How many subnets are possible with a Class B address and a subnet mask of 255.255.255.128?
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255
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256
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254
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512
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510
Class B addresses have a default mask of 255.255.0.0 (/16). The given mask 255.255.255.128 is /25, meaning 9 bits are borrowed for subnetting (25-16=9). With 9 subnet bits, we get 2^9 = 512 possible subnets. However, using traditional subnetting rules that exclude the all-zeros and all-ones subnets, we get 512-2 = 510 usable subnets. Options A, B, C, and D are incorrect because they don't account for the correct number of borrowed bits or the subnet exclusion rule.
A Class B address has a default mask of 255.255.0.0 (/16). The given mask 255.255.255.128 is /25, so 25 - 16 = 9 bits are borrowed for subnetting. Under the classic (older) convention taught in CCNA-era material, the number of usable subnets is 2^9 - 2 = 510 (the all-zeros and all-ones subnets are excluded). Under the modern RFC 1878 / 'ip subnet-zero' convention, all 2^9 = 512 subnets are usable. Both 510 and 512 appear as options, so the answer is convention-dependent. Because this is phrased as a traditional textbook question, the intended and marked answer 510 (subtract-2 convention) is correct within that framework.