The answer is 192.168.200.97 to 192.168.200.110. This is correct because a /28 prefix length means the subnet has 16 total addresses, calculated as 2^(32-28) = 16. The network address 192.168.200.96 starts the block, which ends at 192.168.200.111, but the first address is reserved as the network identifier and the last as the broadcast address, leaving the usable host range between them. On the CCNA 200-301 v2 exam, this question tests your ability to find usable host range from network address and prefix, a core subnetting skill that often appears in both multiple-choice and simulation formats. A common trap is forgetting to exclude both boundary addresses or misidentifying the block size—remember that a /28 gives you 14 usable hosts, not 16. Memory tip: for any prefix, subtract the network and broadcast addresses from the total block size to get the usable range.
CCNA Network Infrastructure and Connectivity Practice Question
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 /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, resulting in 16 total IP addresses per subnet.. 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 subnet uses network address 192.168.200.96/28. Which range contains the usable host addresses?
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
192.168.200.97 to 192.168.200.110
A /28 block contains 16 addresses. In practical terms, the block starting at 192.168.200.96 runs through 192.168.200.111. The first address is the network address and the last address is the broadcast address. That means the usable host range is 192.168.200.97 through 192.168.200.110.
This question checks whether you can calculate the correct block and then exclude the two reserved boundary addresses.
Key principle: A /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, resulting in 16 total IP addresses per subnet.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✓
192.168.200.97 to 192.168.200.110
Why this is correct
This is correct because .96 is the network and .111 is the broadcast.
Related concept
A /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, resulting in 16 total IP addresses per subnet.
✗
192.168.200.96 to 192.168.200.111
Why it's wrong here
This is wrong because it includes the network and broadcast addresses.
When this WOULD be correct
If the question asked for the entire range of addresses in the subnet, including the network and broadcast addresses, then this option would be correct. For example, a question could specify 'list all addresses in the subnet' rather than just usable host addresses.
✗
192.168.200.98 to 192.168.200.111
Why it's wrong here
This is wrong because it excludes one valid host and includes the broadcast.
When this WOULD be correct
If the question asked for the range of usable host addresses for a subnet with a different CIDR notation, such as 192.168.200.96/27, then the range 192.168.200.98 to 192.168.200.111 would be correct, as it would include valid host addresses in that larger subnet.
✗
192.168.200.81 to 192.168.200.94
Why it's wrong here
This is wrong because that range belongs to a different subnet.
When this WOULD be correct
This option would be correct if the subnet were defined as 192.168.200.80/28, where the usable host addresses would indeed range from 192.168.200.81 to 192.168.200.94, making option D valid in that context.
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.
✓192.168.200.97 to 192.168.200.110Correct answer▾
Why this is correct
This is correct because .96 is the network and .111 is the broadcast.
✗192.168.200.96 to 192.168.200.111Wrong answer — click to see why▾
Why this is wrong here
This range includes the network address (.96) and the broadcast address (.111), which cannot be assigned to hosts. Usable host addresses must exclude these two addresses.
★ When this WOULD be the correct answer
If the question asked for the entire range of addresses in the subnet, including the network and broadcast addresses, then this option would be correct. For example, a question could specify 'list all addresses in the subnet' rather than just usable host addresses.
Why candidates choose this
Students often mistakenly include the network and broadcast addresses in the usable range because they forget that these addresses are reserved and not assignable to hosts.
✗192.168.200.98 to 192.168.200.111Wrong answer — click to see why▾
Why this is wrong here
This range starts at .98, which excludes the valid host .97, and ends at .111, which includes the broadcast address. The correct usable range is .97 to .110.
★ When this WOULD be the correct answer
If the question asked for the range of usable host addresses for a subnet with a different CIDR notation, such as 192.168.200.96/27, then the range 192.168.200.98 to 192.168.200.111 would be correct, as it would include valid host addresses in that larger subnet.
