Exam 200-301 topic 1 question 1209 discussion

Seems people are confused. Let's break it down: There's only one subnet that potentially matches (the /23); the other two are host routes and since they don't match the destination they're automatically excluded. First, break the last 2 octets of the IPv4 address into binary: 172.16.3.254 -> 172.16.[00000011].[11111110] Then, do the same for the network address: 172.16.2.0 -> 172.16.[00000010].[00000000] The prefix-length is /23, meaning the first 2 octets + the first 7 bits of the third octet need to match, which they do. So, the packet takes this route. A /23 prefix-length is equivalent to a 255.255.254.0 subnet mask, which is B.

B is correct. Plug in 172.16.2.0/23 into a subnet calculator and you can see the usable IP ranges: 172.16.2.1 - 172.16.3.254 Ref: https://www.calculator.net/ip-subnet-calculator.html?cclass=any&csubnet=23&cip=172.16.2.0&ctype=ipv4&x=Calculate

The 172.168.2.0 /23 network is the longest prefix match and the subnet of /23 is 255.255.254.0

B is correct

Connected has a AD value of 0. In addition, the subnet mask covers the range. Provided answer is correct.

The Answer is D

There is no route for the destination so it would be the gateway of last resort.

the usable host ip range for 172.16.2.0/23: 172.16.2.1 - 172.16.3.254 The answer is B

Address: 172.16.3.254 10101100.00010000.0000001 1.11111110 Netmask: 255.255.254.0 = 23 11111111.11111111.1111111 0.00000000 Wildcard: 0.0.1.255 00000000.00000000.0000000 1.11111111 => Network: 172.16.2.0/23 10101100.00010000.0000001 0.00000000 (Class B) Broadcast: 172.16.3.255 10101100.00010000.0000001 1.11111111 HostMin: 172.16.2.1 10101100.00010000.0000001 0.00000001 HostMax: 172.16.3.254 10101100.00010000.0000001 1.11111110 Hosts/Net: 510 (Private Internet)

It will send via default gateway