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in Computer Networks by Active (2.4k points)
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here first we find network Id  means that all host bit zero and net id all 1's  and if  see id  of first octet is 192 so its belongs to class C which range( 192-223) so  

 Net id -->  192.168.1.187

mask----> 255 .255 .255.224

---------------------------------------------------------

AND operation-->  192.168.1.160  ( net id)

and  directed broadcast --> all host bit 1.  so ---->  192.168. 1. 10111111

                                                                 ------> 192.168.1.191

so option D ia correct
by Loyal (6.5k points)
edited by
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what you are saying? is what you said corresponds to the answer?

I know that whatever you said , i am just curious how they deduced option 4?
0
i update my answer.........plz check option D is correct
0
thank you.

i was also sure that options are wrong but just want to ask others also.
+1
option 4) is correct.

192.168.1.187 ANDed with 255.255.255.224 gives 192.168.1.160 (Network id)

by making last 5 bits(host bits) of last octet '1' we get 192.168.1.10111111 (broadcast id)

hence option 4)

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