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 /28 subnet mask allocates 28 bits to the network portion and leaves 4 bits for host addresses, creating subnets with 16 IP addresses each.. 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 192.168.22.145/28. Which subnet contains that host?
192.168.22.128/28
Why wrong: This is wrong because that block ends at .143.
192.168.22.144/28
This is correct because .145 falls within the .144 through .159 block.
192.168.22.160/28
Why wrong: This is wrong because that block begins above the host address.
192.168.22.148/28
Why wrong: This is wrong because /28 boundaries do not start at .148.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
192.168.22.144/28
A /28 prefix creates address blocks of 16. In practical terms, the fourth-octet ranges are 0–15, 16–31, 32–47, and so on. Because 145 falls inside the 144–159 range, the network address of the containing subnet is 192.168.22.144/28. This type of question checks whether you can move from prefix length to block size and then place a host into the correct interval. The common mistake is choosing a nearby familiar number instead of calculating the actual block boundary.
Key principle: A /28 subnet mask allocates 28 bits to the network portion and leaves 4 bits for host addresses, creating subnets with 16 IP addresses each.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
192.168.22.128/28
Why it's wrong here
This is wrong because that block ends at .143.
When this WOULD be correct
If the question asked for the subnet containing the address 192.168.22.128/28, then option A would be correct, as it directly represents the subnet range from 192.168.22.128 to 192.168.22.143.
192.168.22.144/28
Why this is correct
This is correct because .145 falls within the .144 through .159 block.
Related concept
A /28 subnet mask allocates 28 bits to the network portion and leaves 4 bits for host addresses, creating subnets with 16 IP addresses each.
192.168.22.160/28
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 for the subnet that contains a host address of 192.168.22.160 with a subnet mask of /28, then option C would be the correct answer. This scenario would involve identifying the subnet for a host at the upper end of the range.
192.168.22.148/28
Why it's wrong here
This is wrong because /28 boundaries do not start at .148.
When this WOULD be correct
If the question were to ask for a specific host address within the subnet 192.168.22.144/28, then option D (192.168.22.148/28) would be correct as it is a valid host address within that 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 .145 falls within the .144 through .159 block.
Why this is wrong here
The subnet 192.168.22.128/28 includes addresses 192.168.22.128 to 192.168.22.143. The host address 192.168.22.145 is outside this range, so it does not belong to this subnet.
★ When this WOULD be the correct answer
If the question asked for the subnet containing the address 192.168.22.128/28, then option A would be correct, as it directly represents the subnet range from 192.168.22.128 to 192.168.22.143.
Why candidates choose this
Students might mistakenly think that any address starting with 192.168.22.1xx could be in the .128 subnet, not realizing that the /28 mask creates a block of 16 addresses ending at .143.
Why this is wrong here
The subnet 192.168.22.160/28 includes addresses 192.168.22.160 to 192.168.22.175. The host address 192.168.22.145 is below this range, so it cannot be in this subnet.
★ When this WOULD be the correct answer
If the question asked for the subnet that contains a host address of 192.168.22.160 with a subnet mask of /28, then option C would be the correct answer. This scenario would involve identifying the subnet for a host at the upper end of the range.
Why candidates choose this
A student might confuse the subnet boundaries and think that .145 is close to .160, or they might incorrectly calculate the subnet size and assume .145 falls into the next subnet.
Why this is wrong here
The subnet 192.168.22.148/28 is not a valid subnet because the network address must be a multiple of the subnet size (16). Valid network addresses for /28 are 0, 16, 32, 48, etc. 148 is not a multiple of 16, so this is not a valid subnet.
★ When this WOULD be the correct answer
If the question were to ask for a specific host address within the subnet 192.168.22.144/28, then option D (192.168.22.148/28) would be correct as it is a valid host address within that subnet.
Why candidates choose this
Students might think that any address can be a network address, or they might incorrectly calculate the subnet boundaries and assume .148 is a valid starting point because it is close to .145.
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
Avoid assuming a host belongs to a subnet without calculating the exact range. Always verify the block size and boundaries.
Detailed technical explanation
Subnetting divides an IP network into smaller, manageable segments by borrowing bits from the host portion of the address to create subnetworks. A /28 prefix means the subnet mask is 255.255.255.240, which allocates 28 bits for the network and leaves 4 bits for host addresses. This results in subnets with 16 IP addresses each, including network and broadcast addresses. To determine which subnet contains a given host IP, calculate the subnet block size using the prefix length. For /28, the block size is 16 addresses. Starting from zero, subnets increment by 16 in the fourth octet: 0–15, 16–31, 32–47, and so forth. Since 192.168.22.145 falls between 144 and 159, it belongs to the 192.168.22.144/28 subnet. A common exam trap is selecting a subnet based on a close but incorrect boundary, such as 192.168.22.128/28, which ends at .143 and excludes .145. Understanding subnet boundaries and block sizes is critical in Cisco CCNA subnetting questions to avoid misclassification of host addresses and ensure accurate network segmentation in practical Cisco environments.
Key takeaway
A /28 subnet mask allocates 28 bits to the network portion and leaves 4 bits for host addresses, creating subnets with 16 IP addresses each.
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 /28 subnet mask allocates 28 bits to the network portion and leaves 4 bits for host addresses, creating subnets with 16 IP addresses each., 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 /28 subnet mask allocates 28 bits to the network portion and leaves 4 bits for host addresses, creating subnets with 16 IP addresses each..
The correct answer is: 192.168.22.144/28 — A /28 prefix creates address blocks of 16. In practical terms, the fourth-octet ranges are 0–15, 16–31, 32–47, and so on. Because 145 falls inside the 144–159 range, the network address of the containing subnet is 192.168.22.144/28. This type of question checks whether you can move from prefix length to block size and then place a host into the correct interval. The common mistake is choosing a nearby familiar number instead of calculating the actual block boundary.
Review a /28 subnet mask allocates 28 bits to the network portion and leaves 4 bits for host addresses, creating subnets with 16 IP addresses each., then practise related 200-301 questions on the same topic to reinforce the concept.
A /28 subnet mask allocates 28 bits to the network portion and leaves 4 bits for host addresses, creating subnets with 16 IP addresses each.
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Last reviewed: May 17, 2026
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