Question from examsim at work..

p0etp0et Member Posts: 48 ■■□□□□□□□□
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
B) 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
B) 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
B) 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!

Comments

  • EdTheLadEdTheLad Member Posts: 2,111 ■■■■□□□□□□
    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.
    Networking, sometimes i love it, mostly i hate it.Its all about the $$$$
  • p0etp0et Member Posts: 48 ■■□□□□□□□□
    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. icon_redface.gif
  • EdTheLadEdTheLad Member Posts: 2,111 ■■■■□□□□□□
    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.
    Networking, sometimes i love it, mostly i hate it.Its all about the $$$$
  • p0etp0et Member Posts: 48 ■■□□□□□□□□
    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. icon_wink.gif
  • p0etp0et Member Posts: 48 ■■□□□□□□□□
    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!
  • georgemcgeorgemc Member Posts: 429
    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. :)
    WGU BS: Business - Information Technology Management
    Start Date: 01 October 2012
    QFT1,PFIT in progress.
    TRANSFERRED/COMPLETED: AGC1,BBC1,LAE1,QBT1,LUT1,QLC1,QMC1,QLT1,IWC1,INC1,INT1,BVC1,CLC1,MGC1, CWV1 BNC1, LIT1,LWC1,QAT1,WFV1,EST1,EGC1,EGT1,IWT1,MKC1,MKT1,RWT1,FNT1,FNC1, BDC1,TPV1 REQUIRED:
  • malcyboodmalcybood Member Posts: 900 ■■■□□□□□□□
    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
  • georgemcgeorgemc Member Posts: 429
    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. icon_wink.gif


    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.
    WGU BS: Business - Information Technology Management
    Start Date: 01 October 2012
    QFT1,PFIT in progress.
    TRANSFERRED/COMPLETED: AGC1,BBC1,LAE1,QBT1,LUT1,QLC1,QMC1,QLT1,IWC1,INC1,INT1,BVC1,CLC1,MGC1, CWV1 BNC1, LIT1,LWC1,QAT1,WFV1,EST1,EGC1,EGT1,IWT1,MKC1,MKT1,RWT1,FNT1,FNC1, BDC1,TPV1 REQUIRED:
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