Which of the following are valid host addresses for the subnet 203.0.113.64/28? (Choose two.)
Which of the following are valid host addresses for the subnet 203.0.113.64/28? (Choose two.)
For the subnet 203.0.113.64/28, the subnet mask is 255.255.255.240, which means there are 16 addresses in total (2^4 = 16). This range includes addresses from 203.0.113.64 to 203.0.113.79. The first address (203.0.113.64) is reserved for the network address, and the last address (203.0.113.79) is reserved for the broadcast address. Hence, valid host addresses must fall between 203.0.113.65 and 203.0.113.78. Therefore, the valid host addresses in the provided options are 203.0.113.65 and 203.0.113.78.
/28 is counting in 16. So it will range from .64 ~ .79 You need to use valid host addresses so the actual range is: .65 ~ .78 Why you ask? .64 is used to call the actual subnet. .79 is used for a Broadcast Address. Answer is B and C because .78 and .65 fall in the ranges mentioned above.
Network: 203.0.113.64/28 11001011.00000000.01110001.0100 0000 (Class C) Broadcast: 203.0.113.79 11001011.00000000.01110001.0100 1111 HostMin: 203.0.113.65 11001011.00000000.01110001.0100 0001 HostMax: 203.0.113.78 11001011.00000000.01110001.0100 1110 Hosts/Net: 14
64+14=78 (sennò me scordo)
sto ad albano un paesino prima di genzano
jump= 2^4=16=>16 ip/subnet => ip from .64 to .79 except .64 because it is ip subnet
.65 .78
203.0.113.64/28 203.0.113.0100|0000 <-- 28 bits .64 is the network address so the range would be the remaining 15 (00001111), hence B & C are correct
only B and C please