![Nielit Scentist-B [02-12-2018]](https://solutionsadda.in/wp-content/uploads/2019/05/green-new-logo.png){kind=logo}
Database-Management-System
October 4, 2023KVS 30-12-2018 Part-A
October 4, 2023Nielit Scentist-B [02-12-2018]
| Question 6 |
Identify the subnet mask for the given direct broadcast address of subnet of subnet is 201.15.16.31.
| 255.255.192.192 | |
| 255.255.255.198 | |
| 255.255.255.240 | |
| 255.255.257.240 | |
| None of the above |
Question 6 Explanation:
Subnet mask ID: 201.15.16.16/28 and subnet broadcast ID is 201.15.16.31.
The last octet of given DBA is 0001 1111. So, in Subnet mask address all should be 1’s except last 5 digits,
i.e., 27 bit for NID which is 255.255.255.224.
The last octet of given DBA is 0001 1111. So, in Subnet mask address all should be 1’s except last 5 digits,
i.e., 27 bit for NID which is 255.255.255.224.
The correct answer would be none of these.
Correct Answer: E
Question 6 Explanation:
Subnet mask ID: 201.15.16.16/28 and subnet broadcast ID is 201.15.16.31.
The last octet of given DBA is 0001 1111. So, in Subnet mask address all should be 1’s except last 5 digits,
i.e., 27 bit for NID which is 255.255.255.224.
The last octet of given DBA is 0001 1111. So, in Subnet mask address all should be 1’s except last 5 digits,
i.e., 27 bit for NID which is 255.255.255.224.
The correct answer would be none of these.
Subscribe
Login
0 Comments