Class C Subnetting for Multiple Branches
Short answer
B 255.255.255.224 provides 30 usable hosts per subnet and creates 8 subnets, satisfying both the 26-host maximum and 6-branch requirement.
Question
Key Takeaways
The trap is focusing solely on host capacity; you must also ensure the borrowed bits create enough subnets for all physical locations.
Determine the correct Class C subnet mask for a network with 6 branches needing up to 26 hosts each. The solution involves balancing subnet count and host capacity using VLSM principles.
Common mistake: Choosing Option D (255.255.255.192) because it offers plenty of hosts (62), but forgetting that it only yields 4 subnets, which is insufficient for 6 branches.
Expert Analysis
The question requires selecting a subnet mask that satisfies two constraints simultaneously: supporting at least six branches (subnets) and accommodating at least 26 usable hosts in the largest branch. This is a classic Variable Length Subnet Masking (VLSM) or fixed-subnet scenario where one Class C network must be divided.
First, analyze the host requirement. To support 26 usable hosts, we need enough host bits such that 2^h - 2 ≥ 26. If h = 4, 2^4 - 2 = 14 (insufficient). If h = 5, 2^5 - 2 = 30 (sufficient). Therefore, the subnet mask must leave at least 5 bits for hosts. A /27 mask (255.255.255.224) leaves 5 host bits, providing 30 usable addresses, which meets the host constraint.
Next, verify the subnet count. Using a /27 mask on a Class C address (/24) borrows 3 bits from the host portion (24 + 3 = 27). The number of subnets created is 2^3 = 8. Since the company has 6 branches, 8 subnets are sufficient to cover all locations.
Evaluate the other options:
- Option A (255.255.255.128 or /25) provides 126 hosts per subnet but only creates 2 subnets. Two subnets cannot support 6 branches.
- Option C (255.255.255.0 or /24) provides 254 hosts but creates only 1 subnet, failing the branch count.
- Option D (255.255.255.192 or /26) creates 4 subnets (insufficient for 6 branches) and provides 62 hosts per subnet. While the host count is adequate, the subnet count is not.
Thus, 255.255.255.224 is the only option that satisfies both the subnet quantity and host capacity requirements.
Exam Strategy
When solving subnetting questions with multiple constraints, always check the 'tightest' constraint first. In this case, calculate the minimum host bits required for the largest segment, then determine how many bits remain for subnetting. Verify that the remaining bits can generate enough subnets for all segments. If not, you may need a larger block or different addressing scheme.
Frequently Asked Questions
Why is 255.255.255.192 incorrect if it supports 62 hosts?
How do I calculate usable hosts quickly?
Community Votes
100% of anonymous learners picked answer B. Votes are pick records left by other test-takers — they are not the verified answer.
Discussion 4 contributions
Compiled from public community threads and reader submissions, lightly edited for clarity. Handles and vote tallies are shown as recorded, and comments reflect the views of individual contributors.
From public community threads 4
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Fmoura17 👍 2 Answered BCorrect Answer: /27 = 255.255.255.224 (Option B) actually satisfies both: 8 subnets (≥ 6 branches ✓) 30 usable hosts (≥ 26 hosts ✓) The correct answer is B. 255.255.255.224, as it provides enough subnets for 6 branches and enough hosts (30) for the largest branch.
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Heidorn 👍 2 Answered B/27 with 30 hosts to fit 27 users
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amiie00 👍 1 Answered B255.255.255.224 = /27 can use 30 hosts and /24 can separate /27 = 8 network. -- correct // 255.255.255.128 = /25 can use 126 hosts and /24 can separate /25 = 2 network. -- incorrect
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ButcherCapybara 👍 1 Answered B255.255.255.224 gives us 30 hosts, which is the most suitable from host and netwrok management
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