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Certified Ethical Hacker CEH (CEH) — Questions 76150

870 questions total · 12pages · All types, answers revealed

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76
MCQmedium

After a security incident, an analyst retrieves a suspicious file. The analyst runs the 'strings' command on it and sees references to 'CreateRemoteThread' and 'WriteProcessMemory'. Which technique does this indicate?

A.DLL hijacking
B.Privilege escalation
C.Process injection
D.Buffer overflow
AnswerC

Process injection is a sophisticated technique where an attacker writes malicious code into the address space of a legitimate, running process and then forces that process to execute it. The `WriteProcessMemory` API is crucial for writing the attacker's shellcode or payload into the target process's memory. Subsequently, `CreateRemoteThread` is commonly used to create a new thread within the remote process, directing its execution flow to the starting address of the newly injected code, thereby achieving execution within the target's context.

Why this answer

The presence of 'CreateRemoteThread' and 'WriteProcessMemory' in the output of the 'strings' command is a strong indicator of process injection. These Windows API functions are commonly used together to allocate memory in a target process (VirtualAllocEx), write malicious code into that memory (WriteProcessMemory), and then execute it in the context of the remote process (CreateRemoteThread). This technique allows an attacker to run arbitrary code within a legitimate process, bypassing security controls.

Exam trap

The EC-CEH exam often tests the distinction between process injection and DLL hijacking. Candidates mistakenly associate any DLL-related API call with DLL hijacking, but the key differentiator is that process injection explicitly uses WriteProcessMemory and CreateRemoteThread to write and execute code in a remote process, whereas DLL hijacking relies on search order manipulation without direct memory writing.

How to eliminate wrong answers

Option A is wrong because DLL hijacking involves tricking a legitimate application into loading a malicious DLL by placing it in a directory where the application searches first, not by using API calls to inject code into a remote process. Option B is wrong because privilege escalation typically exploits vulnerabilities or misconfigurations to gain higher-level access rights (e.g., SeBackupPrivilege abuse, token manipulation), and does not inherently rely on WriteProcessMemory and CreateRemoteThread. Option D is wrong because a buffer overflow exploits memory corruption to overwrite adjacent data or control flow (e.g., overwriting a return address on the stack), not by explicitly calling WriteProcessMemory and CreateRemoteThread to inject code into another process.

77
MCQeasy

Which type of password cracking attack uses a precomputed table of hash chains to reverse hashes quickly?

A.Rainbow table attack
B.Dictionary attack
C.Brute-force attack
D.Hybrid attack
AnswerA

A rainbow table attack leverages precomputed tables that store chains of hash-to-plaintext reductions. This method significantly reduces the time required to crack passwords by performing a time-memory trade-off, avoiding the need to recompute every possible hash during the attack. Instead of brute-forcing, it looks up the target hash within the table to find the corresponding original password or a chain leading to it.

Why this answer

A rainbow table attack is correct because it uses a precomputed table of hash chains to reverse hashes quickly. Rainbow tables reduce the time needed for cracking by storing chains of hash values that allow for efficient lookup, trading off storage space for computational speed. This technique is specifically designed to reverse cryptographic hash functions like LM, NTLM, or MD5 without performing brute-force or dictionary lookups for each attempt.

Exam trap

The trap here is that candidates confuse rainbow tables with dictionary attacks because both involve precomputed data, but rainbow tables specifically use hash chains with reduction functions to enable efficient reversal, not just a list of plaintext-to-hash mappings.

How to eliminate wrong answers

Option B is wrong because a dictionary attack uses a list of likely passwords (words from a dictionary) and hashes each one for comparison, not a precomputed table of hash chains. Option C is wrong because a brute-force attack systematically tries every possible combination of characters until the correct password is found, which is computationally intensive and does not rely on precomputed tables. Option D is wrong because a hybrid attack combines dictionary words with variations (e.g., appending numbers or symbols) but still hashes each candidate on the fly rather than using precomputed hash chains.

78
Multi-Selecthard

A web application is vulnerable to SQL injection. Which THREE of the following techniques can be used to extract data from the database using blind SQL injection?

Select 3 answers
A.Time-based
B.Error-based
C.Boolean-based
D.Out-of-band
E.Union-based
AnswersA, C, D

Time-based blind SQL injection involves injecting queries that cause a measurable time delay on the database server if a specific condition evaluates to true. By observing the server's response time, an attacker can infer the truthfulness of the injected statement, character by character. This method is crucial when no direct output or error messages are returned by the application, making it a viable technique for data exfiltration in blind scenarios.

Why this answer

Boolean-based, time-based, and out-of-band are all types of blind SQL injection. Error-based and union-based are in-band techniques, not blind.

79
Multi-Selectmedium

Which TWO of the following are examples of application layer DDoS attacks? (Select two.)

Select 2 answers
A.Slowloris
B.UDP flood
C.Smurf attack
D.HTTP flood
E.SYN flood
AnswersA, D

Correct. Slowloris keeps many connections open to exhaust server resources.

