Refer to the exhibit.
The internetwork is using subnets of the address 192.168.1.0 with a subnet mask of
255.255.255.224. The routing protocol in use is RIP version 1. Which address could be assigned
to the FastEthernet interface on RouterA?
A.
192.168.1.31
B.
192.168.1.64
C.
192.168.1.127
D.
192.168.1.190
E.
192.168.1.192
/27
Network Hosts Broadcast Address
from to
192.168.1.0 192.168.1.1 192.168.1.30 192.168.1.31
192.168.1.32 192.168.1.33 192.168.1.62 192.168.1.63
192.168.1.64 192.168.1.65 192.168.1.94 192.168.1.95
192.168.1.96 192.168.1.97 192.168.1.126 192.168.1.127
192.168.1.128 192.168.1.129 192.168.1.158 192.168.1.159
192.168.1.160 192.168.1.161 192.168.1.190 192.168.1.191
192.168.1.192 192.168.1.193 192.168.1.222 192.168.1.223
192.168.1.224 192.168.1.225 192.168.1.254 192.168.1.255
128+64+32=224 which means there is 3 mask bits and gives a total of 8 subnets
256-224=32 is the block size for the subnets.
Then you figure out which are usable ip addresses for the answer.