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. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. 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 192.168.90.33/28. Which address is the last usable host in the subnet?
192.168.90.46
This is correct because .46 is the last usable address before the .47 broadcast in the 32-47 block.
192.168.90.47
Why wrong: This is wrong because .47 is the broadcast address.
192.168.90.33
Why wrong: This is wrong because .33 is usable, but not the last usable host.
192.168.90.48
Why wrong: This is wrong because .48 is in 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
192.168.90.46
The /28 subnet mask (255.255.255.240) gives a block size of 16 addresses. The network address for 192.168.90.33 is 192.168.90.32, so the broadcast address is 192.168.90.47. The last usable host is the broadcast address minus one, which is 192.168.90.46.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
192.168.90.46
Why this is correct
This is correct because .46 is the last usable address before the .47 broadcast in the 32-47 block.
Related concept
Read the scenario before looking for a memorised answer.
192.168.90.47
Why it's wrong here
This is wrong because .47 is the broadcast address.
When this WOULD be correct
If the question asked for the last address in the subnet instead of the last usable host, option B would be correct, as the broadcast address for the subnet 192.168.90.32/28 is indeed 192.168.90.47.
192.168.90.33
Why it's wrong here
This is wrong because .33 is usable, but not the last usable host.
When this WOULD be correct
If the question asked for the first usable host address in the subnet 192.168.90.32/28, then option C would be correct, as it represents the first address available for assignment to a host.
192.168.90.48
Why it's wrong here
This is wrong because .48 is in the next subnet.
When this WOULD be correct
If the question asked for the first address in the next subnet after 192.168.90.32/28, then 192.168.90.48 would be the correct answer, as it represents the start of the subnet 192.168.90.48/28.
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 .46 is the last usable address before the .47 broadcast in the 32-47 block.
Why this is wrong here
192.168.90.47 is the broadcast address for the subnet 192.168.90.32/28. Broadcast addresses cannot be assigned to hosts; they are used to send traffic to all hosts in the subnet.
★ When this WOULD be the correct answer
If the question asked for the last address in the subnet instead of the last usable host, option B would be correct, as the broadcast address for the subnet 192.168.90.32/28 is indeed 192.168.90.47.
Why candidates choose this
Students often confuse the last usable host with the broadcast address, especially when the subnet boundary is not obvious. The broadcast is one less than the next subnet's network address, and it's easy to mistakenly think it's usable.
Why this is wrong here
192.168.90.33 is the first usable host in the subnet (network address .32 + 1). The question asks for the last usable host, not the first.
★ When this WOULD be the correct answer
If the question asked for the first usable host address in the subnet 192.168.90.32/28, then option C would be correct, as it represents the first address available for assignment to a host.
Why candidates choose this
A test-taker might see that .33 is a valid host and assume it is the last because it is close to the given address, without calculating the subnet range correctly.
Why this is wrong here
192.168.90.48 is the network address of the next subnet (192.168.90.48/28). It is not part of the current subnet and cannot be used as a host address in the subnet containing .33.
★ When this WOULD be the correct answer
If the question asked for the first address in the next subnet after 192.168.90.32/28, then 192.168.90.48 would be the correct answer, as it represents the start of the subnet 192.168.90.48/28.
Why candidates choose this
Students may think that .48 is the last usable because it is numerically higher than .46, not realizing that subnet boundaries restrict the range. They might also confuse it with the broadcast address of the previous 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
Cisco often tests the distinction between the broadcast address and the last usable host, tricking candidates who forget to subtract one from the broadcast address.
Detailed technical explanation
With a /28 mask, the subnet has 16 addresses (2^(32-28) = 16). The network address is the first address in the block (all host bits 0), and the broadcast address is the last (all host bits 1). The usable host range is from network+1 to broadcast-1, so for subnet 192.168.90.32/28, the usable hosts are 192.168.90.33 through 192.168.90.46. This concept is critical for VLSM and subnetting in real-world network design, where misidentifying the broadcast address can cause IP conflicts or routing issues.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
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
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FAQ
Network Infrastructure and Connectivity — This question tests Network Infrastructure and Connectivity — Read the scenario before looking for a memorised answer..
The correct answer is: 192.168.90.46 — The /28 subnet mask (255.255.255.240) gives a block size of 16 addresses. The network address for 192.168.90.33 is 192.168.90.32, so the broadcast address is 192.168.90.47. The last usable host is the broadcast address minus one, which is 192.168.90.46.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
Read the scenario before looking for a memorised answer.
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Last reviewed: Jun 11, 2026
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