Cidr

Question

Comments

Given an IP address of block: 210.69.92.39/26

Range of the block:
210.69.92.00 000000....................210.69.92. 00 111111

Last subnet is 210.69.92.11 000000
Second last subnet is 210.69.92.10 000000

Last host of second last subnet is 210.69.92.10111110 = 210.69.92.192

The subnet term seems confusing to me. What is the given solution?

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He don't mention in question how many subnet

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from where you get this question?

Answer 1

Given an IP address of block: 210.69.92.39/26
  so we are left with 6 bits for host id,whose 1st address is all six 0s and last address is all six ones.

so in one subnet maximum 2^6 = 64 addresses are possible.

now i am writing the blocks of addresses as per subnet.

1st block --   210.69.92.0  to 210.69.92.63 (the address given in the qs is resides in this block)

2nd block--    210.69.92.64  to 210.69.92.127

3rd block-- 210.69.92.128 to 210.69.92.191

4th and last block-- 210.69.92.192 to 210.69.92.255

so 2nd last block is 3rd block.. and the last host is which usable 210.69.92.190 /26. 

so answer is  A