Subnetting IP Address Quiz Questions And Answers

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Subnetting IP Address Quiz Questions And Answers - Quiz

Do you have knowledge of subnetting? To test your understanding, you can take this subnetting IP address quiz. Computer networks can be confusing, especially to people without the knowledge and understanding. Among the many concepts, subnetting IP addresses is one that is extensively studied. While taking the quiz, you will get some extra knowledge also. Go for this quiz, and gauge your knowledge on subnetting IP addresses. Try to ace the quiz, and get a perfect score. All the best!


Questions and Answers
  • 1. 

    Which of the following answers is the default subnet mask for a class A network?

    • A.

      255.128.0.0

    • B.

      255.0.0.0

    • C.

      255.224.0.0

    • D.

      224.0.0.0

    Correct Answer
    B. 255.0.0.0
    Explanation
    The default subnet mask for a class A network is 255.0.0.0. In classful networking, class A networks have a default subnet mask of 255.0.0.0, which means that the first octet represents the network portion of the IP address and the remaining three octets represent the host portion. This allows for a large number of hosts to be connected to the network. The other subnet masks listed in the options are not the default subnet masks for a class A network.

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  • 2. 

    Subnet the IP Address 203.10.93.0 /24 into 30 Subnets.  Is 203.10.93.30 a valid Host ID after subnetting?

    • A.

      Yes

    • B.

      No

    Correct Answer
    A. Yes
    Explanation
    203.10.93.0 Need 30 Networks 128 64 32 16 8 4 2 1 0 0 0 1 1 1 1 0 = 30 in binary Borrow 5 Bits for # of Networks New Subnet Mask = 255.255.255.11111000 (248) or /29 ^ Increment = 8 2^5 = # of Networks = 32 2^3-2 = # of Hosts per Network = 6 First 5 IP Ranges 223.10.93.0 - 223.10.93.7 93.8 – 93.15 93.16 – 93.23 93.24 – 93.31 93.32

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  • 3. 

    You are given the IP Address of 193.103.20.0 /24 and need 50 Subnets. How many hosts per network and what total networks do you get once subnetted.

    • A.

      20 Hosts and 50 Subnets

    • B.

      2 Hosts and 64 Subnets

    • C.

      4 Hosts and 50 Subnets

    • D.

      6 Hosts and 64 Subnets

    Correct Answer
    B. 2 Hosts and 64 Subnets
    Explanation
    193.103.20.0 Need 50 Networks

    128 64 32 16 8 4 2 1

    0 0 1 1 0 0 1 0 = 50 in binary
    Need to borrow 6 bits for # of Networks

    New Subnet Mask = 255.255.255.11111100 (252) or /30
    ^ Increment = 4

    2^6 = # of Networks = 64
    2^2-2 = # of Hosts per Network = 2

    First 5 IP Ranges
    193.103.20.0 - 193.103.20.3
    20.4 – 20.7
    20.8 – 20.11
    20.12 – 20.15
    20.16 – 20.19

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  • 4. 

    Your company has been given the IP Address of 199.2.1.0 /24 to 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 your subnet.

    • A.

      199.2.1.224 - 199.2.1.255

    • B.

      199.2.1.0 - 199.2.1.255

    • C.

      199.2.1.252 - 199.2.1.255

    • D.

      199.2.1.32 - 199.2.1.64

    Correct Answer
    A. 199.2.1.224 - 199.2.1.255
    Explanation
    199.2.1.0 Need 5 Networks

    128 64 32 16 8 4 2 1

    0 0 0 0 0 0 1 0 1 = 5 in binary
    Need to borrow 3 bits for # of Networks

    New Subnet Mask = 255.255.255.11100000 (224) or /27
    ^ Increment = 32

    2^3 = # of Networks = 8
    2^5-2 = # of Hosts per Network = 30

    All 8 IP Ranges
    199.2.1.0 - 199.2.1.31
    1.32 – 1.63
    1.64 – 1.95
    1.96 – 1.127
    1.128 – 1.159
    1.160 – 1.191
    1.192 – 1.223
    1.224 – 1.255

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  • 5. 

    Refer to the exhibit. In this VLSM addressing scheme, what summary address would be sent from router A?

    • A.

      172.16.0.0/16

    • B.

      172.16.0.0/20

    • C.

      172.16.0.0/24

    • D.

      172.32.0.0/16

    Correct Answer
    A. 172.16.0.0/16
    Explanation
    Router A receives 3 subnets: 172.16.64.0/18, 172.16.32.0/24 and 172.16.128.0/18.

    All these 3 subnets have the same form of 172.16.x.x so our summarized subnet must be also in that form -> Only A, B or C is correct.

    The smallest subnet mask of these 3 subnets is /18 so our summarized subnet must also have its subnet mask equal or smaller than /18.

    -> Only answer A has these 2 conditions -> A is correct.

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  • 6. 

    Subnet the Class B IP Address 130.13.0.0 into 500 Subnets.  What is the new Subnet Mask and what is the Increment?

    • A.

      Subnet Mask 255.255.255.0 with an Increment of 2

    • B.

      Subnet Mask 255.255.255.192 with an Increment of 128

    • C.

      Subnet Mask 255.255.255.128 with an Increment of 128

    • D.

