For 115.64.4.0/22 why only b and d are correct I know it means the network has assigned the first 22 bits are assigned for host.. Can somoene explain me why only those 2 are part of it. Please help me.
a. 115.64.8.32 b 115.64.6.255 c 115.64.8.31 d 115.64.5.128
In the slash notation for IPs, the number after thes slash denotes the length of the subnet bitmask in bits. Therefore, "/22" is analogous to a subnet mask of 255.255.252.0, or put differently, the IPs in your subnet may vary only in the last 10 bits. The first 22 bits are fixed.
This leaves all 8 bits of the last number and also the two least significant bits of the number before. Thus, 115.64.4.0/22 covers the following range of IPs:
115.64.{4...7}.{0...255}