Subnetting IP Address Quiz Questions And Answers

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

255.128.0.0

255.0.0.0

255.224.0.0

224.0.0.0

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.

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.

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

Yes

No

To subnet a /24 network into 30 subnets, we need to borrow 5 bits from the host portion, giving us a new subnet mask of /29 (255.255.255.248). Each /29 subnet will have 8 IP addresses (6 usable host addresses after excluding the network and broadcast addresses).

Let's consider the first few subnets:

203.10.93.0/29: Host range 203.10.93.1 - 203.10.93.6

203.10.93.8/29: Host range 203.10.93.9 - 203.10.93.14

203.10.93.16/29: Host range 203.10.93.17 - 203.10.93.22

203.10.93.24/29: Host range 203.10.93.25 - 203.10.93.30

In this subnetting scheme, 203.10.93.30 is the broadcast address for the subnet 203.10.93.24/29, making it not a valid host address.

Explanation

To subnet a /24 network into 30 subnets, we need to borrow 5 bits from the host portion, giving us a new subnet mask of /29 (255.255.255.248). Each /29 subnet will have 8 IP addresses (6 usable host addresses after excluding the network and broadcast addresses).

Let's consider the first few subnets:

203.10.93.0/29: Host range 203.10.93.1 - 203.10.93.6

203.10.93.8/29: Host range 203.10.93.9 - 203.10.93.14

203.10.93.16/29: Host range 203.10.93.17 - 203.10.93.22

203.10.93.24/29: Host range 203.10.93.25 - 203.10.93.30

In this subnetting scheme, 203.10.93.30 is the broadcast address for the subnet 203.10.93.24/29, making it not a valid host address.

3. 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?

8 Networks, First usable from second Network range = 192.168.1.33

3 Networks, First usable from second Network range = 192.168.2.33

5 Networks, First usable from second Network range = 192.168.1.1

8 Networks, First usable from second Network range = 192.168.1.1

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

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

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

210.30.12.63, 127, 191, and 255

210.30.12.64, 128,192, and 255

210.30.12.255

210.30.12.0

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

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

5. 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.

199.2.1.224 - 199.2.1.255

199.2.1.0 - 199.2.1.255

199.2.1.252 - 199.2.1.255

199.2.1.32 - 199.2.1.64

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

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

6. 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.

20 Hosts and 50 Subnets

2 Hosts and 64 Subnets

4 Hosts and 50 Subnets

6 Hosts and 64 Subnets

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

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

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

172.16.0.0/16

172.16.0.0/20

172.16.0.0/24

172.32.0.0/16

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.

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.

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

Subnet Mask 255.255.255.0 with an Increment of 2

Subnet Mask 255.255.255.192 with an Increment of 128

Subnet Mask 255.255.255.128 with an Increment of 128

Subnet Mask 255.255.255.128 with an Increment of 64

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

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

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?

Subnet Mask 255.255.255.0, Range 150.20.0.0 - 150.20.0.255

Subnet Mask 255.255.0.0, Range 150.20.0.0 - 150.20.1.255

Subnet Mask 255.255.255.254, Range 150.20.0.0 - 150.20.255.255

Subnet Mask 255.255.254.0, Range 150.20.0.0 - 150.20.1.255

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

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

10. 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)

The subnet mask in use is 255.255.255.192.

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

The subnet mask in use is 255.255.255.128.

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

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

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.

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.

Submit