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A network advertises the $CIDR$ network number $50.1.56.0/22$ (and no other  numbers).  Which of the following $\text{IP addresses}$ could the network own?

1. $50.1.57.0$
2. $50.1.59.1$
3. $50.1.60.0$
4. $50.1.120.0$

1. $\text{Only 1}$
2. $\text{Only 1 and 2}$
3. $\text{Only 1,2 and 3}$
4. $\text{All of 1,2,3 and 4 }$

Given IP = $50.1.56.0/22$

Subnet mask = $255.255.252.0$

$252.0 = 1111 \enspace 1100 . 0000 \enspace 0000$

which means to find the range In given IP address put all last 10 bits 0 once which will give the initial address and keep those 10 bits as 1 which will give the last address. So

$56.0 = 0011 \enspace 1000 . 0000 \enspace 0000$

$59.255 = 0011 \enspace 1011. 1111 \enspace 1111$

Thus range is :- $50.1.56.0/22 \text{ - } 50.1.59.255/22$

Thus Only suitable addresses which comes under the defined range is $1$ and $2$.

Hence Answer is option B) Only 1 and 2.
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+1 vote
$50.1.56.0/22$ for this $32-22=10,2^{10}$ address possible

First address =  $50.1.00111000.00000000 \rightarrow 50.1.56.0$

Last address = $50.1.00111011.11111111 \rightarrow 50.1.59.255$

now we can easily pick options $1,2$ belong to network

Hence $Option \space B$
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