nielit 2018 q52

Question

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

Answer 1

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 $(D)$.

Comments

This is only when class is C..

But in the question, class is not given.

Hence, If class is A the subnet bits = 21

class B then subnet bits = 13

class C  then subnet bits = 5

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Yes, you are correct. But in the options, numbers below 8 are given. So it is fair to assume that it is class C I guess. Good that you pointed it out.

So let me say, the $\textit{minimum}$ number of subnet bits required for the given mask is, $5$.

Answer 2

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

Answer 3

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

Answer 4

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