GATE IT 2008 | Question: 85

Question

Host $X$ has $IP$ address $192.168.1.97$ and is connected through two routers $R1$ and $R2$ to an­other host $Y$ with $IP$ address $192.168.1.80$. Router $R1$ has $IP$ addresses $192.168.1.135$ and $192.168.1.110$. $R2$ has $IP$ addresses $192.168.1.67$ and $192.168.1.155$. The netmask used in the network is $255.255.255.224$.

Which $IP$ address should $X$ configure its gateway as?

• A. $192.168.1.67$
• B. $192.168.1.110$
• C. $192.168.1.135$
• D. $192.168.1.155$

Comments

net mask is .224 => 11100000
so ip address should have first three bits as same as that of X   .97 =>  011
writing only the first three bits
.67  =>  010
.110 => 011
.135 => 100
.155 => 100

.110  (B) 

 

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The question is straightforward if you know one key fact: the host and the gateway must share the same network ID (netid), as they reside on the same network.

This turns the problem into a classic routing question, where we need to determine which of the given addresses belong to the same network.

Here, it is clear that 192.168.1.110 and 192.168.1.97 belong to the same network.

Therefore, the correct gateway for host X will be 192.168.1.110 (option B).

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Source : https://en.wikipedia.org/wiki/Gateway_address#:~:text=The%20gateway%20operates,facilitate%20network%20planning.

Answer 1

$X$ must be able to reach the gateway using the net mask.
Subnet number of host $X = 192.168.1.97 \;\&\; 255.255.255.224 = 192.168.1.96$

Now, the gateway must also have the same subnet number. Lets take $\text{IP}\;192.168.1.110$ of $R1. \; 192.168.1.110 \;\&\; 255.255.255.224 = 192.168.1.96$ and hence this can be used by $X$. 

$($To quickly identify the matching mask divide the last part of mask $(224\;\text{here})$ into powers of $2$. So, $224 = 128 + 64+ 32$. Now, our host $X$ has $97$ as the last part of $\text{IP} = 64 + 32 + 1$. So, the last part of subnet number becomes $64 +32 = 96$. Now, we need to consider only those $\text{IPs}$ whose last part will contain $64$ as well as $32)$

http://courses.washington.edu/css432/joemcc/slides/03_cidr.ppt
Correct Answer: $B$

Comments

if anyone is still confused then check @Hemant Parihar comment in this reply section only its the most important thing I learned today.

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There are more simple method but this one explain well

 

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updated link 

https://courses.washington.edu/css432/joemcc/syllabi/2012au.html

Answer 2

The concept for this question and all other questions like this is as follows.

The router takes 1 IP address for every interface that it is connected to and the IP address is taken from the same subnet in which the whole network on that interface is.

So here every router is connected to 2 interfaces. Like R1 on the one side is connected to X and on the other side connected to R2. So similarly R2 is connected to R1 and Y.

So both the routers will be having 2 IP addresses. Both IPs in the interface should be in the same network.
So one side of the wire is X and on the second side is R2, So The default gateway for X will be the Ip address that belongs to the same subnet out of 2 IP which R1 has.
So you can get the Ip address router is using on the X side, remaining will be used to communicate by R2. Similarly, R2 has 2 IPs, check which is in the same subnet with the remaining IP of R1.

Comments

@Tendua
Thanks for explaining this.

Answer 3

Network mask:255.255.255.224
Subnet bit =3 bit
For: 192.168.1.97
Network address = 192.168.1.01100000 =>192.168.1.96
For:192.168.1.110
Network address=192.168.1.01101110
=>192.168.1.96

Comments

@heena Singh here we have 3 bits in subnet ID part so 2^3 = 8 combinations can be used so we will have 8 subnets but last and first subnet id will not be used hence total 6 subnets are possible,But how it is 3 subnets ?

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@sethi  you are right but here question is about gatway not how many subnet possible???

Subnet number of host X = 192.168.1.97 & 255.255.255.224 =192.168.1.96

the gateway must also have the same subnet number.

Take ip addresses and net mask find subnet number if it have same subnet no. then it is used as gatway

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nice explanation.

Answer 4

.

Answer 5

COMMENT BELOW IF YOU HAVE ANY DOUBT

Answer 6

Method 1 - Without checking options,

A host's default gateway must be the router interface that is in the SAME subnet as the host.

Mask - 255.255.255.224  =>  256 - 224 = 32

X = 192.168.1.97          So subnets are, 

Network             Host Range           Broadcast
     0                         1–30                     31
    32                       33–62                    63
    64                       65–94                    95
    96                       97–126                 127
   128                    129–158                 159
     ...                         ...                          ...

Since, 97 lies in 96–127,

X = 192.168.1.96/27 

 

Now, let's see router interfaces, 

R1 : 192.168.1.135     ............falls in 128–159   (different subnet)
        192.168.1.110     ............falls in 96–127     (same subnet)

R2 :  192.168.1.67      ............falls in 64–95       (different subnet)
         192.168.1.155    ............falls in 128–159   (different subnet)

Only, one interface is same as of X's interface
Ans 192.168.1.110

 

Method 2 - With checking options, 

Host =>  .97
Mask => 255.255.255.224   => /27    =>   96–127

Now, check options

    Ip                      Subnet

    67                      64–95           .....different subnet
    110                    96–127         ......same subnet
    135                    128–159       .....different subnet
    155                    128–159       .....different subnet

Hence, Ans (B)

Method 3 - Even faster method 

X =             .97  =  64 + 32 + 1
Subnet  =  .224 =  128 + 64 + 32 
-----------------------------------------------
              =   64 + 32   i.e 011 (starting 3 bits)

Check options, 

A - 67    =    010    (not matches)
B - 110  =    011     (matches)
C - 135  =   100     (not matches)
D - 155  =   100     (not matches)

Hence, Ans (B)