For the class C network 192.168.10.0, which subnet mask yields exactly four subnets while maximizing hosts per subnet?

Introduction / Context:
Subnetting a class C network (default /24) to obtain a required number of subnets while maximizing hosts per subnet is a core IP addressing skill. Here we need four subnets with as many hosts as possible per subnet on 192.168.10.0.

Given Data / Assumptions:

• Base network: 192.168.10.0/24.
• Required subnets: 4.
• Goal: maximize hosts per subnet.

Concept / Approach:
Number of subnets with borrowed bits: subnets = 2^n. Number of hosts per subnet = 2^(host_bits) - 2. For a class C (/24), there are 8 host bits before borrowing. Borrow as few bits as necessary to reach 4 subnets to keep more host bits (thus maximizing hosts per subnet).

Step-by-Step Solution:

Verification / Alternative check:
Subnets would start at 192.168.10.0, .64, .128, .192 with broadcast addresses .63, .127, .191, .255 respectively, each supporting 62 usable host IPs. This satisfies the requirement of exactly 4 subnets with the maximum host count per subnet.

Why Other Options Are Wrong:

• 255.255.255.224 (/27): 8 subnets, only 30 hosts each.
• 255.255.255.240 (/28): 16 subnets, 14 hosts each.
• 255.255.255.248 (/29): 32 subnets, 6 hosts each.
• 255.255.255.252 (/30): 64 subnets, 2 hosts each (point-to-point).

Common Pitfalls:
Borrowing more bits than necessary (reduces hosts); forgetting to subtract 2 for network and broadcast; mixing up /26 vs /27 boundaries.

Final Answer:
255.255.255.192