You have an interface on a router with the IP address of 192.168.192.10/29. What is the broadcast address the hosts will use on this LAN?
A. 192.168.192.15
B. 192.168.192.31
C. 192.168.192.63
D. 192.168.192.127
Answer: Option A
Solution(By Examveda Team)
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0, 8, 16, 24, etc. 10 is in the 8 subnet. The next subnet is 16, so 15 is the broadcast address.Related Questions on Subnetting
A. 1 only
B. 2 and 3 only
C. 3 and 4 only
D. None of the above
E. 2 and 5 only
A. 1 and 3
B. 2 and 4
C. 1, 2 and 4
D. 2, 3 and 4
E. 2 and 5 only
A. 1 only
B. 2 only
C. 3 and 4 only
D. 1 and 5 only
E. 2 and 5 only
Join The Discussion