CCNA Network Infrastructure and Connectivity Practice Question
A host address is 172.22.14.99/27. Which address is the broadcast address of the subnet?
⚠ Common exam trap
Be careful not to confuse network addresses with broadcast addresses, and ensure you are calculating the correct subnet 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
✓
172.22.14.127
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.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
172.22.14.95
Why it's wrong here
172.22.14.95 is the broadcast address of the previous /27 subnet, which spans 172.22.14.64 through 172.22.14.95. Because 172.22.14.99 lies in the next higher block starting at .96, this address is not the broadcast for the host's subnet. The /27 boundary means .95 and .99 fall in different subnets, so .95 cannot serve as the broadcast for the network that contains .99.
When this WOULD be correct
In a different question where the subnet mask was /26 and the host address was 172.22.14.99, option A (172.22.14.95) could be the broadcast address for that subnet. This would occur if the address range was defined such that 172.22.14.64/26 had 172.22.14.95 as its broadcast address.
- ✓
172.22.14.127
Why this is correct
172.22.14.127 is the broadcast address for the /27 subnet that contains host 172.22.14.99. Since the /27 mask creates 32-address blocks, the relevant block runs from 172.22.14.96 through 172.22.14.127, and the broadcast is always the final address in that block. The host bits in .127 are all ones, which identifies this as the directed broadcast for this subnet.
- ✗
172.22.14.96
Why it's wrong here
172.22.14.96 is the network address of the /27 subnet that contains host 172.22.14.99, not the broadcast. The network address always has all host bits set to 0, whereas the broadcast address is the very next last address in the block, which is 172.22.14.127. Since .96 identifies the subnet itself, it is never used as a broadcast or host address.
When this WOULD be correct
In a different scenario where the subnet mask is changed to /26, the address 172.22.14.96 could be the broadcast address for the subnet 172.22.14.64/26. In this case, the range would include addresses from 172.22.14.65 to 172.22.14.126, making 172.22.14.126 the broadcast address.
- ✗
172.22.14.128
Why it's wrong here
172.22.14.128 is the network address of the next /27 subnet, which begins immediately after the current block ends at 172.22.14.127. A broadcast address must have all host bits set to 1, but .128 has all host bits set to 0, making it the network identifier for the 172.22.14.128/27 range. As a result, it cannot be the broadcast for the subnet containing .99.
When this WOULD be correct
In a different question where the subnet mask is /25, the broadcast address for the subnet containing 172.22.14.128 would be 172.22.14.255. In this scenario, option D would be the correct answer as it would represent the broadcast address for that specific 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.
✓172.22.14.127Correct answer▾
Why this is correct
172.22.14.127 is the broadcast address for the /27 subnet that contains host 172.22.14.99. Since the /27 mask creates 32-address blocks, the relevant block runs from 172.22.14.96 through 172.22.14.127, and the broadcast is always the final address in that block. The host bits in .127 are all ones, which identifies this as the directed broadcast for this subnet.
✗172.22.14.95Wrong answer — click to see why▾
Why this is wrong here
Option A is incorrect because the broadcast address for the subnet 172.22.14.99/27 is 172.22.14.127, not 172.22.14.95. The /27 subnet mask indicates that the last 5 bits are for host addresses, allowing for a range of addresses that ends at 172.22.14.127.
★ When this WOULD be the correct answer
In a different question where the subnet mask was /26 and the host address was 172.22.14.99, option A (172.22.14.95) could be the broadcast address for that subnet. This would occur if the address range was defined such that 172.22.14.64/26 had 172.22.14.95 as its broadcast address.
Why candidates choose this
Candidates may choose option A due to a miscalculation of the subnet range, mistakenly believing that the lower end of the host address range is the broadcast address. This can happen if they confuse the calculation of usable addresses in a subnet.
✗172.22.14.96Wrong answer — click to see why▾
Why this is wrong here
Option C (172.22.14.96) is incorrect because it falls within the usable host range of the subnet and is not the broadcast address. The broadcast address for the subnet 172.22.14.96/27 is actually 172.22.14.127.
★ When this WOULD be the correct answer
In a different scenario where the subnet mask is changed to /26, the address 172.22.14.96 could be the broadcast address for the subnet 172.22.14.64/26. In this case, the range would include addresses from 172.22.14.65 to 172.22.14.126, making 172.22.14.126 the broadcast address.
Why candidates choose this
Candidates may choose this option due to confusion between the subnet address and the broadcast address, as they might miscalculate the range of usable addresses and mistakenly identify 172.22.14.96 as a boundary address.
✗172.22.14.128Wrong answer — click to see why▾
Why this is wrong here
Option D, 172.22.14.128, is incorrect because it does not fall within the subnet defined by 172.22.14.99/27, which has a valid range of 172.22.14.96 to 172.22.14.127. The broadcast address for this subnet is 172.22.14.127.
★ When this WOULD be the correct answer
In a different question where the subnet mask is /25, the broadcast address for the subnet containing 172.22.14.128 would be 172.22.14.255. In this scenario, option D would be the correct answer as it would represent the broadcast address for that specific subnet.
Why candidates choose this
Candidates may be tempted by option D because it is a higher address within the same octet, leading them to mistakenly associate it with the broadcast address due to common subnetting patterns.
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?”
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Related to this question
Learn chapter
IPv4 Addressing and Address Classes
Key term
Block size
Block size is the amount of data, measured in bytes, that a system reads or writes in a single operation, affecting storage efficiency and network performance.
Key term
Cisco Discovery Protocol
A proprietary Layer 2 network protocol used by Cisco devices to discover information about directly connected neighbor devices.
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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.