CCNA Network Infrastructure and Connectivity Practice Question
A host address is 192.168.22.145/28. Which subnet contains that host?
⚠ Common exam trap
Avoid assuming a host belongs to a subnet without calculating the exact range. Always verify the block size and boundaries.
Answer choices
Why each option matters
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.
Answer analysis
Option-by-option breakdown
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
The /28 subnet mask creates a block of 16 addresses, with network boundaries at multiples of 16. 192.168.22.145 falls within the range 192.168.22.144–159, making 192.168.22.144/28 the correct network address. 192.168.22.128/28 covers 192.168.22.128–143, so it excludes .145. This option is tempting because 192.168.22.128 is a common boundary for a /25 or /27 subnet, and one might mistakenly apply a different prefix length that aligns with .128 as the network address.
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
A /28 prefix length uses subnet mask 255.255.255.240, which creates blocks of 16 addresses. Dividing the fourth octet 145 by 16 gives 9.0625, so the network boundary is 9 × 16 = 144. Therefore, the subnet containing 192.168.22.145 is 192.168.22.144/28, which spans usable hosts .145 through .158 and has a broadcast address of .159. This matches the host being the first usable address in that range.
- ✗
192.168.22.160/28
Why it's wrong here
192.168.22.160/28 is the next consecutive /28 subnet, starting at the network address .160 and ending at the broadcast address .175. Since the host .145 is numerically lower than the .160 network boundary, it falls entirely outside this range; it resides in the preceding .144 subnet. A common error is to select the next boundary instead of the one that actually contains the host, but a /28 subnet always increments in fixed blocks of 16, so .145 is not in the .160 block.
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
192.168.22.148/28 is incorrect because it represents a host address within the subnet, not the subnet's network address itself. The question asks to identify the subnet that contains the host 192.168.22.145. For a /28 mask, subnets increment by 16; the subnet containing 192.168.22.145 starts at 192.168.22.144. This option is tempting as 192.168.22.148 is a valid assignable IP address within the 192.168.22.144/28 subnet, and would be a correct choice if the question requested a valid host IP.
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
Why each answer is right or wrong
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.
✓192.168.22.144/28Correct answer▾
Why this is correct
A /28 prefix length uses subnet mask 255.255.255.240, which creates blocks of 16 addresses. Dividing the fourth octet 145 by 16 gives 9.0625, so the network boundary is 9 × 16 = 144. Therefore, the subnet containing 192.168.22.145 is 192.168.22.144/28, which spans usable hosts .145 through .158 and has a broadcast address of .159. This matches the host being the first usable address in that range.
✗192.168.22.128/28Wrong answer — click to see why▾
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.
✗192.168.22.160/28Wrong answer — click to see why▾
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.
✗192.168.22.148/28Wrong answer — click to see why▾
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?”
Visual reference
Quick reference
IPv4 Address Class Summary
| Class | First Octet Range | Default Mask | Networks | Hosts per Network |
|---|---|---|---|---|
| A | 1–126 | /8 (255.0.0.0) | 126 | 16,777,214 |
| B | 128–191 | /16 (255.255.0.0) | 16,384 | 65,534 |
| C | 192–223 | /24 (255.255.255.0) | 2,097,152 | 254 |
| D | 224–239 | N/A | Multicast groups | — |
| E | 240–255 | N/A | Reserved / experimental | — |
127.x.x.x is reserved for loopback. Modern networks use CIDR (classless) rather than classful addressing.
Go deeper
Related to this question
Learn chapter
IPv4 Addressing and Address Classes
Key term
Prefix length
A prefix length is the number of bits in a subnet mask that identifies the network portion of an IP address, written after a slash (e.g., /24).
Key term
Block size
Block size is the amount of data, measured in bytes, that a system reads or writes in a single operation, affecting storage efficiency and network performance.
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JA
Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
This 200-301 practice question is part of Courseiva's free Cisco certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the 200-301 exam.