• # How many subnets were created from the original Class B addressing space

Can anyone explain to me why the answer to this question is 1024?

How many subnets were created from the original Class B addressing space?

• I'm thinking there's some information missing from that question. I thought I would try a few different things to try to get to that number but I've arrive at many more than 1024.

If I just do 128.0.x.x -191.0.x.x/16 =63 or 64 subnets

If I try to do something like 128.1.x.x, 128.2,x.x/16 ... I could do 128,129.130,131 and that would give me close to the 1024 subnets. but that doesn't account for 132.1.x.x - 191.255.x.x/16

if we just count bits from the original class B numbers (128-191) to the first 16 bits that's 13 bits after the 1st 3 bits 2^13= 8192 possible...

So there is some critical piece of information missing or we need to see the other possible answers to this question. If you're getting this question from some internet sources, you may find the question answer may have just been simply mistaken. Or you'll need to contact the author of the question. I'm stumped on this one but looking forward to seeing the rest of the answers to this question.

• Thanks for confirming Ronnie. I was having the same struggle. I test tomorrow for CCENT, figners crossed.

• Prepared. Get good rest.

Read the question and every answer before trying to answer.

Do not hit next until you're sure you've answered what you want.

Pace yourself about 1 minute per question.

• if we just count bits from the original class B numbers (128-191) to the first 16 bits that's 13 bits after the 1st 3 bits 2^13= 8192 possible...

Aren't there 16384 possible Class B Networks? You've got 16-bits in the native Class B mask and 2 of those bits '10' (10000000-10111111) [128-191]) are fixed, leaving 2^14 bits for addressing?

• @Jason-Leschnik,
you're right, I miscounted the bits.

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