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The subnet mask for a particular network is $255.255.31.0.$ Which of the following pairs of $\text{IP}$ addresses could belong to this network?

1. $172.57.88.62$ and $172.56.87.23$

2. $10.35.28.2$ and $10.35.29.4$

3. $191.203.31.87$ and $191.234.31.88$

4. $128.8.129.43$ and $128.8.161.55$

Need not be @s_dr_13.

Theoretically we can set the bits in anywhere in the remaining unset bits. But practically we don't do it though.
The question is not valid as the subnet mask given is not valid. The subnet mask is required to be contiguous. Try putting this for any NIC and you will not be able to.
Which one?

(A) and (C) are not the answers as the second byte of IP differs and subnet mast has $255$ for second byte.

Consider (B), (& for bitwise AND)

$10.35.28.2 \ \& \ 255.255.31.0 = 10.35.28.0 (28 = 11100_2)$
$10.35.29.4 \ \& \ 255.255.31.0 = 10.35.29.0 (29 = 11101_2)$
So, we get different subnet numbers

Consider (D).

$128.8.129.43 \ \& \ 255.255.31.0 = 128.8.1.0 (129 = 10000001_2)$
$128.8.161.55 \ \& \ 255.255.31.0 = 128.8.1.0 (161 = 10100001_2)$
The subnet number matches. So, (D) is the answer.

by

sir, we can only check the bits of third octet to conclude whether the ids belong to the same network or not, right?

Also, one more thing to note: If I have subnet masks with not contiguous 1’s. It is not possible to divide the subnets into explicit ranges. Such subnet masks are not allowed in CIDR notation as the subnets will not have continous IP addresses. See allowed masks in CIDR notation here.

Ex. Subnet Mask is 255.255.255.1, In this network, we will have two subnets as the last bit is subnet Id. One subnet will have all odd IP addresses (ex. 192.168.2.15)  and the other subnet with all even IP addresses (ex. 192.168.2.16). So we cannot represent the subnets by ranges (by ranges I mean say 192.168.2.0 – 192.168.2.127, we cannot do this here) when we have subnet masks with not contiguous 1’s.

@Arjun sir, why are you trying to make same subnet?? The question says subnet mask is 255.255.31.0 and later it is asked that the hosts given in option whether they belong to the same network ( since network is subnetted so we must check whether the host bit doesn't have all 0's and all 1's and also subnet bits are not all 0's and all 1's )

it's nowhere mentioned that they must belong to same subnet, it's just same network????

I am solving here through elimination type procedure

The subnet mask 255.255.x.x belongs to class B type(IP address range will be from 128.1.0.1 to 191.255.255.254) so we can eliminate option B

Now we understand A, C, D belongs to class B network.

we will do a bit by bit AND operation of network and subnet mask to find the subnet.

A)172.57.88.62 and 172.56.87.23

C)191.203.31.87 and 191.234.31.88

D)128.8.129.43 and 128.8.161.55

clearly, we see the second octet of option A and C (57,56       203,234      are different.   57 & 255, 56 & 255 will give different values       203 & 255, 234 & 255 will give different values) are different so this implies that they don't belong to the same subnet.

Now we eliminated A B C then option D will be the answer.

still, if you want to check the answer for being safe side then follow below procedure

now for first option D the first two octets are same so you don't need to do  AND with 255 and waste time so now go to the third octet

129 =1000 0001     161=1010 0001

31=  0000 11111      31= 0000 11111

AND.............................................................

0000 0001             0000 0001

The third octet of subnets are same(first 2 octets are also same).

still not confident abt D then do fourth octet of network AND with fourth octet subnet mask.

Here anyhow fourth octet of mask is 0 so anything AND with 0 is zero

So the conclusion is, D is the correct answer.

There is no carry in doing AND operation. it is just bit by bit AND.

Awesome! :)

Please explain this line The subnet mask 255.255.x.x belongs to class B type(IP address range will be from 128.1.0.1 to 191.255.255.254) so we can eliminate option B

How did you wrote the range of ip addresses? @saiteja chekka

but they didn’t mentioned that it is classful IP addressing or classless IP Addressing

The subnet 255.255.31.0  belongs to Class B, since its net id is 255.255 .

So, we can conclude that either option could be (A) or (D).

Now, the question asks that the the IP should be  same network, so, to test that the IP are in same network we check that whether their netwrok address is same, by using AND operation.

In case of option A, we can see that when we apply AND operation to first byte of IP and first byte of MASK both the IP will give same, but when we apply AND operation is applied on second byte then both the IP gives different. So, option A is eliminated and hence Option D is selected because both IP has first and second IP as same.

ANS: D

Class A can also have Subnet Mask starting with 255.255

Please do not post wrong explanation!
Wrong explanation. They have given subnet mask, not network mask. Don't just reach the answer by giving any random explanation.

Here,

255 is having 11111111 LSB is 1,

31 is having 00011111 that means LSB 1, which means When we AND anything it will depend on the LSB of data we are ANDing.

Now in option a) is 57(odd) LSB 1 and 56(even) LSB 0 that means when we AND this with above SUBNET Mask we will get different result for sure..So, Wrong

b) 28 & 29 again same odd and even, Wrong

c)  203 & 234 odd and even same again, Wrong

d) 129 & 161 both are odd, Hence Option left so Answer