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 /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.. 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 10.100.12.94/26. Which address is the broadcast address for that subnet?
10.100.12.63
Why wrong: This is wrong because .63 is the broadcast address of the previous /26 block.
10.100.12.127
This is correct because .94 is in the 64-127 /26 range.
10.100.12.64
Why wrong: This is wrong because .64 is the network address of that subnet.
10.100.12.128
Why wrong: This is wrong because .128 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
10.100.12.127
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).
Key principle: A /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
10.100.12.63
Why it's wrong here
This is wrong because .63 is the broadcast address of the previous /26 block.
When this WOULD be correct
In a question where the subnet mask is /26 and the network address is 10.100.12.0, if the question mistakenly states that the subnet ends at 10.100.12.63, then 10.100.12.63 would be the correct broadcast address for that defined range.
10.100.12.127
Why this is correct
This is correct because .94 is in the 64-127 /26 range.
Related concept
A /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
10.100.12.64
Why it's wrong here
This is wrong because .64 is the network address of that subnet.
When this WOULD be correct
In a different question setup where the subnet mask is /26 and the network address is 10.100.12.64, option C would be the correct answer if the question asked for the first usable host address instead of the broadcast address.
10.100.12.128
Why it's wrong here
This is wrong because .128 starts the next subnet.
When this WOULD be correct
If the question were to ask for the broadcast address of a subnet defined by 10.100.12.128/26, then option D would be correct, as the broadcast address for that subnet would be 10.100.12.191.
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 .94 is in the 64-127 /26 range.
Why this is wrong here
The address 10.100.12.63 is incorrect as the broadcast address for the subnet 10.100.12.0/26, which actually has a range from 10.100.12.0 to 10.100.12.63. The correct broadcast address for the subnet is 10.100.12.127.
★ When this WOULD be the correct answer
In a question where the subnet mask is /26 and the network address is 10.100.12.0, if the question mistakenly states that the subnet ends at 10.100.12.63, then 10.100.12.63 would be the correct broadcast address for that defined range.
Why candidates choose this
Candidates may choose this option due to a misunderstanding of subnetting, where they incorrectly calculate the broadcast address by only considering the last octet and not accounting for the subnet mask properly.
Why this is wrong here
Option C (10.100.12.64) is incorrect because it is the first usable host address in the subnet 10.100.12.64/26, not the broadcast address. The broadcast address for this subnet is 10.100.12.127.
★ When this WOULD be the correct answer
In a different question setup where the subnet mask is /26 and the network address is 10.100.12.64, option C would be the correct answer if the question asked for the first usable host address instead of the broadcast address.
Why candidates choose this
Candidates might choose option C due to confusion between the broadcast address and the first usable address in a subnet, especially if they miscalculate the range of addresses available in a /26 subnet.
Why this is wrong here
Option D (10.100.12.128) is incorrect because it does not fall within the range of addresses for the subnet defined by 10.100.12.94/26, which spans from 10.100.12.64 to 10.100.12.127.
★ When this WOULD be the correct answer
If the question were to ask for the broadcast address of a subnet defined by 10.100.12.128/26, then option D would be correct, as the broadcast address for that subnet would be 10.100.12.191.
Why candidates choose this
Candidates may be tempted by option D because it is a common mistake to miscalculate subnet boundaries, especially when dealing with larger subnets or when they misinterpret the CIDR notation.
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
Be careful not to confuse host addresses or network addresses with the broadcast address. Remember, the broadcast address is the last address in the subnet range.
Detailed technical explanation
Subnetting is a fundamental concept in IP networking that divides a larger network into smaller, manageable subnetworks. A /26 subnet mask means the first 26 bits are fixed for the network portion, leaving 6 bits for host addresses. This results in 64 IP addresses per subnet, including network and broadcast addresses. The subnet mask 255.255.255.192 corresponds to /26, which splits the IP address range into blocks of 64 addresses. To find the broadcast address for a given host IP and subnet mask, you first identify the subnet block the IP belongs to by calculating the block size (256 minus the last octet of the subnet mask). For /26, the block size is 64, so the subnets are 0-63, 64-127, 128-191, and 192-255. Since 10.100.12.94 falls within 64-127, the broadcast address is the last address in that block, 10.100.12.127. A common exam trap is confusing the broadcast address with the network address or the broadcast address of an adjacent subnet. For example, 10.100.12.63 is the broadcast for the previous subnet (0-63), and 10.100.12.64 is the network address of the current subnet. Understanding the block boundaries and correctly identifying the subnet range prevents this mistake. In practical Cisco networking, accurate broadcast address calculation is essential for proper subnet communication and avoiding IP conflicts.
Key takeaway
A /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
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 /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses., 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 /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses..
The correct answer is: 10.100.12.127 — 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).
Review a /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses., then practise related 200-301 questions on the same topic to reinforce the concept.
A /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
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Last reviewed: May 17, 2026
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