Why this answer

Slowloris is an application layer DDoS attack that targets HTTP servers by opening multiple connections and sending partial HTTP requests, keeping them open as long as possible. It exploits the server's connection handling by sending incomplete headers, preventing the server from timing out the connection and exhausting its connection pool. This attack operates at Layer 7 and does not require high bandwidth, making it effective against web servers.

Exam trap

The CEH exam often tests the distinction between Layer 4 (transport) and Layer 7 (application) attacks, and the trap here is that candidates may confuse SYN flood (a TCP-based Layer 4 attack) with an application layer attack because it targets web servers, but it operates at a lower layer of the OSI model.

80
MCQmedium

A security analyst reviews the iptables firewall configuration on a Linux server acting as a gateway for a small office. The server has two interfaces: eth0 (external) and eth1 (internal, 192.168.1.0/24). Based on the exhibit, which of the following is a valid security concern?

A.All traffic to the loopback interface is accepted, which could allow local attacks to bypass firewall rules.
B.The OUTPUT chain policy is set to ACCEPT, which allows any outbound traffic.
C.The FORWARD chain only allows traffic from 192.168.1.0/24 to any destination, which is too permissive.
D.UDP port 53 is allowed, which could permit DNS tunneling attacks.
AnswerA

While allowing loopback traffic (lo) is common for inter-process communication, an `ACCEPT` rule for all traffic on the loopback interface can be a significant security vulnerability. Malicious local processes or compromised applications could exploit this permissive rule to communicate with other local services, potentially bypassing more restrictive firewall rules designed for external interfaces. This allows attackers to pivot internally, escalating privileges or exfiltrating data without network-level firewall intervention.

Why this answer

The iptables rules show that all traffic to the loopback interface (lo) is accepted in the INPUT chain. This means any process on the local host can send packets to 127.0.0.1 without being filtered, potentially allowing local privilege escalation or local attacks to bypass firewall restrictions. In a gateway configuration, this can be exploited if an attacker gains local access and uses the loopback to communicate with services that should be protected.

Exam trap

The trap here is that candidates often overlook the loopback interface rules and focus on external-facing chains, assuming that only external interfaces matter for security, while the question specifically tests awareness of local attack vectors through the loopback interface.

How to eliminate wrong answers

Option B is wrong because the OUTPUT chain policy being set to ACCEPT is not inherently a security concern; it is a common default that allows the gateway itself to initiate outbound connections, which is expected for normal operation. Option C is wrong because the FORWARD chain rule only allowing traffic from 192.168.1.0/24 to any destination is a typical and correct configuration for a gateway that forwards internal traffic to the internet; it is not 'too permissive' as it restricts forwarding to the internal subnet only. Option D is wrong because allowing UDP port 53 (DNS) is necessary for name resolution and does not inherently permit DNS tunneling; tunneling requires additional exploitation and is not a direct consequence of allowing standard DNS traffic.

81
Multi-Selecthard

Which THREE of the following are techniques used in session hijacking? (Select three.)

Select 3 answers
A.ARP poisoning
B.DNS amplification
C.TCP sequence prediction
D.Cookie theft
E.MAC flooding
AnswersA, C, D

Correct. ARP poisoning enables MITM, which can be used to hijack sessions.

Why this answer

ARP poisoning is correct because it allows an attacker on the same local network to associate their MAC address with the IP address of a legitimate host (e.g., the default gateway). This redirects traffic through the attacker's machine, enabling them to intercept and hijack active sessions by capturing or modifying packets in transit, a classic man-in-the-middle (MITM) technique used in session hijacking.

Exam trap

The trap here is that candidates confuse network reconnaissance or DoS techniques (like DNS amplification or MAC flooding) with active session hijacking methods, which require direct manipulation of session tokens or TCP state.

82
MCQeasy

A penetration tester wants to perform a stealthy TCP scan that does not complete the three-way handshake. Which Nmap flag should be used?

A.-sU
B.-sS
C.-sV
D.-sT
AnswerB

The -sS option initiates a SYN scan, often referred to as a 'stealth scan,' because it does not complete the full TCP three-way handshake. Nmap sends a SYN packet and, if a SYN/ACK is received, immediately responds with an RST packet, preventing the target system from logging a fully established connection. This technique allows the penetration tester to identify open TCP ports while minimizing the footprint and avoiding detection by many intrusion detection systems and application logs.

Why this answer

(-sS) is correct because it performs a SYN scan, which sends a TCP SYN packet and waits for a SYN-ACK response without completing the three-way handshake (i.e., it sends a RST instead of an ACK). This makes the scan stealthy as it avoids establishing a full TCP connection, reducing the chance of being logged by the target.

Exam trap

The trap here is that candidates often confuse -sS (SYN scan) with -sT (TCP connect scan), mistakenly thinking that -sT is stealthy because it uses TCP, but -sT actually completes the full handshake and is easily logged, while -sS is the true stealthy option.

