Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 13)
13.
You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
Answer: Option
Explanation:
A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host.
Discussion:
13 comments Page 2 of 2.
Usha said:
6 years ago
As per question, we need valid host address and the correct option is D, can you explain how 172.16.18.255 is valid host address?
Gio said:
4 years ago
Let me explain.
We learned that subnets with CIDR /22 can skip at 4 size block.
So, how is possible that the question subnet is 17.0, when the only ones available are 16.0, 20.0 etc?
Please explain.
We learned that subnets with CIDR /22 can skip at 4 size block.
So, how is possible that the question subnet is 17.0, when the only ones available are 16.0, 20.0 etc?
Please explain.
(1)
Abera said:
3 years ago
D is the correct answer in the case of /22 =255.255.252.0 but A not.
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