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 /29 subnet mask allocates 29 bits for the network and 3 bits for hosts, resulting in 8 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 10.10.10.14/29. Which address is the broadcast address for its subnet?
10.10.10.7
Why wrong: This is wrong because .7 is the broadcast of the previous /29 block.
10.10.10.15
This is correct because .14 is in the 8–15 /29 subnet.
10.10.10.8
Why wrong: This is wrong because .8 is the network address of the block.
10.10.10.16
Why wrong: This is wrong because .16 starts the next 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
10.10.10.15
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.
Key principle: A /29 subnet mask allocates 29 bits for the network and 3 bits for hosts, resulting in 8 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.
10.10.10.7
Why it's wrong here
This is wrong because .7 is the broadcast of the previous /29 block.
When this WOULD be correct
In a different question, if the subnet mask were /28 and the host address were 10.10.10.0, then 10.10.10.7 could be the broadcast address for a subnet that includes 10.10.10.0/28, where the valid host range is 10.10.10.1 to 10.10.10.14.
10.10.10.15
Why this is correct
This is correct because .14 is in the 8–15 /29 subnet.
Related concept
A /29 subnet mask allocates 29 bits for the network and 3 bits for hosts, resulting in 8 IP addresses per subnet block.
10.10.10.8
Why it's wrong here
This is wrong because .8 is the network address of the block.
When this WOULD be correct
If the question asked for the network address of the subnet that includes 10.10.10.14/29, then 10.10.10.8 would be the correct answer, as it represents the first address in that subnet range.
10.10.10.16
Why it's wrong here
This is wrong because .16 starts the next subnet.
When this WOULD be correct
If the question specified a different subnet, such as 10.10.10.16/29, then 10.10.10.16 would be the broadcast address for that subnet. This would change the range of valid addresses and make D the correct answer.
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 .14 is in the 8–15 /29 subnet.
Why this is wrong here
Option A, 10.10.10.7, is incorrect because it does not represent the broadcast address for the subnet defined by 10.10.10.14/29. The correct broadcast address is 10.10.10.15, which is the highest address in the subnet range.
★ When this WOULD be the correct answer
In a different question, if the subnet mask were /28 and the host address were 10.10.10.0, then 10.10.10.7 could be the broadcast address for a subnet that includes 10.10.10.0/28, where the valid host range is 10.10.10.1 to 10.10.10.14.
Why candidates choose this
Candidates may choose option A because they might misinterpret the subnetting rules and mistakenly calculate the broadcast address based on an incorrect understanding of subnet ranges.
Why this is wrong here
The address 10.10.10.8 is not the broadcast address for the subnet 10.10.10.14/29. The correct broadcast address is 10.10.10.15, which is the highest address in the subnet range from 10.10.10.8 to 10.10.10.15.
★ When this WOULD be the correct answer
If the question asked for the network address of the subnet that includes 10.10.10.14/29, then 10.10.10.8 would be the correct answer, as it represents the first address in that subnet range.
Why candidates choose this
Candidates may choose this option due to confusion between the concepts of broadcast and network addresses, mistakenly associating the address with a common subnet boundary.
Why this is wrong here
The address 10.10.10.16 is outside the subnet defined by 10.10.10.14/29, which includes addresses from 10.10.10.8 to 10.10.10.15. Therefore, it cannot be the broadcast address for this subnet.
★ When this WOULD be the correct answer
If the question specified a different subnet, such as 10.10.10.16/29, then 10.10.10.16 would be the broadcast address for that subnet. This would change the range of valid addresses and make D the correct answer.
Why candidates choose this
Candidates might confuse the subnetting boundaries and mistakenly think that 10.10.10.16 could be a valid broadcast address due to its proximity to the range of addresses in the original subnet.
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 confusing the network address or a host address with the broadcast address. Remember, the broadcast address is the last address in the subnet range.
Detailed technical explanation
Subnetting divides an IP network into smaller logical segments called subnets, each with its own network address, broadcast address, and host range. The subnet mask determines the size of each subnet by defining which portion of the IP address is the network and which is the host. A /29 subnet mask means 29 bits are fixed for the network, leaving 3 bits for host addresses, resulting in 8 IP addresses per subnet (2^3 = 8). These 8 addresses include the network address, broadcast address, and usable host addresses. To find the broadcast address for a /29 subnet, first identify the block size, which is 8 addresses. The subnet ranges increment by 8 in the last octet (e.g., 0–7, 8–15, 16–23). The broadcast address is always the last IP in the subnet block. Since the host IP 10.10.10.14 falls within the 8–15 range, the broadcast address is 10.10.10.15. This method applies consistently across all /29 subnets. A common exam trap is confusing the broadcast address with the network address or the broadcast of an adjacent subnet. For example, 10.10.10.7 is the broadcast for the previous subnet (0–7), not the one containing 10.10.10.14. Understanding the block size and correctly identifying the subnet range prevents this mistake. In practical Cisco networking, correctly identifying broadcast addresses is crucial for configuring ACLs, DHCP scopes, and troubleshooting network issues.
Key takeaway
A /29 subnet mask allocates 29 bits for the network and 3 bits for hosts, resulting in 8 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 /29 subnet mask allocates 29 bits for the network and 3 bits for hosts, resulting in 8 IP addresses per subnet block., 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 — A /29 subnet mask allocates 29 bits for the network and 3 bits for hosts, resulting in 8 IP addresses per subnet block..
The correct answer is: 10.10.10.15 — 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.
Review a /29 subnet mask allocates 29 bits for the network and 3 bits for hosts, resulting in 8 IP addresses per subnet block., then practise related 200-301 questions on the same topic to reinforce the concept.
A /29 subnet mask allocates 29 bits for the network and 3 bits for hosts, resulting in 8 IP addresses per subnet block.
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Last reviewed: May 17, 2026
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