Q1. A device has IP address 192.168.1.130 with subnet mask 255.255.255.192. What is the network address?
- A.192.168.1.0
- B.192.168.1.64
- C.192.168.1.128✓ Correct
- D.192.168.1.192
Explanation
A /26 mask (255.255.255.192) gives 64-host blocks: .0, .64, .128, .192. The address .130 falls into the .128 block, so the network address is 192.168.1.128. The last valid host in this subnet is .190 and the broadcast is .191.