Subnetting(Networking ) Questions and Answers

Question 1. 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?
1. 172.16.1.100
2. 172.16.1.198
3. 172.16.2.255
4. 172.16.3.0
  1.    1 only
  2.    2 and 3 only
  3.    3 and 4 only
  4.    None of the above
  5.    2 and 5 only
Explanation:-
Answer: Option C. -> 3 and 4 only
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.
Question 2. Which two statements describe the IP address 10.16.3.65/23?
1. The subnet address is 10.16.3.0 255.255.254.0.
2. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
3. The last valid host address in the subnet is 10.16.2.254 255.255.254.0.
4. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
  1.    1 and 3
  2.    2 and 4
  3.    1, 2 and 4
  4.    2, 3 and 4
  5.    2 and 5 only
Explanation:-
Answer: Option B. -> 2 and 4
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.
Question 3. 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?
  1.    14
  2.    15
  3.    16
  4.    30
Explanation:-
Answer: Option D. -> 30
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.
Question 4. A network administrator is connecting hosts A and B directly through their Ethernet interfaces, as shown in the illustration. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts?
1. A crossover cable should be used in place of the straight-through cable.
2. A rollover cable should be used in place of the straight-through cable.
3. The subnet masks should be set to 255.255.255.192.
4. A default gateway needs to be set on each host.
5. The subnet masks should be set to 255.255.255.0.
  1.    1 only
  2.    2 only
  3.    3 and 4 only
  4.    1 and 5 only
  5.    2 and 5 only
Explanation:-
Answer: Option D. -> 1 and 5 only
First, if you have two hosts directly connected, as shown in the graphic, then you need a crossover cable. A straight-through cable won't work. Second, the hosts have different masks, which puts them in different subnets. The easy solution is just to set both masks to 255.255.255.0 (/24).
Question 5. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
  1.    255.255.255.192
  2.    255.255.255.224
  3.    255.255.255.240
  4.    255.255.255.248
Explanation:-
Answer: Option B. -> 255.255.255.224
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.
Question 6. 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?
  1.    192.168.19.0 255.255.255.0
  2.    192.168.19.33 255.255.255.240
  3.    192.168.19.26 255.255.255.248
  4.    192.168.19.31 255.255.255.248
Explanation:-
Answer: Option C. -> 192.168.19.26 255.255.255.248
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.
Question 7. You have an interface on a router with the IP address of 192.168.192.10/29. What is the broadcast address the hosts will use on this LAN?
  1.    192.168.192.15
  2.    192.168.192.31
  3.    192.168.192.63
  4.    192.168.192.127
Explanation:-
Answer: Option A. -> 192.168.192.15
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0, 8, 16, 24, etc. 10 is in the 8 subnet. The next subnet is 16, so 15 is the broadcast address.
Question 8. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
  1.    172.16.17.1 255.255.255.252
  2.    172.16.0.1 255.255.240.0
  3.    172.16.20.1 255.255.254.0
  4.    172.16.18.255 255.255.252.0
Explanation:-
Answer: Option D. -> 172.16.18.255 255.255.252.0
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.
Question 9. On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
  1.    /27
  2.    /28
  3.    /29
  4.    /30
Explanation:-
Answer: Option D. -> /30
A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.
Question 10. To test the IP stack on your local host, which IP address would you ping?
  1.    127.0.0.0
  2.    1.0.0.127
  3.    127.0.0.1
  4.    127.0.0.255
Explanation:-
Answer: Option C. -> 127.0.0.1
To test the local stack on your host, ping the loopback interface of 127.0.0.1.
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