Refer to the exhibit. Host B has just been added to the network and must acquire an IP address.
Which two addresses are possible addresses that will allow host B to communicate with other devices in the network? (Choose two.)
A. 192.168.10.32
B. 192.168.10.38
C. 192.168.10.46
D. 192.168.10.47
E. 192.168.10.49
F. 192.168.10.51
Can you please explain why ??
WIll try…
subnet mask = 255.255.255.240 ..revert that to the wildcard: 0.0.0.15 ->
192.168.10.32
0 . 0 . 0.16
15 is the last address for the last network ID. 16 starts a new range – every 16 bits is a new range …. And if you put the wildcard below the IP, the octets where the 0’s are, won’t change. Combine these 2 things to create:
192.168.10.0
192.168.10.16
192.168.10.32
192.168.10.48
etc.. – We have found our range. Gateway IP = 192.168.10.33 – To communicate with the GW. You need to be in that IP range between .32 and .47 – .32 and .47 are the subnet ID and Broadcast address of this range. And .48 starts a new range
Hope this helps a bit.