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EC-Council

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.

Scanning Networks

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)
  • A
    5% (1)
  • B
    14% (3)
  • C
    77% (17)
  • D
    5% (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.

A190.86.168.255

190.86.168.255 is the broadcast for the smaller 190.86.168.0/24 subnet, not the larger /22 block.

B190.86.255.255

190.86.255.255 would be the broadcast for a /16 block starting at 190.86.0.0, which is a much larger range.

C190.86.171.255Correct

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.

D190.86.169.255

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

#subnetting#broadcast address#CIDR notation#IPv4

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