N10-005 Question #622: Real Exam Question with Answer & Explanation

The correct answer is A: 255.255.255.0. Only a /24 subnet mask (255.255.255.0) places both 192.168.0.3 and 192.168.0.240 in the same network range (192.168.0.0 to 192.168.0.255).

Question

Which of the following subnet masks would allow for 192.168.0.3 and 192.168.0.240 to be in the same subnet?

Options

A255.255.255.0
B255.255.255.128
C255.255.255.192
D255.255.255.248

Explanation

Only a /24 subnet mask (255.255.255.0) places both 192.168.0.3 and 192.168.0.240 in the same network range (192.168.0.0 to 192.168.0.255).

Common mistakes.

B. A 255.255.255.128 (/25) mask divides the space into 192.168.0.0-127 and 192.168.0.128-255, placing .3 and .240 in separate subnets.
C. A 255.255.255.192 (/26) mask creates four /26 blocks - .3 falls in the 0-63 block and .240 falls in the 192-255 block, so they are in different subnets.
D. A 255.255.255.248 (/29) mask creates very small 8-host blocks - .3 is in the 0-7 block and .240 is in the 240-247 block, placing them in entirely different subnets.

Concept tested. Subnet mask selection to include hosts in same network

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

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