Subnetting question

Would someone like to give this question a go. Its taken from Que certification's Exam Cram 2

Identify three valid host addresses in the 201.168.27.0 network with a subnet mask of 255.255.255.240 (choose three).

A. 201.168.27.33
B. 201.168.27.112
C. 201.168.27.119
D. 201.168.27.126
E. 201.168.27.175
F. 201.168.27.208

If you give it a go, a brief explanation as to how you reached your answer would be appreciated.

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Comments

georgemc

I'm not sure I like how this question is worded. The valid hosts for 201.168.27.0 /28 (255.255.255.240) are 201.168.27.1 - 201.168.27.14.

The answers I believe they are looking for are: A, C, & D since 201.168.27.112 & .208 are subnet addresses and .175 is a broadcast address.

Cucumber

First, figure out the netmask, you must automatically tell it from the number 240:

11110000

Notice the last four bits are zeroes, that means you have to look after the last four bits of each given IP address to tell whether the address is a valid host IP, a subnet IP (0000) or a directed broadcast IP address (1111)

Next, translate each number into digital format (we only care about the last octet)

A. 33
0010 0001
Look at the last four bits, 0001, neither 0000 nor 1111, so this one is a valid host ip

B. 112
0111 0000
Look at the last four bits, 0000, this is a subnet, not a valid host addy

C. 119
0111 0111
Look at the last four bits, 0111, neither 0000 nor 1111, so this one is a valid host ip

D. 126
0111 1110
Look at the last four bits, 1110, neither 0000 nor 1111, so this one is a valid host ip

E. 175
1010 1111
Look at the last four bits, 1111, its a directed broadcast addy, not valid as a host ip

F. 208
1101 0000
Look at the last four bits, 0000, this is a subnet, not a valid host addy

kafifi13

They way this was worded threw me off. Is that how they had it in the text?

Pash

You have the mask there with a prefix of /28 correct?

interesting octet is the last octet

256 is your magic number

take 240 from 256 and you get ?

16

Your network go's up in 16's, remember just like triangle houses with one up two down (refering to rooms) this applies to your useable addresses, lets not confuse anything.

useable IP's is?

A: looks funky dory to me, .32 being the network address (remember your ranges)

B: super fast mega rush maths would make you do 2x 64 which is easy peasy, - 16 you get .112, this one is not funky dory, its a network address chief.

C: looks fan dabby dousy to me chum, .112 being your network address, whats your last useable IP chief?

Well this ties in with the question i asked you in C

E: This looks like an all the F's to me chief, whats the network address here? i say .160 you?

F: again super fast fann dabby dousy maths says that 128 +64 is 192, add 16 to that? .208 is a network address.

It's all about patterns, seriously you will be able to do it in your head soon, if im honest i took a while coz i had no sleep last night, but the principal doesnt change.

And damn you who showed him in binary, its too early in the day for that.

Cucumber

sorry Im a binary freak

dtlokee

Yeah the wording is a bit tricky, keep in mind what the difference is between "network" and "subnet", when they say "network" they are referring to the classful network number (the mask is the default mask for the class), a "subnet" is a smaller piece of that (anytime the mask is longer than the default mask), and a "block of networks" or "supernet" is a group of networks that have been summarized (the mask is shorter than the default mask for the class)

so when they say 201.168.27.0 network they are impling the classful mask, then by saying a mask of 255.255.255.240 they are impling there are subnets of that network. Cucumber and pash already did the math but there would be 16 subnets with 14 host addresses on each.

Pash

Cucumber wrote:

sorry Im a binary freak

It's alright cucumber, but you will be going in my pastrami and ham sandwhich later. Sorry

faisal79

as everyone said

i worked out that correct aswers are A,C and D this is how

255.255.255.240 means 4bits has been borrowed which makes \28 so 201.168.27.0\28

in binary when you borrow 4, you turn the first 4 bits as follows 11110000

i have taken the 256-240 =16 so the first subent is 0, 16 32 48 and so on you increment by 16 every time
hare 32,112 and 208 are subets is the subnet

A is the first host address of the subnet 32,
C is seventh host address of subnet of 112
D is last host address of the subnet 112
those are the correct answers

E is broadcast address of the subnet 160

i hope this hellps

totts

faisal79 wrote:

as everyone said

i worked out that correct aswers are A,C and D this is how

255.255.255.240 means 4bits has been borrowed which makes \28 so 201.168.27.0\28

in binary when you borrow 4, you turn the first 4 bits as follows 11110000

i have taken the 256-240 =16 so the first subent is 0, 16 32 48 and so on you increment by 16 every time
hare 32,112 and 208 are subets is the subnet

A is the first host address of the subnet 32,
C is seventh host address of subnet of 112
D is last host address of the subnet 112
those are the correct answers

E is broadcast address of the subnet 160

i hope this hellps

Yes, A, C and D are the correct answers. Thanks for everyone who made and input. I think the wording can throw you off a bit on these questions. The distinction is being aware that they're giving you the 'network' number rather than the 'subnet'. This was the mistake I kept making. BTW cucumber... the binary explanation was good and this is the way they detail it in the book. If you know the binary then you know the nuts n bolts!