Question 5595 – NIELIT Junior Teachnical Assistant_2016_march
December 9, 2023
Question 5594 – NIELIT Junior Teachnical Assistant_2016_march
December 9, 2023
Question 5595 – NIELIT Junior Teachnical Assistant_2016_march
December 9, 2023
Question 5594 – NIELIT Junior Teachnical Assistant_2016_march
December 9, 2023

Computer-Networks

Question 56

A subnetted Class B network has the following broadcast address: 144.16.95.255. Its subnet mask

A
is necessarily 255.255.224.0
B
is necessarily 255.255.240.0
C
is necessarily 255.255.248.0
D
could be any one of 255.255.224.0, 255.255.240.0, 255.255.248.0
Question 56 Explanation: 
In the broadcast address for a subnet, all the host bits are set to 1. So as long as all the bits to the right are 1, bits left to it can be taken as possible subnet.
Broadcast address for subnet is
.95.255 or .01011111.11111111
(as in class B, 16 bits each are used for network and host)
So, we can take minimum 3 bits (from left) as subnet and make rest as host bits (as they are 1)
.224.0 → 11100000.00000000 (leftmost 3 bits for subnet)
.240.0 → 11110000.00000000 (leftmost 4 bits for subnet)
.248.0 → 11111000.00000000 (leftmost 5 bits for subnet)
Correct Answer: D
Question 56 Explanation: 
In the broadcast address for a subnet, all the host bits are set to 1. So as long as all the bits to the right are 1, bits left to it can be taken as possible subnet.
Broadcast address for subnet is
.95.255 or .01011111.11111111
(as in class B, 16 bits each are used for network and host)
So, we can take minimum 3 bits (from left) as subnet and make rest as host bits (as they are 1)
.224.0 → 11100000.00000000 (leftmost 3 bits for subnet)
.240.0 → 11110000.00000000 (leftmost 4 bits for subnet)
.248.0 → 11111000.00000000 (leftmost 5 bits for subnet)
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