+1 vote
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Plz any one solve this is answer is B?

asked
edited | 62 views
0
AND the subnet mask with given IP to get the Network ID as 172.60.32.0 and OR the complemented subnet mask with the IP to get the direct broadcast address as 172.60.32.255.Hence the assignable range of IP addresses would be 172.60.32.1-172.60.63.254.

Hence,answer is (B).
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Logically if we see the question asked assignable IP address.

As we cannot assign X.X.X.0(N/W address) and X.X.X.255(DBA), the remaining option is 'B' :).

## 2 Answers

+2 votes
Best answer

Given subnet Mask :- 255.255.224.0 = 1111 1111 . 11111111 . 1110 0000 . 0000 0000

host bits = zero bits in Subnet Mask represent HOST bits ===> no.of host bits = 13

if you want Network IP address:- put all zeros in host bits in the given ip.

Given IP Address = 172.60.50.2 = 172.60.0010 0000 . 0000 0000 = 172.60.32.0

if you want first host address:- put all zeros in host bits in the given ip except LSB, make LSB as 1

Given IP Address = 172.60.50.2 = 172.60.0010 0000 . 0000 0001 = 172.60.32.1

if you want last host address:- put all ones in host bits in the given ip except LSB, make LSB as 0

Given IP Address = 172.60.50.2 = 172.60.0011 1111 . 1111 111= 172.60.63.254

if you want Broadcast address of this network :- put all ones in host bits in the given ip.

Given IP Address = 172.60.50.2 = 172.60.0011 1111 . 1111 1111 = 172.60.63.255

answered by Veteran (60.8k points)
edited
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I guess you need to modify the last line.. The last host address which can't be an assignable IP address...

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@the_bob

Thanks for the correction.

it's a typing mistake ( it occurred due to copy and paste), i will update it.

0 votes

Ans  is B answered by Junior (619 points)

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