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
Explanation:
Subnet mask 255.255.255.224 with CIDR of /27 which results in 32 hosts per.
192.168.1.31 is the broadcast address for sunbet ‘0’
192.168.1.64 is the network address for subnet ‘2’
192.168.1.127 is the broadcast address for subnet ‘3’
192.168.1.192 is the network address for subnet ‘6’
the only correct answer is D
The increment for these subnets is 32 (256 – 224).
This results in the following subnets:
192.168.1.0
192.168.1.32
192.168.1.64
192.168.1.96
…
192.168.1.160
192.168.1.192
answer A is the broadcast address for the first subnet
answer B is the network address for the third subnet
answer C is the broadcast address for the fourth subnet
answer E is the network address for the 192.168.1.192 subnet
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