns the first 128 addresses to .0/25 and the second 128 to .128/25.
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
A /25 network must begin at a multiple of 128 in the last octet, which only yields .0 and .128 within this /24. 192.168.10.64 is not a valid network address for a /25 prefix because its host bits are not all zero; it would be a host address inside .0/25, not a subnet boundary. Therefore no host can be said to reside in a ".64/25" subnet.
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
A /25 mask uses a 128-bit block size in the fourth octet, so the valid subnets within 192.168.10.0/24 are 192.168.10.0/25 and 192.168.10.128/25. The address .129 is within the upper block's range .128–.255, making it a host in that subnet. Its usable host range is .129–.254 with broadcast .255.
- ✗
192.168.10.192/25
Why it's wrong here
192.168.10.192/25 would, if it existed, cover .192–.255, which excludes .129 entirely. Moreover, /25 network boundaries within a /24 are limited to .0 and .128; .192 is not a legal starting address because it violates the alignment requirement of the 25-bit mask. The address .129 is already accounted for by the .128/25 subnet.
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
Why each answer is right or wrong
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.
✓192.168.10.128/25Correct answer▾
Why this is correct
A /25 mask uses a 128-bit block size in the fourth octet, so the valid subnets within 192.168.10.0/24 are 192.168.10.0/25 and 192.168.10.128/25. The address .129 is within the upper block's range .128–.255, making it a host in that subnet. Its usable host range is .129–.254 with broadcast .255.
✗192.168.10.0/25Wrong answer — click to see why▾
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.
✗192.168.10.64/25Wrong answer — click to see why▾
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.
✗192.168.10.192/25Wrong answer — click to see why▾
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?”
Visual reference
Quick reference
Routing Protocol Comparison
| Protocol | Metric | Max Hops | Algorithm | Type |
|---|---|---|---|---|
| RIP v2 | Hop count | 15 | Bellman-Ford | Distance vector |
| OSPF | Cost (bandwidth) | Unlimited | Dijkstra (SPF) | Link state |
| EIGRP | Composite metric | Unlimited | DUAL | Hybrid |
| IS-IS | Cost | Unlimited | Dijkstra | Link state |
| BGP | Policy / attributes | Unlimited | Path vector | Path vector |
RIP's 15-hop limit makes it unsuitable for large networks. OSPF and EIGRP dominate modern enterprise deployments.
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Last reviewed: May 17, 2026
This 200-301 practice question is part of Courseiva's free Cisco 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 200-301 exam.
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