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20 Comments on “What two address and subnet combinations are properly matched?

    1. Willyx says:

      172.16.255.255/14
      assuming that we can use a /14 subnet in a B class (we shouldn’t):
      /14=255.252.0.0 , the magic number is 4. We have
      172.0.0.0 as first subnet ID
      172.4.0.0 as first subnet ID
      172.8.0.0 as first subnet ID
      then we have a range of 172.4.0.1 to 172.7.255.254 and 172.7.255.255 is the broadcast address.
      No, 172.16.255.255/14 isn’t a broadcast address, just the mask is wrong in a B class 🙂




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  1. Galante says:

    B/D are correct, they are both valid IP address within their subnet

    A: 10.230.33.15/28 is the broadcast address for the network with subnet mask 255.255.255.240
    D: 172.25.44.96/26 is the Network ID based on the subnet mask 255.255.255.192
    E: 192.168.73.223/29 is the broadcast address for the network with subnet mask 255.255.255.248




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  2. Galante says:

    Let me expand on my last comment ..

    B/D are correct, they are both are valid HOST IP addresses within their subnet, whereas A,D & E are either the subnet IP or broadcast IP addresses.

    Hope they helps?




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