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Network Infrastructure and ConnectivityhardMultiple ChoiceObjective-mapped

CCNA Network Infrastructure and Connectivity Practice Question

A host address is 192.168.50.158/27. Which address is the network address of its subnet?

⚠ Common exam trap

Ensure you calculate the correct block range for the subnet mask given, not just any multiple of the block size.

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

192.168.50.128

A /27 subnet has a block size of 32. In practical terms, the relevant ranges in the last octet are 0-31, 32-63, 64-95, 96-127, 128-159, and so on. Because 158 falls in the 128-159 block, the network address is 192.168.50.128. This is a block-identification question. Once you identify the correct /27 block, the network address is the first address in that 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.

  • 192.168.50.128

    Why this is correct

    With a /27 prefix, the subnet block size is 32 addresses. The host .158 falls within the range .128–.159, and the network address is always the first address of that block, so it is 192.168.50.128. This is the lowest address that identifies this subnet.

  • 192.168.50.159

    Why it's wrong here

    192.168.50.159 is actually the directed broadcast address for the subnet that starts at .128, not its network address. In a /27 block, the first address is the network and the last is the broadcast, so .158 is a valid host while .159 is reserved for broadcast traffic. Choosing .159 as the network address ignores the distinction between network and broadcast addresses.

    When this WOULD be correct

    If the question asked for the last usable host address in the subnet 192.168.50.128/27, then 192.168.50.159 would be the correct answer, as it is the highest address before the broadcast address.

  • 192.168.50.160

    Why it's wrong here

    A /27 subnet with a block size of 32 puts 192.168.50.160 at the very beginning of the next network segment, which covers .160–.191. Because the host .158 does not fall in that range, .160 cannot be its network address. This answer confuses the boundary of the next subnet with the one actually containing the host.

    When this WOULD be correct

    In a different question where the subnet mask is /26 and the host address is 192.168.50.158, the network address would be 192.168.50.128, making 192.168.50.160 a valid host address within that subnet.

  • 192.168.50.96

    Why it's wrong here

    This address, 192.168.50.96, is the network address for the /27 subnet spanning .96–.127. However, the host .158 lies well beyond that range, so this is an earlier subnet entirely. Mistaking it for the network would mean miscounting the subnet boundaries, since the correct block is the one that numerically contains the host.

    When this WOULD be correct

    If the question asked for a network address within a different subnet using a /25 mask, such as 192.168.50.96/25, then option D would be correct as it would represent the network address 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.

192.168.50.128Correct answer

Why this is correct

With a /27 prefix, the subnet block size is 32 addresses. The host .158 falls within the range .128–.159, and the network address is always the first address of that block, so it is 192.168.50.128. This is the lowest address that identifies this subnet.

192.168.50.159Wrong answer — click to see why

Why this is wrong here

192.168.50.159 is the broadcast address for the 192.168.50.128/27 subnet, not the network address. The broadcast address is used to send packets to all hosts in the subnet and cannot be assigned to a host.

★ When this WOULD be the correct answer

If the question asked for the last usable host address in the subnet 192.168.50.128/27, then 192.168.50.159 would be the correct answer, as it is the highest address before the broadcast address.

Why candidates choose this

Students might confuse the last usable host address (158) with the broadcast address (159) or think that the network address is the highest number in the range.

192.168.50.160Wrong answer — click to see why

Why this is wrong here

192.168.50.160 is the network address of the next /27 subnet (160-191), not the subnet containing 158. The /27 mask creates subnets with a block size of 32, so the subnet boundaries are multiples of 32.

★ When this WOULD be the correct answer

In a different question where the subnet mask is /26 and the host address is 192.168.50.158, the network address would be 192.168.50.128, making 192.168.50.160 a valid host address within that subnet.

Why candidates choose this

A student might miscalculate the subnet boundaries by using a block size of 32 but starting from 0 incorrectly, or they might think 158 is closer to 160 than to 128.

192.168.50.96Wrong answer — click to see why

Why this is wrong here

192.168.50.96 is the network address of the 96-127 /27 subnet, which is an earlier subnet. The address 158 falls in the 128-159 range, not the 96-127 range.

★ When this WOULD be the correct answer

If the question asked for a network address within a different subnet using a /25 mask, such as 192.168.50.96/25, then option D would be correct as it would represent the network address for that subnet.

Why candidates choose this

Students might incorrectly calculate the subnet by using a block size of 32 but starting from 96 instead of 128, or they might confuse the subnet mask and think the network address is 96.

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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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.