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5 Written questions

5 Matching questions

  1. /25
  2. What is the broadcast address of 192.168.192.10/29?
  3. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
  4. 192.168.100.17/29
  5. You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?
  1. a A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host.
  2. b /29 is 255.255.255.248. The fourth octet is a block size of 8. 0, 8, 16, 24. The host is in the 16 subnet, broadcast of 23. Valid host 17-22
  3. c This is a pretty simple question. A /28 is 255.255.255.240, which means that our block size is 16 in the fourth octet. 0, 16, 32, 48, 64, 80, etc. The host is in the 64 subnet.
  4. d /29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0, 8, 16. The host is in the 8 subnet, broadcast is 15
  5. e 10000000 = 128

5 Multiple choice questions

  1. /28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0, 16, 32, 48. The host is in the 32 subnet, with a broadcast address of 47. Valid hosts 33-46
  2. The subnet is 172.16.32.0, and the broadcast must be 172.16.63.25
  3. 4
  4. This is 5 bits of subnetting, which provides 32 subnets. This is our best answer, a /21
  5. Tracert, Show ip arp

5 True/False questions

  1. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. Again, the zero subnet should not be considered valid for this questionA /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32, so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.

          

  2. 192.168.100.99/26/26 is 255.255.255.192. The fourth octet has a block size of 64. 0, 64, 128. The host is in the 64 subnet, broadcast of 127. Valid host 65-126

          

  3. To test the IP stack on your local host, which IP address would you ping?The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254.

          

  4. What are the valid subnets?256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached

          

  5. What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 (/27) a member of?The subnet is 172.16.45.12, with a broadcast of 172.16.45.15

          

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