2 Comments on “What would be the IP address of Ethernet 0/0 using the first available address from the sixth subnet of the ne”

1. Tamer Ruby says:

   September 8, 2012 at 3:38 pm

   quoted:

   “From the subnet mask, we deduce the increment is 8

   First subnet: 192.168.8.0/29
   Second subnet: 192.168.8.8/29
   3rd subnet: 192.168.8.16/29
   4th subnet: 192.168.8.24/2
   5th subnet: 192.168.8.32/29
   6th subnet: 192.168.8.40/29
   7th subnet: 192.168.8.48/29

   But the question says that the ip subnet zero is not used, so we can’t use the first subnet. Therefore the “the sixth subnet” will become the seventh subnet, that is 192.168.8.48/29 and its range is 192.168.8.48/29 (network address) to 192.168.8.55/29 (broadcast address) and the first usable host address of this subnet is 192.168.8.49/29.

   
   
   
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2. Ashraf Salah says:

   January 30, 2013 at 10:01 pm

   Hi,
   the sub net is /29=248
   255-248=7
   using the first available address from the sixth subnet of the network
   192.168.8.0/29?
   1- 192.168.8.0—–network address
   2- 192.168.8.1 to 7
   3- 192.168.8.14 to 21
   4- 192.168.8.28 to 35
   5- 192.168.8.35 to 43
   6- 192.168.8.49
   the answer is C ……..

   
   
   
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