Q151. What are the three phases involved in security testing?
A. Reconnaissance, Conduct, Report
B. Reconnaissance, Scanning, Conclusion
C. Preparation, Conduct, Conclusion
D. Preparation, Conduct, Billing
Answer: C
Explanation: Preparation phase - A formal contract is executed containing non-disclosure of the client's data and legal protection for the tester. At a minimum, it also lists the IP addresses to be tested and time to test. Conduct phase - In this phase the penetration test is executed, with the tester looking for potential vulnerabilities. Conclusion phase - The results of the evaluation are communicated to the pre-defined organizational contact, and corrective action is advised.
Q152. While probing an organization you discover that they have a wireless network. From your attempts to connect to the WLAN you determine that they have deployed MAC filtering by using ACL on the access points. What would be the easiest way to circumvent and communicate on the WLAN?
A. Attempt to crack the WEP key using Airsnort.
B. Attempt to brute force the access point and update or delete the MAC ACL.
C. Steel a client computer and use it to access the wireless network.
D. Sniff traffic if the WLAN and spoof your MAC address to one that you captured.
Answer: D
Explanation: The easiest way to gain access to the WLAN would be to spoof your MAC address to one that already exists on the network.
Q153. nn would like to perform a reliable scan against a remote target. She is not concerned about being stealth at this point.
Which of the following type of scans would be the most accurate and reliable option?
A. A half-scan
B. A UDP scan
C. A TCP Connect scan
D. A FIN scan
Answer: C
Explanation: A TCP Connect scan, named after the Unix connect() system call is the most accurate scanning method. If a port is open the operating system completes the TCP three-way handshake, and the port scanner immediately closes the connection. Otherwise an error code is returned. Example of a three-way handshake followed by a reset: Source Destination Summary
[192.168.0.8] [192.168.0.10] TCP: D=80 S=49389 SYN SEQ=3362197786 LEN=0 WIN=5840
[192.168.0.10] [192.168.0.8] TCP: D=49389 S=80 SYN ACK=3362197787 SEQ=58695210 LEN=0 WIN=65535
[192.168.0.8]
[192.168.0.10] TCP: D=80 S=49389 ACK=58695211 WIN<<2=5840
[192.168.0.8]
[192.168.0.10] TCP: D=80 S=49389 RST ACK=58695211 WIN<<2=5840
Q154. One of the most common and the best way of cracking RSA encryption is to being to derive the two prime numbers, which are used in the RSA PKI mathematical process. If the two numbers p and q are discovered through a _________________ process, then the private key can be derived.
A. Factorization
B. Prime Detection
C. Hashing
D. Brute-forcing
Answer: A
Explanation: In April 1994, an international cooperative group of mathematicians and computer scientists solved a 17-year-old challenge problem, the factoring of a 129-digit number, called RSA-129, into two primes. That is, RSA-129 = 1143816257578888676692357799761466120102182 9672124236256256184293570693524573389783059 7123563958705058989075147599290026879543541 = 34905295108476509491478496199038 98133417764638493387843990820577 times 32769132993266709549961988190834 461413177642967992942539798288533. Se more at http://en.wikipedia.org/wiki/RSA_Factoring_Challenge
Q155. A zone file consists of which of the following Resource Records (RRs)?
A. DNS, NS, AXFR, and MX records
B. DNS, NS, PTR, and MX records
C. SOA, NS, AXFR, and MX records
D. SOA, NS, A, and MX records
Answer: D
Explanation: The zone file typically contains the following records:
SOA – Start Of Authority NS – Name Server record MX – Mail eXchange record A – Address record
Q156. As a securing consultant, what are some of the things you would recommend to a company to ensure DNS security?
Select the best answers.
A. Use the same machines for DNS and other applications
B. Harden DNS servers
C. Use split-horizon operation for DNS servers
D. Restrict Zone transfers
E. Have subnet diversity between DNS servers
Answer: BCDE
Explanations:
A is not a correct answer as it is never recommended to use a DNS server for any other application. Hardening of the DNS servers makes them less vulnerable to attack. It is recommended to split internal and external DNS servers (called split-horizon operation). Zone transfers should only be accepted from authorized DNS servers. By having DNS servers on different subnets, you may prevent both from going down, even if one of your networks goes down.
