CCNA Subnetting Quiz Questions And Answers

1. Subnetting allows a network administrator to divide a large network into smaller, more efficient sub-networks.

Yes

No

Subnetting is the process of dividing a large network into smaller sub-networks, known as subnets. This practice helps in improving network performance, managing network traffic more effectively, and enhancing security by logically grouping devices based on their functions or geographic locations. Each subnet operates as a separate network with its own subnet mask, allowing efficient use of IP addresses and reducing the size of broadcast domains. Therefore, subnetting is a fundamental technique used in network administration to optimize network resources and organization.

Explanation

Subnetting is the process of dividing a large network into smaller sub-networks, known as subnets. This practice helps in improving network performance, managing network traffic more effectively, and enhancing security by logically grouping devices based on their functions or geographic locations. Each subnet operates as a separate network with its own subnet mask, allowing efficient use of IP addresses and reducing the size of broadcast domains. Therefore, subnetting is a fundamental technique used in network administration to optimize network resources and organization.

2. 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

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

20 Hosts and 50 Subnets

6 Hosts and 64 Subnets

4 Hosts and 50 Subnets

2 Hosts and 64 Subnets

To create 50 subnets from a 193.103.20.0 /24 address, you need to borrow bits from the host part to increase the number of subnets. Borrowing 6 bits from the original 24-bit subnet mask gives you a /30 subnet mask. This allows for 64 subnets in total, with each subnet having 4 IP addresses. However, because one address is used as the network address and another as the broadcast address in each subnet, only 2 addresses in each subnet are available for hosts. So, you end up with 64 subnets, each capable of hosting 2 devices.

Explanation

To create 50 subnets from a 193.103.20.0 /24 address, you need to borrow bits from the host part to increase the number of subnets. Borrowing 6 bits from the original 24-bit subnet mask gives you a /30 subnet mask. This allows for 64 subnets in total, with each subnet having 4 IP addresses. However, because one address is used as the network address and another as the broadcast address in each subnet, only 2 addresses in each subnet are available for hosts. So, you end up with 64 subnets, each capable of hosting 2 devices.

5. 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 your subnet?

199.2.1.0 - 199.2.1.255

199.2.1.32 - 199.2.1.64

199.2.1.224 - 199.2.1.255

199.2.1.128 - 199.2.1.159

1st subnet: 199.2.1.0 - 199.2.1.31

2nd subnet: 199.2.1.32 - 199.2.1.63

3rd subnet: 199.2.1.64 - 199.2.1.95

4th subnet: 199.2.1.96 - 199.2.1.127

5th subnet: 199.2.1.128 - 199.2.1.159 (This is the last available network)

So, the IP range of the last available network is from 199.2.1.128 to 199.2.1.159.

Explanation

1st subnet: 199.2.1.0 - 199.2.1.31

2nd subnet: 199.2.1.32 - 199.2.1.63

3rd subnet: 199.2.1.64 - 199.2.1.95

4th subnet: 199.2.1.96 - 199.2.1.127

5th subnet: 199.2.1.128 - 199.2.1.159 (This is the last available network)

So, the IP range of the last available network is from 199.2.1.128 to 199.2.1.159.

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?

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

To subnet the Class B IP address 130.13.0.0 into 500 subnets, you need to use a subnet mask of 255.255.255.128, and each subnet will have an increment of 128 host addresses. This allows you to create 500 smaller subnetworks from the original Class B address while providing a sufficient number of host addresses in each subnet.

Explanation

To subnet the Class B IP address 130.13.0.0 into 500 subnets, you need to use a subnet mask of 255.255.255.128, and each subnet will have an increment of 128 host addresses. This allows you to create 500 smaller subnetworks from the original Class B address while providing a sufficient number of host addresses in each subnet.

7. 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: 150.20.0.1 to 150.20.1.254

To subnet the address 150.20.0.0 into networks supporting 500 hosts each, we need to determine the appropriate subnet mask and calculate the IP address range for the first network.