How to eliminate wrong answers

Option A (-sU) is wrong because it performs a UDP scan, not a TCP scan, and UDP is connectionless, so it does not involve a three-way handshake at all. Option C (-sV) is wrong because it is used for version detection, which requires completing the three-way handshake to probe services, not for stealthy scanning. Option D (-sT) is wrong because it performs a full TCP connect scan, which completes the three-way handshake and is not stealthy, as it is more likely to be logged by the target system.

83
Multi-Selectmedium

A security analyst wants to perform passive reconnaissance on a target domain. Which TWO of the following methods are considered passive? (Choose 2)

Select 2 answers
A.WHOIS lookup
B.Shodan search
C.Telnet banner grab
D.Ping sweep
E.Nmap SYN scan
AnswersA, B

A WHOIS lookup involves querying publicly accessible databases maintained by domain registrars and regional internet registries (RIRs) to retrieve information about a domain name or IP address. This process is entirely passive because the queries are directed at third-party databases, not the target's systems, ensuring no direct network traffic is sent to the target organization. Information gathered can include registrant contact details, registration and expiration dates, and nameservers.

Why this answer

WHOIS lookup is passive because it queries public registration databases (e.g., ARIN, RIPE) via the WHOIS protocol (RFC 3912) to retrieve domain ownership, registrar, and name server information without sending any packets to the target's own infrastructure. This data is publicly available and does not interact with the target's servers or network, making it a classic passive reconnaissance technique.

Exam trap

EC-Council often tests the distinction between passive and active reconnaissance by including tools like Shodan (which is passive) alongside active scanning tools like Nmap, leading candidates to mistakenly classify Shodan as active because it involves a search engine rather than direct network interaction.

84
Multi-Selectmedium

Which TWO of the following are types of SQL injection? (Select 2)

Select 2 answers
A.Stored
B.DOM-based
C.Union-based
D.Blind boolean-based
E.Reflected
AnswersC, D

Union-based SQL injection is an in-band technique where an attacker leverages the UNION SELECT SQL operator to combine the results of the original legitimate query with a malicious query. This allows the attacker to retrieve data from other tables or databases within the same database server, and have it returned directly in the application's HTTP response. The attacker can then extract sensitive information by carefully crafting the injected SELECT statement.

Why this answer

In-band SQL injection includes union-based and error-based. Blind SQL injection includes boolean-based and time-based. Out-of-band is another type.

85
MCQeasy

Which of the following best describes the difference between active and passive reconnaissance?

A.Active reconnaissance is legal, while passive reconnaissance is not
B.Active reconnaissance involves direct interaction with the target, whereas passive reconnaissance does not
C.Passive reconnaissance uses tools like Nmap, while active reconnaissance uses Google dorks
D.Passive reconnaissance is used only during the exploitation phase
AnswerB

This is the core difference between the two approaches.

Why this answer

Active reconnaissance involves direct interaction with the target system, such as sending packets, probes, or connection requests (e.g., using Nmap scans, ping sweeps, or banner grabbing) that can be logged or detected by the target. Passive reconnaissance, in contrast, gathers information without engaging the target directly, relying on publicly available sources (e.g., WHOIS lookups, DNS records, social media, or search engines) and does not generate traffic that reaches the target's network. This distinction is fundamental in the CEH methodology because active techniques carry a higher risk of alerting the target, while passive techniques are stealthier and often used first to avoid detection.

Exam trap

EC-Council often tests the misconception that passive reconnaissance is 'safer' or 'always legal,' but the trap here is confusing the method of interaction (direct vs. indirect) with legality or tool assignment, leading candidates to pick Option A or C instead of the correct definition based on target interaction.

How to eliminate wrong answers

Option A is wrong because legality is not the defining difference; both active and passive reconnaissance can be legal or illegal depending on authorization and jurisdiction—active reconnaissance is not inherently legal, and passive reconnaissance is not inherently illegal. Option C is wrong because it reverses the typical tool usage: Nmap is a primary tool for active reconnaissance (sending packets to discover hosts and services), while Google dorks are a form of passive reconnaissance (searching publicly indexed data without direct interaction). Option D is wrong because passive reconnaissance is primarily used during the footprinting and reconnaissance phase, not the exploitation phase; exploitation occurs after reconnaissance and scanning are complete.

86
MCQeasy

Which of the following tools is specifically designed for ARP poisoning and can be used to perform man-in-the-middle attacks on a local network?

A.Nmap
B.Wireshark
C.Metasploit
D.Ettercap
AnswerD

Ettercap is a comprehensive, open-source suite specifically designed for man-in-the-middle (MITM) attacks on local area networks, alongside network sniffing, content filtering, and active protocol dissection. It natively supports various MITM techniques, including robust ARP poisoning (ARP spoofing), which allows it to intercept traffic between two hosts by sending forged ARP replies. This capability makes Ettercap a primary and highly effective tool for manipulating network traffic flows and performing session hijacking or data interception.

Why this answer

Ettercap is specifically designed for ARP poisoning and man-in-the-middle (MITM) attacks on a local network. It exploits the Address Resolution Protocol (ARP) by sending forged ARP replies to associate the attacker's MAC address with the IP address of a legitimate host, thereby intercepting traffic between two hosts on the same subnet.

