Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?

Question

Options

• APermit 10.0.0.0 0.0.0.1
• BPermit 10.0.0.1 0.0.0.0
• CPermit 10.0.0.1 0.0.0.254
• DPermit 10.0.0.0 255.255.255.254

• A
  5% (1)
• B
  15% (3)
• C
  75% (15)
• D
  5% (1)

Explanation

Explanation

Option C (permit 10.0.0.1 0.0.0.254) is correct because in a standard ACE, the wildcard mask 0.0.0.254 in binary is 11111110, meaning the last bit is fixed at 0 while all other host bits are free to vary - but since the source address ends in .1 (binary 00000001), the fixed bit forces matching only addresses where the last bit is 1, which is the definition of an odd number.

Why the distractors are wrong:

• A (10.0.0.0 0.0.0.1) matches even hosts (.0, .2, .4…) because the source address .0 has a last bit of 0, and the wildcard 0.0.0.1 allows that last bit to vary - anchoring matches on the even pattern.
• B (10.0.0.1 0.0.0.0) is an exact host match for only one address (10.0.0.1), not all odd hosts.
• D uses a subnet mask format (255.255.255.254) instead of a wildcard mask, which is invalid syntax for a standard ACE wildcard.

Memory Tip: Think of the wildcard mask as a "don't care" mask - wherever there's a 0, the bit must match the source address. Since odd numbers always have a 1 in the last bit, pair source .1 with wildcard .254 (which locks that last bit) to catch all odd hosts.

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