# Question 123 of 530 from exam 400-251: CCIE Security written exam

### Question

In what subnet does address 192.168.23.197/27 reside?

### Answers

### Explanations

D.

To determine the subnet in which the given address 192.168.23.197/27 resides, we need to apply the subnetting rules.

The /27 in the address represents the subnet mask, which consists of 27 ones followed by 5 zeros in binary form:

`11111111.11111111.11111111.11100000`

To find the subnet address, we perform a bitwise AND operation between the given address and the subnet mask:

`makefile``Address: 11000000.10101000.00010111.11000101 (192.168.23.197) Mask: 11111111.11111111.11111111.11100000 (/27) Result: 11000000.10101000.00010111.11000000 (192.168.23.192) `

Therefore, the subnet address for the given IP address is 192.168.23.192.

Now we need to determine the range of IP addresses in the subnet. Since the subnet mask is /27, we know that there are 32 - 27 = 5 bits available for host addresses. This gives us a total of 2^5 = 32 possible host addresses in the subnet.

The first address in the subnet is the subnet address itself (192.168.23.192), and the last address is the broadcast address (192.168.23.223). Therefore, the valid host range for the subnet is 192.168.23.193 - 192.168.23.222.

We can see that the given address 192.168.23.197 falls within this range. Therefore, the subnet in which the address 192.168.23.197/27 resides is 192.168.23.192/27, which is option D in the list of answers.