Exam trap

The trap here is that candidates may confuse Metasploit's broad exploit capabilities with Ettercap's specialized ARP poisoning functionality, or assume that Wireshark's packet capture implies active attack capabilities, when in fact Ettercap is the quintessential tool for ARP-based MITM attacks on a LAN.

How to eliminate wrong answers

Option A is wrong because Nmap is a network scanning and discovery tool used for port scanning, OS detection, and service enumeration; it does not perform ARP poisoning or MITM attacks. Option B is wrong because Wireshark is a packet analyzer and network protocol sniffer used for passive traffic capture and analysis; it lacks active ARP spoofing capabilities. Option C is wrong because Metasploit is a penetration testing framework that includes many exploit modules, but it is not specifically designed for ARP poisoning; while it may have auxiliary modules for ARP spoofing, Ettercap is the dedicated tool for this purpose.

87
MCQmedium

A penetration tester is scanning a target and receives the output: 'PORT STATE SERVICE 22/tcp open ssh 80/tcp open http 443/tcp open https'. Which Nmap flag was MOST likely used to obtain this output?

A.-sS
B.-O
C.-A
D.-sV
AnswerA

The -sS option performs a SYN scan, which is Nmap's default and most common scanning method. This "half-open" scan sends a SYN packet to each target port and, if a SYN/ACK is received, the port is marked as open before a full TCP connection is established. This efficient technique quickly identifies open ports and infers service names based on standard port assignments, making it ideal for initial reconnaissance without completing the three-way handshake.

Why this answer

The output shows open ports with their service names (ssh, http, https) but no version information. The -sS flag performs a SYN stealth scan, which by default probes common ports and uses the /etc/services file to map port numbers to service names. This matches the output format exactly, as -sS does not perform version detection or OS fingerprinting.

Exam trap

The trap here is that candidates often confuse the service name mapping (from -sS or default scan) with version detection (-sV), assuming that seeing 'ssh' or 'http' implies version probing occurred, when in fact Nmap simply maps the port number to a common service name from its database.

How to eliminate wrong answers

Option B is wrong because -O is used for OS detection, which would add OS fingerprinting details (e.g., 'OS: Linux 2.6.32') to the output, not just port states and service names. Option C is wrong because -A enables aggressive scanning (OS detection, version detection, script scanning, traceroute), which would produce far more verbose output including version strings and script results. Option D is wrong because -sV enables version detection, which would append version information (e.g., 'Apache httpd 2.4.41') to each service line, not just the service name from the port mapping.

88
MCQeasy

Which of the following Burp Suite tools is used to automatically fuzz web application inputs and identify common vulnerabilities like SQL injection and XSS?

A.Proxy
B.Repeater
C.Intruder
D.Scanner
AnswerC

Burp Suite's Intruder tool is purpose-built for performing automated, highly configurable attacks against web applications, including fuzzing, brute-forcing, and enumeration. It allows users to define specific insertion points within a request and then systematically iterate through a list of payloads, observing the server's responses to identify vulnerabilities or weak points. Its advanced payload generation, attack types (e.g., Sniper, Battering Ram, Pitchfork, Cluster Bomb), and result analysis features make it the ideal choice for automated input testing.

Why this answer

Burp Intruder is a tool for automating customized attacks against web applications, including fuzzing for vulnerabilities.

89
MCQhard

After gaining initial access to a Linux server, an attacker runs `find / -perm -4000 -o -perm -2000 2>/dev/null`. What is the primary objective of this command?

A.Locate world-writable files for data exfiltration
B.List all files owned by the root user
C.Find configuration files containing passwords
D.Identify files with SUID or GUID bits set for privilege escalation
AnswerD

Identifying files with SUID (Set User ID) or GUID (Set Group ID) bits set is a critical step for privilege escalation on Linux systems. When a program with the SUID bit is executed, it runs with the permissions of its owner, typically root, regardless of the user who initiated it. This allows a low-privileged attacker to execute specific binaries with elevated privileges, potentially exploiting vulnerabilities within those programs to gain a root shell or execute arbitrary commands as root.

Why this answer

The command `find / -perm -4000 -o -perm -2000 2>/dev/null` searches the entire filesystem for files with the SUID (setuid, permission 4000) or SGID (setgid, permission 2000) bits set. These special permission bits allow a file to execute with the privileges of its owner (often root) or group, respectively. An attacker's primary objective is to identify such files because they can be exploited to escalate privileges from a low-privileged user to a higher-privileged user (e.g., root) by running a vulnerable SUID/SGID binary.

Exam trap

The trap here is that candidates confuse the permission masks for SUID/SGID (4000/2000) with world-writable (0002) or setuid root ownership, leading them to incorrectly select options about data exfiltration or root-owned files instead of recognizing the command's specific purpose for privilege escalation enumeration.

