Quick Answer
The answer is 172.16.100.223. This is the correct broadcast address because the /27 subnet mask uses a block size of 32 in the fourth octet, meaning the subnet boundaries fall at 0, 32, 64, 96, 128, 160, 192, and 224. Since the host address 172.16.100.222 falls within the 192–223 block, the broadcast is the last address in that block, which is 172.16.100.223. On the CCNA 200-301 v2 exam, this classic subnet-boundary question tests your ability to quickly place an IP into the correct subnet block and identify the broadcast as the final address in that range—a skill essential for troubleshooting and routing. A common trap is forgetting that the broadcast address is not the next network’s first address but the last usable address within the current subnet. For a quick memory tip, remember that the broadcast address is always the subnet’s last address, and you can find it by adding the block size minus one to the subnet’s starting octet value.
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: subnetting divides an IP network into smaller address blocks by borrowing bits from the host portion to create subnet masks.. 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 with IP address 172.16.100.222/27. Which address is the broadcast address for its subnet?
172.16.100.191
Why wrong: This is wrong because .191 is the broadcast of the previous /27 block.
172.16.100.223
This is correct because .222 is in the 192–223 /27 block.
172.16.100.224
Why wrong: This is wrong because .224 is the network address of the next block.
172.16.100.255
Why wrong: This is wrong because .255 is not the broadcast of this /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
172.16.100.223
A /27 uses address blocks of 32. In practical terms, the fourth-octet ranges are 0–31, 32–63, 64–95, 96–127, 128–159, 160–191, 192–223, and 224–255. Since 222 falls inside the 192–223 block, the broadcast address is the last address in that block, which is 172.16.100.223. This is a classic subnet-boundary question because it tests whether you can place a host in the correct block and then identify the final address in that block as the broadcast.
Key principle: Subnetting divides an IP network into smaller address blocks by borrowing bits from the host portion to create subnet masks.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
172.16.100.191
Why it's wrong here
This is wrong because .191 is the broadcast of the previous /27 block.
When this WOULD be correct
In a different question setup where the subnet mask was /26, the subnet would be 172.16.100.128/26, and the broadcast address would be 172.16.100.191. This would make option A the correct answer in that context.
172.16.100.223
Why this is correct
This is correct because .222 is in the 192–223 /27 block.
Related concept
Subnetting divides an IP network into smaller address blocks by borrowing bits from the host portion to create subnet masks.
172.16.100.224
Why it's wrong here
This is wrong because .224 is the network address of the next block.
When this WOULD be correct
In a different question setup where the subnet mask is /28 (255.255.255.240) and the host IP is 172.16.100.224, the broadcast address would be 172.16.100.239. In this case, option C would be the correct answer.
172.16.100.255
Why it's wrong here
This is wrong because .255 is not the broadcast of this /27 subnet.
When this WOULD be correct
If the question specified a subnet mask of /24 instead of /27, then 172.16.100.255 would be the broadcast address for the subnet 172.16.100.0/24, making this option correct in that context.
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 .222 is in the 192–223 /27 block.
Why this is wrong here
172.16.100.191 is the broadcast address of the previous /27 subnet (172.16.100.160/27), not the subnet containing .222. The host .222 is in the 172.16.100.192/27 subnet, so its broadcast is .223.
★ When this WOULD be the correct answer
In a different question setup where the subnet mask was /26, the subnet would be 172.16.100.128/26, and the broadcast address would be 172.16.100.191. This would make option A the correct answer in that context.
Why candidates choose this
Students often miscalculate subnet boundaries or confuse the broadcast of a different subnet. They might think .191 is the broadcast because it is a common broadcast address in a /27 starting at .160.
Why this is wrong here
172.16.100.224 is the network address of the next /27 subnet (172.16.100.224/27), not a broadcast address. Broadcast addresses are always the last address in a subnet, not the first.
★ When this WOULD be the correct answer
In a different question setup where the subnet mask is /28 (255.255.255.240) and the host IP is 172.16.100.224, the broadcast address would be 172.16.100.239. In this case, option C would be the correct answer.
Why candidates choose this
A student might mistakenly think that .224 is the broadcast because it is the next multiple of 32, but they forget that the broadcast is one less than the next network address.
Why this is wrong here
172.16.100.255 is the broadcast address of the entire /24 subnet (172.16.100.0/24), not the /27 subnet containing .222. The /27 subnet has a smaller range, so its broadcast is .223.
★ When this WOULD be the correct answer
If the question specified a subnet mask of /24 instead of /27, then 172.16.100.255 would be the broadcast address for the subnet 172.16.100.0/24, making this option correct in that context.
Why candidates choose this
Students often default to the classful broadcast address (ending in .255) without considering the subnet mask. They may assume a /27 still uses the classful broadcast.
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
Avoid assuming the broadcast address is always .255 or miscalculating subnet ranges.
