GO Classes DPP | CN | SUBNETTING

Answer 1

Valid is A,  C

Explanation:- 150 = 1001 0110

                        175 = 1010 1111

So till 2nd bit we can take subnet mask M , i.e 128 or 192

Answer 2

Answer is b,d 

• Both x and y are from class c.
• So now to keep them in the same subnet we need the bits which are same in both x and y ip addresses host part so that we can differentiate the subnet part and host part. 
• Look only the host part of both x and y  i,e. 150 and 175 we can have common upto first 2 bits msb (01).
• So now to keep them in the same subetwork we can subnet masks as option A, C which are valid.
• But in the last part of the question they asked for invalid M. So options B,D are correct.

Answer 3

Correct answer is B,D 

Answer 4

• A. 255.255.255.192 (/26): Block size = 64 → subnet range 128–191. Both 192.168.10.150 and 192.168.10.175 belong to this subnet, so this mask is valid.

• B. 255.255.255.224 (/27): Block size = 32 → subnets 128–159, 160–191, ... Here, 150 belongs to 128–159 while 175 belongs to 160–191. Hence, they are in different subnets, making this mask invalid.

• C. 255.255.255.128 (/25): Block size = 128 → subnet range 128–255. Both IP addresses lie in the same subnet, so this mask is valid.

• D. 255.255.255.240 (/28): Block size = 16 → subnets 144–159, 160–175, ... Here, 150 belongs to 144–159 and 175 belongs to 160–175, so they are in different subnets. Hence, this mask is invalid.

Therefore, the invalid subnet masks are B and D.
 

Silly mistake from my side 😅. The question asks for the invalid subnet masks, but while solving I identified the valid ones instead.