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 /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, creating subnets with 16 IP addresses each.. 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 router interface is configured with 192.0.2.97/28. What is the network address of the subnet?
192.0.2.80
Why wrong: That is part of a different /28 block.
192.0.2.96
Correct. It is the first address in the 192.0.2.96/28 subnet.
192.0.2.111
Why wrong: That is the broadcast address of this subnet.
192.0.2.112
Why wrong: That is the next /28 network.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
192.0.2.96
A /28 uses blocks of 16 addresses. The block containing .97 runs from .96 through .111, so .96 is the network address.
Key principle: A /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, creating subnets with 16 IP addresses each.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
192.0.2.80
Why it's wrong here
That is part of a different /28 block.
When this WOULD be correct
If the question were to ask for the network address of a subnet with a different subnet mask, such as /26, which has a block size of 64, then 192.0.2.80 could be the correct answer as it would represent the network address for the range 192.0.2.64 to 192.0.2.127.
192.0.2.96
Why this is correct
Correct. It is the first address in the 192.0.2.96/28 subnet.
Related concept
A /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, creating subnets with 16 IP addresses each.
192.0.2.111
Why it's wrong here
That is the broadcast address of this subnet.
When this WOULD be correct
If the question were to ask for the broadcast address of the subnet 192.0.2.96/28, then 192.0.2.111 would be the correct answer, as it is the last address in that subnet range.
192.0.2.112
Why it's wrong here
That is the next /28 network.
When this WOULD be correct
If the question asked for the broadcast address of the subnet instead of the network address, then 192.0.2.112 would be correct, as it is the last address in the subnet range for 192.0.2.96/28.
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
Correct. It is the first address in the 192.0.2.96/28 subnet.
Why this is wrong here
Option A (192.0.2.80) is incorrect because the subnet mask /28 indicates a block size of 16, making the valid network addresses range from 192.0.2.96 to 192.0.2.111. The network address for this subnet is 192.0.2.96.
★ When this WOULD be the correct answer
If the question were to ask for the network address of a subnet with a different subnet mask, such as /26, which has a block size of 64, then 192.0.2.80 could be the correct answer as it would represent the network address for the range 192.0.2.64 to 192.0.2.127.
Why candidates choose this
Candidates might choose this option due to a misunderstanding of subnetting, mistakenly calculating the network address based on an incorrect block size or misinterpreting the range of addresses within the subnet.
Why this is wrong here
The address 192.0.2.111 is not a valid network address for the subnet defined by 192.0.2.97/28, as it falls outside the range of usable addresses for that subnet, which spans from 192.0.2.96 to 192.0.2.111.
★ When this WOULD be the correct answer
If the question were to ask for the broadcast address of the subnet 192.0.2.96/28, then 192.0.2.111 would be the correct answer, as it is the last address in that subnet range.
Why candidates choose this
Candidates may confuse the network address with the last usable address in the subnet, leading them to mistakenly select 192.0.2.111, which is the highest usable IP address in that range.
Why this is wrong here
Option D, 192.0.2.112, is wrong because it does not represent the network address for the subnet defined by the IP address 192.0.2.97/28. The correct network address is 192.0.2.96, which is the first address in the subnet range.
★ When this WOULD be the correct answer
If the question asked for the broadcast address of the subnet instead of the network address, then 192.0.2.112 would be correct, as it is the last address in the subnet range for 192.0.2.96/28.
Why candidates choose this
Candidates may choose this option due to confusion between network and broadcast addresses, as they might remember that the last address in a subnet is often significant and mistakenly associate it with the question.
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
Be careful not to confuse the given IP address or the broadcast address with the network address.
Detailed technical explanation
Subnetting divides an IP network into smaller, manageable segments called subnets, each identified by a network address. A /28 subnet mask means 28 bits are fixed for the network portion, leaving 4 bits for host addresses, resulting in 16 IP addresses per subnet (2^4 = 16). The subnet mask for /28 is 255.255.255.240, which helps determine the range of addresses in each subnet block. To find the network address for 192.0.2.97/28, you calculate the subnet block size (16 addresses) and identify which block contains the IP. The blocks start at multiples of 16: .0, .16, .32, .48, .64, .80, .96, .112, etc. Since 97 falls between 96 and 111, the network address is 192.0.2.96. This address represents the subnet's base, with the first usable host starting at .97 and the broadcast at .111. A common exam trap is confusing the network address with the broadcast or next subnet address. For example, .111 is the broadcast address for the 192.0.2.96/28 subnet, not the network address. Misidentifying these can lead to incorrect subnetting decisions. Practically, Cisco devices use the network address to route traffic correctly and avoid IP conflicts within subnets.
Key takeaway
A /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, creating subnets with 16 IP addresses each.
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 /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, creating subnets with 16 IP addresses each., 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 /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, creating subnets with 16 IP addresses each..
The correct answer is: 192.0.2.96 — A /28 uses blocks of 16 addresses. The block containing .97 runs from .96 through .111, so .96 is the network address.
Review a /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, creating subnets with 16 IP addresses each., then practise related 200-301 questions on the same topic to reinforce the concept.
A /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, creating subnets with 16 IP addresses each.
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Last reviewed: May 17, 2026
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