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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.
Ashu said:   8 years ago
I think option D is correct.
(1)
Snehal said:   1 decade ago
As we need to adjust 3 hosts, so 2^2=4 will suffice that,as per binary chart for subnet we have 252 in block of 4, hence 255.255.252.0.
Asis said:   1 decade ago
Please help me while writing more.

192 is class C address, then how 255.255.252.0 is possible as 255.255.255.0 is default class C subnet mask.
Krishna said:   1 decade ago
How the answer is possible ?
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.
Shiva said:   9 years ago
I think 255.255.255.252 should be the answer as it is a class c network.
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.
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.
Ravikumarmaradana said:   6 years ago
Correct @Saddam.
HarshavardHan said:   4 years ago
Thank You @AK Saddam.
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