Difficulty: Easy
Correct Answer: 255.255.255.0
Explanation:
Introduction / Context: In classical IPv4 addressing (pre–Classless Inter-Domain Routing), each address fell into a class with a default subnet mask. Recognizing the default mask for a given dotted-decimal address helps with quick network/broadcast calculations and legacy exam questions. Here, we analyze address 201.142.23.12 and identify its default mask prior to any subnetting changes.
Given Data / Assumptions:
Concept / Approach: Classful ranges by first octet are: Class A = 1–126 → default mask 255.0.0.0; Class B = 128–191 → default mask 255.255.0.0; Class C = 192–223 → default mask 255.255.255.0. Since 201 falls within 192–223, 201.x.y.z is Class C, so the default mask is 255.255.255.0.
Step-by-Step Solution:
Identify first octet: 201.Map 201 to class ranges: 192–223 → Class C.Recall default mask for Class C: 255.255.255.0.Select 255.255.255.0 as the correct answer.Verification / Alternative check: Convert 201 to binary (11001001). The leading bits for Class C are 110, which matches the 192–223 range and therefore the /24 (255.255.255.0) default network mask in classful notation.
Why Other Options Are Wrong:
0.0.0.0: not a default mask; often represents a default route or unspecified address context.255.0.0.0: Class A default; does not apply to 201.x.x.x.255.255.0.0: Class B default; does not apply to 201.x.x.x.None of the above: incorrect because 255.255.255.0 is correct.Common Pitfalls: Confusing classful defaults with designed subnet masks; mixing up Class B and C boundaries (191 vs 192). Remember, 201 clearly lies in Class C.
Final Answer: 255.255.255.0.
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