To support 500 hosts, we need to find the smallest subnet that accommodates at least 500 host addresses. This requires 9 bits for host addresses (2^9 = 512, which is more than 500), leaving the remaining bits for the network portion.

Since the original address 150.20.0.0 is a Class B address, it has a default subnet mask of /16 (255.255.0.0). To support 500 hosts, we need to borrow 9 bits for host addresses, leaving 7 bits for the network portion in the subnet mask.

The new subnet mask will be /23, which corresponds to 255.255.254.0 (because 8 bits in the third octet plus the first bit of the fourth octet make up 9 bits in total).

For the IP address range of the first network, we'll consider the subnet address and the range of host addresses.

Subnet Address: 150.20.0.0 (Given)

First IP Address in Range: This will be the subnet address + 1, which is 150.20.0.1.

Last IP Address in Range: We calculate the last IP address by subtracting 1 from the broadcast address of the subnet. The broadcast address is the highest address in the subnet range. Since we borrowed 9 bits for hosts, the broadcast address will end in 11111111, or 255 in decimal, in the third octet and 254 in the fourth octet. So, the broadcast address is 150.20.1.255. Therefore, the last usable IP address in the range is 150.20.1.254.

So, for the first network:

Subnet mask: 255.255.254.0 (/23)

IP address range: 150.20.0.1 to 150.20.1.254

Explanation

To subnet the address 150.20.0.0 into networks supporting 500 hosts each, we need to determine the appropriate subnet mask and calculate the IP address range for the first network.

To support 500 hosts, we need to find the smallest subnet that accommodates at least 500 host addresses. This requires 9 bits for host addresses (2^9 = 512, which is more than 500), leaving the remaining bits for the network portion.

Since the original address 150.20.0.0 is a Class B address, it has a default subnet mask of /16 (255.255.0.0). To support 500 hosts, we need to borrow 9 bits for host addresses, leaving 7 bits for the network portion in the subnet mask.

The new subnet mask will be /23, which corresponds to 255.255.254.0 (because 8 bits in the third octet plus the first bit of the fourth octet make up 9 bits in total).

For the IP address range of the first network, we'll consider the subnet address and the range of host addresses.

Subnet Address: 150.20.0.0 (Given)

First IP Address in Range: This will be the subnet address + 1, which is 150.20.0.1.

Last IP Address in Range: We calculate the last IP address by subtracting 1 from the broadcast address of the subnet. The broadcast address is the highest address in the subnet range. Since we borrowed 9 bits for hosts, the broadcast address will end in 11111111, or 255 in decimal, in the third octet and 254 in the fourth octet. So, the broadcast address is 150.20.1.255. Therefore, the last usable IP address in the range is 150.20.1.254.

So, for the first network:

Subnet mask: 255.255.254.0 (/23)

IP address range: 150.20.0.1 to 150.20.1.254

8. Your ISP has given you the address 223.5.14.6/29 to assign to your router's interface. They have also given you the default gateway address of 223.5.14.7. After you have configured the address, the router is unable to ping any remote devices. What is preventing the router from pinging remote devices?

The default gateway is not an address on this subnet.

The default gateway is the broadcast address for this subnet.

The IP address is the broadcast address for this subnet.

The IP address is an invalid class D multicast address.

To resolve the issue described, we must first understand the addressing and subnetting based on the information provided:

IP Address Assigned: 223.5.14.6/29

Subnet Mask (/29): This subnet mask indicates that 29 bits are used for the network part, leaving 3 bits for the host part within the subnet. This means the subnet supports up to 23−2=623−2=6 usable host IP addresses.

Calculating the Subnet:

The subnet /29 means the subnet mask is 255.255.255.248.

To find the range, consider the last octet of the subnet mask (248): 256 - 248 = 8, which means each subnet increments by 8 in the last octet.

So, the subnet 223.5.14.0/29 covers addresses from 223.5.14.0 to 223.5.14.7.