How to eliminate wrong answers

Option A is wrong because world-writable files are found using `-perm -0002` or `-o -perm -0002`, not the `-4000` (SUID) or `-2000` (SGID) permission masks; this command specifically targets setuid/setgid bits, not write permissions. Option B is wrong because the command does not filter by owner; it finds files based on permission bits, not the `-user root` predicate, so it will list SUID/SGID files owned by any user, not just root. Option C is wrong because configuration files containing passwords are typically found by searching for specific filenames (e.g., `*.conf`, `*.cfg`) or content with `grep`, not by checking for the SUID/SGID permission bits; this command has no string or pattern matching.

90
MCQmedium

A penetration tester executes the following command: 'reaver -i wlan0mon -b 00:11:22:33:44:55 -vv'. Which attack is being performed?

A.WEP IV attack
B.Evil twin attack
C.De-authentication attack
D.WPS PIN brute-force attack
AnswerD

Reaver is purpose-built to execute a brute-force attack against the Wi-Fi Protected Setup (WPS) PIN, exploiting a significant design flaw in the protocol. This vulnerability allows an attacker to test the first four digits and the subsequent three digits of the 8-digit PIN independently, drastically reducing the number of attempts required. Upon successfully guessing the correct PIN, Reaver can then extract the network's Pre-Shared Key (PSK), thereby compromising the Wi-Fi network's security.

Why this answer

Reaver is a tool used for brute-forcing WPS PINs to recover the WPA/WPA2 pre-shared key. The command specifies the interface and BSSID, indicating a WPS attack.

91
Multi-Selectmedium

Which TWO of the following are common attack vectors against IoT devices? (Select 2)

Select 2 answers
A.Default credentials
B.Firmware extraction via JTAG
C.Insecure protocols like MQTT without encryption
D.Replay attacks on encrypted sessions
E.SQL injection
AnswersA, C

Many IoT devices are deployed with factory-set usernames and passwords that are publicly known or easily guessable. Attackers leverage extensive databases of these default credentials to gain unauthorized access to devices, often en masse. This vulnerability allows for full device compromise, enabling actions like data exfiltration, device manipulation, or recruitment into botnets without requiring complex exploits.

Why this answer

Default credentials and insecure protocols (e.g., MQTT without TLS) are common IoT attack vectors; firmware extraction is a technique, not a vector; SQL injection is more common in web apps; replay attacks can occur but are not specific to IoT.

92
MCQhard

During a penetration test, you run the following Nmap command: nmap -sS -sV -O -A -T4 --script=default 10.0.0.1. The scan results show that port 443 is open and the service is 'Apache httpd 2.4.29'. However, banner grabbing with Netcat shows 'Apache/2.4.41 (Ubuntu)'. What is the MOST likely explanation for the discrepancy?

A.The server is using a reverse proxy that presents a different version to Nmap
B.Netcat banner grabbing is more reliable because it reads the actual server response
C.The discrepancy is due to Nmap's OS fingerprinting conflicting with version detection
D.Nmap is more accurate because it uses deep packet inspection
AnswerB

Netcat performs a direct banner grab by establishing a raw TCP connection to the target port and simply displaying the initial data sent by the service. This method directly reads the service's self-reported version string, which is typically the most authoritative source. In contrast, Nmap's -sV (service version detection) relies on a database of signatures and probes, which can sometimes be outdated, incomplete, or misinterpret non-standard banners, leading to discrepancies compared to the service's actual response.

Why this answer

Netcat performs a direct TCP connection to the service and reads the raw banner as sent by the application, which is the most immediate and unfiltered version information. Nmap's version detection (-sV) relies on probe-response matching against its signature database, which can be outdated or misinterpret the service if the server uses banner obfuscation or if the Nmap database does not have an exact match for the newer version. In this case, Netcat reveals the actual server version (2.4.41), while Nmap's database may only have a signature for 2.4.29, leading to a false lower version.

Exam trap

The trap here is that candidates assume Nmap is always more accurate because it is a sophisticated scanning tool, but in version detection, a direct banner grab with Netcat is often more reliable when the service banner is not suppressed.

How to eliminate wrong answers

Option A is wrong because a reverse proxy would typically present the same version to both Nmap and Netcat, or could mask the backend version entirely; it would not cause Nmap to report a lower version than the actual banner. Option C is wrong because OS fingerprinting (-O) is a separate function that does not interfere with version detection; the discrepancy is between two version detection methods, not OS fingerprinting. Option D is wrong because Nmap's version detection does not use deep packet inspection; it sends specific probes and matches responses to a signature database, which can be less accurate than a direct banner grab if the database is outdated or the service responds differently to probes.

93
MCQeasy

Which of the following is a characteristic of a polymorphic virus?

A.It changes its code pattern with each infection to evade detection
B.It remains dormant until a specific date
C.It spreads without user interaction
D.It attaches to the boot sector of a hard drive
AnswerA

Polymorphic malware, such as a polymorphic virus, possesses the sophisticated ability to alter its internal code structure and signature with every new infection or replication. This mutation typically involves encrypting its payload with a different key and using a varying decryption routine, making each instance appear unique. This constant code transformation is a primary tactic to bypass traditional signature-based antivirus software, which relies on identifying fixed patterns, thereby significantly increasing its stealth and persistence within a system.

