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 /27 subnet mask uses 27 bits for the network portion, leaving 5 bits for host addresses, resulting in 32 IP addresses per subnet block.. 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 address is 192.168.50.158/27. Which address is the network address of its subnet?
192.168.50.128
This is correct because 158 belongs to the 128-159 /27 subnet.
192.168.50.159
Why wrong: This is wrong because .159 is the broadcast address of that subnet.
192.168.50.160
Why wrong: This is wrong because .160 starts the next /27 subnet.
192.168.50.96
Why wrong: This is wrong because that is the start of an earlier /27 subnet.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
192.168.50.128
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.
Key principle: A /27 subnet mask uses 27 bits for the network portion, leaving 5 bits for host addresses, resulting in 32 IP addresses per subnet block.
Answer analysis
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
This is correct because 158 belongs to the 128-159 /27 subnet.
Related concept
A /27 subnet mask uses 27 bits for the network portion, leaving 5 bits for host addresses, resulting in 32 IP addresses per subnet block.
192.168.50.159
Why it's wrong here
This is wrong because .159 is the broadcast address of that subnet.
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
This is wrong because .160 starts the next /27 subnet.
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 is wrong because that is the start of an earlier /27 subnet.
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
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
This is correct because 158 belongs to the 128-159 /27 subnet.
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.
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.
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?”
Common exam traps
Ensure you calculate the correct block range for the subnet mask given, not just any multiple of the block size.
Detailed technical explanation
Subnetting is a fundamental concept in IP networking that divides a larger network into smaller, manageable segments called subnets. Each subnet has a network address, a range of usable host addresses, and a broadcast address. The subnet mask determines the size of each subnet by defining how many bits are used for the network portion versus the host portion of the IP address. In this question, the /27 mask means 27 bits are fixed for the network, leaving 5 bits for host addresses, resulting in subnets with 32 IP addresses each. To find the network address of a host IP within a /27 subnet, you calculate the block size as 2^(32-27) = 32 addresses per subnet. The subnets increment in blocks of 32 in the last octet: 0-31, 32-63, 64-95, 96-127, 128-159, and so forth. Since the host IP 192.168.50.158 falls between 128 and 159, the network address is the first IP in that range, 192.168.50.128. This address identifies the subnet and is used by routers and switches to route traffic correctly. A common exam trap is confusing the network address with the broadcast or next subnet address. For example, 192.168.50.159 is the broadcast address for the 192.168.50.128/27 subnet, not the network address. Similarly, 192.168.50.160 is the start of the next subnet. Understanding the block size and correctly identifying the first address in the block as the network address is critical for subnetting questions on the CCNA exam and practical network design.
Key takeaway
A /27 subnet mask uses 27 bits for the network portion, leaving 5 bits for host addresses, resulting in 32 IP addresses per subnet block.
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 /27 subnet mask uses 27 bits for the network portion, leaving 5 bits for host addresses, resulting in 32 IP addresses per subnet block., then practise related 200-301 questions on the same topic to reinforce the concept.
Network Infrastructure and Connectivity — study guide chapter
Learn the concepts, then practise the questions
Network Infrastructure and Connectivity practice questions
Targeted practice on this topic area only
All 200-301 questions
1,819 questions across all exam domains
CCNA 200-301 v2 study guide
Full concept coverage aligned to exam objectives
200-301 practice test guide
How to use practice tests most effectively before exam day
Related practice questions
Use these pages to review the topic behind this question. This is how one missed question becomes focused revision.
Practise 200-301 questions linked to Network Infrastructure and Connectivity.
Practise 200-301 questions linked to Switching and Network Access.
Practise 200-301 questions linked to IP Routing.
Practise 200-301 questions linked to Network Services and Security.
Practise 200-301 questions linked to AI and Network Operations.
Practise IPv4 subnetting, CIDR, masks, host ranges and subnet selection.
Practise OSPF neighbours, router IDs, metrics, areas and routing-table interpretation.
Practise VLANs, access ports, trunks, allowed VLANs and switching scenarios.
Practise spanning tree, root bridge election, port roles and STP troubleshooting.
Practise LACP, PAgP, port-channel behaviour and bundle requirements.
Practise standard and extended ACLs, permit/deny logic and traffic filtering.
Practise static NAT, dynamic NAT, PAT and inside/outside address translation.
Practice this exam
Short sessions build daily habit. Longer sessions build exam-day stamina. Try a timed session to simulate real conditions.
FAQ
Network Infrastructure and Connectivity — This question tests Network Infrastructure and Connectivity — A /27 subnet mask uses 27 bits for the network portion, leaving 5 bits for host addresses, resulting in 32 IP addresses per subnet block..
The correct answer is: 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.
Review a /27 subnet mask uses 27 bits for the network portion, leaving 5 bits for host addresses, resulting in 32 IP addresses per subnet block., then practise related 200-301 questions on the same topic to reinforce the concept.
A /27 subnet mask uses 27 bits for the network portion, leaving 5 bits for host addresses, resulting in 32 IP addresses per subnet block.
About these practice questions
Courseiva creates original exam-style practice questions with explanations and wrong-answer analysis. It does not publish real exam questions, exam dumps, or protected exam content. Learn why practice questions differ from exam dumps →
Last reviewed: May 17, 2026
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.
Question Discussion
Share a tip, memory trick, or ask about the reasoning behind this question. Do not post real exam questions, leaked content, braindumps, or copyrighted exam material. Comments are moderated and may be removed without notice.
Sign in to join the discussion.