4A0-102 · Question #28
Which of the following prefix-lists is the most specific match for prefixes 192.168.64.1/20 and 192.168.32.5/19?
The correct answer is A. Prefix 192.168.0.0/17 longer. Option A (192.168.0.0/17) is the most specific supernet that contains both target prefixes. A /17 mask fixes the first 17 bits, covering 192.168.0.0–192.168.127.255 (third octet 0–127). Since 192.168.64.x (64 < 128) and 192.168.32.x (32 < 128) both fall inside this range, the…
Question
Options
- APrefix 192.168.0.0/17 longer.
- BPrefix 192.168.0.0/19 longer.
- CPrefix 192.168.0.0/10 longer.
- DPrefix 192.168.0.0/18 longer.
How the community answered
(18 responses)- A78% (14)
- B6% (1)
- C6% (1)
- D11% (2)
Explanation
Option A (192.168.0.0/17) is the most specific supernet that contains both target prefixes. A /17 mask fixes the first 17 bits, covering 192.168.0.0–192.168.127.255 (third octet 0–127). Since 192.168.64.x (64 < 128) and 192.168.32.x (32 < 128) both fall inside this range, the longer keyword then matches any more-specific routes (like /19 and /20) within it - making this the tightest possible supernet umbrella for both.
Why the distractors fail:
- B (/19): Covers only 192.168.0.0–192.168.31.255; neither 192.168.32.x nor 192.168.64.x falls in that range (32 and 64 are both ≥ 32, outside the /19 block starting at 0).
- C (/10): Does technically contain both prefixes, but /10 is far less specific than /17 - it spans a massive 192.128.0.0–192.191.255.255 range, so it's the least specific option, not the most.
- D (/18): Covers 192.168.0.0–192.168.63.255; 192.168.32.x fits, but 192.168.64.x does not (64 > 63).
Memory tip: Convert the third octet to binary and count shared leading bits across all target prefixes - that bit count is your most-specific common prefix length. Here, 64 = 01000000 and 32 = 00100000 share only the first bit (0), placing them both under a /17 boundary.
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