Networking - Networking Basics - Discussion
Discussion Forum : Networking Basics - Networking Basics (Q.No. 7)
7.
Which of the following is the valid host range for the subnet on which the IP address 192.168.168.188 255.255.255.192 resides?
Answer: Option
Explanation:
256 - 192 = 64. 64 + 64 = 128. 128 + 64 = 192. The subnet is 128, the broadcast address
is 191, and the valid host range is the numbers in between, or 129-190.
Discussion:
46 comments Page 5 of 5.
Harish said:
9 years ago
64 - 2 = 62 then b is the answer.
Shyamnarayan2003@gmail.com said:
9 years ago
Hi, @All.
Subtract 192 from 256 = 64,
Means, the subnetting has been done by dividing into four parts, 64 * 4 = 256.
Now see the option.
64 * 2 = 128,
And 64 * 3 = 192.
So IP add should be in between.
192.168.168.129 to 192.168.168.191.
Subtract 192 from 256 = 64,
Means, the subnetting has been done by dividing into four parts, 64 * 4 = 256.
Now see the option.
64 * 2 = 128,
And 64 * 3 = 192.
So IP add should be in between.
192.168.168.129 to 192.168.168.191.
Punith said:
9 years ago
@Shyamnarayan.
Your answer is perfect with the simple solution.
Your answer is perfect with the simple solution.
Robel said:
9 years ago
128 is called a network address and 191 is called broadcast address.
Thanks it is easy to understand.
Thanks it is easy to understand.
Vivek singh rathore said:
9 years ago
Not getting the solution, please explain me in detail.
Sumit said:
8 years ago
The correct answer will be 192.168.168.128 - 191.
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