Host X has IP address 192.168.1.97 and is connected through two routers R1 and…
2008
Host X has IP address 192.168.1.97 and is connected through two routers R1 and R2 to another host Y with IP address 192.168.1.80. Router R1 has IP addresses 192.168.1.135 and 192.168.1.110. R2 has IP addresses 192.168.1.67 and 192.168.1.155. The netmask used in the network is 255.255.255.224.
Given the information above, how many distinct subnets are guaranteed to already exist in the network?
- A.
6
- B.
3
- C.
2
- D.
1
Attempted by 155 students.
Show answer & explanation
Correct answer: B
Key idea: with netmask 255.255.255.224 (which is /27), each subnet contains 32 addresses.
192.168.1.97 falls in the subnet 192.168.1.96/27 (range .96–.127).
192.168.1.110 also falls in 192.168.1.96/27 (same as 192.168.1.97).
192.168.1.80 falls in the subnet 192.168.1.64/27 (range .64–.95).
192.168.1.67 also falls in 192.168.1.64/27 (same as 192.168.1.80).
192.168.1.135 falls in the subnet 192.168.1.128/27 (range .128–.159).
192.168.1.155 also falls in 192.168.1.128/27 (same as 192.168.1.135).
Conclusion: The listed addresses occupy three distinct /27 subnets: 192.168.1.64/27, 192.168.1.96/27, and 192.168.1.128/27. Therefore, three distinct subnets are guaranteed to already exist.
Note: You cannot infer any additional distinct subnets beyond these from the given addresses, because some interfaces share the same /27 block.