ace test series 2018

Question

Comments

In each networks you have 2^8 IP addresses possible so in total you will require 2^8 *4 ip addresses means in aggregated network 10 bits required for host id part and 22 for Super-net id part.

So supernet address will be 128.56.24.0/22

Answer 1

Just analyze binary representation of third octet of given IP addresses:

128.56.24.0 = 128.56.00011000.0

128.56.25.0 = 128.56.00011001.0

128.56.26.0 = 128.56.00011010.0

128.56.27.0 = 128.56.00011011.0

It is clear that first 22 bits are common for all.

The answer is C) 128.56.24.0/22