Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 11)
11.
You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?
Answer: Option
Explanation:
A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.
Discussion:
13 comments Page 2 of 2.
Nick said:
8 years ago
I think D is the right answer.
192.168.19.24/29 --> 11000000.10110000.00010011.00011000 with the first 5 bits of the 4th octet defining the sub net 24 and last 3 bits defining the host number. So if the next sub net is 32--> 11000000.10110000.00010011.00100000 then the broadcast address for sub net 24 would be 11000000.10110000.00010011.00011111.
Am I missing something?
192.168.19.24/29 --> 11000000.10110000.00010011.00011000 with the first 5 bits of the 4th octet defining the sub net 24 and last 3 bits defining the host number. So if the next sub net is 32--> 11000000.10110000.00010011.00100000 then the broadcast address for sub net 24 would be 11000000.10110000.00010011.00011111.
Am I missing something?
Nani said:
9 years ago
How the block size is 8 in the fourth octet?
Bhadana said:
9 years ago
Subnet id is 192.168.19.24/29 or mask is 255.255.255.248.
First host of the subnet is 11000000.10110000.00010011.00011001 that is 192.168.19.25/29.
First host of the subnet is 11000000.10110000.00010011.00011001 that is 192.168.19.25/29.
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