Which one of the following IP addresses is the last valid host in the subnet using mask
255.255.255.224?
A.
192.168.2.63
B.
192.168.2.62
C.
192.168.2.61
D.
192.168.2.60
E.
192.168.2.32
Explanation:
With the 224 there are 8 networks with increments of 32
One of these is 32 33 62 63 where 63 is broadcast so 62 is last valid host out
of given choices.
B
The increment for these subnets is 32 (256 – 224)
Therefore we can create the following subnets:
192.168.2.0
192.168.2.32
192.168.2.64
answer A is incorrect because this is the broadcast address for the 192.168.2.32 subnet
answer C is incorrect because this is not the LAST usable host address for the 192.168.2.32 subnet
answer D is incorrect because of the same reason
answer E is the network adddress for the 192.168.2.32 subnet
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Here is the network range 192.168.2.32 – 192.168.2.63; .32 is network and .63 is broadcast. Therefore last valid host ip is .62
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