If IP address: 199.141.27.23 Subnet mask: 255.255.255.240 Which of the following addresses can be allocated to hosts on the resulting subnet?Question 1Answera.199.141.27.30b.199.141.27.8c.199.141.27.13d.199.141.27.12

To determine which of the given addresses can be allocated to hosts on the resulting subnet, we need to perform a bitwise AND operation between the IP address and the subnet mask.

First, let's convert the IP address and subnet mask to binary:

IP address: 199.141.27.23 -> 11000111.10001101.00011011.00010111 Subnet mask: 255.255.255.240 -> 11111111.11111111.11111111.11110000

Next, we perform the bitwise AND operation:

11000111.10001101.00011011.00010111 AND 11111111.11111111.11111111.11110000

11000111.10001101.00011011.00010000

The resulting binary value is 11000111.10001101.00011011.00010000, which is equivalent to 199.141.27.16 in decimal notation.

Now, we need to determine the range of valid host addresses within this subnet. Since the subnet mask is 255.255.255.240, it means that the last 4 bits of the IP address are used for host addresses.

The range of valid host addresses is from the network address + 1 to the broadcast address - 1.

The network address is 199.141.27.16, so the first valid host address is 199.141.27.17.

The broadcast address is obtained by setting all the host bits to 1. In this case, the broadcast address is 199.141.27.31.

Therefore, the valid host addresses on the resulting subnet are:

a. 199.141.27.30 b. 199.141.27.8 c. 199.141.27.13 d. 199.141.27.12

So, the correct answer is a. 199.141.27.30, b. 199.141.27.8, c. 199.141.27.13, and d. 199.141.27.12.