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Which of the following is an invalid host address using a
netmask
of 255.255.255.192?
A.) 10.1.1.1
B.) 10.1.1.66
C.) 10.1.1.130
D.) 10.1.1.127
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Answer Posted / subramani
D.) 10.1.1.127
No of valued hosts is
2^6-2=64-2=62
Therefore subnets are 0,64,128,256
since third subnet starts with 128,
127 will become the broadcast address for the second subnet
which is starting from 64.
| Is This Answer Correct ? | 2 Yes | 0 No |
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