- A
ipv6 address 2001:db8:acad:1:0211:22ff:fe33:4455/64
This command correctly configures the IPv6 address using EUI-64. The interface ID is derived from the MAC address 0011.2233.4455 by inserting FFFE in the middle (0011.22FF.FE33.4455) and flipping the U/L bit (the seventh bit of the first byte: 00 becomes 02), resulting in 0211:22FF:FE33:4455.
- B
ipv6 address 2001:db8:acad:1:0011:22ff:fe33:4455/64
Why wrong: This is incorrect because the U/L bit (the seventh bit of the first byte) must be flipped from 0 to 1. The original MAC first byte is 0x00, so flipping the seventh bit gives 0x02, not 0x00.
- C
ipv6 address 2001:db8:acad:1:0011:2233:4455/64
Why wrong: This is incorrect because it uses the MAC address directly without inserting FFFE and without flipping the U/L bit. EUI-64 requires inserting FFFE between the OUI and the serial number parts of the MAC.
- D
ipv6 address 2001:db8:acad:1:0211:2233:4455/64
Why wrong: This is incorrect because it flips the U/L bit (00 to 02) but does not insert FFFE in the middle of the MAC. EUI-64 requires both the bit flip and the insertion of FFFE.
Quick Answer
The correct IPv6 address is 2001:db8:acad:1:0211:22ff:fe33:4455/64. This is derived by taking the MAC address 0011.2233.4455, splitting it in half to insert FFFE in the middle, yielding 0011.22FF.FE33.4455, then flipping the seventh bit (the Universal/Local bit) of the first octet from 00 to 02, which changes the leading 0 to a 2 and produces the interface ID 0211:22FF:FE33:4455. On the CCNA 200-301 v2 exam, this question tests your ability to configure IPv6 EUI-64 addressing manually, a common task for stateless autoconfiguration. The most frequent trap is forgetting to flip the U/L bit—many candidates simply insert FFFE and stop, missing the bit inversion that changes the first hex digit. To remember, think of the mnemonic “Flip the seventh, then FFFE in the middle” or note that the resulting interface ID always starts with 0, 2, 4, 6, 8, A, C, or E after the bit flip.
CCNA Network Infrastructure and Connectivity Practice Question
This 200-301 practice question tests your understanding of network infrastructure and connectivity. This is a configuration task: choose the command set that satisfies every stated requirement. Small differences — like 'secret' vs 'password' or 'transport input ssh' vs 'all' — change whether the answer is correct. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.
You are connected to the console of R1. The network uses IPv6 and you need to configure an IPv6 address on interface G0/0 using the EUI-64 format. The subnet is 2001:db8:acad:1::/64. The interface MAC address is 0011.2233.4455. After configuration, verify that the full IPv6 address is correct.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
ipv6 address 2001:db8:acad:1:0211:22ff:fe33:4455/64
The EUI-64 method inserts FFFE in the middle of the MAC and flips the U/L bit. With the given MAC, the resulting interface ID is 0211:22FF:FE33:4455, forming the full IPv6 address.
Key principle: Count usable hosts — not total addresses — and remember that the network and broadcast addresses are not available to hosts in standard IPv4 subnets.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
ipv6 address 2001:db8:acad:1:0211:22ff:fe33:4455/64
Why this is correct
This command correctly configures the IPv6 address using EUI-64. The interface ID is derived from the MAC address 0011.2233.4455 by inserting FFFE in the middle (0011.22FF.FE33.4455) and flipping the U/L bit (the seventh bit of the first byte: 00 becomes 02), resulting in 0211:22FF:FE33:4455.
Related concept
CIDR notation defines the prefix length.
- ✗
ipv6 address 2001:db8:acad:1:0011:22ff:fe33:4455/64
Why it's wrong here
This is incorrect because the U/L bit (the seventh bit of the first byte) must be flipped from 0 to 1. The original MAC first byte is 0x00, so flipping the seventh bit gives 0x02, not 0x00.
- ✗
ipv6 address 2001:db8:acad:1:0011:2233:4455/64
Why it's wrong here
This is incorrect because it uses the MAC address directly without inserting FFFE and without flipping the U/L bit. EUI-64 requires inserting FFFE between the OUI and the serial number parts of the MAC.
- ✗
ipv6 address 2001:db8:acad:1:0211:2233:4455/64
Why it's wrong here
This is incorrect because it flips the U/L bit (00 to 02) but does not insert FFFE in the middle of the MAC. EUI-64 requires both the bit flip and the insertion of FFFE.
