CCNA Network Infrastructure and Connectivity Practice Question
A host with address 10.0.0.130/25 needs to identify its subnet. Which subnet is correct?
⚠ Common exam trap
Be careful not to confuse the subnet mask with the number of addresses it covers. Always calculate the address range based on the subnet mask.
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
✓
10.0.0.128/25
A /25 uses blocks of 128 addresses. In plain language, that means the fourth-octet ranges are 0–127 and 128–255. Since the host address ends in 130, it belongs to the upper block, which means the subnet is 10.0.0.128/25. This kind of question is a staple of subnetting because it tests whether you can identify the correct subnet boundary from the prefix and host address. Once you recognize the /25 split, the answer becomes straightforward.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
10.0.0.0/25
Why it's wrong here
With a /25 mask, the subnet beginning at 10.0.0.0 covers only addresses 10.0.0.0 through 10.0.0.127, and its broadcast address is 10.0.0.127. Since 130 is greater than 127, it clearly lies outside this range, so the correct network cannot be 10.0.0.0/25. The only two /25 subnets in this octet are 10.0.0.0/25 and 10.0.0.128/25, and the host belongs to the latter.
When this WOULD be correct
If the question asked for the base address of the subnet that includes all addresses from 10.0.0.0 to 10.0.0.127, then option A would be correct as it represents the starting address of that subnet.
- ✗
10.0.0.64/25
Why it's wrong here
A /25 prefix defines a subnet block size of 128, so valid network addresses must increment by 128 (0, 128, 256), never by 64. The address 10.0.0.64/25 would imply a range from 64 through 191, but that range is not aligned to a /25 boundary because it crosses the 128 boundary. In a /25 addressing scheme, 64 is not a valid network address, and host 130 must be assigned to the subnet starting at 10.0.0.128.
When this WOULD be correct
In a question where the subnet mask is changed to /26 and the host address is specified as 10.0.0.130, option B (10.0.0.64/25) could be correct if the question asks for the subnet that includes addresses from 10.0.0.64 to 10.0.0.127. This would make it a valid subnet for a different host range.
- ✓
10.0.0.128/25
Why this is correct
The /25 prefix length creates subnets with a block size of 128 addresses. The host address 10.0.0.130 has a last octet binary of 10000010, which lies within the 128-255 range, so it belongs to the subnet that starts at 10.0.0.128. The network address is therefore 10.0.0.128, with usable host addresses 10.0.0.129 through 10.0.0.254 and a broadcast address of 10.0.0.255.
- ✗
10.0.0.192/25
Why it's wrong here
This choice fails because a valid /25 network must begin at a multiple of 128, and 192 is not one of those boundaries. The address 10.0.0.130 falls inside the 128-255 block, whose network address is 10.0.0.128, not a separate 192-255 /25 sub-block. Treating 192 as a /25 start would misalign the subnet boundaries and incorrectly include addresses that belong to the 10.0.0.128/25 network.
When this WOULD be correct
This option would be correct if the question asked for the subnet of a host with an address in the range of 10.0.0.192 to 10.0.0.255, such as 10.0.0.200/25. In that case, 10.0.0.192/25 would accurately represent the subnet for that host.
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.
✓10.0.0.128/25Correct answer▾
Why this is correct
The /25 prefix length creates subnets with a block size of 128 addresses. The host address 10.0.0.130 has a last octet binary of 10000010, which lies within the 128-255 range, so it belongs to the subnet that starts at 10.0.0.128. The network address is therefore 10.0.0.128, with usable host addresses 10.0.0.129 through 10.0.0.254 and a broadcast address of 10.0.0.255.
✗10.0.0.0/25Wrong answer — click to see why▾
Why this is wrong here
The subnet 10.0.0.0/25 covers addresses 10.0.0.0 through 10.0.0.127. Since 10.0.0.130 is outside this range, it cannot belong to this subnet. The host's address must be within the subnet's range.
★ When this WOULD be the correct answer
If the question asked for the base address of the subnet that includes all addresses from 10.0.0.0 to 10.0.0.127, then option A would be correct as it represents the starting address of that subnet.
Why candidates choose this
A student might assume that any address starting with 10.0.0.x belongs to the 10.0.0.0/24 network and forget that the /25 mask splits the /24 into two subnets. They may incorrectly think 10.0.0.0/25 includes all addresses from 0 to 255.
✗10.0.0.64/25Wrong answer — click to see why▾
Why this is wrong here
The /25 prefix length has a subnet size of 128, so valid network addresses are multiples of 128 (0, 128, 256, etc.). 10.0.0.64 is not a multiple of 128, so it is not a valid /25 network address.
★ When this WOULD be the correct answer
In a question where the subnet mask is changed to /26 and the host address is specified as 10.0.0.130, option B (10.0.0.64/25) could be correct if the question asks for the subnet that includes addresses from 10.0.0.64 to 10.0.0.127. This would make it a valid subnet for a different host range.
Why candidates choose this
Students often confuse /25 with /26 (which has a 64-address block size) or /27 (32-address block). The number 64 is a common boundary for /26 subnets, leading to the mistaken belief that 10.0.0.64/25 is valid.
✗10.0.0.192/25Wrong answer — click to see why▾
Why this is wrong here
The /25 prefix length creates subnets with a block size of 128, starting at 0, 128, 256, etc. 10.0.0.192/25 would be a valid subnet if the network started at 192, but 192 is not a valid /25 boundary because 192 is not a multiple of 128.
★ When this WOULD be the correct answer
This option would be correct if the question asked for the subnet of a host with an address in the range of 10.0.0.192 to 10.0.0.255, such as 10.0.0.200/25. In that case, 10.0.0.192/25 would accurately represent the subnet for that host.
Why candidates choose this
Students often confuse /25 boundaries with /26 boundaries (which increment by 64) or mistakenly think that any address ending in .192 could be a subnet start. The number 192 is a common subnet boundary for /26 or /27, leading to this error.
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
Go deeper
Related to this question
Learn chapter
IPv4 Addressing and Address Classes
Key term
Classless Inter-domain Routing
Classless Inter-domain Routing (CIDR) is a method for allocating IP addresses and routing IP packets more efficiently by replacing the old class-based system with variable-length subnet masks.
Key term
CAN
A CAN (Controller Area Network) is a robust vehicle bus standard designed to allow microcontrollers and devices to communicate with each other without a host computer.
About these practice questions
This 200-301 question is part of Courseiva's 1,389-question bank — original exam-style content with full explanations and wrong-answer analysis, never real exam questions or exam dumps. Learn why practice questions differ from exam dumps →
JA
Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
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.