Why this answer

A polymorphic virus is designed to change its code pattern—often by using a mutation engine that generates new decryption routines or alters the payload's signature—each time it infects a new file or system. This constant mutation makes it difficult for signature-based antivirus solutions to detect it because the virus's binary fingerprint is never the same across infections.

Exam trap

The trap here is that candidates confuse 'polymorphic' with other malware types like worms or boot sector viruses, focusing on propagation methods or triggers instead of the defining characteristic of code mutation to evade signature-based detection.

How to eliminate wrong answers

Option B is wrong because a virus that remains dormant until a specific date is a logic bomb or time bomb, not a polymorphic virus; polymorphic viruses are defined by their code-changing ability, not by a trigger condition. Option C is wrong because spreading without user interaction describes a worm, which self-propagates across networks, whereas a virus typically requires some form of user action (e.g., opening a file) to execute and spread. Option D is wrong because attaching to the boot sector of a hard drive defines a boot sector virus, which infects the Master Boot Record (MBR) or Volume Boot Record (VBR), not a polymorphic virus that focuses on altering its own code to evade detection.

94
MCQmedium

During the reconnaissance phase, a tester discovers that the target company's email server is configured to automatically respond to delivery status notifications (DSNs). Which type of attack could this information facilitate?

A.DNS cache poisoning
B.Email enumeration
C.Man-in-the-middle attack
D.Phishing attack
AnswerB

Email enumeration is a reconnaissance technique used to discover valid email addresses within an organization. When an email is sent to a non-existent address, the mail server often returns a Delivery Status Notification (DSN) or Non-Delivery Report (NDR) indicating the address was invalid. Conversely, the *absence* of such a bounce, or a specific DSN indicating a deferred delivery rather than an immediate rejection, can confirm the validity of an email address, making DSN responses a valuable tool for identifying active accounts.

Why this answer

Email servers that automatically respond to Delivery Status Notifications (DSNs) as defined in RFC 1891/3464 can be exploited for email enumeration. By sending a message to a non-existent address, the DSN response will indicate the address is invalid, while a valid address may generate no DSN or a different response. This allows an attacker to systematically verify valid email addresses on the target domain without triggering a full bounce-back to the original sender.

Exam trap

EC-Council often tests the distinction between passive reconnaissance (like email enumeration via DSN) and active attacks (like MITM or phishing), so candidates mistakenly choose 'Phishing attack' because they associate email servers with phishing, but the question specifically asks what the DSN behavior facilitates during reconnaissance.

How to eliminate wrong answers

Option A is wrong because DNS cache poisoning targets the DNS resolver's cache with forged records, not email server DSN behavior. Option C is wrong because a man-in-the-middle attack requires intercepting and relaying communications between two parties, which is unrelated to DSN responses. Option D is wrong because phishing is a social engineering attack that uses deceptive messages to steal credentials, not a reconnaissance technique to enumerate valid email addresses.

95
MCQhard

A security analyst runs a vulnerability scan with Nessus and receives a report indicating that multiple hosts have the 'MS17-010' vulnerability. What is the MOST likely impact of this vulnerability if exploited?

A.Remote code execution on Windows systems
B.SQL injection
C.Cross-site scripting
D.DNS cache poisoning
AnswerA

Nessus scans are highly effective at identifying critical operating system vulnerabilities, such as the MS17-010 vulnerability, famously known as EternalBlue. This flaw in the Server Message Block (SMB) protocol allows unauthenticated remote code execution on vulnerable Windows systems. A successful scan would flag this specific vulnerability, indicating a severe risk of compromise and potential for widespread malware infection like WannaCry or NotPetya, making it the most relevant finding.

Why this answer

MS17-010 is a critical remote code execution vulnerability in the Microsoft Server Message Block (SMB) protocol. Exploitation allows an unauthenticated attacker to send specially crafted packets to an SMB server, enabling arbitrary code execution with system privileges. This is the same vulnerability leveraged by the EternalBlue exploit used in the WannaCry ransomware attacks.

Exam trap

The trap here is that candidates may confuse MS17-010 with a general network vulnerability, but the CEH exam specifically tests that it is a remote code execution flaw in Windows SMB, not a web or DNS attack.

How to eliminate wrong answers

Option B is wrong because SQL injection targets database query layers (e.g., SQL statements) and is unrelated to SMB protocol vulnerabilities. Option C is wrong because cross-site scripting (XSS) exploits web application input validation to inject client-side scripts, not SMB remote code execution. Option D is wrong because DNS cache poising manipulates DNS resolver caches via forged responses, which is a network-layer attack distinct from the SMB-based MS17-010 flaw.

96
MCQeasy

During a penetration test, a security analyst discovers that an organization's web application uses HTTP for login forms, potentially exposing credentials to interception. Which of the following is the BEST cryptographic control to implement to protect credentials in transit?

