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Network Infrastructure and ConnectivityhardMultiple ChoiceObjective-mapped

CCNA Network Infrastructure and Connectivity Practice Question

A host is configured as 192.168.100.65/26. What is the valid host range for its subnet?

⚠ Common exam trap

Remember to exclude the network and broadcast addresses when determining the valid host 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

192.168.100.65 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.

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.

  • 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

    A /26 subnet mask (255.255.255.192) creates a block size of 64 addresses, so the subnet containing 192.168.100.65 starts at 192.168.100.64, not 192.168.100.0. The valid host range is therefore 192.168.100.65 to 192.168.100.126, because .64 is the network address and .127 is the broadcast address. This option is tempting because it describes the valid host range for a /26 subnet that begins at 192.168.100.0, which would be correct if the host address were between .1 and .62, but the given host .65 falls into the next subnet block.

    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?”

Visual reference

192.168.1.0 /24 256 addresses (254 usable) 192.168.1.0 /25 Subnet A 128 addr (126 usable) 192.168.1.128 /25 Subnet B 128 addr (126 usable) Borrowing 1 bit from host portion creates 2 subnets (/25)

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JA

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.