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 defined range of IP addresses determined by the subnet mask. A /27 subnet mask means 27 bits are used for the network portion, leaving 5 bits for host addresses. This results in 32 IP addresses per subnet block, including the network and broadcast addresses. The subnet mask 255.255.255.224 reflects this division, where the last octet increments in steps of 32 (0, 32, 64, etc.). To determine the broadcast address for a host IP like 172.16.100.222/27, you first identify which subnet block the IP belongs to. The blocks for /27 subnets in the fourth octet are 0–31, 32–63, 64–95, 96–127, 128–159, 160–191, 192–223, and 224–255. Since 222 falls within the 192–223 block, the broadcast address is the last IP in that block, 172.16.100.223. This address is reserved for broadcast traffic to all hosts in that subnet. A common exam trap is confusing the broadcast address with the network address or the broadcast address of adjacent subnets. For example, 172.16.100.191 is the broadcast for the previous subnet block (160–191), and 172.16.100.224 is the network address of the next block. Misidentifying these addresses can lead to incorrect subnetting answers and network misconfigurations. Cisco routers and switches rely on accurate subnet and broadcast address calculations to forward traffic correctly and maintain network segmentation.
Key takeaway
Subnetting divides an IP network into smaller address blocks by borrowing bits from the host portion to create subnet masks.
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 subnetting divides an IP network into smaller address blocks by borrowing bits from the host portion to create subnet masks., then practise related 200-301 questions on the same topic to reinforce the concept.
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FAQ
Network Infrastructure and Connectivity — This question tests Network Infrastructure and Connectivity — Subnetting divides an IP network into smaller address blocks by borrowing bits from the host portion to create subnet masks..
The correct answer is: 172.16.100.223 — A /27 uses address blocks of 32. In practical terms, the fourth-octet ranges are 0–31, 32–63, 64–95, 96–127, 128–159, 160–191, 192–223, and 224–255. Since 222 falls inside the 192–223 block, the broadcast address is the last address in that block, which is 172.16.100.223. This is a classic subnet-boundary question because it tests whether you can place a host in the correct block and then identify the final address in that block as the broadcast.
Review subnetting divides an IP network into smaller address blocks by borrowing bits from the host portion to create subnet masks., then practise related 200-301 questions on the same topic to reinforce the concept.
Subnetting divides an IP network into smaller address blocks by borrowing bits from the host portion to create subnet masks.
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Same concept, more angles
These questions test the same concept from different angles. Work through them to make sure you can recognise it however the exam phrases it.
Variation 1. A host address is 192.168.1.14/29. Which address is the broadcast address for that host’s subnet?
hardWhy C: A /29 uses blocks of 8 addresses. In plain language, the subnets in the last octet move in increments of 8: 0–7, 8–15, 16–23, and so on. Since the host address ends in 14, it belongs to the 8–15 block. In that block, the last address is the broadcast address, so the broadcast is 192.168.1.15. This is a classic subnetting pattern because it requires you to place the host inside the correct block and then identify the last address in that block rather than guessing based on the host value itself.
Variation 2. A host address is 10.100.12.94/26. Which address is the broadcast address for that subnet?
hardWhy B: A /26 subnet has a block size of 64, so the fourth-octet ranges are 0–63, 64–127, 128–191, and 192–255. The host address 10.100.12.94 lies in the 64–127 range, making the broadcast address the last address in that range: 10.100.12.127. Option A (10.100.12.63) is the broadcast of the previous subnet (0–63). Option C (10.100.12.64) is the network address of the subnet containing the host. Option D (10.100.12.128) is the network address of the next subnet (128–191).
Variation 3. A host uses address 192.168.5.126/25. Which address is the broadcast address for its subnet?
hardWhy B: A /25 divides the /24 into two halves: 0–127 and 128–255. In plain language, the host address 192.168.5.126 is in the lower half, which runs from .0 through .127. The last address in that block is the broadcast address, so the broadcast is 192.168.5.127. This is a classic subnetting question because it checks whether you can identify the correct block and then select the last address in that block as the broadcast address.
Variation 4. A host address is 192.168.14.222/28. Which address is the broadcast address of its subnet?
hardWhy B: A /28 subnet has a block size of 16. In practical terms, the fourth-octet blocks are 0-15, 16-31, and so on. Because 222 falls within the 208-223 block, the broadcast address is the last address in that block: 192.168.14.223. This is a subnet-boundary question that depends on identifying the correct /28 block before choosing the broadcast address.
Variation 5. A host address is 172.22.14.99/27. Which address is the broadcast address of the subnet?
hardWhy B: A /27 subnet has a block size of 32. In practical terms, the relevant blocks are 0-31, 32-63, 64-95, 96-127, and so on. Because 99 falls within the 96-127 block, the broadcast address is the last address in that block: 172.22.14.127. This is a classic subnet-boundary question and remains important because addressing precision appears throughout the CCNA blueprint.
Variation 6. A host address is 10.10.10.14/29. Which address is the broadcast address for its subnet?
hardWhy B: A /29 subnet has a block size of 8. In practical terms, the fourth-octet ranges are 0–7, 8–15, 16–23, and so on. Since 14 falls in the 8–15 block, the broadcast address is the last address in that block, which is 10.10.10.15. This is a classic subnetting question that checks whether you can find the block first and then identify the final address in that block.
Variation 7. A host address is 192.168.50.158/27. Which address is the network address of its subnet?
hardWhy A: 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.
Variation 8. A host address is 10.55.8.117/29. Which address is the network address of the subnet?
hardWhy A: A /29 subnet has a block size of 8. In practical terms, the relevant last-octet blocks are 112-119 for this host. That means the network address is 10.55.8.112. Once you identify the correct block, the first address in the block is the network address. This is a useful addressing-boundary question because it checks careful block calculation, not memorized guesses.
Last reviewed: May 17, 2026
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