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Certified Ethical Hacker 312-49 .
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312-49 PDF DEMO:
QUESTION NO: 1
If you come across a sheepdip machine at your client site, what would you infer?
A. A sheepdip coordinates several honeypots
B. A sheepdip computer is another name for a honeypot
C. A sheepdip computer is used only for virus-checking.
D. A sheepdip computer defers a denial of service attack
Answer: C
QUESTION NO: 2
In a computer forensics investigation, what describes the route that evidence takes from the time you find it until the case is closed or goes to court?
A. rules of evidence
B. law of probability
C. chain of custody
D. policy of separation
Answer: C
QUESTION NO: 3
How many characters long is the fixed-length MD5 algorithm checksum of a critical system file?
A. 128
B. 64
C. 32
D. 16
Answer: C
QUESTION NO: 4
What does the superblock in Linux define?
A. file system names
B. available space
C. location of the first inode
D. disk geometry
Answer: B, C, D
QUESTION NO: 5
A honey pot deployed with the IP 172.16.1.108 was compromised by an attacker . Given below is an excerpt from a Snort binary capture of the attack. Decipher the activity carried out by the attacker by studying the log. Please note that you are required to infer only what is explicit in the excerpt. (Note: The student is being tested on concepts learnt during passive OS fingerprinting, basic TCP/IP connection concepts and the ability to read packet signatures from a sniff dump.)
03/15-20:21:24.107053 211.185.125.124:3500 -> 172.16.1.108:111
TCP TTL:43 TOS:0x0 ID:29726 IpLen:20 DgmLen:52 DF
***A**** Seq: 0x9B6338C5 Ack: 0x5820ADD0 Win: 0x7D78 TcpLen: 32
TCP Options (3) => NOP NOP TS: 23678634 2878772
=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=
03/15-20:21:24.452051 211.185.125.124:789 -> 172.16.1.103:111
UDP TTL:43 TOS:0x0 ID:29733 IpLen:20 DgmLen:84
Len: 64
01 0A 8A 0A 00 00 00 00 00 00 00 02 00 01 86 A0 ................
00 00 00 02 00 00 00 03 00 00 00 00 00 00 00 00 ................
00 00 00 00 00 00 00 00 00 01 86 B8 00 00 00 01 ................
00 00 00 11 00 00 00 00 ........
=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=
03/15-20:21:24.730436 211.185.125.124:790 -> 172.16.1.103:32773
UDP TTL:43 TOS:0x0 ID:29781 IpLen:20 DgmLen:1104
Len: 1084
47 F7 9F 63 00 00 00 00 00 00 00 02 00 01 86 B8 G..c............
00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 20 ...............
3A B1 5E E5 00 00 00 09 6C 6F 63 61 6C 68 6F 73 :......localhost
=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=
+
03/15-20:21:36.539731 211.185.125.124:4450 -> 172.16.1.108:39168
TCP TTL:43 TOS:0x0 ID:31660 IpLen:20 DgmLen:71 DF
***AP*** Seq: 0x9C6D2BFF Ack: 0x59606333 Win: 0x7D78 TcpLen: 32
TCP Options (3) => NOP NOP TS: 23679878 2880015
63 64 20 2F 3B 20 75 6E 61 6D 65 20 2D 61 3B 20 cd /; uname -a;
69 64 3B id;
A. The attacker has conducted a network sweep on port 111
B. The attacker has scanned and exploited the system using Buffer Overflow
C. The attacker has used a Trojan on port 32773
D. The attacker has installed a backdoor
Answer: A
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Updated: May 27, 2022