subnets

Question

The subnet mask for a particular network is 255.255.252.0. Which of the following pairs of
IP addresses could belong to this network?
A. 172.57.88.62 and 172.57.87.233
B. 10.35.28.2 and 10.35.29.4
C. 191.203.31.87 and 191.234.31.88
D. 128.8.129.43 and 128.8.131.42

Answer 1

Option b as in this pairs both have the same network id as 10.35.28.0

Comments

Yes may be,

128.8.129.43 & 255.255.252.0 = 128.8.128.0

128.8.131.42 & 255.255.252.0 = 128.8.128.0

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and above it it belongs to class B.

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I think you are right brother.

Answer 2

Answer (B) and (D)

Host ID - 10 bit

B) Both IP have Same Subnet (Initial 22 bit are same)

D) Both IP Have Same Subnet (Initial 22 bit are same) 

Answer 3

Ans is B and D .

Here the Subnet mask is 255.255.252.0 which is equal to  /22 , so easy way to check is to see if for both the IP address in pairs first 22 network bits are exactly same then they belongs to same network

For A part  :
 172.57.88.62 in binary is 172.57.01011000.62 and 172.57.87.33 is 172.57.01010111.233 first 22 bits are different

For B part :
10.35.28.2 in binary is 10.35.00011100.2  and 10.35.29.4 is 10.35.00011101.4 first 22 bits are exactly same therefore this is one of the pair

For C part :
The second octet is different in 191.203.31.87 and 191.234.31.88 therefore first 22 bits cant be same , hence this IP address don;t belongs to one network with SM /22

For D part:
128.8.129.43 in binary form is 128.8. 1000001.43 and 128.8.131.42 in binary form is 128.8.10000011.42 , first 22 bits are same , therefore both the IP address belongs to same subnet with SM/22 i.e 255.255.252.0

Hence B and D are correct answers.

Correct me If I am wrong , if anyone can suggest any other fast method please do advise