Welcome to CCNA Exam 200-301 Practice Quiz on Subnetting. This quiz section provides you with multiple-choice questions with a single answer or multiple answers. The MCQ in this section is based on IP subnetting.

This MCQ quiz may be helpful for the preparation of CCNA Exam 200-301.

CCNA Exam 300-201 MCQ Quiz- Subnetting

1. Which of the following statements are true about the subnetting of IPv4 address? (choose any three)

  1. Subnetting breaks down a single large network into a smaller network.
  2. Subnetting in IPv4 is a permanent solution to control the rapid exhaustion of IPv4 addresses.
  3. Host bits are borrowed by the network to create extra subnets.
  4. Subnetting controls broadcast traffic, hence reduces network congestion.

A, C, and D are the correct answers.

2. You are a network administrator and you are asked to develop an IP addressing plan with 192.20.20.0/24 to allow the maximum number of subnets with as many as 50 hosts each. Which IP address range meets these requirements?

  1. 192.20.20.0/27
  2. 192.20.20.0/26
  3. 192.20.20.0/29
  4. 192.20.20.0/28
  5. 192.20.20.0/25

B is the correct answer.

Explanation:

To create a subnet with 50 hosts, we need 2h – 2 >= 50 (h = number of host bits required to create a subnet for 50 hosts). If h = 6, we have ; 62 >= 25. Hence, 32 – 6 = 26 (network bits).

 

3. Using the VLSM techniques which is the most suitable subnet mask for the point to point serial link?

  1. 255.255.255.254
  2. 255.255.255.252
  3. 255.255.255.248
  4. 255.255.255.0

B is the correct answer.

Explanation:

For point to point link, we need only two IP. So, if h = 2 ( h = host bits) we have, 22 – 2 = 2. Thus, using VLASM we can take 2 host bits and network bits = 32 -2 = 30; will give the subnet mask 255.255.255.252.

 

4. In the following exhibit, assuming the routing configuration of both the router is properly done. But, HostA cannot ping HostB. What could be the problem?

  1. HostA and its gateway are not on the same subnet.
  2. Host B and its gateway are not on the same subnet.
  3. The serial link between the two routers is not on the same subnet.
  4. The G 0/0 interface on RouterB is using a broadcast address.
  5. The G 0/0 interface on RouterA is using a broadcast address.

C is the correct answer.

Explanation:

The correct answer is C because the exhibit is using the CIDR prefix /27 for all the subnets. Hence, the subnet range will be 0-31, 32-63,64-95, 96-127, and so on. Router 1 s0/0/0 IP 192.168.5.62/27 belongs to the subnet block 32-63 and the router 2 s0/0/0 IP 192.168.5.65/27 belongs to the subnet block 64-95. They are not on the same subnet, that is why host A cannot ping Host B.

 

5. The gigabit Ethernet port G0/0 is configured with IP address 172.20.10.1/23 to connect subnetworks. What will be the maximum number of valid hosts allowed to this subnet?

  1. 512
  2. 254
  3. 510
  4. 1022

C is the correct answer.

Explanation:

172.20.10.1/23 contains network bits 23, so host bit = 32 – 23 = 9. Thus, 9 bits are available for hosts. Therefore, the number of valid hosts is given by 29 – 2 = 510. Hence, option C is the correct answer.

 

6. As a network administrator, you are trying to configure the serial interface of a router with IP address 120.10.10.28/30. But the error message is displayed. What could be the possible cause for it?

A. The address is a broadcast address.
B. The Ethernet interface is faulty.
C. The router does not support VLSM.
D. This address is a network address.

D is the correct answer.

Explanation:

Subnet mask for 120.10.10.28/30 is 255.255.255.252 (because, 2³²³⁰ – 2 =2² – 2 = 2. Therefore , the block size for each subnet = 2+1+1 (each 1 denotes network and broadcast address) = 4.

Then, the subnet range will be 0-3, 4-7, 8-11, 12-15, 16-19, 20-23, 24-27, 28-31 and so on. Thus, 120.10.10.28/30 is the network id of the 28-30 block of IP.

