Subnetting Questions and Answers updated daily – Networking



13 Subnetting Questions and answers section with explanation for various online exam preparation, various interviews, Networking Subnetting online test. Subnetting Questions with detailed description, explanation will help you to master the topic.

Subnetting Questions

1.
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?



SHOW ANSWER
Correct Ans:Router(config)#ip subnet-zero
Explanation:
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets, each with 30 hosts. However, if the command ip subnet-zero is not used, then only 6 subnets would be available for use.


2. To test the IP stack on your local host, which IP address would you ping?



SHOW ANSWER
Correct Ans:127.0.0.1
Explanation:
To test the local stack on your host, ping the loopback interface of 127.0.0.1.


3. On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?



SHOW ANSWER
Correct Ans:/30
Explanation:
A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.


4. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?



SHOW ANSWER
Correct Ans:192.168.19.26 255.255.255.248
Explanation:
A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.


5. The network address of 172.16.0.0/19 provides how many subnets and hosts?



SHOW ANSWER
Correct Ans:8 subnets, 8,190 hosts each
Explanation:
A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.


6. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?



SHOW ANSWER
Correct Ans:172.16.64.0
Explanation:
A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255


7. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts can have IP addresses on the LAN attached to the router interface?



SHOW ANSWER
Correct Ans:6
Explanation:
A /29 (255.255.255.248), regardless of the class of address, has only 3 host bits. Six hosts is the maximum number of hosts on this LAN, including the router interface.


8. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?



SHOW ANSWER
Correct Ans:172.16.112.0
Explanation:
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.


9. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet. How many bits must you borrow from the host field to provide the correct subnet mask?



SHOW ANSWER
Correct Ans:5
Explanation:
A 240 mask is 4 subnet bits and provides 16 subnets, each with 14 hosts. We need more subnets, so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 host bits (6 hosts per subnet). This is the best answer.


10. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?



SHOW ANSWER
Correct Ans:255.255.255.224
Explanation:
You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts-this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts. This is the best answer.


11. What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?



SHOW ANSWER
Correct Ans:30
Explanation:
A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets, each with 30 hosts. Does it matter if this mask is used with a Class A, B, or C network address? Not at all. The number of host bits would never change.


12.
Which two statements describe the IP address 10.16.3.65/23?

The subnet address is 10.16.3.0 255.255.254.0.
The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
The last valid host address in the subnet is 10.16.2.254 255.255.254.0.
The broadcast address of the subnet is 10.16.3.255 255.255.254.0.



SHOW ANSWER
Correct Ans:2 and 4
Explanation:
The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host 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.


13.
Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host IDs on the LAN interface attached to the router?

172.16.1.100
172.16.1.198
172.16.2.255
172.16.3.0



SHOW ANSWER
Correct Ans:3 and 4 only
Explanation:
The router's IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. This makes the third octet a block size of 2. The router's interface is in the 2.0 subnet, and 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.


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