A.Implement password hashing with bcrypt on the server side.
B.Use digital signatures to sign the login request.
C.Enforce HTTPS using TLS 1.2 or higher.
D.Encrypt the password field using AES-256 before sending via HTTP.
AnswerC

Enforcing HTTPS using TLS 1.2 or higher is the most effective solution for protecting credentials during transmission. TLS establishes an encrypted tunnel between the client and the server, ensuring confidentiality, integrity, and authenticity for all data exchanged, including login credentials. This robust encryption prevents eavesdropping, tampering, and man-in-the-middle attacks by encrypting the entire communication channel end-to-end.

Why this answer

HTTPS with TLS 1.2 or higher encrypts the entire HTTP session, including login credentials, preventing interception and man-in-the-middle attacks. This is the standard cryptographic control for protecting data in transit, as mandated by RFC 2818 and PCI DSS. TLS 1.2+ uses strong cipher suites like ECDHE-RSA-AES256-GCM-SHA384 to ensure forward secrecy and confidentiality.

Exam trap

The trap here is that candidates confuse encryption at rest (hashing) or partial encryption (AES on password field) with full-session encryption (TLS), or they think digital signatures provide confidentiality, when in fact they only ensure authenticity and integrity.

How to eliminate wrong answers

Option A is wrong because password hashing with bcrypt protects credentials at rest on the server, not during transit over the network. Option B is wrong because digital signatures provide integrity and non-repudiation but do not encrypt the login request, leaving the credentials visible to an interceptor. Option D is wrong because encrypting only the password field with AES-256 before sending over HTTP still leaves the rest of the request (e.g., session tokens, form data) in plaintext, and the encryption key must be shared insecurely, defeating the purpose.

97
Multi-Selecteasy

Which TWO of the following are passive reconnaissance techniques? (Select 2)

Select 2 answers
A.Performing a WHOIS lookup
B.Running an Nmap version scan
C.Performing a ping sweep
D.Using Shodan to find exposed devices
E.Banner grabbing with Netcat
AnswersA, D

Performing a WHOIS lookup is a classic passive reconnaissance technique because it queries publicly available databases maintained by domain registrars and registries. This action retrieves information such as domain owner, administrative contacts, nameservers, and registration dates without sending any packets directly to the target's servers or network. The data is simply pulled from a third-party repository, making it entirely non-intrusive and undetectable by the target.

Why this answer

A WHOIS lookup queries public databases (e.g., ARIN, RIPE) to retrieve domain registration details such as registrar, creation date, and name server records. This is passive because it relies on publicly available information without sending any packets to the target network or interacting with its live systems.

Exam trap

EC-Council often tests the distinction between passive and active reconnaissance by making candidates confuse techniques that use public databases (passive) with those that send packets to the target (active); the trap here is that banner grabbing with Netcat feels passive because it only reads a response, but it still requires initiating a TCP connection to the target.

98
MCQeasy

Which tool is specifically designed to enumerate SMB shares and user accounts on a Windows target by leveraging the SMB protocol?

A.Enum4linux
B.Wireshark
C.Nmap
D.Hydra
AnswerA

Enum4linux is a specialized command-line utility explicitly engineered for enumerating information from Windows and Samba hosts. It leverages NetBIOS and SMB protocols to extract details such as user lists, group memberships, share names, operating system versions, and security policies. This tool is invaluable during the reconnaissance phase of a penetration test, providing critical insights into potential attack surfaces.

Why this answer

Enum4linux is a tool specifically designed to enumerate SMB shares and user accounts on Windows targets by leveraging the SMB protocol. It uses the SMB/CIFS protocol to query services like the SAMR (Security Account Manager Remote) and LSARPC (Local Security Authority Remote Procedure Call) interfaces to extract user lists, share listings, and OS information. This makes it the correct choice for targeted SMB enumeration.

Exam trap

The trap here is that candidates may confuse Nmap's SMB enumeration scripts (like smb-enum-shares) with a dedicated tool, but Enum4linux is the specific tool designed for comprehensive SMB share and user enumeration, not just scanning for open ports.

How to eliminate wrong answers

Option B (Wireshark) is wrong because it is a network protocol analyzer that captures and inspects packets, not a tool that actively enumerates SMB shares or user accounts by sending SMB-specific queries. Option C (Nmap) is wrong because while it can scan for open SMB ports (139, 445) and run some SMB scripts via NSE, its primary purpose is port scanning and service discovery, not dedicated enumeration of SMB shares and user accounts. Option D (Hydra) is wrong because it is a password-cracking tool that performs brute-force attacks against authentication services, not an enumeration tool for listing SMB shares or user accounts.

99
MCQmedium

A security analyst observes the following in Apache access logs: 'GET /cgi-bin/test.cgi?cmd=id HTTP/1.1' 200. This is most likely an attempt at which attack?

