This 200-301 practice question tests your understanding of network infrastructure and connectivity. This is a configuration task: choose the command set that satisfies every stated requirement. Small differences — like 'secret' vs 'password' or 'transport input ssh' vs 'all' — change whether the answer is correct. A key principle to apply: a /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP 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 is configured as 192.168.50.130/25. Which address is the broadcast address for its subnet?
192.168.50.127
Why wrong: This is wrong because .127 is the broadcast of the lower /25 block.
192.168.50.128
Why wrong: This is wrong because .128 is the network address of the upper /25 block.
192.168.50.255
This is correct because the upper /25 block runs through .255, which is the broadcast address.
192.168.50.254
Why wrong: This is wrong because .254 is a usable host address in the upper /25 block.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
192.168.50.255
A /25 divides the /24 into two blocks: 0–127 and 128–255. In plain language, because the host ends in 130, it belongs to the upper half, which starts at 128 and ends at 255. The last address in that block is the broadcast address, so the broadcast is 192.168.50.255. This is a classic subnetting pattern because it tests whether you can identify not just the subnet, but also the reserved last address in that subnet.
Key principle: A /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP 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.
192.168.50.127
Why it's wrong here
This is wrong because .127 is the broadcast of the lower /25 block.
When this WOULD be correct
In a different scenario where the subnet mask is /25 and the network address is 192.168.50.0, the broadcast address for the subnet would be 192.168.50.127. A question could ask for the broadcast address of the subnet 192.168.50.0/25, making this option correct.
192.168.50.128
Why it's wrong here
This is wrong because .128 is the network address of the upper /25 block.
When this WOULD be correct
If the question specified a subnet mask of /25 for the address 192.168.50.128, then option B would be correct as the broadcast address for that subnet would be 192.168.50.255, and the first usable address would be 192.168.50.129.
192.168.50.255
Why this is correct
This is correct because the upper /25 block runs through .255, which is the broadcast address.
Related concept
A /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP addresses including network and broadcast addresses.
192.168.50.254
Why it's wrong here
This is wrong because .254 is a usable host address in the upper /25 block.
When this WOULD be correct
If the question specified a subnet mask of /24 instead of /25, then 192.168.50.254 would be the broadcast address for the subnet 192.168.50.0/24, as it would cover the range from 192.168.50.0 to 192.168.50.255.
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 the upper /25 block runs through .255, which is the broadcast address.
Why this is wrong here
The /25 subnet mask (255.255.255.128) divides the 192.168.50.0/24 network into two subnets: 192.168.50.0/25 (range .0-.127) and 192.168.50.128/25 (range .128-.255). The broadcast address for the lower subnet is 192.168.50.127, but the host 192.168.50.130 belongs to the upper subnet, so this is not its broadcast.
★ When this WOULD be the correct answer
In a different scenario where the subnet mask is /25 and the network address is 192.168.50.0, the broadcast address for the subnet would be 192.168.50.127. A question could ask for the broadcast address of the subnet 192.168.50.0/25, making this option correct.
Why candidates choose this
Students often confuse the broadcast address of the lower subnet with that of the upper subnet, especially when the host IP is close to the subnet boundary.
Why this is wrong here
192.168.50.128 is the network address of the upper /25 subnet (192.168.50.128/25). It is not a broadcast address; it identifies the subnet itself and cannot be assigned to a host.
★ When this WOULD be the correct answer
If the question specified a subnet mask of /25 for the address 192.168.50.128, then option B would be correct as the broadcast address for that subnet would be 192.168.50.255, and the first usable address would be 192.168.50.129.
Why candidates choose this
Some might think that the first address after the subnet boundary is the broadcast, but it is actually the network address. The broadcast is the last address in the range.
Why this is wrong here
192.168.50.254 is a valid host address within the 192.168.50.128/25 subnet (usable range .129-.254). It is not the broadcast address; the broadcast is .255.
★ When this WOULD be the correct answer
If the question specified a subnet mask of /24 instead of /25, then 192.168.50.254 would be the broadcast address for the subnet 192.168.50.0/24, as it would cover the range from 192.168.50.0 to 192.168.50.255.
Why candidates choose this
Students might assume that the last usable host address (.254) is the broadcast, but the broadcast is actually the very last address (.255) in the 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
A frequent exam trap is mistaking the network address or a high usable host address for the broadcast address. Candidates often select 192.168.50.128, confusing it as the broadcast because it is the start of the upper subnet, or 192.168.50.254, assuming it is the broadcast since it is near the subnet's end. The trap lies in not recognizing that the broadcast address is always the highest address in the subnet, which in this case is 192.168.50.255. Misidentifying these addresses leads to incorrect subnet calculations and can cause network communication failures in real scenarios.
Detailed technical explanation
Subnetting is a fundamental concept in IP networking that divides a larger network into smaller, manageable segments called subnets. Each subnet has a network address, a range of usable host addresses, and a broadcast address. The subnet mask determines the size of each subnet by defining how many bits are used for the network portion versus the host portion of the IP address. In this question, a /25 mask means 25 bits are fixed for the network, leaving 7 bits for host addressing, effectively splitting a standard Class C /24 network into two equal subnets. The /25 subnet mask (255.255.255.128) divides the 192.168.50.0/24 network into two subnets: 192.168.50.0 to 192.168.50.127 and 192.168.50.128 to 192.168.50.255. The first subnet uses 0 as the network address and 127 as the broadcast address, while the second subnet uses 128 as the network address and 255 as the broadcast address. Since the host IP 192.168.50.130 falls within the second subnet range, the broadcast address for this subnet is 192.168.50.255. A common exam trap is confusing the broadcast address with the network address or a usable host address within the subnet. For example, 192.168.50.128 is the network address of the second subnet, not the broadcast, and 192.168.50.254 is a valid host address, not a broadcast. Understanding the binary division of the subnet and recognizing the last address in the subnet as the broadcast address is critical for accurate subnetting in Cisco exams and practical network design.
Key takeaway
A /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP addresses including network and broadcast addresses.
Real-world example
A small business has 20 workstations on the 192.168.1.0/24 network and one public IP from its ISP. The router uses PAT (NAT overload) so all 20 devices share one public address using different source ports. NAT questions test whether you understand the four address terms and which direction each translation applies.
What to study next
Review a /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP 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 /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP addresses including network and broadcast addresses..
The correct answer is: 192.168.50.255 — A /25 divides the /24 into two blocks: 0–127 and 128–255. In plain language, because the host ends in 130, it belongs to the upper half, which starts at 128 and ends at 255. The last address in that block is the broadcast address, so the broadcast is 192.168.50.255. This is a classic subnetting pattern because it tests whether you can identify not just the subnet, but also the reserved last address in that subnet.
Review a /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP addresses including network and broadcast addresses., then practise related 200-301 questions on the same topic to reinforce the concept.
A /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP addresses including network and broadcast addresses.
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Last reviewed: May 17, 2026
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