CCNA Network Infrastructure and Connectivity Practice Question
A host address is 192.168.88.66/27. Which address is the network address of the subnet?
⚠ Common exam trap
Avoid assuming the host address is in the first or last subnet without calculating the correct range.
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
✓
192.168.88.64
A /27 subnet has a block size of 32. In practical terms, the fourth-octet ranges are 0-31, 32-63, 64-95, and so on. Because 66 falls within the 64-95 block, the network address is 192.168.88.64. This is a classic subnet-boundary calculation. The key step is identifying the correct block first.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
192.168.88.32
Why it's wrong here
With a /27 prefix (255.255.255.224), each subnet contains 32 addresses. The subnets are sequential: 0-31, 32-63, 64-95, etc. Because the host address .66 falls in the 64-95 block, the network address must be .64. The address .32 is the network identifier for the previous block (32-63), which does not encompass .66, so it is incorrect.
When this WOULD be correct
In a different question where the subnet mask is /26 (255.255.255.192) and the host address is 192.168.88.66, the network address would be 192.168.88.0, and if the question asked for the network address of a subnet that starts at 192.168.88.32, then option A would be correct.
- ✓
192.168.88.64
Why this is correct
The /27 mask creates 32-address subnets, and the range from .64 to .95 is one such subnet. Host .66 is within this range, so the network address is the first address of the block, 192.168.88.64. All hosts in this subnet share the same network bits, and .64 is the all-zero host portion for that range.
- ✗
192.168.88.95
Why it's wrong here
In a /27 subnet, the final address in each 32-address block is the directed broadcast address. The block 64-95 ends with .95, which is reserved as the broadcast for that subnet; it cannot be assigned to a host nor represent the network. Since .66 is a host, its network address is the initial address .64, not the broadcast .95.
When this WOULD be correct
In a different question where the subnet mask is /25 and the host address is 192.168.88.95, this address could be the correct answer as it would then represent the last usable host address in that subnet, which ranges from 192.168.88.64 to 192.168.88.95.
- ✗
192.168.88.96
Why it's wrong here
The next /27 subnet after 64-95 starts at .96 and spans .96-127. A host with address .66 resides in the 64-95 subnet, so its network address is .64. The address .96 is the network identifier of a different, higher subnet, and does not share the same network portion as .66. Therefore, it is not the correct network address.
When this WOULD be correct
In a different scenario where the subnet mask was changed to /26 (255.255.255.192), the network address for the range starting at 192.168.88.64 would be 192.168.88.64, and the next subnet would start at 192.168.88.64 + 64 = 192.168.88.128. In this case, 192.168.88.96 could be a valid address within a different subnet.
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.
✓192.168.88.64Correct answer▾
Why this is correct
The /27 mask creates 32-address subnets, and the range from .64 to .95 is one such subnet. Host .66 is within this range, so the network address is the first address of the block, 192.168.88.64. All hosts in this subnet share the same network bits, and .64 is the all-zero host portion for that range.
✗192.168.88.32Wrong answer — click to see why▾
Why this is wrong here
192.168.88.32 is the network address of the previous /27 subnet (32-63). Since the host address 192.168.88.66 falls in the 64-95 range, the correct network address is 192.168.88.64, not 192.168.88.32.
★ When this WOULD be the correct answer
In a different question where the subnet mask is /26 (255.255.255.192) and the host address is 192.168.88.66, the network address would be 192.168.88.0, and if the question asked for the network address of a subnet that starts at 192.168.88.32, then option A would be correct.
Why candidates choose this
A student might miscalculate the subnet boundaries by using a wrong block size or misidentifying the subnet increment. For example, they might think the block size is 16 instead of 32, leading them to choose 32 as the network address.
✗192.168.88.95Wrong answer — click to see why▾
Why this is wrong here
192.168.88.95 is the broadcast address for the /27 subnet that starts at 192.168.88.64. The broadcast address is the last address in the subnet, used to send packets to all hosts in that subnet, not the network address.
★ When this WOULD be the correct answer
In a different question where the subnet mask is /25 and the host address is 192.168.88.95, this address could be the correct answer as it would then represent the last usable host address in that subnet, which ranges from 192.168.88.64 to 192.168.88.95.
Why candidates choose this
Students often confuse the broadcast address with the network address because both are boundary addresses. They might think the last address is the network address, especially when they remember that the subnet includes addresses from 64 to 95.
✗192.168.88.96Wrong answer — click to see why▾
Why this is wrong here
192.168.88.96 is the network address of the next /27 subnet (96-127). The host 192.168.88.66 is not in that range; it belongs to the subnet starting at 192.168.88.64.
★ When this WOULD be the correct answer
In a different scenario where the subnet mask was changed to /26 (255.255.255.192), the network address for the range starting at 192.168.88.64 would be 192.168.88.64, and the next subnet would start at 192.168.88.64 + 64 = 192.168.88.128. In this case, 192.168.88.96 could be a valid address within a different subnet.
Why candidates choose this
A student might incorrectly round up the host address to the next multiple of 32 (96) instead of rounding down to the previous multiple (64). This is a common mistake when calculating network addresses without careful division.
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?”
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Related to this question
Learn chapter
IPv4 Addressing and Address Classes
Key term
Block size
Block size is the amount of data, measured in bytes, that a system reads or writes in a single operation, affecting storage efficiency and network performance.
Key term
Network address
A network address is a unique identifier that allows devices to find and communicate with each other on a computer network.
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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.