312-50V10 · Question #344
What is the broadcast address for the subnet 190.86.168.0/22?
The correct answer is C. 190.86.171.255. A /22 subnet covers 4 consecutive /24 blocks, so 190.86.168.0/22 spans .168.0 through .171.255, making the broadcast address 190.86.171.255.
Question
What is the broadcast address for the subnet 190.86.168.0/22?
Options
- A190.86.168.255
- B190.86.255.255
- C190.86.171.255
- D190.86.169.255
How the community answered
(22 responses)- A5% (1)
- B14% (3)
- C77% (17)
- D5% (1)
Why each option
A /22 subnet covers 4 consecutive /24 blocks, so 190.86.168.0/22 spans .168.0 through .171.255, making the broadcast address 190.86.171.255.
190.86.168.255 is the broadcast for the smaller 190.86.168.0/24 subnet, not the larger /22 block.
190.86.255.255 would be the broadcast for a /16 block starting at 190.86.0.0, which is a much larger range.
A /22 prefix leaves 10 host bits (32 - 22 = 10) and a subnet mask of 255.255.252.0. Starting at 190.86.168.0, the host portion spans the last 2 bits of the third octet (allowing values 168-171) and all 8 bits of the fourth octet. Setting all host bits to 1 yields a third octet of 171 (10101011) and fourth octet of 255, producing broadcast address 190.86.171.255.
190.86.169.255 is the broadcast for 190.86.169.0/24, a subnet that falls inside the /22 range but is not the final broadcast address.
Concept tested: IPv4 CIDR broadcast address calculation
Source: https://learn.microsoft.com/en-us/troubleshoot/windows-server/networking/tcpip-addressing-and-subnetting
Topics
Community Discussion
No community discussion yet for this question.