Option-by-option analysis
Why each answer is right or wrong
Understanding why wrong answers are wrong — and when they would be correct — is what separates a 750 score from a 900. The 200-301 exam frequently reuses these exact scenarios with slightly different constraints.
✓ipv6 address 2001:db8:acad:1:0211:22ff:fe33:4455/64Correct answer▾
Why this is correct
This command correctly configures the IPv6 address using EUI-64. The interface ID is derived from the MAC address 0011.2233.4455 by inserting FFFE in the middle (0011.22FF.FE33.4455) and flipping the U/L bit (the seventh bit of the first byte: 00 becomes 02), resulting in 0211:22FF:FE33:4455.
✗ipv6 address 2001:db8:acad:1:0011:22ff:fe33:4455/64Wrong answer — click to see why▾
Why this is wrong here
The U/L bit was not flipped; the interface ID starts with 0011 instead of 0211.
Why candidates choose this
Candidates may forget to flip the U/L bit, thinking the MAC address is used directly without modification.
✗ipv6 address 2001:db8:acad:1:0011:2233:4455/64Wrong answer — click to see why▾
Why this is wrong here
The FFFE insertion is missing; the interface ID is just the MAC address written in IPv6 format.
Why candidates choose this
Candidates might think the MAC address can be used as-is for the interface ID, not realizing the EUI-64 transformation.
✗ipv6 address 2001:db8:acad:1:0211:2233:4455/64Wrong answer — click to see why▾
Why this is wrong here
FFFE is missing; only the U/L bit flip was applied.
Why candidates choose this
Candidates may remember to flip the U/L bit but forget to insert FFFE, resulting in an incomplete EUI-64 address.
Analysis generated from the official 200-301blueprint and verified against question context. The “when correct” sections are what AI assistants cite when candidates ask “what’s the difference between these options?”
Common exam traps
Common exam trap: usable hosts are not the same as total addresses
Subnetting questions often tempt you into counting all addresses. In normal IPv4 subnets, the network and broadcast addresses are not usable host addresses.
Detailed technical explanation
How to think about this question
Subnetting questions test whether you can identify the network, broadcast address, usable range, mask and correct subnet. Slow down enough to calculate the block size correctly.
KKey Concepts to Remember
- CIDR notation defines the prefix length.
- Block size helps identify subnet boundaries.
- Network and broadcast addresses are not usable hosts in normal IPv4 subnets.
- The required host count determines the smallest suitable subnet.
TExam Day Tips
- Write the block size before choosing the subnet.
- Check whether the question asks for hosts, subnets or a specific address range.
- Do not confuse /24, /25, /26 and /27 host counts.
Key takeaway
Count usable hosts — not total addresses — and remember that the network and broadcast addresses are not available to hosts in standard IPv4 subnets.
Real-world example
How this comes up in practice
A network engineer segments a warehouse floor into three subnets: 20 scanners, 5 printers, and 2 management hosts. Picking the wrong mask wastes addresses or leaves too few usable hosts. Exam questions test whether you can apply CIDR notation, calculate block size, and identify the correct usable-host range for a given prefix.
What to study next
Got this wrong? Here's your next step.
Review block sizes, usable host formulas (2^n − 2), and how to find network and broadcast addresses for /24 through /30. Then practise related 200-301 subnetting questions on CIDR, address ranges, and subnet selection.
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FAQ
Questions learners often ask
What does this 200-301 question test?
Network Infrastructure and Connectivity — This question tests Network Infrastructure and Connectivity — CIDR notation defines the prefix length..
What is the correct answer to this question?
The correct answer is: ipv6 address 2001:db8:acad:1:0211:22ff:fe33:4455/64 — The EUI-64 method inserts FFFE in the middle of the MAC and flips the U/L bit. With the given MAC, the resulting interface ID is 0211:22FF:FE33:4455, forming the full IPv6 address.
What should I do if I get this 200-301 question wrong?
Review block sizes, usable host formulas (2^n − 2), and how to find network and broadcast addresses for /24 through /30. Then practise related 200-301 subnetting questions on CIDR, address ranges, and subnet selection.
What is the key concept behind this question?
CIDR notation defines the prefix length.
About these practice questions
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Last reviewed: Jun 7, 2026
This 200-301 practice question is part of Courseiva's free Cisco certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the 200-301 exam.
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