 

7. Which of the following statements are true about 170.20.30.70/23? (Choose any three)

  1. The network address is 170.20.30.0
  2. The broadcast address is 170.20.30.255
  3. The number of subnets is 128.
  4. It belongs to the subnet range from 170.20.30.0 to 170.20.31.255

A, C, and D are the correct answers.

Explanation:

For 170.20.30.70/23, Netmask = 255.255.254.0, net id = 170.20.30.0, broadcast id = 170.20.31.255

Number of subnets = 27 = 128.

170.20.30.70/23 falls under the range 170.20.30.0 to 170.20.31.255.

 

8. Which of the following statements are true about 192.168.5.128/28 and 192.168.5.129/28? (Choose any two)

  1. Both of them belong to the same subnet.
  2. 192.168.5.128/28 is not the valid host IP address.
  3. The subnet mask for both of them is 255.255.255.192.
  4. All of the above statements are correct.

A and B  are the correct answers.

Explanation:

The subnet mask for /28 is 255.255.255.240. The Block size of the given subnet is 256 – 240 = 16. Therefore, the IP range are 0-15, 16-31, 32-63, 64-79, 80-95, 96-111, 112-127, 128-159 and so on. Both IP 192.168.5.128/28 and 192.168.5.129/28 belong to the subnet block 128-159. 192.168.5.128/28 is the subnet id, hence cannot be assigned to the host.

 

9. You are given the IP Address of 193.100.100.0/24 and need 50 hosts per subnetwork. How total sub-networks, valid host per subnet and subnet mask do you get after subnetting?

  1. 4 subnets, 64 hosts per subnets, and 255.255.255.128
  2. 2 subnets, 62 hosts per subnets, and 255.255.255.192
  3. 4 subnets, 62 hosts per subnets, and 255.255.255.192
  4. 4 subnets, 62 hosts per subnets, and 255.255.255.128

C is the correct answer.

 

Explanation:

For 50 hosts, the numbers of host bits required are 6 i.e. each subnet will have 26 – 2 = 62 host which is enough to accommodate 5o hosts.

Number of subnets = 22 = 4 ( since two subnet bits are borrowed from the host portion)

Thus, the new subnet mask = 255.255.255.192 or /26.

10. Your company has been given the IP Address of 199.2.1.0 /24 to the subnet. You plan to put each of the 5 floors in your building on its own subnet. What is the IP range of the LAST available network once subnetted?

  1. 199.2.1.0 – 199.2.1.255
  2. 199.2.1.128 – 199.2.1.255
  3. 199.2.1.224 – 199.2.1.255
  4. 199.2.1.223 – 199.2.1.255

C is the correct answer.

Explanation:

Your office has 5 floors and each floor will have different subnets. So, you have to take 23 = 8 subnets. Thus, 3 bits must be borrowed from the host to create subnets. The new subnet mask will be 255.255.255.224. the block size of each subnet = 256-224 = 32. Hence, the corresponding subnet range will be 0-31, 32-63, 64-95, 95-127, 128-159, 160-191, 192-223 and 224-255. The IP range for the last subnet is 199.2.1.224 to 199.2.1.255

 

11. How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many usable host addresses will there be per subnet?

  1. 8 subnets, 30 hosts
  2. 16 subnets, 32 hosts
  3. 16 subnets, 30 hosts
  4. 8 subnets, 32 hosts

C is the correct answer.

Explanation:
We have a network address 172.17.32.0/23. Out of this network, we further need to create the subnets with mask /27. For this, we have to borrow 4 extra subnets from the host portion to create the subnets. The number of new subnets created is given by 24 = 16.
Again, the remaining host bit = 32-27 = 5. Therefore, number of usable host addresses per subnet is given by 25 – 2 = 30.

 

12. What could be the maximum numbers of usable host IP for 20.20.20.20/20?

  1. 4096 usable hosts
  2. 4094 usable hosts
  3. 4098 usable hosts
  4. 4092 usable hosts

B is the correct answer.

