New CCNA – IPv6 Questions 2

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Who can explain why in Q3 anwer B is incorrect?

IPv6 addresses are written in eight groups of four hexadecimal digits separated by colons,

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IPV6 calculation is hexadecimal format.In that XYZ aplaphets will not come.ABCDEF will come.

Hi Admin,

Will you be setting up composite Questions for new CCNA??

@Student
B is incorrect because it is missing a few Zeros in the address field. you can abbreviate multiple fields of Zero’s with a :: but can only use it ONCE. So if you have a second spacing of a field of zeros you can have just a single 0 in the address space and it is still valid. The 61 in the address space however can be taken as 0061 or 6100 so it needs to be clearly defined or it is not a valid address.

the 8 grouping as well but IOS I think just adds 0′s at the end if you end early

plz help me to understand in Q3 how D is correct

because :: contains 4 times leading 0 ‘s so that its 8 hexadecimal

WXYZ is not a valid hex #. its starts from 0-9 and A-F (10-15).

For question:
“What are three features of the IPv6 protocol? (Choose three)”
the answer “E. plug-and-play”
is incorrect?!

What is “plug-and-play” for IPv6?

“Only IPv6 Supports Auto-Configuration”
Another common claim is that only IPv6 offers “plug-and-play” auto-configuration. Again this is an overenthusiastic statement, given the widespread use of the Dynamic Host Configuration Protocol (DHCP) in IPv4 networks these days. Both protocol environments support some level of “plug-and-play” auto-configuration capability, and in this respect the situation is pretty much the same for both IPv4 and IPv6.
(from Opinion: The Mythology of IP Version 6 by Geoff Huston, Telstra)

May be “Optional IPsec” is the right choice?

Ok, IPsec is not optional any more.

“pnp = marketing hype” for auto-configuration. And it’s the correct answer.

IPSec is a part of IPv6 now, not an option

please explame me Q3
why not the right answer is B
i think both of B and D are correct.
please someone tell me…

B is incorrect. IPv6 should have 8 groups of 4 hexadecimal characters. Letter B only has 6 groups and it doesn’t contain double colons (::). If we convert it back to the original address, it will be 2002:7654:A1AD:0061:81AF:CCC1, still it only has 6 groups

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Q6 was in today’s exam.

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Hi
Who can explain why in Q5 anwer A is incorrect?
Help me please….

@MORNY
router elected as root bridge because he has lowest mac-address between 2 lowest priority A and B

ps: * switch A

Question3 (My answer ?)
Which option is a valid IPv6 address?

A. 2001:0000:130F::099a::12a (:: appears twice. Not valid)
B. 2002:7654:A1AD:61:81AF:CCC1 (only 6*16 bits. Need 8*16 bits to make ipV6)
C. FEC0:ABCD:WXYZ:0067::2A4 ( Contains letters above “F”. Not valid)
D. 2004:1:25A4:886F::1 (Valid IP)

Dear All,

In Q4 they have mentioned -

What is the alternative notation for the IPV6 address B514:82C3:0000:0000:0029:EC7A:0000:EC72?

Ans : C. B514:82C3::0029:EC7A:0:EC72

But IPv6 don’t encourage leading 0′s so the correct format should be -
B514:82C3::29:EC7A:0:EC72 (so none of the option matches out here).

Am I wrong?? Looking forward for your valuable reply to the earliest since I have my exam on coming Monday.

Amartya,
you’re right but C B514:82C3::0029:EC7A:0:EC72 is also correct (the only one in the options)

Thanks Mac

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Question 2: The statement “Every IPv6 interface contains at least one loopback address” is incorrect. It should say “Every IPv6 interface contains one link-local address”.

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Reply Q3 (B) incorrect

an IPv6 address is made of 128 bits and it is represented using exadecimal digits.

Each hex digits represents 4 binary digits = 1 hex = 4 bits

So an IPv6 address is made of 128/4 = 32 hexadecimal digits

the usual hex dotted representation groups 4 hex and separate these groups using a ‘:’

So an IPv6 address is made of 8 4hex groups divided by :

the notation allows to represent one of more 4 hex consecutive groups made of all zeroes with a ::

This can appear only once in the number.

In your case the proposed hexadecimal sequence is:

2002:7654:a1ad:61:81af:ccc1

2002:7654:a1ad:0061:81af:ccc1

Here we see only 6 4-hex groups, no :: appears in the middle to represent all zero groups so it is too short.

an IPv6 prefix is represented with a final ::/prefix length

this is not a valid IPv6 address because it is too short, it could be an IPv6 prefix if written as:

2002:7654:a1ad:61:81af:ccc1::/96

Hope to help

Hi all

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Clóvis Madri (Brasil): the answer to Q3 states D not B. Maybe you were just giving explanation why B is not correct?

Can someone explain me the Q5?

@Steve Row:
Question 5
Which switch would STP choose to become the root bridge in the selection process?

A. 32768: 11-22-33-44-55-66
B. 32768: 22-33-44-55-66-77
C. 32769: 11-22-33-44-55-65
D. 32769: 22-33-44-55-66-78

Hi, For the root bridge selection process it chooses the least BID(least priority-32768 least MAC-11-22-33-44-55-66)……So which has the least BID will become the root bridge!!!!

Question 6 was in the exam recently

Please can someone explain how
A. 32768: 11-22-33-44-55-66 is the least BID…. And
C. 32768: 11-22-33-44-55-65 isn’t the least….
Or is it impossible to have an BID like “C’s” ??

Please notice: A and C. the first octet of A. is 32768 while C. is 32769. its very obvious answer is A!
5.

A. 32768: 11-22-33-44-55-66
B. 32768: 22-33-44-55-66-77
C. 32769: 11-22-33-44-55-65
D. 32769: 22-33-44-55-66-78

Hope it will help……

@Padreskey

the answer is A because its bridge priority is lower than C and D but the comparison to letter B is its MAC ADDRESS..

to electing rootbridge in STP you have to choose the lowest bridge priority and lowest mac address

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In question 3 the answers B is not correct because in IPv6 we need at least 8 hexadecimal groups. Look that we can´t identified any ::

got confused in question 5