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Computer Science - Networking - Discussion

Discussion Forum : Networking - Section 2 (Q.No. 40)
Networking
40.
You are working with three networks that have the network IDs 192.168.5.0, 192.168.6.0, and 192.168.7.0. What subnet mask can you use to combine these addresses into one?
255.255.252.0
225.255.254.0
255.255.255.240
255.255.255.252
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.
Ekta said:   3 years ago
Agree, Option A is correct.
Wasim sajjad said:   4 years ago
11000000.10101000.00000101.000000 00 (192.168.5.0) =255.255.255.255-3.
11000000.10101000.00000110.000000 00 (192.168.6.0) = 255.255.255.255-3.
11000000.10101000.00000111.000000 00 (192.168.7.0). = 255.255.255.255-3.
Rahmat said:   4 years ago
D is correct, as default subnet mask for class C is 255.255.255. , as 3 hosts are required, hence 252 is used.
HarshavardHan said:   4 years ago
Thank You @AK Saddam.
Ravikumarmaradana said:   6 years ago
Correct @Saddam.
AK Saddam said:   7 years ago
Binary of 3 hosts are:

11000000.10101000.00000101.00000000 (192.168.5.0)
11000000.10101000.00000110.00000000 (192.168.6.0)
11000000.10101000.00000111.00000000 (192.168.7.0).

Here, 22 bits are common for 3 hosts.
So in the 3rd octet, we will count 6 bit for the subnet mask. so it will be 255.255.252.0.
Ashu said:   8 years ago
I think option D is correct.
(1)
Shewangizaw Bogale said:   9 years ago
I think 255.255.255.252 should be the answer as it is a class C network.

Thank you @Shiva and @Sehal.
Shiva said:   9 years ago
I think 255.255.255.252 should be the answer as it is a class c network.
Sehal said:   1 decade ago
I think right answer is C because we are using 3 bits for sub net we will skip 3 bits and then add all other bits then we got answer C.

Anyone explain please.
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