The routers in this network are running RIPv2. Which addressing scheme would satisfy the needs of this network yet waste the fewest addresses?

A TestKing network is shown in the exhibit below:

A. Network 1: 192.168.10.0/26Network 2: 192.168.10.64/26Network 3:
192.168.10.128/26Serial link 1: 192.168.20.0/24Serial link 2: 192.168.30.0/24
B. Network 1: 192.168.10.0/26Network 2: 192.168.10.64/28Network 3:
192.168.10.80/29Serial link 1: 192.168.10.88/30Serial link 2: 192.168.10.96/30
C. Network 1: 192.168.10.0/26Network 2: 192.168.10.64/27Network 3:
192.168.10.96/28Serial link 1: 192.168.10.112/30Serial link 2: 192.168.10.116/30
D. Network 1: 192.168.10.0/27Network 2: 192.168.10.64/28Network 3:
192.168.10.96/29Serial link 1: 192.168.10.112/30Serial link 2: 192.168.10.116/30

Answer: C

Explanation:

Network 1

Required Number of hosts :50
When We use the 26 bits for Network : 11111111.11111111.11111111.11000000 so 62
usable host can be in one network. 50 host for now and remaining hosts address for
further growth.

Network 2

Required Number of Hosts: 20

When we use the 27 bits for Network: 11111111.11111111.1111111.11100000 so 30
usable hosts can be in one network.

Network 3

Required Number of Hosts: 10
When we use the 28 bits for Network: 11111111.11111111.11111111.11110000 so 14
usable hosts can be in one network.

Connection between TK1, TK2 and TK3 is WAN so when you use 30 bits network IP,
you will not lose any IP addresses from network since this subnet allows for only 2 host
addresses.