Borrowing subnet bits from a Class C network for 20 hosts
Short answer
B Borrow three bits from the fourth octet, leaving five host bits for 30 usable hosts per subnet, which meets the requirement of at least 20 hosts.
Question
Key Takeaways
It tests the host-bit formula 2^h − 2 ≥ 20 and the trap of stopping at two borrowed bits, which also leaves enough hosts but does not maximize subnetting.
This subnetting question asks how many bits to borrow from a Class C fourth octet to provide at least 20 hosts. The correct answer is B: borrow three bits, leaving five host bits and 30 usable addresses per subnet.
Common mistake: Choosing A because 2^6 − 2 = 62 hosts meets the minimum, but it leaves only two borrowed bits and fewer subnets; the question expects the minimum host bits that satisfy 20 hosts, which requires borrowing three bits.
Expert Analysis
Start with the fourth octet's role A Class C network has a 24-bit default mask, so the fourth octet supplies all eight host bits. The scenario requires subdividing that space while keeping at least 20 usable hosts per subnet. That means some of the eight bits become subnet bits, and the remaining bits must be enough to produce 20 usable addresses after excluding the network and broadcast values.
Apply the usable-host formula
For h remaining host bits, usable hosts = 2^h − 2. Testing the relevant values: h = 5 gives 2^5 − 2 = 30, while h = 4 gives 2^4 − 2 = 14. Because 30 meets the 20-host minimum and 14 does not, the design needs at least five host bits. On a Class C fourth octet with 8 total bits, that leaves 8 − 5 = 3 bits to borrow for subnetting. Borrowing three bits creates 2^3 = 8 subnets, each with 30 usable host addresses.
Why option B is the intended answer Option B, three subnet bits, is the standard exam interpretation: leave the minimum number of host bits that satisfies the 20-host requirement, thereby maximizing subnet bits. Option A (two borrowed bits) leaves six host bits and 62 usable hosts; that also clears the raw host minimum, but it creates only four subnets and leaves more host space than required. Options C and D fail because four borrowed bits leave only four host bits (14 usable hosts), and five borrowed bits leave three host bits (6 usable hosts), both below the 20-host requirement.
Exam takeaway Work from the host requirement first, not from a desired subnet count. Once the minimum host bits are identified, subtract that value from the Class C host width. The trap is that “at least 20” can be satisfied by more than one host-bit count; the exam expects the efficient design that meets the host minimum while maximizing subnetting. That path points to B.
Official Reference
https://www.rfc-editor.org/rfc/rfc950.html
This link opens the official vendor documentation in a new tab.
Exam Strategy
When a subnetting item gives a minimum host count, convert it to host bits first: find the smallest h where 2^h − 2 meets the requirement, then subtract h from the host portion. For a Class C fourth octet, that subtraction is 8 − h. Watch for options that also satisfy the host count with more host bits, because exam wording usually asks for the efficient subnet design that maximizes subnets while meeting the host minimum.
Frequently Asked Questions
Why isn't two subnet bits correct if 2^6 − 2 gives 62 hosts?
How many usable hosts remain after borrowing three bits from a Class C /24?
Do the network and broadcast addresses count toward the 20-host requirement?
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 1 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 1
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ERB14 👍 1 Answered BThe correct answer is B. Three.
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