An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets. Find first and last addresses of last subnet.

1.

130.56.0.0

130.56.254.255

2.

130.56.255.0

130.56.255.255

3.

130.56.0.192

130.255.255.255

4.

130.56.255.192

130.56.255.255

Question

Question

This question was previously asked in

UGC NET Computer Science (Paper 2) 11 March 2023 Official Paper

Answer 1

The correct answer is 4

Key Points

Based on the previous question where we established the new subnet mask as /16 resulting in 64 addresses in each subnet:
The first address of each subnet is a multiple of 64 in the last octet, and the last address is 63 greater than the first address (because we start at 0).
The last subnet can be calculated by subtracting 64 (one subnet's worth of addresses) from the entire address space. So the last subnet's first address can be calculated as 256 - 64 = 192 (the "256" being representative of the total number of addresses accessible by 8 bits in the last section of the IP address), with the "192" being the starting point of the final subnet in the /26 partitioning.
Hence, the first address of the last subnet is: 130.56.255.192
The last address in the subnet is the first address plus 63, because there are 64 addresses in each subnet. That would be: 130.56.255.255

So, the answer is: 4) First address = 130.56.255.192, Last address = 130.56.255.255.