1,885 views
1 votes
1 votes
Given a mask, M=255.255.255.248. How many subnet bits are required for given mask M?

(A) 2 (B) 3 (C) 4 (D)5

6 Answers

2 votes
2 votes
Given mask is $255.255.255.248$. That is, $255.255.255.11111000$.

Here, number of mask bits are $24 + 5 = 29$.

To find the number of subnet bits, look at the $last \space 8 \space bits$ of the subnet mask. The number of subnet bits are the number of $1$'s in the final byte. Here, we have $five$ $1$'s.

So, the answer is $(D)$.
1 votes
1 votes
In this question the mask is 255.255.255.248

It means it belongs to the Class C Network.

24 bits+ 5 bits=29 bits

no of subnets= 2^5=32

no of host= 2^3=8
0 votes
0 votes
3 bits are required as when you write the binary format the last 3 digits will be 0 hence referring to 8 hosts per network
0 votes
0 votes
subnet mask - 255.255.255.248 which means 11111111.11111111.11111111,11111000

we have 32 bit addressing system and network ID part is represented by 1 and host ID is by 0

so therefore NID part is 29 bit so, hence remaining is  HID part i.e.32-29= 3

so answer is 3 bits

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