Difficulty: Medium
Correct Answer: 255.255.255.224
Explanation:
Introduction / Context:
Subnetting a Class C (default /24) network requires balancing two goals: (1) ensuring each subnet has enough usable host addresses and (2) producing at least the required number of subnets. This question asks for a subnet mask that yields at least five subnets while providing a minimum of 25 usable host addresses per subnet.
Given Data / Assumptions:
Concept / Approach:
For a mask /n, each subnet contains 2^(32 − n) total addresses and 2^(32 − n) − 2 usable addresses. The smallest subnet that still supports ≥ 25 hosts maximizes how many subnets can be carved from the /24. We will evaluate /27, /28, and finer masks and count how many such subnets fit within /24.
Step-by-Step Solution:
Check /28 (255.255.255.240): total 16, usable 14 → not enough for 25.Check /27 (255.255.255.224): total 32, usable 30 → meets the ≥ 25 requirement.How many /27 subnets fit into a /24? 2^(27 − 24) = 2^3 = 8 subnets → satisfies the need for five.Finer masks (/29, /30, etc.) reduce hosts below 25 and are invalid; larger masks like /26 waste capacity and reduce the number of available subnets compared with /27.
Verification / Alternative check:
The /27 subnets in a /24 start at .0, .32, .64, .96, .128, .160, .192, .224. Each has 30 usable hosts, which comfortably exceeds the target of 25. You can assign any five of these ranges to satisfy the requirement and still have room for growth.
Why Other Options Are Wrong:
255.255.255.0 (/24): No subnetting performed; only one subnet.
Common Pitfalls:
Forgetting to subtract 2 for network/broadcast, mixing total vs usable counts, or choosing an oversized subnet (e.g., /26) that unnecessarily reduces the number of available subnets.
Final Answer:
255.255.255.224
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