最新EC-COUNCIL 312-49試験の練習問題と解答

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試験番号：312-49問題集
試験科目：EC-COUNCIL 「Computer Hacking Forensic Investigator 」
最近更新時間：2014-03-18
問題と解答：150

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NO.1 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

EC-COUNCIL問題集   312-49問題集   312-49認定証   312-49

NO.2 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

EC-COUNCIL過去問   312-49   312-49認定資格   312-49   312-49   312-49過去問

NO.3 With the standard Linux second extended file system (Ext2fs), a file is deleted when the inode
internal link count reaches ________.
A. 0
B. 10
C. 100
D. 1
Answer: A

EC-COUNCIL   312-49   312-49練習問題   312-49   312-49

NO.4 In the context of file deletion process, which of the following statement holds true?
A. When files are deleted, the data is overwritten and the cluster marked as available
B. The longer a disk is in use, the less likely it is that deleted files will be overwritten
C. While booting, the machine may create temporary files that can delete evidence
D. Secure delete programs work by completely overwriting the file in one go
Answer: C

EC-COUNCIL過去問   312-49   312-49認証試験   312-49   312-49

NO.5 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過去問   312-49練習問題   312-49認定証

NO.6 You are contracted to work as a computer forensics investigator for a regional bank that has four
30 TB storage area networks that store customer data. What method would be most efficient for
you to acquire digital evidence from this network?
A. create a compressed copy of the file with DoubleSpace
B. create a sparse data copy of a folder or file
C. make a bit-stream disk-to-image fileC
D. make a bit-stream disk-to-disk file
Answer: C

EC-COUNCIL   312-49参考書   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認定証   312-49   312-49   312-49

NO.8 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練習問題   312-49過去問   312-49   312-49

NO.9 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

EC-COUNCIL認証試験   312-49認証試験   312-49参考書   312-49認定証   312-49練習問題

NO.10 When examining a file with a Hex Editor, what space does the file header occupy?
A. the last several bytes of the file
B. the first several bytes of the file
C. none, file headers are contained in the FAT
D. one byte at the beginning of the file
Answer: D

EC-COUNCIL   312-49   312-49

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 :.