The 172.16.25.0/25 subnet provides 128 IP addresses (2^7 = 128), including network and broadcast.
To subnet a /25 network (128 addresses), we can break it into:
* 2 x /26 # each with 64 addresses (62 usable) ##Valid
* 4 x /27 # each with 32 addresses # Also valid, but let's verify options Let's analyze the options:
A: one /23 and one /27 ##Invalid. /23 is larger than /25 - can't derive a larger subnet from a smaller one.
B: four times /27 ##Possible. 4 × 32 = 128. But not the best or only answer - and option D is more precise.
C: two times /24 ##Invalid. /24 = 256 addresses; you can't divide a 128-address block into 2 larger ones.
D: two times /26 ##Each /26 = 64 addresses. Two such subnets exactly fit into a /25 network.
MTCNA Course Manual - Subnetting and Address Planning:
"To divide a /25, you may use two /26s, four /27s, or other equal parts as long as they total no more than the parent subnet." Rene Meneses MTCNA Guide - Subnet Design:
"A /25 subnet can be split into 2 /26s (64 IPs each) or 4 /27s (32 IPs each), depending on host requirements." Terry Combs Notes - IP Subnetting Exercises:
"Always check if the proposed subnet sizes logically fit within the assigned block."