CCNA Subnetting Quiz Questions And Answers

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

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

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.

 

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.

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.

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

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.

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.

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.