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Code d'Examen: 312-49
Nom d'Examen: EC-COUNCIL (Computer Hacking Forensic Investigator )
Questions et réponses: 150 Q&As
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NO.1 The newer Macintosh Operating System is based on:
A. OS/2
B. BSD Unix
C. Linux
D. Microsoft Windows
Answer: B
EC-COUNCIL examen 312-49 312-49 312-49 examen
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
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NO.3 You are working for a large clothing manufacturer as a computer forensics investigator and are
called in to investigate an unusual case of an employee possibly stealing clothing designs from
the company and selling them under a different brand name for a different company. What you
discover during the course of the investigation is that the clothing designs are actually original
products of the employee and the company has no policy against an employee selling his own
designs on his own time. The only thing that you can find that the employee is doing wrong is that
his clothing design incorporates the same graphic symbol as that of the company with only the
wording in the graphic being different. What area of the law is the employee violating?
A. trademark law
B. copyright law
C. printright law
D. brandmark law
Answer: A
EC-COUNCIL certification 312-49 312-49 examen 312-49
NO.4 The offset in a hexadecimal code is:
A. The last byte after the colon
B. The 0x at the beginning of the code
C. The 0x at the end of the code
D. The first byte after the colon
Answer: B
EC-COUNCIL certification 312-49 312-49 examen 312-49
NO.5 Before you are called to testify as an expert, what must an attorney do first?
A. engage in damage control
B. prove that the tools you used to conduct your examination are perfect
C. read your curriculum vitae to the jury
D. qualify you as an expert witness
Answer: D
EC-COUNCIL 312-49 312-49 examen 312-49 312-49
NO.6 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
EC-COUNCIL 312-49 312-49 312-49
NO.7 A suspect is accused of violating the acceptable use of computing resources, as he has visited
adult websites and downloaded images. The investigator wants to demonstrate that the suspect
did indeed visit these sites. However, the suspect has cleared the search history and emptied the
cookie cache. Moreover, he has removed any images he might have downloaded. What can the
investigator do to prove the violation? Choose the most feasible option.
A. Image the disk and try to recover deleted files
B. Seek the help of co-workers who are eye-witnesses
C. Check the Windows registry for connection data (You may or may not recover)
D. Approach the websites for evidence
Answer: A
EC-COUNCIL certification 312-49 312-49 312-49 examen
NO.8 A(n) _____________________ is one that's performed by a computer program rather than the
attacker manually performing the steps in the attack sequence.
A. blackout attack
B. automated attack
C. distributed attack
D. central processing attack
Answer: B
EC-COUNCIL examen 312-49 certification 312-49
NO.9 What file structure database would you expect to find on floppy disks?
A. NTFS
B. FAT32
C. FAT16
D. FAT12
Answer: D
EC-COUNCIL 312-49 312-49 312-49 examen
NO.10 It takes _____________ mismanaged case/s to ruin your professional reputation as a computer
forensics examiner?
A. by law, three
B. quite a few
C. only one
D. at least two
Answer: C
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NO.11 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 :.
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