Quick Answer
The correct ACL entry is deny ip 192.168.1.0 0.0.0.255 any, because in Cisco ACL syntax, the wildcard mask 0.0.0.255 acts as an inverse of a subnet mask, where zeros indicate bits that must match exactly and ones indicate bits that can vary. This specific wildcard mask matches the entire 192.168.1.0/24 subnet, blocking all IP traffic from 192.168.1.1 through 192.168.1.255 to any destination. On the CompTIA SecurityX CAS-004 exam, this tests your ability to translate subnet ranges into ACL wildcard masks, a common task in firewall configuration scenarios. A frequent trap is confusing the wildcard mask with a standard subnet mask—remember that a /24 subnet uses a wildcard mask of 0.0.0.255, not 255.255.255.0. A helpful memory tip: think of the wildcard mask as telling the ACL “which bits to ignore,” so zeros mean “check this bit,” and ones mean “skip it.”
This CAS-004 practice question tests your understanding of security engineering. 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 network administrator is configuring a firewall to block traffic from a specific IP address range. The firewall uses ACLs. Which ACL entry would deny traffic from 192.168.1.0/24?
deny ip 192.168.1.0 0.0.0.255 any
Wildcard 0.0.0.255 matches the entire /24 subnet.
deny ip 192.168.1.0 0.0.0.0 any
Why wrong: Wildcard 0.0.0.0 matches only the specific host 192.168.1.0.
deny ip 192.168.1.0 0.0.0.127 any
Why wrong: Wildcard 0.0.0.127 matches a /25 subnet, not /24.
deny ip 192.168.1.0 255.255.255.0 any
Why wrong: ACLs use wildcard masks, not subnet masks.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
deny ip 192.168.1.0 0.0.0.255 any
Option A is correct because in Cisco ACL syntax, the wildcard mask 0.0.0.255 matches all addresses in the 192.168.1.0/24 network. The 'deny ip 192.168.1.0 0.0.0.255 any' entry blocks any IP traffic from the source subnet 192.168.1.0 through 192.168.1.255 to any destination.
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.
deny ip 192.168.1.0 0.0.0.255 any
Why this is correct
Wildcard 0.0.0.255 matches the entire /24 subnet.
Related concept
Read the scenario before looking for a memorised answer.
deny ip 192.168.1.0 0.0.0.0 any
Why it's wrong here
Wildcard 0.0.0.0 matches only the specific host 192.168.1.0.
deny ip 192.168.1.0 0.0.0.127 any
Why it's wrong here
Wildcard 0.0.0.127 matches a /25 subnet, not /24.
deny ip 192.168.1.0 255.255.255.0 any
Why it's wrong here
ACLs use wildcard masks, not subnet masks.
Common exam traps
The trap here is that candidates often confuse subnet masks with wildcard masks, selecting option D (255.255.255.0) instead of the correct wildcard mask 0.0.0.255.
Detailed technical explanation
Cisco ACLs use wildcard masks where 0 means 'must match' and 1 means 'ignore'. For a /24 network, the wildcard mask is 0.0.0.255 (binary: 00000000.00000000.00000000.11111111). A common real-world pitfall is confusing subnet masks with wildcard masks, especially when migrating from other firewall platforms that use CIDR notation directly.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
A small business has 20 workstations on the 192.168.1.0/24 network and one public IP from its ISP. The router uses PAT (NAT overload) so all 20 devices share one public address using different source ports. NAT questions test whether you understand the four address terms and which direction each translation applies.
What to study next
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FAQ
Security Engineering — This question tests Security Engineering — Read the scenario before looking for a memorised answer..
The correct answer is: deny ip 192.168.1.0 0.0.0.255 any — Option A is correct because in Cisco ACL syntax, the wildcard mask 0.0.0.255 matches all addresses in the 192.168.1.0/24 network. The 'deny ip 192.168.1.0 0.0.0.255 any' entry blocks any IP traffic from the source subnet 192.168.1.0 through 192.168.1.255 to any destination.
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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