A small manufacturing company has a Class C network address on the plant floor and needs to create five subnets, each accommodating 25 endpoints. Which subnet mask needs to be configured?

Question

Options

• A255.255.240.0
• B255.255.255.128
• C255.255.255.192
• D255.255.255.224
• E255.255.255.240
• F255.255.255.248

• A
  2% (1)
• B
  2% (1)
• D
  73% (30)
• E
  15% (6)
• F
  7% (3)

Explanation

Option D (255.255.255.224) works because borrowing 3 bits from the host portion of a Class C address yields 2³ = 8 subnets (satisfying the requirement of 5) while leaving 5 host bits, providing 2⁵ − 2 = 30 usable hosts per subnet (satisfying the requirement of 25).

Why the distractors fail:

• A (255.255.240.0) is a Class B-range mask that modifies the third octet - invalid for a Class C network.
• B (255.255.255.128) borrows only 1 bit, giving just 2 subnets - not enough for 5.
• C (255.255.255.192) borrows 2 bits, giving only 4 subnets - still one short.
• E (255.255.255.240) borrows 4 bits, giving 16 subnets but only 14 hosts per subnet - fails the 25-host requirement.
• F (255.255.255.248) borrows 5 bits, giving 32 subnets but only 6 hosts per subnet - far too few.

Memory tip: Use the "two-constraint squeeze" - you must satisfy both the subnet count (2^n ≥ required subnets) and the host count (2^h − 2 ≥ required hosts) simultaneously; in an 8-bit host field, n + h = 8, so adding more subnets always means fewer hosts, and 3+5 is the only split that clears both bars here.

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