      Subnet Mask 255.255.255.128 with an Increment of 64

    Correct Answer
    C. Subnet Mask 255.255.255.128 with an Increment of 128
    Explanation
    130.13.0.0 Need 500 Networks

    128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

    0 0 0 0 0 0 0 1 . 1 1 1 1 0 1 0 0 = 500 in binary
    Need to borrow 9 bits for # of Networks

    New Subnet Mask = 255.255.11111111.10000000 (255.128) or /25
    ^ Increment = 128

    2^9 = # of Networks = 512
    2^7-2 = # of Hosts per Network = 126

    First 5 IP Ranges
    130.13.0.0 - 130.13.0.127
    13.0.128 – 13.0.255
    13.1.0 – 13.1.127
    13.1.128 – 13.1.255
    13.2.0 – 13.2.127

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  • 7. 

    Your company wants to utilize the private Class C IP Address of 192.168.1.0.  You are tasked with Subnetting the Address to get the most networks with at least 30 Hosts per Subnet. How many Networks will be created after you subnet?  Also, what is the first usable IP Address in the Second Network range?

    • A.

      8 Networks, First usable from second Network range = 192.168.1.33

    • B.

      3 Networks, First usable from second Network range = 192.168.2.33

    • C.

      5 Networks, First usable from second Network range = 192.168.1.1

    • D.

      8 Networks, First usable from second Network range = 192.168.1.1

    Correct Answer
    A. 8 Networks, First usable from second Network range = 192.168.1.33
    Explanation
    192.168.1.0 Need 30 Hosts per Network

    128 64 32 16 8 4 2 1

    0 0 0 1 1 1 1 0 = 30 in binary
    Need to save 5 bits for # of Hosts

    New Subnet Mask = 255.255.255.11100000 (224) or /27
    ^ Increment = 32

    2^3 = # of Networks = 8
    2^5-2 = # of Hosts per Network = 30

    First 5 IP Ranges
    192.168.1.0 - 192.168.1.31
    1.32 – 1.63
    1.64 – 1.95
    1.96 – 1.127
    1.128 – 1.159

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  • 8. 

    Refer to the diagram. All hosts have connectivity with one another. Which statements describe the addressing scheme that is in use in the network? (Choose three)

    • A.

      The subnet mask in use is 255.255.255.192.

    • B.

      The LAN interface of the router is configured with one IP address.

    • C.

      The subnet mask in use is 255.255.255.128.

    • D.

      The IP address 172.16.1.25 can be assigned to hosts in VLAN1.

    • E.

      The LAN interface of the router is configured with multiple IP addresses.

    Correct Answer(s)
    C. The subnet mask in use is 255.255.255.128.
    D. The IP address 172.16.1.25 can be assigned to hosts in VLAN1.
    E. The LAN interface of the router is configured with multiple IP addresses.
    Explanation
    VLAN 2 has 114 hosts so we need to leave 7 bits 0 for the host addresses (27 – 2 = 126 > 114). Notice that we are working with class B (both Host A and Host B belong to class B) and the default subnet mask of class B is /16 so we need to use 16 – 7 = 9 bits 1 for the sub-network mask, that means the subnet mask should be 255.255.255.128 -> B is correct.

    By using above scheme, C is correct because the IP 172.16.1.25 belongs to the sub-network of VLAN 1 (172.16.1.0/25) and can be assigned to hosts in VLAN 1.

    For communication between VLAN 1 and VLAN 2, the LAN interface of the router should be divided into multiple sub-interfaces with multiple IP addresses -> F is correct.

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  • 9. 

    Subnet the Address 150.20.0.0 into networks supporting 500 Hosts each.  What is the New Subnet Mask and the IP Address Range of the first Network?

    • A.

      Subnet Mask 255.255.255.0, Range 150.20.0.0 - 150.20.0.255

    • B.

      Subnet Mask 255.255.0.0, Range 150.20.0.0 - 150.20.1.255

    • C.

      Subnet Mask 255.255.255.254, Range 150.20.0.0 - 150.20.255.255

    • D.

      Subnet Mask 255.255.254.0, Range 150.20.0.0 - 150.20.1.255

    Correct Answer
    D. Subnet Mask 255.255.254.0, Range 150.20.0.0 - 150.20.1.255
    Explanation
    150.20.0.0 Need 500 Hosts per Network

    128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1

    0 0 0 0 0 0 0 1 . 1 1 1 1 0 1 0 0 = 500 in binary
    Need to save 9 bits for # of Hosts

    New Subnet Mask = 255.255.11111110.00000000 (254.0) or /23
    ^ Increment = 2 in the 3rd Octet

    2^7 = # of Networks = 128
    2^9-2 = # of Hosts per Network = 510

    First 5 IP Ranges
    150.20.0.0 - 150.20.1.255
    20.2.0 – 20.3.255
    20.4.0 – 20.5.255
    20.6.0 – 20.7.255
    20.8.0 – 20.9.255

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  • 10. 

    Subnet the IP Address 210.30.12.0, so there are 60 Hosts in each network.  What are the Broadcast Addresses of each Network?

    • A.

      210.30.12.63, 127, 191, and 255

    • B.

      210.30.12.64, 128,192, and 255

    • C.

      210.30.12.255

    • D.

      210.30.12.0

    Correct Answer
    A. 210.30.12.63, 127, 191, and 255
    Explanation
    210.30.12.0 Need 60 Hosts per Network

    128 64 32 16 8 4 2 1

    0 0 1 1 1 1 0 0 = 60 in binary
    Need to save 6 bits for # of Hosts

    New Subnet Mask = 255.255.255.11000000 (192) or /26
    ^ Increment = 64

    2^2 = # of Networks = 4
    2^6-2 = # of Hosts per Network = 62

    All 4 IP Ranges
    210.30.12.0 - 210.30.12.63
    12.64 – 12.127
    12.128 – 12.191
    12.192 – 12.255

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