Addresses in the Subnet:

Network Address: 223.5.14.0

Usable Addresses: 223.5.14.1 to 223.5.14.6

Broadcast Address: 223.5.14.7

Analyzing the Problem:

The router's IP address given is 223.5.14.6, which falls within the usable range.

The default gateway address provided is 223.5.14.7, which is the broadcast address for the subnet.

Answer: The problem is that the default gateway is the broadcast address for this subnet. Routers or devices in a network cannot use the broadcast address as a gateway because it is reserved for broadcasting to all devices in the subnet. Thus, any packets intended for external networks sent to this gateway address would not be properly routed.

Explanation

To resolve the issue described, we must first understand the addressing and subnetting based on the information provided:

IP Address Assigned: 223.5.14.6/29

Subnet Mask (/29): This subnet mask indicates that 29 bits are used for the network part, leaving 3 bits for the host part within the subnet. This means the subnet supports up to 23−2=623−2=6 usable host IP addresses.

Calculating the Subnet:

The subnet /29 means the subnet mask is 255.255.255.248.

To find the range, consider the last octet of the subnet mask (248): 256 - 248 = 8, which means each subnet increments by 8 in the last octet.

So, the subnet 223.5.14.0/29 covers addresses from 223.5.14.0 to 223.5.14.7.

Addresses in the Subnet:

Network Address: 223.5.14.0

Usable Addresses: 223.5.14.1 to 223.5.14.6

Broadcast Address: 223.5.14.7

Analyzing the Problem:

The router's IP address given is 223.5.14.6, which falls within the usable range.

The default gateway address provided is 223.5.14.7, which is the broadcast address for the subnet.

Answer: The problem is that the default gateway is the broadcast address for this subnet. Routers or devices in a network cannot use the broadcast address as a gateway because it is reserved for broadcasting to all devices in the subnet. Thus, any packets intended for external networks sent to this gateway address would not be properly routed.

9. Given the address 192.168.20.19/28, which of the following are valid host addresses on this subnet? (Choose two)

192.168.20.29

192.168.20.16

192.168.20.17

192.168.20.31

The given address 192.168.20.19/28 has a subnet mask of 28 bits, which means that the first 28 bits of the IP address are used for the network portion and the remaining 4 bits are used for the host portion.

The valid host addresses on this subnet are those that have the same network portion (first 28 bits) as the given address, but different host portions.

Out of the given options, 192.168.20.29 and 192.168.20.17 have the same network portion as 192.168.20.19 but different host portions, making them valid host addresses on this subnet.

Therefore, the correct answers are 192.168.20.29 and 192.168.20.17.

Explanation

The given address 192.168.20.19/28 has a subnet mask of 28 bits, which means that the first 28 bits of the IP address are used for the network portion and the remaining 4 bits are used for the host portion.

The valid host addresses on this subnet are those that have the same network portion (first 28 bits) as the given address, but different host portions.

Out of the given options, 192.168.20.29 and 192.168.20.17 have the same network portion as 192.168.20.19 but different host portions, making them valid host addresses on this subnet.

Therefore, the correct answers are 192.168.20.29 and 192.168.20.17.

Submit

10. 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?

8 subnets, 31 hosts

8 subnets, 32 hosts

16 subnets, 30 hosts

16 subnets, 32 hosts

By subnetting the network 172.17.32.0/23 into a /27 mask, we are borrowing 4 bits from the host portion to create subnets. With 4 borrowed bits, we can create 2^4 = 16 subnets. However, since the /27 mask only allows for 5 host bits, we have 2^5 - 2 = 30 usable host addresses per subnet (subtracting 2 for the network and broadcast addresses). Therefore, the correct answer is 16 subnets, 30 hosts.

Explanation

By subnetting the network 172.17.32.0/23 into a /27 mask, we are borrowing 4 bits from the host portion to create subnets. With 4 borrowed bits, we can create 2^4 = 16 subnets. However, since the /27 mask only allows for 5 host bits, we have 2^5 - 2 = 30 usable host addresses per subnet (subtracting 2 for the network and broadcast addresses). Therefore, the correct answer is 16 subnets, 30 hosts.