Building the EUI-64 Interface Identifier for Router R2

200-301 By islever Editorial Last updated

Configure and verify IPv6 addressing and prefix

Short answer

D Applying EUI-64 (split MAC, insert FFFE, invert 7th bit) to R2's MAC yields 2001:db8:9bb6:6bb9:C801:B6FF:FEB8:1.

Question

Refer to the exhibit. The IPv6 address for the LAN segment on router R2 must be configured using the EUI-64 format. When configured which ipv6 address is produced by the router? - Exhibit: Building the EUI-64 Interface Identifier for Router R2
  1. 2001:db8:9bb6:6bb9:C081:B6FF:FF4B:1
  2. 2001:db8:9bb6:6bb9:C001:6BFE:FF01:1
  3. 2001:db8:9bb6:6bb9:4679:824F:FE88:1
  4. 2001:db8:9bb6:6bb9:C801:B6FF:FEB8:1 Correct Answer

Key Takeaways

As with the other EUI-64 questions, the answer shows the inverted seventh bit (C8, not C0/C2) and the FFFE insertion; D is the only option consistent with that transformation.

EUI-64 builds the interface ID from the MAC: split at the OUI/host boundary, insert FFFE, and flip the seventh bit (U/L) of the first byte. For R2 the inverted first byte becomes C8, placing C801 in the third hextet, matching option D.

Common mistake: Picking an option whose first hextet lacks the C8 (inverted-bit) result. The 7th-bit flip is the step most often missed.

Expert Analysis

Why the Answer Is Correct

D is correct. EUI-64 on R2's MAC: the first byte CA (1100 1010) flips its seventh bit to 1100 1000 = C8, so the interface ID begins with C801, and the FFFE insertion produces the remainder shown in D (2001:db8:9bb6:6bb9:C801:B6FF:FEB8:1).

Why the Other Options Are Wrong

A (C081) and B (C001) miss the seventh-bit inversion, so their first hextet is wrong. C does not follow the EUI-64 transformation of the exhibited MAC at all.

Community Comment Notes

The vote is D (100). Commenters derive CA -> C8 by flipping the seventh bit, confirming C801 as the correct start and noting D is the only option that follows the EUI-64 rules.

Official Reference

https://learningnetwork.cisco.com/s/ccna-exam-topics

This link opens the official vendor documentation in a new tab.

Community Votes

D
100%

100% of anonymous learners picked answer D. Votes are pick records left by other test-takers — they are not the verified answer.

Discussion 9 contributions

Compiled from public community threads and reader submissions, lightly edited for clarity. Handles and vote tallies are shown as recorded, and comments reflect the views of individual contributors.

From public community threads 9

  1. JulesAZ 👍 3 Answered D
    Can be the only answer using the rules for these questions. The fact that the question / solution, is wrong just does my head in!! DDDDD
  2. picho707 👍 2 Answered D
    Correct answer is D. CA = 1100 1010 so flipping the 7 bit => 1100 1000 = C8
  3. ladarius01 👍 1
    1. Split MAC in half and insert FF FE. 2. Invert 7th bit: a = 1010 - > 1000 = 8
  4. KuyaErik101 👍 1
    c801:bfff:fea4:1 - the correct answer nearest is D
  5. bymrdas 👍 2 Answered D
    D is Correct
Show 4 more comments
  1. exiledwl 👍 2 Answered D
    D for sure
  2. 1625b6d 👍 1 Answered D
    Answer is D
  3. d5c03de 👍 1
    It's D fam
  4. askar430 👍 1 Answered D
    its D , C8

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