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A subnetted Class $B$ network has the following broadcast address: $144.16.95.255$

Its subnet mask

  1. is necessarily $255.255.224.0$
  2. is necessarily $255.255.240.0$
  3. is necessarily $255.255.248.0$
  4. could be any one of $255.255.224.0$, $255.255.240.0$,$255.255.248.0$
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Best answer
63 votes
63 votes
Option ($D$) is correct. 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 $.95.255$   $.0101 1111. 1111 1111$ (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 $1$).

$.224.0$     $1110 0000. 0000 0000$   (leftmost $3$ $bits$ for subnet)

$.240.0$     $1111 0000. 0000 0000$  (leftmost $4$ $bits$ for subnet)

$.248.0$     $1111 1000. 0000 0000$  (...          $5$ $bits$ for subnet )
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2 votes

The below image clearly explains it can be any of those IDs

As you can see we dont have same network ID for all 3.

Had the option d been only 224 and 240 in the third octet then , both A and B are correct. But 

As we have different nid for all, the SM can be different for all and based on how many subnets you want to create

With 224 you can create 6 different sub networks

With 240 you can create 14 different subnetworks

With 248 you can create 30 different subnetworks.

1 votes
1 votes
As it is a DBA, so there must be continuous 1s at the end.

So here are 13 continuous 1s.

And to get the first ID, i.e., NID, we need to replace all these 1s with 0s.

So NID for 144.16.95.255 = 144.16.01000000.0 = 144.16.64.0

So when we perform given IP Address AND Subnet mask in Options, we get 144.16.64.0 every time as NID.

 

\[
\begin{align*}
144.16.95.255 \text{ AND } 255.255.224.0 &= 144.16.64.0 \\
144.16.95.255 \text{ AND } 255.255.240.0 &= 144.16.64.0 \\
144.16.95.255 \text{ AND } 255.255.248.0 &= 144.16.64.0 \\
\end{align*}
\]
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