Hey why not C? Isn't both B and C right?

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+2 votes

if the subnet mask is

255.255.255.224

which of the following can be DBA of subnet.

a)201.55.66.15

b)201.55.66.31

c)201.55.66.63

d)all

255.255.255.224

which of the following can be DBA of subnet.

a)201.55.66.15

b)201.55.66.31

c)201.55.66.63

d)all

0 votes

0

Given subnet mask is 11111111.11111111.11111111.11100000

The first 27 bits belong to the network here .Free bits are last 5 bits which are Host id bits. To get DBA we need to make the host bits all 1's .

Here host bits are 5 . so make allo of them as 1. This gives 11111=31

The first 27 bits belong to the network here .Free bits are last 5 bits which are Host id bits. To get DBA we need to make the host bits all 1's .

Here host bits are 5 . so make allo of them as 1. This gives 11111=31

0

But doesn't the subnet mask the network in subnets ..in this case into 201.55.66.0 - 31 and 201.55.66.32 - 63 and so on...... ?

0

As u said to get 63 we need --111111 but here the first 3 bits are fixed . you cannot do anything on the first 3 bits .So we can only use the remaining 5 bits

+1

What I mean is if a subnet mask 255.255.255.224 is applied on 201.55.66.0, it means that the ip address can be subnetted into groups of 32 ip's (2^5)

In this case 201.55.66.0/27 will have 201.55.66.31 has DBA

In the case of 201.55.66.32/27, 201.55.66.63 will be the DBA

Correct me if I'm wrong

In this case 201.55.66.0/27 will have 201.55.66.31 has DBA

In the case of 201.55.66.32/27, 201.55.66.63 will be the DBA

Correct me if I'm wrong

0

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