UPPCL AE 2018:40

Question

If in $\text{CIDR}$ notation an $\text{IP}$ address is written as $172.26.17.1/25$, the subnet mask is:

• A. $255.255.255.192$
• B. $255.255.255.0$
• C. $255.255.240.0$
• D. $255.255.255.128$

Answer 1

 CIDR general notation format:$a.b.c.d/n$ where n is the number of bits that are presented in network id.

• NID+SID= n bits
• HID=(32-n) bits

Subnet mask:  it is $32$ bit number in which the number of 1’s indicates network id and subnet id, number of 0’s indicates host id part.

• NID+SID=all 1’s
• HID=all 0’s

Given IP address is $172.26.17.1/25$, here /25 represents the first 25 bits are all 1’s. that is

$172.26.17.1$ can be represented in binary form:

$11111111.11111111.11111111.10000000$, last $7$ 0’s represented as HID.

Therefore SM would be: $255.255.255.128$

So option $(D)$ is correct.

Ref:CIDR

Answer 2

CIDR Representation:    172.26.17.1/25                                                /25 shows no of bit in Network Id

 

Subnet Mask=1 in Network Id Pat and 0 in Host Id Part.

Host ID part=32-25

=7

i.e 25 bit Network Id + 7 bit Host Id

So Subnet Mask= 11111111.11111111.11111111.10000000

                             =255.255.255.128

Hence,D is the correct Option.