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An organization has a class C network and wants to form a subnet for four departments with hosts as follows:

A:72

B:35

C:20

D:18

what is the possible arrangements?

options:

a) for A: 255.255.255.128 for B: 255.255.255.192 for C & D: 255.255.255.223

b)for A 255.255.255.223 for B: 255.255.255.223 for C & D : 255.255.255.128

My approach for solution:

I just eliminated the rest options and i did it as follows,

for option a) 128=10000000 , so there is $2^{7} = 128$ hosts possible, so it can be for A , again 192= 1100000 so there are \$2^{6} = 64 hosts possible , but in 223 = 10110111 means only 4 hosts are possible.

so option A is not possible. But answer given is option A.

I know i am lacking some concept here , please care to correct :)
asked | 66 views
I think instead of 255.255.255.223, it should be 255.255.255.224
May be, is my approach correct?
Approach is correct but binary of 223 is not and even correct form will not satisfy the given condition.