CCNA Network Services and Security Practice Question
An ACL entry reads:
access-list 25 permit 192.168.8.0 0.0.0.15
Which address range does this statement match?
⚠ Common exam trap
Be careful not to confuse the block size determined by the wildcard mask with a full subnet or miscalculate the starting address.
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
✓
192.168.8.0 through 192.168.8.15
A wildcard of 0.0.0.15 means the last 4 bits can vary, which corresponds to a block size of 16 addresses. Starting at 192.168.8.0, the range is 192.168.8.0 through 192.168.8.15.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
192.168.8.0 through 192.168.8.15
Why this is correct
The wildcard mask 0.0.0.15 (binary 00000000.00000000.00000000.00001111) fixes the first 28 bits and allows the last 4 bits of the fourth octet to vary. Because the network portion is 192.168.8.0, these 4 variable bits produce every address from .0 (00000000) to .15 (00001111), inclusive. Thus this ACL entry matches exactly the 16 addresses 192.168.8.0 through 192.168.8.15.
- ✗
192.168.8.0 through 192.168.8.31
Why it's wrong here
A wildcard mask of 0.0.0.31 would set the last five bits of the octet as wildcard bits, yielding the range 192.168.8.0 through 192.168.8.31 — a total of 32 addresses. However, the ACL entry specifies 0.0.0.15, not 0.0.0.31, so only the low four bits are free. The result is that this option broadens the match beyond what the mask actually permits, including addresses that the ACL would not match.
When this WOULD be correct
If the question were modified to read 'access-list 25 permit 192.168.8.0 0.0.0.31', then option B would be correct, as it would match the entire range from 192.168.8.0 to 192.168.8.31.
- ✗
192.168.8.0 through 192.168.8.7
Why it's wrong here
The correct range does indeed start at .0—not at .8—but the wildcard 0.0.0.15 lets the low four bits vary, so it extends all the way to .15. Stopping at .7 would require a wildcard mask of 0.0.0.7, because only the low three bits would be free. This option covers just the lower half of the match set and is therefore incomplete.
When this WOULD be correct
This option would be correct if the question specified that the ACL was intended to match only the addresses within the range of 192.168.8.8 to 192.168.8.15, perhaps in a context where a specific subnet mask was applied to limit the range.
- ✗
Only host 192.168.8.15
Why it's wrong here
A host entry matches exactly one IPv4 address and requires a wildcard mask of 0.0.0.0, as in 192.168.8.15 0.0.0.0 or the shorthand 'host 192.168.8.15'. The ACL given uses wildcard 0.0.0.15, which makes the final octet's low four bits variable, so it matches .15 only as one of sixteen addresses, not exclusively. Even though .15 falls inside the permitted range, the statement 'only host 192.168.8.15' is far too narrow and misstates how wildcard matching works.
When this WOULD be correct
In a different question, if the ACL entry specified 'access-list 25 permit 192.168.8.15 0.0.0.0', then option D would be correct, as it would permit only the host address 192.168.8.15.
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.
✓192.168.8.0 through 192.168.8.15Correct answer▾
Why this is correct
The wildcard mask 0.0.0.15 (binary 00000000.00000000.00000000.00001111) fixes the first 28 bits and allows the last 4 bits of the fourth octet to vary. Because the network portion is 192.168.8.0, these 4 variable bits produce every address from .0 (00000000) to .15 (00001111), inclusive. Thus this ACL entry matches exactly the 16 addresses 192.168.8.0 through 192.168.8.15.
✗192.168.8.0 through 192.168.8.31Wrong answer — click to see why▾
Why this is wrong here
The wildcard mask 0.0.0.15 matches only the last 4 bits, resulting in a range of 16 addresses (0-15). A wildcard of 0.0.0.31 would match 32 addresses (0-31), which is incorrect for this ACL.
★ When this WOULD be the correct answer
If the question were modified to read 'access-list 25 permit 192.168.8.0 0.0.0.31', then option B would be correct, as it would match the entire range from 192.168.8.0 to 192.168.8.31.
Why candidates choose this
Students often confuse wildcard masks with subnet masks, thinking that 0.0.0.15 corresponds to a /27 subnet (255.255.255.224) which has 32 addresses, but wildcard masks are the inverse of subnet masks.
✗192.168.8.0 through 192.168.8.7Wrong answer — click to see why▾
Why this is wrong here
This range corresponds to a wildcard mask of 0.0.0.7 (block size 8), not 0.0.0.15 (block size 16).
★ When this WOULD be the correct answer
This option would be correct if the question specified that the ACL was intended to match only the addresses within the range of 192.168.8.8 to 192.168.8.15, perhaps in a context where a specific subnet mask was applied to limit the range.
Why candidates choose this
A test-taker might mistakenly think the wildcard mask indicates the starting point or that the range is centered around the given address, leading to an incorrect starting offset.
✗Only host 192.168.8.15Wrong answer — click to see why▾
Why this is wrong here
A host match requires a wildcard mask of 0.0.0.0, not 0.0.0.15. The given ACL matches a range of 16 addresses, not a single host.
★ When this WOULD be the correct answer
In a different question, if the ACL entry specified 'access-list 25 permit 192.168.8.15 0.0.0.0', then option D would be correct, as it would permit only the host address 192.168.8.15.
Why candidates choose this
Students may see the last octet .15 in the wildcard mask and incorrectly assume it matches only the host with that specific value, confusing the wildcard mask with a host address.
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
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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.