A host is configured as 192.168.50.130/25. Which address is the broadcast address for its subnet?
Answer choices
Why each option matters
Good practice is not just finding the correct option. The wrong answers often show the exact trap the exam wants you to fall into.
Distractor review
192.168.50.127
This is wrong because .127 is the broadcast of the lower /25 block.
Distractor review
192.168.50.128
This is wrong because .128 is the network address of the upper /25 block.
Best answer
192.168.50.255
This is correct because the upper /25 block runs through .255, which is the broadcast address.
Distractor review
192.168.50.254
This is wrong because .254 is a usable host address in the upper /25 block.
Common exam trap
Common exam trap: answer the scenario, not the keyword
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.
Technical deep dive
How to think about this question
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.
KKey Concepts to Remember
- A /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP addresses including network and broadcast addresses.
- The broadcast address is the highest IP address in a subnet, where all host bits are set to one in binary notation.
- The network address is the lowest IP address in a subnet, where all host bits are set to zero in binary notation.
- Hosts with IP addresses within a subnet range share the same network and broadcast addresses defined by the subnet mask.
- Subnetting requires understanding binary math to correctly identify network, broadcast, and usable host addresses for each subnet.
- In Cisco networking, the broadcast address is used to send packets to all hosts within a subnet simultaneously.
- Confusing the network address or a high usable host address with the broadcast address is a common subnetting error on the CCNA exam.
- The /25 mask (255.255.255.128) splits a /24 network into two subnets: 0–127 and 128–255, each with distinct network and broadcast addresses.
TExam Day Tips
- Watch for words such as best, first, most likely and least administrative effort.
- Review why wrong options are wrong, not only why the correct option is correct.
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More questions from this exam
Keep practising from the same exam bank, or move into a focused topic page if this question exposed a weak area.
Question 1
A router learns the same prefix from both OSPF and EIGRP. Which route is installed by default?
Question 2
A router shows this output: R1#show ip ospf neighbor Neighbor ID Pri State Dead Time Address Interface 10.1.1.2 1 FULL/DR 00:00:34 192.168.12.2 GigabitEthernet0/0 10.1.1.3 1 2WAY/DROTHER 00:00:39 192.168.12.3 GigabitEthernet0/0 Which statement is correct?
Question 3
What is the OSPF metric called?
Question 4
A non-root switch has two uplinks toward the root bridge. One path has a lower total STP cost than the other. What role will the lower-cost uplink have?
Question 5
A router interface applies this ACL inbound: 10 deny tcp any any eq 80 20 permit ip any any A user reports that web browsing to a server by IP address fails, but ping works. Which statement best explains the behavior?
Question 6
A router learns route 198.51.100.0/24 from OSPF with AD 110 and also has a static route to the same prefix configured with AD 150. Which route is installed?
FAQ
Questions learners often ask
What does this 200-301 question test?
A /25 subnet mask divides a Class C network into two equal subnets, each with 128 IP addresses including network and broadcast addresses.
What is the correct answer to this question?
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.
What should I do if I get this 200-301 question wrong?
Then try more questions from the same exam bank and focus on understanding why the wrong options are tempting.
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