An administrator is working with the 192.168.4.0 netwrok, which has been subnetted with a /26 mask. Which two addresses can be assigned to hosts within the same subnet? (Choose two.)
The question has specified for network 192.168.4.0 and subnet /26 the range of usable addresses are 192.168.4.1 to 192.168.4.62. Answer A and E would be correct for the next network 192.168.4.63 and /26 where the usable addresses range from 192.168.4.64 to 192.168.4.125. I think the answer B and F cannot be right because 192.168.4.63 is the last address for broadcast and cannot be a usable address. My view the closest answer is A and E
Broadcast addres can't be assigned to hosts bro. You're gonna fail your exam if you don't clarifiy that. A-E are correct because Network 192.168.4.64 contains "Available Hosts addreses" from 192.168.4.65 to 192.168.4.126, where 192.168.4.127 is the broadcast.
But how you will assign the broadcast address to hosts? 192.168.4.63 is the broadcast address for subnet 192.168.4.0. It is especially mentioned that the address should be assigned to hosts. The question is not specifically for subnet 192.168.4.0 , but for any networks which are derived with this subnet. The important thing is to be in the same subnet (no matter which 192.168.4.0 or 192.168.4.64 or 192.168.4.128 etc.) and to be addresses which can be assigned to hosts.
So A and E are the correct answers. B and F are not.
A and E are wrong, as they fall in with the 192.168.4.64 subnet. This subnet mask /26 (in CIDR) is basically 255.255.255.192. Same with answer E. You need to turn the /26 to binary and the IP addresses themselves as well and apply the AND logic operator.
Binary for the subnet mask is 11111111 11111111 11111111 11000000
The answers should be B and F, since the last host is 192.168.4.62 (first host is 192.168.4.1) and the broadcast is 192.168.4.63, given that you have 6 host bits (2 ^ 6) = 64, but you only have 62 usable hosts since the network and broadcast addresses are NOT assigned to hosts
you really believe the rubbish you are uttering A and E are correct, /26 means the network going up by 64
192.168.4.0 range valid host 192.168.4.1 - 192.168.4.62
192.168.4.64 range valid host 192.168.4.65 - 192.168.4.126
check over your work
upvoted 3 times
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