Explanation:

For 20.20.20.20/20, the number of host bits is 32 – 20 = 12.

Therefore, numbers of usable host = 212 – 2 = 4094

13. Which of the following is not the valid host IP that does not belong to the subnet 100.100.1.128/25.

  1. 100.100.1.127
  2. 100.100.1.129
  3. 100.100.1.254
  4. 100.100.1.130

A is the correct answer.

Explanation:

The valid host range for subnet 100.100.1.128/25 is 100.100.1.129 to 100.100.1.254.

Only 100.100.1.127 doesn’t belong to the given subnet. Hence, the correct answer is A.

 

14. You have a network address 170.25.0.0/16. You are going to create 64 subnets and 1000 hosts for each subnet. What would be the correct subnet mask for each subnet?

  1. 255.255.254.0
  2. 255.255.252.0
  3. 255.255.240.0
  4. 255.255.248.0

B is the correct answer.

Explanation:

For creating 1000 host subnet for 170.25.0.0/16, we need to take the number of host bits as given by the following calculation:

2h – 2 =1000; where h = host bits required

0r, 210 – 2 >= 1000

or, 1024 – 2 >= 1000

Therefore, host bit = 10. then new subnet mask =

11111111.11111111.11111100.00000000

255.255.252.0

 

15. The correct usable host range for 180.110.0.0/26 is ______________.

  1. 180.110.0.0 – 180.110.0.63
  2. 180.110.0.1 – 180.110.0.63
  3. 180.110.0.0 – 180.110.0.62
  4. 180.110.0.1 – 180.110.0.62

D is the correct answer.

 

Explanation:

For 180.110.0.0/26, the subnet mask is

11111111.11111111.11111111.11000000

0r, 255.255.255.192.

hence, the block size ( 4th octet) = 256 – 192 = 64

Ip range is from 180.110.0.0 to 180.110.0.63

Therefore, the usable host range is 180.110.0.1 to 180.110.0.62

 

16. A station with 10.10.46.33/21 and gateway address of 10.10.46.1. What would be the destination mac address of the packet destined for 10.10.47.2?

  1. Mac address of 10.10.46.1
  2. Mac address of 10.10.47.2
  3. Mac address of 10.10.44.0/21
  4. None of the above
A is the correct answer.

17. Which of the following devices share the same network? (Choose any two)

  1. 170.20.20.23/29
  2. 170.20.20.25/29
  3. 170.20.20.33/29
  4. 170.20.20.38/29

C and D are the correct answers.

Explanation:

170.20.20.33/29 and 170.20.20.38/29 belong to the same subnet block 32-39.

18. The best subnetting plan for point to point link is _________.

  1. /30
  2. /31
  3. /32
  4. /29

C is the correct answer.

Explanation:

For point to point link, we need only 2 addresses. Hence, 2² – 2 = 2 . Keeping two bits for hosts, we will have the subnet mask /30.

 

19. The subnet address of 150.100.100.100/22 is ________.

  1. 150.100.101.0
  2. 150.100.100.0
  3. 150.100.103.0
  4. 150.100.102.0

B is the correct answer.

Explanation:

The subnet mask for 150.100.100.100/22 will be 255.255.252.0. Using AND operation for finding net id, we get;

Binary of 3rd octet of IP: 01100100

Binary of 3rd octet of mask: 11111100

By ANDing, we get 01100100. Therefore, net id = 150.100.100.0

 


Also Read: How to calculate network Id and broadcast Id

20. Refer to the exhibit. Which of the following IP address do you assign to serial interface s0/0/0 of Router A so that Host A can ping Host B successfully?

  1. 192.168.5.63/27
  2. 192.168.5.64/27
  3. 192.168.5.65/27
  4. 192.168.5.66/27

D is the correct answer.

Explanation:

Router B is configured with IP 192.168.5.65/27 that belongs to the subnet range 192.168.5.64 to 192.168.5.95.

192.168.5.66/27 is the valid IP that can be assigned to the serial interface s0/0/0 of Router A.

 



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