ISRO CS 2025 Q.41: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. The mask $\mathrm{255.255.255.192}$ has last octet $11000000$, so it has $26$ network bits (a $/26$ mask). The host part is $6$ bits, so each subnet spans $2^6=64$ addresses in the last octet.

2. Subnet boundaries in the last octet are $0,64,128,192$. Two hosts are in the same subnet only if their addresses agree in all $26$ network bits.

3. Computer A: last octet $126$ lies in $64$ to $127$, so A is in subnet $10.0.0.64/26$ (broadcast address $\mathrm{10.0.0.127}$).

4. Computer B: last octet $131$ lies in $128$ to $191$, so B is in subnet $10.0.0.128/26$.

5. The two hosts are in different subnets. Traffic between different IP subnets must be forwarded by a router, so option (c) is true.

6. Option (a) is false: $10.0.0.0/8$ is a private address block meant for local use. Hosts can use these addresses freely, and no allocation from IANA is needed for each user.

7. Option (b) is false: a class A address has the default mask $\mathrm{255.0.0.0}$, but a longer mask is allowed. Subnetting lets a class A address carry a $/26$ mask.

8. Option (d) is false: multicast addresses lie in $\mathrm{224.0.0.0}$ to $\mathrm{239.255.255.255}$, and $\mathrm{10.0.0.}x$ is a unicast private address.