CCNA Network Infrastructure and Connectivity Practice Question
A host address is 192.168.1.14/29. Which address is the broadcast address for that host’s subnet?
⚠ Common exam trap
Be careful not to confuse the network address or the next subnet's network address with the broadcast address.
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
✓
192.168.1.15
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.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
192.168.1.7
Why it's wrong here
192.168.1.7 is the broadcast address of the previous /29 subnet (192.168.1.0–192.168.1.7), not the subnet containing 192.168.1.14. Since the /29 block size is 8, the host's subnet starts at 192.168.1.8 and ends at 192.168.1.15. Choosing .7 would indicate a subnet boundary that does not include the host address in question.
When this WOULD be correct
If the question asked for the broadcast address of the subnet 192.168.1.0/29 instead, then 192.168.1.7 would be the correct answer, as it would be the highest address in that specific subnet range.
- ✗
192.168.1.14
Why it's wrong here
192.168.1.14 is the host address itself, which is the last usable address in the /29 subnet, not the broadcast. In a /29 block sized 8, addresses are assigned as network (.8), hosts (.9–.14), and broadcast (.15). The broadcast is always the final address in the block, so .14 cannot be the broadcast while also being the host address.
When this WOULD be correct
In a different question setup where the question asks for the host address of a specific device within a subnet, and the subnet is defined as 192.168.1.14/29, option B could be correct if the question specifically inquires about the address assigned to that device.
- ✓
192.168.1.15
Why this is correct
A /29 prefix (255.255.255.248) creates subnets with 8 addresses each. The host 192.168.1.14 falls in the subnet from 192.168.1.8 to 192.168.1.15, where the first address is the network ID and the last is the directed broadcast. Therefore, 192.168.1.15 is the broadcast address for this subnet, and .14 is the last usable host address.
- ✗
192.168.1.16
Why it's wrong here
192.168.1.16 is not a broadcast address at all; it is the network address of the next /29 subnet (192.168.1.16–192.168.1.23). For the subnet containing 192.168.1.14, the broadcast is .15, which is one less than .16. This option confuses the start of the following block with the end of the current block.
When this WOULD be correct
In a different scenario where the subnet mask is changed to /28, the subnet would range from 192.168.1.0 to 192.168.1.15. In this case, if the host address was 192.168.1.14, the broadcast address would indeed be 192.168.1.16.
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.
✓192.168.1.15Correct answer▾
Why this is correct
A /29 prefix (255.255.255.248) creates subnets with 8 addresses each. The host 192.168.1.14 falls in the subnet from 192.168.1.8 to 192.168.1.15, where the first address is the network ID and the last is the directed broadcast. Therefore, 192.168.1.15 is the broadcast address for this subnet, and .14 is the last usable host address.
✗192.168.1.7Wrong answer — click to see why▾
Why this is wrong here
192.168.1.7 is the broadcast address for the /29 block 0–7, which does not contain host .14. The host .14 is in the block 8–15, so its broadcast is .15.
★ When this WOULD be the correct answer
If the question asked for the broadcast address of the subnet 192.168.1.0/29 instead, then 192.168.1.7 would be the correct answer, as it would be the highest address in that specific subnet range.
Why candidates choose this
Students often miscalculate the block size or confuse the network address with the broadcast address. They might think .7 is the broadcast because it is the last address in the first block, but they forget to check which block contains .14.
✗192.168.1.14Wrong answer — click to see why▾
Why this is wrong here
192.168.1.14 is the host address itself, not the broadcast address. The broadcast address is always the last address in the subnet, which is .15 for the block 8–15.
★ When this WOULD be the correct answer
In a different question setup where the question asks for the host address of a specific device within a subnet, and the subnet is defined as 192.168.1.14/29, option B could be correct if the question specifically inquires about the address assigned to that device.
Why candidates choose this
Some students think the broadcast address is the same as the host address or that the host address itself can be used for broadcasting. They may not understand that the broadcast address is a special reserved address.
✗192.168.1.16Wrong answer — click to see why▾
Why this is wrong here
192.168.1.16 is the network address of the next /29 block (16–23), not the broadcast address for the block containing .14. The broadcast address must be the last address in the same block as the host.
★ When this WOULD be the correct answer
In a different scenario where the subnet mask is changed to /28, the subnet would range from 192.168.1.0 to 192.168.1.15. In this case, if the host address was 192.168.1.14, the broadcast address would indeed be 192.168.1.16.
Why candidates choose this
A common mistake is to add the block size (8) to the host address and assume that is the broadcast. For example, 14 + 8 = 22, but .16 is not the broadcast; it is the next network address. Students may also confuse broadcast with network address.
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
Go deeper
Related to this question
Learn chapter
IPv4 Addressing and Address Classes
Key term
Subnet
A subnet is a logical subdivision of an IP network, created by partitioning a larger network address space using subnet masks.
Key term
Classless Inter-domain Routing
Classless Inter-domain Routing (CIDR) is a method for allocating IP addresses and routing IP packets more efficiently by replacing the old class-based system with variable-length subnet masks.
About these practice questions
One of 1,389 original 200-301 practice questions on Courseiva, each with a full explanation and wrong-answer analysis — not exam dumps or protected exam content. Learn why practice questions differ from exam dumps →
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.