Q157. Bill successfully executed a buffer overflow against a Windows IIS web server. He has been able to spawn in interactive shell and plans to deface the main web page. He fist attempts to use the “Echo” command to simply overwrite index.html and remains unsuccessful. He then attempts to delete the page and achieves no progress. Finally, he tires to overwrite it with another page in which also he remains unsuccessful. What is the probable cause of Bill’s problem?
A. The system is a honeypot
B. The HTML file has permissions of read only
C. You can’t use a buffer overflow to deface a web page
D. There is a problem with the shell and he needs to run the attack again
Answer: B
Explanation: A honeypot has no interest in stopping an intruder from altering the “target” files. A buffer overflow is a way to gain access to the target computer. Once he has spawned a shell it is unlikely that it will not work as intended, but the user context that the shell is spawned in might stop him from altering the index.html file incase he doesn’t have sufficient rights.
Q158. You are writing an antivirus bypassing Trojan using C++ code wrapped into chess.c to create an executable file chess.exe. This Trojan when executed on the victim machine, scans the entire system (c:\) for data with the following text “Credit Card” and “password”. It then zips all the scanned files and sends an email to a predefined hotmail address.
You want to make this Trojan persistent so that it survives computer reboots. Which registry entry will you add a key to make it persistent?
A. HKEY_LOCAL_MACHINE\SOFTWARE\MICROOSFT\Windows\CurrentVersion\RunServices
B. HKEY_LOCAL_USER\SOFTWARE\MICROOSFT\Windows\CurrentVersion\RunServices
C. HKEY_LOCAL_SYSTEM\SOFTWARE\MICROOSFT\Windows\CurrentVersion\RunServices
D. HKEY_CURRENT_USER\SOFTWARE\MICROOSFT\Windows\CurrentVersion\RunServices
Answer: A
Explanation: HKEY_LOCAL_MACHINE would be the natural place for a registry entry that starts services when the MACHINE is rebooted.
Topic 7, Sniffers
248. Exhibit:
ettercap –NCLzs --quiet
What does the command in the exhibit do in “Ettercap”?
A. This command will provide you the entire list of hosts in the LAN
B. This command will check if someone is poisoning you and will report its IP.
C. This command will detach from console and log all the collected passwords from the network to a file.
D. This command broadcasts ping to scan the LAN instead of ARP request of all the subnet IPs.
Answer: C
Explanation: -N = NON interactive mode (without ncurses)
-C = collect all users and passwords
-L = if used with -C (collector) it creates a file with all the password sniffed in the session in the
form "YYYYMMDD-collected-pass.log"
-z = start in silent mode (no arp storm on start up)
-s = IP BASED sniffing
--quiet = "demonize" ettercap. Useful if you want to log all data in background.
Q159. While scanning a network you observe that all of the web servers in the DMZ are responding to ACK packets on port 80.
What can you infer from this observation?
A. They are using Windows based web servers.
B. They are using UNIX based web servers.
C. They are not using an intrusion detection system.
D. They are not using a stateful inspection firewall.
Answer: D
Explanation: If they used a stateful inspection firewall this firewall would know if there has been a SYN-ACK before the ACK.
Q160. You want to use netcat to generate huge amount of useless network data continuously for various performance testing between 2 hosts.
Which of the following commands accomplish this?
A. Machine A #yes AAAAAAAAAAAAAAAAAAAAAA | nc –v –v –l –p 2222 > /dev/null Machine B #yes BBBBBBBBBBBBBBBBBBBBBB | nc machinea 2222 > /dev/null
B. Machine A cat somefile | nc –v –v –l –p 2222 Machine B cat somefile | nc othermachine 2222 C. Machine A nc –l –p 1234 | uncompress –c | tar xvfp Machine B tar cfp - /some/dir | compress –c | nc –w 3 machinea 1234
D. Machine A while true : do nc –v –l –s –p 6000 machineb 2 Machine B while true ; do nc –v –l –s –p 6000 machinea 2 done
Answer: A
Explanation: Machine A is setting up a listener on port 2222 using the nc command and then having the letter A sent an infinite amount of times, when yes is used to send data yes NEVER stops until it recieves a break signal from the terminal (Control+C), on the client end (machine B), nc is being used as a client to connect to machine A, sending the letter B and infinite amount of times, while both clients have established a TCP connection each client is infinitely sending data to each other, this process will run FOREVER until it has been stopped by an administrator or the attacker.