The valid host range for 192.168.100.65/26 is 192.168.100.65 through 192.168.100.126. This is correct because a /26 subnet mask (255.255.255.192) creates blocks of 64 addresses, and the given host falls into the 64–127 block, where 192.168.100.64 is the network address and 192.168.100.127 is the broadcast address, leaving the addresses between them as usable hosts. On the CCNA 200-301 v2 exam, this question tests your ability to quickly identify subnet boundaries and calculate the valid host range, a fundamental skill for IP addressing and routing. A common trap is assuming the host address itself is the network address—here, .65 is the first usable host, not the network ID. To avoid mistakes, remember the magic number method: for a /26, the block size is 64, so subnets increment by 64 in the last octet (0, 64, 128, 192). A quick memory tip: “Subtract the mask from 256 to find the block size, then the network address is the first multiple of that block below the host IP.”
CCNA Network Infrastructure and Connectivity Practice Question
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 /26 subnet mask divides the IP address space into blocks of 64 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 is configured as 192.168.100.65/26. What is the valid host range for its subnet?
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
192.168.100.65 to 192.168.100.126
A /26 creates blocks of 64 addresses. In plain language, the subnets in the last octet are 0–63, 64–127, 128–191, and 192–255. Because the host address is 192.168.100.65, it belongs to the 64–127 block. In that block, 192.168.100.64 is the network address and 192.168.100.127 is the broadcast address. That leaves 192.168.100.65 through 192.168.100.126 as the valid host range.
This checks whether you can identify both the subnet boundary and the usable range.
Key principle: A /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✓
192.168.100.65 to 192.168.100.126
Why this is correct
This is correct because the subnet is 192.168.100.64/26, leaving .65 through .126 as usable hosts.
Related concept
A /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
✗
192.168.100.64 to 192.168.100.127
Why it's wrong here
This is wrong because those are the network and broadcast addresses, not usable endpoints.
When this WOULD be correct
If the question asked for the entire range of IP addresses in the subnet, including the network and broadcast addresses, then this option would be correct. For example, a question could ask for the range of addresses in the 192.168.100.64/26 subnet, which would include 192.168.100.64 to 192.168.100.127.
✗
192.168.100.1 to 192.168.100.62
Why it's wrong here
This is wrong because that range belongs to a different /26 subnet.
When this WOULD be correct
If the question were to ask for the valid host range of a different subnet, such as 192.168.100.0/26, then option C would be correct, as it would encompass the valid host addresses from 192.168.100.1 to 192.168.100.62 within that subnet.
✗
192.168.100.66 to 192.168.100.127
Why it's wrong here
This is wrong because it excludes one valid host and includes the broadcast address.
When this WOULD be correct
If the question specified a different subnet, such as 192.168.100.64/25, then option D would be correct, as it would represent the valid host range from 192.168.100.66 to 192.168.100.127 for that 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.
✓192.168.100.65 to 192.168.100.126Correct answer▾
Why this is correct
This is correct because the subnet is 192.168.100.64/26, leaving .65 through .126 as usable hosts.
✗192.168.100.64 to 192.168.100.127Wrong answer — click to see why▾
Why this is wrong here
This option is incorrect because it includes the network address (192.168.100.64) and the broadcast address (192.168.100.127) for the subnet, which are not valid host addresses. The valid host range for a /26 subnet starts from 192.168.100.65 to 192.168.100.126.
★ When this WOULD be the correct answer
If the question asked for the entire range of IP addresses in the subnet, including the network and broadcast addresses, then this option would be correct. For example, a question could ask for the range of addresses in the 192.168.100.64/26 subnet, which would include 192.168.100.64 to 192.168.100.127.
Why candidates choose this
Candidates may choose this option because it closely resembles the correct range and includes the starting address of the subnet, leading to confusion about the inclusion of network and broadcast addresses.
✗192.168.100.1 to 192.168.100.62Wrong answer — click to see why▾
Why this is wrong here
Option C is incorrect because it suggests a host range that does not align with the subnet mask /26, which allows for a range of 192.168.100.64 to 192.168.100.127, but excludes the actual valid hosts for the specified IP address.
★ When this WOULD be the correct answer
If the question were to ask for the valid host range of a different subnet, such as 192.168.100.0/26, then option C would be correct, as it would encompass the valid host addresses from 192.168.100.1 to 192.168.100.62 within that subnet.
Why candidates choose this
Candidates may choose this option due to confusion between the subnet mask and the valid host range, leading them to mistakenly believe that the lower range of IPs is valid for a different subnet configuration.
✗192.168.100.66 to 192.168.100.127Wrong answer — click to see why▾
Why this is wrong here
Option D is incorrect because it suggests a host range that starts from 192.168.100.66, which is outside the valid range for the subnet 192.168.100.64/26. The correct range should include all hosts from 192.168.100.65 to 192.168.100.126.
★ When this WOULD be the correct answer
If the question specified a different subnet, such as 192.168.100.64/25, then option D would be correct, as it would represent the valid host range from 192.168.100.66 to 192.168.100.127 for that subnet.
Why candidates choose this
Candidates may choose this option due to confusion between the starting address of the subnet and the usable host addresses, mistakenly believing that addresses immediately following the subnet address are valid.
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 the valid host range.
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. In this question, the IP address 192.168.100.65 with a /26 mask means the first 26 bits are network bits, leaving 6 bits for host addresses. This results in subnets with 64 IP addresses each (2^6 = 64), including network and broadcast addresses.
The /26 subnet mask (255.255.255.192) creates subnets with address ranges incrementing by 64 in the last octet: 0-63, 64-127, 128-191, and 192-255. Since the host IP is 192.168.100.65, it falls into the 64-127 subnet. The first address (192.168.100.64) is the network address, and the last (192.168.100.127) is the broadcast address, leaving 192.168.100.65 through 192.168.100.126 as valid host addresses.
A common exam trap is confusing the network and broadcast addresses as valid hosts or miscalculating subnet boundaries by ignoring the subnet mask’s effect on the address range. Practically, Cisco devices use these subnet boundaries to route traffic correctly and prevent IP conflicts. Understanding subnetting ensures accurate IP allocation and efficient network design.
KKey Concepts to Remember
A /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
The network address is always the first IP in the subnet block and cannot be assigned to a host device.
The broadcast address is always the last IP in the subnet block and is reserved for broadcasting to all hosts in that subnet.
Valid host IP addresses fall between the network and broadcast addresses, excluding both.
Subnet boundaries are determined by incrementing the block size defined by the subnet mask in the last octet.
Cisco devices use subnet masks to correctly route traffic and prevent IP address conflicts within subnets.
Misidentifying subnet boundaries or including network/broadcast addresses as hosts leads to configuration errors.
Understanding subnetting is critical for IP address planning and troubleshooting in Cisco network environments.
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 /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
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 /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses., 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 /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses..
What is the correct answer to this question?
The correct answer is: 192.168.100.65 to 192.168.100.126 — A /26 creates blocks of 64 addresses. In plain language, the subnets in the last octet are 0–63, 64–127, 128–191, and 192–255. Because the host address is 192.168.100.65, it belongs to the 64–127 block. In that block, 192.168.100.64 is the network address and 192.168.100.127 is the broadcast address. That leaves 192.168.100.65 through 192.168.100.126 as the valid host range.
This checks whether you can identify both the subnet boundary and the usable range.
What should I do if I get this 200-301 question wrong?
Review a /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses., then practise related 200-301 questions on the same topic to reinforce the concept.
What is the key concept behind this question?
A /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
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