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SY0-501 · Question #42

SY0-501 Question #42: Real Exam Question with Answer & Explanation

The correct answer is B: 192.168.0.16/28. This question tests the ability to select the smallest subnet that can accommodate 14 physical hosts, minimizing wasted address space while meeting requirements.

Submitted by parkjh· Mar 4, 2026

Question

A security administrator is creating a subnet on one of the corporate firewall interfaces to use as a DMZ which is expected to accommodate at most 14 physical hosts. Which of the following subnets would BEST meet the requirements?

Options

  • A192.168.0.16 255.25.255.248
  • B192.168.0.16/28
  • C192.168.1.50 255.255.25.240
  • D192.168.2.32/27

Explanation

This question tests the ability to select the smallest subnet that can accommodate 14 physical hosts, minimizing wasted address space while meeting requirements.

Common mistakes.

  • A. The subnet mask 255.25.255.248 is invalid - the second octet '25' is not a valid subnet mask value; valid octets in a subnet mask must follow a contiguous bit pattern (e.g., 255.255.255.248).
  • C. The subnet mask 255.255.25.240 is also invalid because '25' in the third octet is not a legal subnet mask value, making this an unusable and incorrect subnet definition.
  • D. A /27 prefix provides 32 total addresses with 30 usable hosts, which accommodates the 14-host requirement but wastes significantly more address space than a /28, making it a suboptimal choice when the question asks for the BEST fit.

Concept tested. Subnetting to meet minimum host requirements efficiently

Reference. https://learn.microsoft.com/en-us/troubleshoot/windows-client/networking/tcpip-addressing-and-subnetting

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