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5 Written questions

5 Matching questions

  1. Variable Length Subnet Masks (VLSMs)
  2. How many subnets?
  3. You have a Class B network and need 29 subnets. What is your mask?
  4. Subnet Mask
  5. Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host ID's on the LAN interface attached to the router?
  1. a The routers IP address on the E0 interface is 172.16.2.1/23, which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet, the broadcast address is 3.255 because the next subnet is 4.0. The valid host range is 2.1 through 3.254. The router is using the first valid host address in the range.
  2. b each network segment can use a different subnet mask
  3. c This is 5 bits of subnetting, which provides 32 subnets. This is our best answer, a /21
  4. d 32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address
  5. e 2^x where x is the number of masked bits (or 1's)

5 Multiple choice questions

  1. /27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0, 32, 64. The host is in the 32 subnet, broadcast address of 63. Valid host range of 33-62.
  2. all interfaces within the classful address space have the same subnet mask
  3. /25 is 255.255.255.128. The fourth octet is a block size of 128. 0, 128. The host is in the 0 subnet, broadcast of 127. Valid host 1-126
  4. 127 & 255
  5. /29 is 255.255.255.248, which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet

5 True/False questions

  1. Given 172.16.0.0/18, What are the valid subnets?.0.1 - .63.254, .64.1 - .127.254, .128.1 - .191.254, .192.1 - .255.254

          

  2. Additional Windows TroubleshootingTracert, Show ip arp

          

  3. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224?A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits but provides 13 host bits, or 8 subnets, each with 8,190 hosts.

          

  4. Given 172.16.0.0/18, How many subnets?16,382

          

  5. Given 192.168.10.0/28, What are the valid subnets?0.0, 64.0, 128.0, and 192.0

          

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