Given the host IP address of 192.168.100.60 and a network mask of 255.255.255.224 (or /27), which address represents the correct network address for the referenced host?
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A. B. C. D.D
Network is a class C network: 192.168.100.32/27
Binary: 11000000.10101000.01100100.001 00000
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To find the network address for a given IP address and subnet mask, we need to perform a bitwise AND operation between the IP address and the subnet mask. This will give us the network address for the subnet.
In this case, the IP address is 192.168.100.60 and the subnet mask is 255.255.255.224 (or /27 in CIDR notation).
To perform the bitwise AND operation, we convert both the IP address and subnet mask to binary:
IP address: 11000000.10101000.01100100.00111100 Subnet mask: 11111111.11111111.11111111.11100000
Performing the bitwise AND operation:
11000000.10101000.01100100.00100000 (192.168.100.32)
The resulting binary value is 11000000.10101000.01100100.00100000, which translates to the IP address 192.168.100.32. Therefore, option D (192.168.100.32) represents the correct network address for the referenced host.