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ECCouncil C1000-125 - IBM Cloud Technical Advocate v3 Braindumps

IBM C1000-125 - IBM Cloud Technical Advocate v3 Braindumps

  • Certification Provider:IBM
  • Exam Code:C1000-125
  • Exam Name:IBM Cloud Technical Advocate v3 Exam
  • Total Questions:765 Questions and Answers
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NEW QUESTION: 1
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.1.0.0.0.254
B. Permit 10.0.0.0.255.255.255.254
C. Permit 10.0.0.0.0.0.0.1
D. Permit 10.0.0.1.0.0.0.0
Answer: A
Explanation:
Remember, for the wildcard mask, 1's are I DON'T CARE, and 0's are I CARE. So now let's
analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of
the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care
about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16;
maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we
only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd
numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even
numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with
a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered
hosts in the 10.0.0.0/24 subnet.

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