Assuming a subnet mask of 255.255.248.0, three of the following addresses are valid host addresses. Which are these addresses? (Choose three.)
A. 172.16.9.0
B. 172.16.8.0
C. 172.16.31.0
D. 172.16.20.0
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Considering the ip address 172.16.8.0 and the given subnet mask:
Address: 172.16.8.0 10101100.00010000.00001 000.00000000
Netmask: 255.255.248.0 = 21 11111111.11111111.11111 000.00000000
Wildcard: 0.0.7.255 00000000.00000000.00000 111.11111111
Network: 172.16.8.0/21 10101100.00010000.00001 000.00000000 (Class B)
Broadcast: 172.16.15.255 10101100.00010000.00001 111.11111111
HostMin: 172.16.8.1 10101100.00010000.00001 000.00000001
HostMax: 172.16.15.254 10101100.00010000.00001 111.11111110
Only option A is correct.
Ignore my previous answer! I hadn’t considered all these IP addresses could belong to different networks. So the correct answer is ACD as some of you said. Regards!
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So you need to find your ranges now.
.0 – .7, 1-6 = valid
.8 – .15, 9-14 = valid
.16 – .23, 17-22 = valid
.24 – .31, 25-30 = valid
.32 – .39, 33-38 = valid
Is it 128 = 128 hosts per
192 = 64 …
224 = 32 …
240 = 16…
248 = 8 …
252 = 4…
254 = 2 …
255 = 1…
Yes, a .128 has 128 hosts (126 usable because of the network address and broadcast address) and also a .0 is 256.
248 holds 8 addresses per subnet.
So you need to find your ranges now.
.0 – .7, 1-6 = valid
.8 – .15, 9-14 = valid
.16 – .23, 17-22 = valid
.24 – .31, 26-32 = valid
.32 – .39, 33-39 = valid
Your broadcast address will be always be the last address within your subnet, in the first subnet that would be 172.16.7.255 (don’t forget that last octet when figuring the broadcast)
The range of your first subnet is 172.16.0.0-172.16.7.255 with 172.16.0.0 being the network address and 172.16.7.255 being your broadcast address. So ANY other address within that range would be a usable address for your hosts.
So, with subnet mask 255.255.248.0, 172.16.31.255 would be the broardcast.
A, C, D are correct answers.
So you need to find your ranges now.
.0 – .7, 1-6 = valid
.8 – .15, 9-14 = valid
.16 – .23, 17-22 = valid
.24 – .31, 25-30 = valid
.32 – .39, 33-38 = valid
172.16.31.255 is the broadcast address of this subnet. ACD are correct answers.
31 in binary is 11111 so the three last bits are 111
> Broadcast address
> Is not a valid host address
> C WRONG
> CORRECT ANSWERS: A&D