D.) 10.1.1.127 No of valued hosts is 2^6-2=64-2=62 Therefore subnet

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

Question Posted / subramani

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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.

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