Question from examsim at work..

I'm confused over the answer to a couple questions I had on an examsim at work.

You have been assigned 8 class C addresses taht you need to use as one network. One of the network addresses is 200.2.35.0. What will the network address be for your new network?

A) 200.2.0.0

200.2.32.0
C) 200.2.36.0
D) 200.2.35.0

I chose D, but they said it's wrong. The correct answer is B. Now how is that possible? What am I not seeing?

Anther question I got wrong is below:

"You have been assigned 8 class C addresses taht you need to use as one network. One of the network addresses is 200.2.35.0. What will the broadcast address be for your new network?"

A) 200.2.40.255

200.2.35.255
C) 200.2.39.255
D) 200.2.32.255

I answered B, but the correct answer is C. What?!

Another one is:

You have been assigned 8 class C addresses taht you need to use as one network. One of the network addresses is 200.2.35.0. Which of the following is not one of the networks you were assigned?

A) 200.2.36.0

200.2.32.0
C) 200.2.39.0
D) 200.2.31.0

I had no idea which one to pick. Answer is D.

Any explanations would be great!
Thanks!

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Comments

EdTheLad

The wording of the question is terrible,are you sure you were given no extra info?
Anyway you were assigned a block of 8 class C addresses, this block starts at 200.2.32.0 and goes to 200.2.39.255, as you can see the third octet range 32->39

The first answer is B as this is the first address that represents your subnet.
The second answer is the broadcast address which is the last address in your given range,which is 200.2.39.255.
Third answer is D as 200.2.31.0 is not in your given range.

p0et

Thanks for the reply, but I'm confused as to why you said "this block starts at 200.2.32.0". How do you know that .32.0 is where it starts at when the only info the question provided which included an IP was "One of the network addresses is 200.2.35.0"? From the question, I'd assume the block could also start at .35.0 and end at .42.0... but i'm just learning and not too sure.

EdTheLad

To understand this type of question you need to have a good understanding of subnetting and supernetting.A natural class C address has a /24 bit mask and uses a range from 192up until 223 for the first octet.
So you were given a block of 8 class C addresses.
This is actually supernetting or cidr, you are effectively moving the mask boundary to the left of the natural /24 bit mask.

If you have /23 you have a block of 2
If you have /22 you have a block of 4
If you have /21 you have a block of 8
If you have /20 you have a block of 16 etc..

You are told you have been given a block of 8 addresses, which corresponds to /21
which is a mask of 255.255.248.0
One addresses in the range was 200.2.35.0, and i know the mask 255.255.248.0

This mask tells me i have subnets in the 3rd octet 0,8,16.24.32,40,48.....etc
The address given falls into the 32->39 range.

This is too complex for Net+ so i doubt it would be on the exam.

p0et

Thanks for the informative reply.

I do understand how you get 255.255.248.0 from /21, and I do the different class ranges and default masks. What I don't understand is the definition of a "block". You said:
"If you have /23 you have a block of 2". Wouldn't that mean you'd have only 1 bit left in the 3rd octet? Not sure what the "block of 2" represents....

Maybe if I understand the above, then the rest of what you wrote will become clear.

p0et

I have another question that I'm unsure of.

Dynamic routing is possible only when a router is configured with routed protocols. T/F?
Your answer: A. True
Correct answer: B. False

Explanation:
Dynamic routing is possible only when a router is configured with routing protocols.

That explanation sure looks a lot like the question itself and if so, that would be true, not false. Can someone shed some light on this one?

Thanks!

georgemc

You need to understand the difference between ROUTING and ROUTED protocols.

ROUTING = RIP, IGRP, EIGRP, OSPF, BGP, IS-IS, etc.

ROUTED = IP, IPX, etc.

ROUTED protcols are capable of being routed across ROUTING protocols.

or

ROUTING protocols route ROUTED protocols.

I'm not sure how else to explain it. Be sure to get yourself a good book and keep on studying, you'll get there.

malcybood

p0et wrote:

I have another question that I'm unsure of.

Dynamic routing is possible only when a router is configured with routed protocols. T/F?
Your answer: A. True
Correct answer: B. False

Explanation:
Dynamic routing is possible only when a router is configured with routing protocols.

That explanation sure looks a lot like the question itself and if so, that would be true, not false. Can someone shed some light on this one?

Thanks!

If the answer is false then it is basically asking you to identify the difference between routing and routed

"Dynamic" routing protocols are RIP, OSPF, IGRP, EIGRP

Routed protocols refers to the protocol (IP, IPX, Appletalk) that your network is running hence why the answer is false

I agree with Ed though the questions a pretty badly worded

georgemc

p0et wrote:

Thanks for the informative reply.

I do understand how you get 255.255.248.0 from /21, and I do the different class ranges and default masks. What I don't understand is the definition of a "block". You said:
"If you have /23 you have a block of 2". Wouldn't that mean you'd have only 1 bit left in the 3rd octet? Not sure what the "block of 2" represents....

Maybe if I understand the above, then the rest of what you wrote will become clear.

200.2.35.0 is naturally a class C address with a subnet mask of:

255.255.255.0 or 11111111.11111111.11111111.00000000

If you were to Supernet 8 contiguous class C's that include the above network you would get:

200.2.32.0 - 200.2.39.255 or 11001000.00000010.00100000.00000000 -11001000.00000010.00100111.11111111

with a subnet mask of:

255.255.248.0 or 11111111.11111111.11111000.00000000

you borrowed 3 bits from the network portion of the address in order to supernet your 8 class C's together.

2^N=number of subnets, or in this case, 2^3=8 subnets

Looking at the third octet, the network portion is a minimum of 32 (00100000) and a maximum of 39 (00100111).

Remember the network portion of the address is fixed and the host portion is variable.