What is the correct broadcast address for a host with IP address of 192, 164.50.

Question

What is the correct broadcast address for a host with IP address of 192, 164.50.50 using a subnet mark of 255.255.255.240? Which two hosts are on the same subset given the address of 192.168.50.0277? a. 192.168.50.33 b. 192.168.50.95 c. 192.168.50.44 d. 192.168.50.96 You have been given a class C address of 192.168.10.0/28. Which of the following lists a valid host range? a. 192.168 10.16 to 192.168.10.31 b. 192.168.10.33 to 192, 168.10.47 c. 192.168.10.65 to 192.168 1080 d. 192.168.10.81 to 192.168.1094 You have been given a Class C address 192.168.10.0/29. Which of the of following lists a valid host range? a. 192.168.1023 to 192.168.10.230 b. 192.168 10.233 to 192.168.10.239 c. 192.168.10.240 to 192.168.10.246 d. 192.168.10.249 to 192.168.10.254

Explanation / Answer

Question 32:

Each subnet has a range of addressess to assign them to it's hosts.
The first address is a reserved one called network address.
Broadcast address is the last address in a subnet reserved for it. This address cannot be assigned to any host.
Here, the IP address is 192.168.50.50 and the subnet mask is 255.255.255.240.
With this subnet mask, from the last 8 bits of the IP address, first 4 bits are for network and the last four are hosts. Each subnet can have 16 addressess.
The valid subnets are, 0, 16, 32, 48, 64, ...., 240.
In the question, the IP address is 192.168.50.50 which if AND with the subnet mask, we'll get 192.168.50.48 which denotes the network ID.
Now we just have to find the last address of subnet 192.168.50.48, which is 192.168.50.63
192.168.50.63 is the broadcast address for the given host.

Question 33:

In the network 192.168.50.0/27, 27 MSBs are used for network address and 5 for hosts in each subnet.
From the last 8 bits, 3 bits are for network and 5 bits are for hosts. 32 addressess are available for each subnet.
So, the valid subnets are, 0, 32, 64, 96, 128, 160, 192, 224.
Among all this options, 192.168.50.33 and 192.168.50.44 lies in the same subnet i.e., 192.168.50.32
Answer: Option a) and c)

Question 34:

In a given class C network address of 192.168.10.0/28 i.e., 192.168.10.240, from the last 8 bits of IP address, first 4 are for network address and last 4 are for hosts just like the network in question 32.
The valid subnets are 0, 16, 32, 48, 64 etc.
The ranges are: 0-15, 16-31, 32-47, 48-63, 64-79, 80-95 etc.
In each range, the first is network ID and the last is broadcast ID.
Therefore, the valid host range is 192.168.10.81 to 192.168.10.94 i.e., option d).

Question 35:

Here, in the network 192.168.10.0/29, from the last 8 bits of IP address, first 5 are for network address and last 3 are for hosts.
Each subnet can have 8 addressess among which valid hosts addressess are 6 in each subnet.
The range of addressess are: 0-7, 8-15, 16-23, 24-31, ....., 232-239, 240-247, 248-255.
In each range, the first is network ID and the last is broadcast ID.
Therefore, the valid host range is 192.168.10.249 to 192.168.10.254 i.e., option d).

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