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Exam PT1-002 topic 1 question 30 discussion

Actual exam question from CompTIA's PT1-002
Question #: 30
Topic #: 1
[All PT1-002 Questions]

A penetration tester conducted a discovery scan that generated the following:

Which of the following commands generated the results above and will transform them into a list of active hosts for further analysis?

  • A. nmap ג€"oG list.txt 192.168.0.1-254 , sort
  • B. nmap ג€"sn 192.168.0.1-254 , grep ג€Nmap scanג€ | awk '{print S5}'
  • C. nmap ג€"-open 192.168.0.1-254, uniq
  • D. nmap ג€"o 192.168.0.1-254, cut ג€"f 2
Show Suggested Answer Hide Answer
Suggested Answer: B 🗳️

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Davar39
Highly Voted 3 years, 3 months ago
Selected Answer: B
Only B makes sense, since those results are from host discovery (-sn) and no ports are reported. AWK command is used in combination with grep to manipulate the output.
upvoted 10 times
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BinarySoldier
Highly Voted 3 years, 5 months ago
I think B is the correct answer here. For the options, only B has the NMAP flag (-sn) which is for host discovery and returns that kind of NMAP output. And the AWK command selects column 5 ({print $5}) which obviously carries the returned IP of the host in the NMAP output.
upvoted 7 times
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Anarckii
Most Recent 1 year, 10 months ago
Selected Answer: B
the results show that that it is discovering which host are up from the 192.168.0.1. You would do this through a ping sweep (-sn)
upvoted 1 times
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maps7
2 years, 10 months ago
B IS THE CORRECT ANSWER
upvoted 2 times
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BinarySoldier
3 years, 2 months ago
Selected Answer: B
B it is.
upvoted 5 times
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[Removed]
3 years, 3 months ago
Answer is B here.
upvoted 3 times
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