Option "B".
https://en.wikipedia.org/wiki/Wildcard_mask
Any wildcard bit-pattern can be masked for examination. For example, a wildcard mask of 0.0.0.254 (binary equivalent = 00000000.00000000.00000000.11111110) applied to IP address 10.10.10.2 (00001010.00001010.00001010.00000010) will match even-numbered IP addresses 10.10.10.0, 10.10.10.2, 10.10.10.4, 10.10.10.6 etc. Same mask applied to 10.10.10.1 (00001010.00001010.00001010.00000001) will match odd-numbered IP addresses 10.10.10.1, 10.10.10.3, 10.10.10.5 etc
This entry permits traffic from odd-numbered hosts in the 10.0.0.0/24 subnet.
The odd-numbered hosts (1, 3, 5, ..., 253) will be permitted, while the even-numbered hosts (0, 2, 4, ..., 254) will be denied.
Answer is B.
A is not correct, would allow 10.0.0.0 or 10.0.0.1
C is not correct, would allow only 10.0.0.1 (wildcard 0.0.0.0 same as 'host')
D is not correct, would allow any even IP: x.x.x.0
Thanks for you link, first time I see a wildmask to split between even-IP-addresses and odd-IP-addresses
access-list 101 permit ip any 10.10.10.0 0.0.0.254 -> even
access-list 102 permit ip any 10.10.10.1 0.0.0.254 -> odd
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