This 200-301 practice question tests your understanding of network infrastructure and connectivity. 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 divides an IP network into blocks of 32 addresses, including network and broadcast 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 address is 192.168.88.66/27. Which address is the network address of the subnet?
192.168.88.32
Why wrong: This is wrong because that is the previous /27 block.
192.168.88.64
This is correct because .66 is in the 64-95 /27 subnet.
192.168.88.95
Why wrong: This is wrong because .95 is the broadcast address of the block, not the network address.
192.168.88.96
Why wrong: This is wrong because that is the start of the next /27 block.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
192.168.88.64
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.
Key principle: A /27 subnet mask divides an IP network into blocks of 32 addresses, including network and broadcast addresses.
Answer analysis
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
This is wrong because that is the previous /27 block.
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
This is correct because .66 is in the 64-95 /27 subnet.
Related concept
A /27 subnet mask divides an IP network into blocks of 32 addresses, including network and broadcast addresses.
192.168.88.95
Why it's wrong here
This is wrong because .95 is the broadcast address of the block, not the network address.
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
This is wrong because that is the start of the next /27 block.
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
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 .66 is in the 64-95 /27 subnet.
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.
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.
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?”
Common exam traps
Avoid assuming the host address is in the first or last subnet without calculating the correct range.
Detailed technical explanation
Subnetting is a fundamental concept in IP networking that divides a larger network into smaller, manageable subnetworks. Each subnet has a unique network address, which identifies the subnet itself, and a range of host addresses within that subnet. The subnet mask, such as /27 in this question, determines the size of each subnet and how the IP address space is divided. A /27 mask means 27 bits are fixed for the network portion, leaving 5 bits for host addresses, resulting in 32 IP addresses per subnet block. To find the network address for an IP with a /27 mask, you calculate the block size as 2^(32-27) = 32 addresses. The IP address 192.168.88.66 falls within the range 192.168.88.64 to 192.168.88.95. The network address is always the first address in the block, which is 192.168.88.64. This calculation is critical in Cisco networking for proper subnet design, routing, and access control. A common exam trap is confusing the network address with the broadcast address or the start of adjacent subnets. For example, 192.168.88.95 is the broadcast address for the 64-95 block, not the network address. Misidentifying the network address can lead to incorrect subnetting decisions and routing errors. Practically, Cisco devices use the network address to identify subnets in routing tables and ACLs, so precise calculation is essential.
Key takeaway
A /27 subnet mask divides an IP network into blocks of 32 addresses, including network and broadcast 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 /27 subnet mask divides an IP network into blocks of 32 addresses, including network and broadcast 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 /27 subnet mask divides an IP network into blocks of 32 addresses, including network and broadcast addresses..
The correct answer is: 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.
Review a /27 subnet mask divides an IP network into blocks of 32 addresses, including network and broadcast addresses., then practise related 200-301 questions on the same topic to reinforce the concept.
A /27 subnet mask divides an IP network into blocks of 32 addresses, including network and broadcast addresses.
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Last reviewed: May 17, 2026
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