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5 Written questions

5 Matching questions

  1. Given 192.168.10.0/28, what is the subnet mask?
  2. 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?
  3. Why subnet?
  4. How many hosts are available with a Class C /29 mask?
  5. Classful Routing Protocols
  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 Optimized network performance
  3. c 255.255.255.128
  4. d /29 is 255.255.255.248, which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
  5. e RIPv1 and IGRP

5 Multiple choice questions

  1. Traceroute, Arp -a, Ipconfig /all
  2. 11000000 = 192
  3. .1 - .126 & .129 - .254
  4. /23 is 255.255.254.0. The third octet is a block size of 2. 0, 2, 4. The subnet is in the 16.2.0 subnet, the broadcast address is 16.3.255
  5. A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on—which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.

5 True/False questions

  1. 192.168.100.25/30/28 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. Which two statements describe the IP address 10.16.3.65/23?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.

          

  3. Classless Routing Protocolsall nodes in the network use the same subnet mask

          

  4. Given 172.16.0.0/18, What's the broadcast address for each subnet?63.255, 127.255, 191.255, 255.255

          

  5. What subnet and broadcast address is the IP address 172.16.46.255 255.255.240.0 (/20) a member of?The subnet is 172.16.32.0, and the broadcast must be 172.16.63.25