Ans: Option D) None of the Above
A class B network address given: 130.50.0.0
Class B = 16 Bits NID (Network ID) and 16 bits HID (Host ID).
Given, 6 bits are reserved for subnet number. Therfore, total subnets = 2^6.
Remaining 10 bits for HID. So, _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ (are last 16 bits) , in which the least significant 10 bits will be 0 0. 0 0 0 0 0 0 0 1 (as we need the 1st valid host address of both 1st and 4th subnets), and the most significant 6 bits are for SIDs (Subnet IDs) [not marked in BOLD Places]
Now the weights of last 4 bits in 3rd Octet = 8 , 4, 2 & 1. Out of which, 2 & 1 are in Host Address. So remaining Subnet ID weights to be filled are for 8 and 4 with combinations : (00, 01, 10, 11)
Now filling the SIDs (Subnet IDs) [ →
For subnet 1 : 0 0 0 0 0 0 (and rest bold ones) → 130.50.0.1 (Ans)
For subnet 2 : 0 0 0 0 0 1 (and rest bold ones) → 130.50.4.1
For subnet 3 : 0 0 0 0 1 0 (and rest bold ones) → 130.50.8.1
For subnet 4 : 0 0 0 0 1 1 (and rest bold ones) → 130.50.12.1 (Ans)