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 /27 subnet mask allocates 27 bits to the network portion and leaves 5 bits for host addresses, resulting in 32 IP addresses per subnet.. 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 with IP address 192.168.50.94/27. Which subnet contains that host?
192.168.50.32/27
Why wrong: This is wrong because that subnet ends at .63.
192.168.50.64/27
This is correct because .94 falls within the .64 through .95 block.
192.168.50.96/27
Why wrong: This is wrong because that block begins above the host address.
192.168.50.0/27
Why wrong: This is wrong because that block covers only .0 through .31.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
192.168.50.64/27
A /27 subnet has a block size of 32. In simple terms, the fourth-octet ranges are 0–31, 32–63, 64–95, 96–127, and so on. Because 94 falls inside the 64–95 range, the network address for the host’s subnet is 192.168.50.64/27. This kind of question tests whether you can move from prefix length to block size and then place the host inside the correct interval. The most common mistake is choosing a nearby boundary like 96 or 32 without calculating the actual block that contains the address.
Key principle: A /27 subnet mask allocates 27 bits to the network portion and leaves 5 bits for host addresses, resulting in 32 IP addresses per subnet.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
192.168.50.32/27
Why it's wrong here
This is wrong because that subnet ends at .63.
When this WOULD be correct
If the question asked which subnet contains the IP address 192.168.50.40, then option A (192.168.50.32/27) would be correct, as this subnet would encompass the host IP within its range.
192.168.50.64/27
Why this is correct
This is correct because .94 falls within the .64 through .95 block.
Related concept
A /27 subnet mask allocates 27 bits to the network portion and leaves 5 bits for host addresses, resulting in 32 IP addresses per subnet.
192.168.50.96/27
Why it's wrong here
This is wrong because that block begins above the host address.
When this WOULD be correct
If the question asked which subnet contains the IP address 192.168.50.96, then option C would be correct, as that address would belong to the subnet 192.168.50.96/27, which covers addresses from 192.168.50.96 to 192.168.50.127.
192.168.50.0/27
Why it's wrong here
This is wrong because that block covers only .0 through .31.
When this WOULD be correct
If the question asked which subnet contains the network address of a different host, such as one configured with an IP address in the range of 192.168.50.0 to 192.168.50.31, then option D would be correct as it would represent the subnet for that specific host.
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 falls within the .64 through .95 block.
Why this is wrong here
The subnet 192.168.50.32/27 covers addresses 192.168.50.32 to 192.168.50.63. The host address 192.168.50.94 is outside this range, so it does not belong to this subnet.
★ When this WOULD be the correct answer
If the question asked which subnet contains the IP address 192.168.50.40, then option A (192.168.50.32/27) would be correct, as this subnet would encompass the host IP within its range.
Why candidates choose this
Students might mistakenly think that because 94 is close to 64, it could be in the .32 subnet, but they forget to calculate the broadcast address correctly.
Why this is wrong here
The subnet 192.168.50.96/27 covers addresses 192.168.50.96 to 192.168.50.127. The host address 192.168.50.94 is below this range, so it does not belong to this subnet.
★ When this WOULD be the correct answer
If the question asked which subnet contains the IP address 192.168.50.96, then option C would be correct, as that address would belong to the subnet 192.168.50.96/27, which covers addresses from 192.168.50.96 to 192.168.50.127.
Why candidates choose this
Students may incorrectly think that because 94 is close to 96, it could be in the .96 subnet, but they overlook that the subnet starts at .96, not .95.
Why this is wrong here
The subnet 192.168.50.0/27 covers addresses 192.168.50.0 to 192.168.50.31. The host address 192.168.50.94 is far above this range, so it does not belong to this subnet.
★ When this WOULD be the correct answer
If the question asked which subnet contains the network address of a different host, such as one configured with an IP address in the range of 192.168.50.0 to 192.168.50.31, then option D would be correct as it would represent the subnet for that specific host.
Why candidates choose this
Students might choose this option if they incorrectly calculate the subnet mask or assume the host is in the first subnet without proper calculation.
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
Always calculate the subnet range using the block size derived from the prefix length to avoid choosing incorrect boundaries.
Detailed technical explanation
Subnetting is a fundamental concept in IP networking that divides a larger network into smaller, manageable subnetworks. A subnet mask, such as /27, defines how many bits are used for the network portion and how many for the host portion of an IP address. In the case of a /27 mask, 27 bits are allocated to the network, leaving 5 bits for host addresses, which results in 32 IP addresses per subnet (2^5 = 32). These addresses include network, usable host, and broadcast addresses. To determine which subnet contains a specific host IP address, you calculate the block size by subtracting the prefix length from 32 and raising 2 to that power. For /27, the block size is 32 addresses. Starting from 0, subnets increment by 32 in the last octet: 0–31, 32–63, 64–95, 96–127, etc. Since the host IP 192.168.50.94 falls within the 64–95 range, it belongs to the 192.168.50.64/27 subnet. A common exam trap is selecting a subnet boundary close to the host IP without calculating the block size properly. For example, choosing 192.168.50.96/27 because 94 is near 96 is incorrect since 96–127 is the next subnet block. Understanding how to convert prefix length to block size and then mapping the host IP to the correct subnet range is critical for accurate subnetting in Cisco exams and real-world networking.
Key takeaway
A /27 subnet mask allocates 27 bits to the network portion and leaves 5 bits for host addresses, resulting in 32 IP addresses per subnet.
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 /27 subnet mask allocates 27 bits to the network portion and leaves 5 bits for host addresses, resulting in 32 IP addresses per subnet., 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 /27 subnet mask allocates 27 bits to the network portion and leaves 5 bits for host addresses, resulting in 32 IP addresses per subnet..
The correct answer is: 192.168.50.64/27 — A /27 subnet has a block size of 32. In simple terms, the fourth-octet ranges are 0–31, 32–63, 64–95, 96–127, and so on. Because 94 falls inside the 64–95 range, the network address for the host’s subnet is 192.168.50.64/27. This kind of question tests whether you can move from prefix length to block size and then place the host inside the correct interval. The most common mistake is choosing a nearby boundary like 96 or 32 without calculating the actual block that contains the address.
Review a /27 subnet mask allocates 27 bits to the network portion and leaves 5 bits for host addresses, resulting in 32 IP addresses per subnet., then practise related 200-301 questions on the same topic to reinforce the concept.
A /27 subnet mask allocates 27 bits to the network portion and leaves 5 bits for host addresses, resulting in 32 IP addresses per subnet.
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Last reviewed: May 17, 2026
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