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Exam 312-50v11 All Questions

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Exam 312-50v11 topic 1 question 299 discussion

Actual exam question from ECCouncil's 312-50v11
Question #: 299
Topic #: 1
[All 312-50v11 Questions]

The network in ABC company is using the network address 192.168.1.64 with mask 255.255.255.192. In the network the servers are in the addresses
192.168.1.122, 192.168.1.123 and 192.168.1.124. An attacker is trying to find those servers but he cannot see them in his scanning. The command he is using is: nmap 192.168.1.64/28.
Why he cannot see the servers?

  • A. He needs to add the command ג€ג€ip addressג€ג€ just before the IP address
  • B. He needs to change the address to 192.168.1.0 with the same mask
  • C. He is scanning from 192.168.1.64 to 192.168.1.78 because of the mask /28 and the servers are not in that range
  • D. The network must be dawn and the nmap command and IP address are ok
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Suggested Answer: C 🗳️

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Daniel8660
6 months, 1 week ago
Selected Answer: C
mask 255.255.255.192 = /26, IP Range : 192.168.1.65 - 192.168.1.126 mask 255.255.255.240 = /28, IP Range : 192.168.1.65 - 192.168.1.78
upvoted 3 times
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EngnSu
10 months, 3 weeks ago
/28 can’t scanning /26
upvoted 3 times
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mileke2
11 months, 2 weeks ago
Can someone please give an explanation to this ?
upvoted 1 times
freho
10 months, 3 weeks ago
It´s a simple Subnetting-Question. 192.168.1.64/28 is what the hacker scanned, as per Netmask (255.255.255.192 == /26 ) the network is four times bigger than what the hacker scanned.
upvoted 1 times
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FrozenOne
1 year ago
255.255.255.192 is /26..not /28
upvoted 2 times
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Scryptic
1 year, 7 months ago
Please don't say 'correct', no one knows you
upvoted 3 times
shpunk
1 year, 5 months ago
Scryptic - no one knows you either.
upvoted 1 times
gtlusciak
1 year, 5 months ago
That's beside the point.
upvoted 2 times
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ANDRESCB1988
1 year, 9 months ago
correct
upvoted 3 times
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C (25%)
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