Why candidates choose this
A student might think the first usable address is .98 because they incorrectly calculate the next address after the network address, or they might include the broadcast address due to confusion.
✗192.168.200.81 to 192.168.200.94Wrong answer — click to see why▾
Why this is wrong here
This range (192.168.200.81 to .94) belongs to a different subnet. For a /28 subnet starting at .96, the valid host range is .97 to .110. This range is from a previous subnet (e.g., 192.168.200.80/28).
★ When this WOULD be the correct answer
This option would be correct if the subnet were defined as 192.168.200.80/28, where the usable host addresses would indeed range from 192.168.200.81 to 192.168.200.94, making option D valid in that context.
Why candidates choose this
A student might miscalculate the subnet boundaries or confuse the network address with a different subnet, leading them to select a range from an adjacent subnet.
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
Common exam trap: answer the scenario, not the keyword
Remember to exclude the network and broadcast addresses when determining usable host ranges.
Detailed technical explanation
How to think about this question
Subnetting divides a larger IP network into smaller, manageable segments by borrowing bits from the host portion of the address to create subnetworks. A /28 subnet mask means 28 bits are fixed for the network portion, leaving 4 bits for host addresses. This results in 2^4 = 16 total IP addresses per subnet, including the network and broadcast addresses.
The network address is always the first IP in the subnet block, and the broadcast address is the last IP. For 192.168.200.96/28, the block ranges from 192.168.200.96 to 192.168.200.111. The usable host addresses exclude these two reserved addresses, so they range from 192.168.200.97 to 192.168.200.110. This rule applies universally in IPv4 subnetting and is critical for proper IP address allocation.
A common exam trap is confusing the network and broadcast addresses with usable hosts, leading to off-by-one errors in the range. Practically, Cisco devices rely on this addressing scheme for routing and access control. Misidentifying usable hosts can cause connectivity issues or misconfigured ACLs, so understanding the boundary addresses is essential for both exams and real-world networking.
KKey Concepts to Remember
A /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, resulting in 16 total IP addresses per subnet.
The first IP address in a subnet block is always the network address and cannot be assigned to a host.
The last IP address in a subnet block is always the broadcast address and is reserved for broadcast traffic.
Usable host addresses in a subnet exclude the network and broadcast addresses, reducing the total usable hosts by two.
Calculating the subnet block range requires understanding the subnet mask and incrementing the block size accordingly.
Cisco devices use subnetting rules to determine valid host addresses for routing and access control configurations.
Misidentifying network or broadcast addresses as usable hosts leads to invalid IP assignments and connectivity issues.
Subnetting questions often test the ability to calculate address ranges and exclude reserved addresses accurately.
TExam Day Tips
→Watch for words such as best, first, most likely and least administrative effort.
→Review why wrong options are wrong, not only why the correct option is correct.
Key takeaway
A /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, resulting in 16 total IP addresses per subnet.
Real-world example
How this comes up in practice
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.
Related glossary terms
Concepts from this question explained
These glossary pages explain the core terms tested in this 200-301 question in full detail.
Review a /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, resulting in 16 total IP addresses per subnet., then practise related 200-301 questions on the same topic to reinforce the concept.
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, resulting in 16 total IP addresses per subnet..
What is the correct answer to this question?
The correct answer is: 192.168.200.97 to 192.168.200.110 — A /28 block contains 16 addresses. In practical terms, the block starting at 192.168.200.96 runs through 192.168.200.111. The first address is the network address and the last address is the broadcast address. That means the usable host range is 192.168.200.97 through 192.168.200.110.
This question checks whether you can calculate the correct block and then exclude the two reserved boundary addresses.
What should I do if I get this 200-301 question wrong?
Review a /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, resulting in 16 total IP addresses per subnet., then practise related 200-301 questions on the same topic to reinforce the concept.
What is the key concept behind this question?
A /28 subnet mask allocates 28 bits for the network and 4 bits for host addresses, resulting in 16 total IP addresses per subnet.
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