GATE 2024 CS2 Q.38: solution and answer

Official answer: ASource: Official answer key, GATE 2024 CS2 (IISc Bengaluru)

1. Count the addresses in the range 10.12.2.0 to 10.12.3.255. The third octet takes the values 2 and 3, and the fourth octet runs over 0 to 255 for each. That gives $2\times 256=512=2^9$ addresses.

2. A CIDR block of $2^9$ addresses has 9 host bits, so its prefix length is $32-9=23$. Only option (A) has a /23 prefix.

3. Check that 10.12.2.0 is a valid /23 network address. In binary the third octet is 00000010. A /23 prefix fixes the first 7 bits of the third octet (0000001), and the last bit is a host bit. That host bit is 0, so all host bits of 10.12.2.0 are zero and it is a valid /23 network address.

4. The block 10.12.2.0/23 covers third octets 0000001x, that is 2 and 3, with any fourth octet. It therefore spans 10.12.2.0 to 10.12.3.255, which is exactly the given range.

5. Check the other options. (B) 10.12.2.0/24 covers only 256 addresses (10.12.2.0 to 10.12.2.255), so it is too small. (C) 10.12.0.0/22 covers 1024 addresses (10.12.0.0 to 10.12.3.255), so it is too large. (D) 10.12.2.0/22 is also a 1024-address block, so it is too large. It is not even a proper network address, because a /22 block must start at a third octet that is a multiple of 4 (here 10.12.0.0). Only (A) matches the range exactly.