You have a Class B address block starting with 150.x.x.x and need at least 16 subnets (6 existing sites + 10 planned). Which subnet mask maximizes hosts per subnet while meeting this requirement?

Introduction:
Subnetting trades host capacity for additional network segments. For a Class B network (e.g., 150.x.x.x), choosing the right mask depends on the required number of subnets and the desire to maximize hosts per subnet.

Given Data / Assumptions:

• Base class: Class B (default mask 255.255.0.0, i.e., /16).
• Required subnets: at least 16 (6 current + 10 planned).
• Objective: meet subnet count with the fewest borrowed bits to keep more host addresses per subnet.

Concept / Approach:
In Class B, subnetting borrows bits from the third octet first. To get ≥16 subnets, borrow 4 bits (since 2^4 = 16). A /16 becomes /20. The corresponding third-octet value is 240, yielding the dotted mask 255.255.240.0, maximizing hosts while meeting exactly 16 subnets.

Step-by-Step Solution:
1) Start with /16 (255.255.0.0).2) Need 16 subnets → borrow 4 bits: /16 + 4 = /20.3) Third-octet mask with 4 high bits set is 240.4) Final mask: 255.255.240.0 → correct choice corresponds to “240.”

Verification / Alternative check:
Hosts per /20 subnet = 2^(32-20) - 2 = 4094, which is ample for most sites and larger than alternatives that borrow more bits.

Why Other Options Are Wrong:

• 192 (255.255.192.0): borrows only 2 bits → 4 subnets (insufficient).
• 224 (255.255.224.0): borrows 3 bits → 8 subnets (still insufficient).
• 248 (255.255.248.0): 5 bits → 32 subnets (extra but fewer hosts per subnet than necessary).
• 252 (255.255.252.0): 6 bits → 64 subnets (even smaller host pools).

Common Pitfalls:
Forgetting the goal to maximize host counts leads to borrowing more bits than needed. Always match the minimum required subnet count to minimize host loss.

Final Answer:
240