GATE IT 2008 | Question: 84

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 $255.255.255.224$.

Given the information above, how many distinct subnets are guaranteed to already exist in the network?

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Comments

Yes this method is very good, stating more briefly in shortly, 

Mask = .224 =>  256 - 224 = 32

Write range and check all available 4th octet addresses

0 - 31          ......nothing
32 - 63        ......nothing
64 - 95        ......(67, 80)
96 - 127      ......(97, 110)
128 - 159    ......(135, 155) 

Total distinct subnets = 3

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YT link to understand this question "https://youtu.be/RKa7PDSuteo?si=7LLGLZylFP1ZHARW"

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Total 3 Subnets.

Answer 1

For each interface, router has one entry in routing table. 
Each interface out of a router = 1 subnet

Here the diagram is look like 

X---------R1-------common-------R2--------Y

R1 has 2 interface

R2 has 2 interface

One is common..so total 3 subnet..

For more clarity https://gateoverflow.in/371924/Gate-cse-2022-question-12

Comments

@Arjun sir @Shaik Masthan sir  @Nirupam Das how this approach will work for this question if we take two different IP addresses for each  R1 and R2.I think here ip addresses are given so so should use bitwise AND method only here . by using bitwise method we can get maximum 6 subnet if we change value of R1 and R2 ip adress but using your approach we still getting 3 only. correct me if i am getting it wrong. 

Answer 2

This is how X and Y are connected. So 3 subnets

Answer 3

Ans

Answer 4

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