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 subnet has the network address 192.168.20.128/26. What is the broadcast address?
192.168.20.159
Why wrong: This is wrong because that would be the end of a smaller block, not the /26 block starting at .128.
192.168.20.191
This is correct because .128/26 covers .128 through .191.
192.168.20.192
Why wrong: This is wrong because .192 is the start of the next /26 block.
192.168.20.255
Why wrong: This is wrong because that would be the broadcast of the full /24, not this /26 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
192.168.20.191
A /26 subnet uses blocks of 64 addresses. In plain language, the block that starts at 192.168.20.128 runs through 192.168.20.191. The first address in that block is the network address, and the last address is the broadcast address. That means the broadcast address is 192.168.20.191. This is a standard subnetting calculation. Once the block size is identified, the broadcast address is simply the last address in the block.
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.
192.168.20.159
Why it's wrong here
This is wrong because that would be the end of a smaller block, not the /26 block starting at .128.
When this WOULD be correct
If the question specified a subnet mask of /25 instead of /26, the network address would be 192.168.20.128 and the broadcast address would then be 192.168.20.159, making this option correct.
192.168.20.191
Why this is correct
This is correct because .128/26 covers .128 through .191.
Related concept
A /26 subnet mask divides the IP address space into blocks of 64 addresses, including network and broadcast addresses.
192.168.20.192
Why it's wrong here
This is wrong because .192 is the start of the next /26 block.
When this WOULD be correct
If the question specified a different subnet, such as 192.168.20.192/26, then option C would be the correct broadcast address for that subnet, as it would encompass the range from 192.168.20.192 to 192.168.20.255.
192.168.20.255
Why it's wrong here
This is wrong because that would be the broadcast of the full /24, not this /26 subnet.
When this WOULD be correct
If the question asked for the broadcast address of the entire 192.168.20.0/24 network instead of the specific /26 subnet, then 192.168.20.255 would be the correct answer, as it serves as the broadcast address for that larger subnet.
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 .128/26 covers .128 through .191.
Why this is wrong here
Option A (192.168.20.159) is incorrect because the broadcast address for the subnet 192.168.20.128/26 is 192.168.20.191, calculated as the highest address in the subnet range from 192.168.20.128 to 192.168.20.191.
★ When this WOULD be the correct answer
If the question specified a subnet mask of /25 instead of /26, the network address would be 192.168.20.128 and the broadcast address would then be 192.168.20.159, making this option correct.
Why candidates choose this
Candidates may choose this option due to confusion between subnet ranges, mistakenly calculating the broadcast address by incorrectly interpreting the subnet mask or misaligning the address range.
Why this is wrong here
The option C, 192.168.20.192, is incorrect because it falls outside the range of the subnet defined by 192.168.20.128/26, which has a valid broadcast address of 192.168.20.191.
★ When this WOULD be the correct answer
If the question specified a different subnet, such as 192.168.20.192/26, then option C would be the correct broadcast address for that subnet, as it would encompass the range from 192.168.20.192 to 192.168.20.255.
Why candidates choose this
Candidates might choose this option due to confusion between the subnet address and the broadcast address, especially if they miscalculate the range or overlook the CIDR notation.
Why this is wrong here
The broadcast address for the subnet 192.168.20.128/26 is 192.168.20.191, not 192.168.20.255. The address 192.168.20.255 is the broadcast address for the entire 192.168.20.0/24 network, which is not relevant to the specified subnet.
★ When this WOULD be the correct answer
If the question asked for the broadcast address of the entire 192.168.20.0/24 network instead of the specific /26 subnet, then 192.168.20.255 would be the correct answer, as it serves as the broadcast address for that larger subnet.
Why candidates choose this
Candidates may choose this option due to familiarity with the concept of broadcast addresses, often associating the highest address in a subnet with the broadcast address without considering the specific subnet mask provided.
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 the broadcast address with the first address of the next subnet or with the broadcast address of a different subnet mask.
Detailed technical explanation
Subnetting divides a larger IP network into smaller, manageable segments called subnets, each with its own network and broadcast addresses. The subnet mask determines the size of each subnet by specifying how many bits are used for the network portion versus the host portion. In this question, the /26 mask means 26 bits are fixed for the network, leaving 6 bits for host addresses, resulting in 64 IP addresses per subnet block. To find the broadcast address for a subnet, identify the range of IP addresses covered by the subnet mask. For 192.168.20.128/26, the subnet block starts at 192.168.20.128 and includes 64 addresses, ending at 192.168.20.191. The first address is the network address, and the last address is the broadcast address, which is used to send packets to all hosts within that subnet. A common exam trap is confusing the broadcast address with the last address of a smaller or larger subnet block or the broadcast of the entire classful network. For example, mistaking 192.168.20.159 as the broadcast address confuses a smaller block ending at .159 with the /26 block starting at .128. Understanding subnet boundaries and block sizes prevents this error and ensures accurate network design and troubleshooting.
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: 192.168.20.191 — A /26 subnet uses blocks of 64 addresses. In plain language, the block that starts at 192.168.20.128 runs through 192.168.20.191. The first address in that block is the network address, and the last address is the broadcast address. That means the broadcast address is 192.168.20.191. This is a standard subnetting calculation. Once the block size is identified, the broadcast address is simply the last address in the block.
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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