A.Command injection
B.Local File Inclusion (LFI)
C.SQL injection
D.Directory traversal
AnswerA

The presence of a 'cmd' parameter in the URL, especially when followed by a system command like 'id', strongly indicates that the web application is passing user-supplied input directly to an underlying operating system shell. This vulnerability arises when the application fails to properly sanitize or validate this input, allowing an attacker to append arbitrary shell commands using special characters such as semicolons, pipes, or ampersands. Consequently, the server executes these injected commands with the privileges of the web server process, potentially leading to remote code execution.

Why this answer

The 'cmd' parameter in a CGI script is a common indicator of command injection, where the attacker tries to execute system commands.

100
Multi-Selectmedium

Which TWO of the following Nmap scans are considered 'stealth' scans that do not complete a full TCP three-way handshake?

Select 2 answers
A.FIN scan (-sF)
B.TCP connect scan (-sT)
C.UDP scan (-sU)
D.SYN scan (-sS)
E.ACK scan (-sA)
AnswersA, D

The FIN scan (-sF) is considered stealthy because it sends only a FIN packet to the target port without initiating a full TCP three-way handshake. If the port is open, it typically ignores the packet, while a closed port will respond with an RST/ACK. This technique often bypasses stateless firewalls and avoids logging on the target system, as no connection is ever established, making it less detectable.

Why this answer

A FIN scan (-sF) sends a TCP packet with only the FIN flag set. According to RFC 793, if the port is closed, the target responds with an RST packet; if open, the packet is ignored. This avoids completing a full TCP three-way handshake, making it a stealth scan.

Exam trap

The trap here is that candidates often confuse 'stealth' with 'invisible' and incorrectly assume that any scan not completing a handshake qualifies, but the CEH defines stealth scans specifically as those that avoid the full three-way handshake (SYN, FIN, Xmas, Null) and are designed to evade detection, not just any non-handshake scan like ACK scan.

101
MCQmedium

A security analyst wants to check if a web application is vulnerable to Server-Side Request Forgery (SSRF). Which of the following actions would be most effective?

A.Submit a base64-encoded payload in a cookie
B.Use SQLMap with a time-based payload
C.Modify the Host header to point to localhost
D.Send a request with a URL parameter pointing to an internal IP address
AnswerD

Sending a request with a URL parameter pointing to an internal IP address is the correct method to test for Server-Side Request Forgery (SSRF). SSRF exploits occur when a web application fetches a remote resource based on user-supplied input. By providing an internal IP address (e.g., `127.0.0.1`, `10.0.0.1`) in a parameter that the server is expected to process and fetch, an attacker can determine if the server attempts to connect to that internal resource. Successful connection attempts, even if resulting in an error, indicate the presence of an SSRF vulnerability.

Why this answer

Crafting a request that makes the server fetch an internal IP address (like 127.0.0.1) and observing if the response includes data from that internal resource is a good test for SSRF.

102
MCQmedium

A penetration tester runs `snmpwalk -c public -v2c 192.168.1.50 1.3.6.1.2.1.1` and receives a list of system descriptions, uptime, and contact information. Which type of information is the tester primarily gathering?

A.SMB share names and permissions
B.System information and version details
C.Network topology and routing tables
D.Active directory users and groups
AnswerB

The snmpwalk command, especially when targeting the default MIB tree or the "system" group (OID 1.3.6.1.2.1.1), is highly effective at retrieving fundamental device information. This includes critical details like the operating system description (sysDescr), device hostname (sysName), system uptime (sysUpTime), and administrative contact information (sysContact). This makes it a primary method for initial reconnaissance to identify the type and version of the target system.

Why this answer

The `snmpwalk` command with the OID `1.3.6.1.2.1.1` (the system group in MIB-II, defined in RFC 1213) queries the SNMP agent for system-level information. The output includes system description, uptime, contact, and version details, which are all part of the system group. This is a classic enumeration technique to gather system information and version details from a target device using SNMP with the default public community string.

Exam trap

The trap here is that candidates often confuse the system group OID (1.3.6.1.2.1.1) with other MIB branches like the interfaces group or IP group, leading them to incorrectly select network topology or routing tables, but the system group specifically returns device identity and version information.

How to eliminate wrong answers

Option A is wrong because SMB share names and permissions are enumerated using tools like `smbclient` or `enum4linux`, not via SNMP OID 1.3.6.1.2.1.1, which is the system group. Option C is wrong because network topology and routing tables are obtained from OIDs under 1.3.6.1.2.1.4 (IP group) and 1.3.6.1.2.1.4.21 (ipRouteTable), not the system group. Option D is wrong because Active Directory users and groups are typically enumerated via LDAP queries or tools like `ldapsearch`, not through SNMP, which does not expose AD object data via the system group OID.

103
MCQmedium

During a penetration test, you successfully gain access to a web server with a low-privileged shell. You want to escalate privileges to root. Which of the following techniques is MOST likely to achieve privilege escalation on a misconfigured Linux system?

A.Use the `netcat` tool to establish a reverse shell back to the attacker
B.Search for and exploit a SUID binary that allows privilege escalation
C.Use a password cracking tool like John the Ripper on the system's shadow file
D.Perform a brute force attack on the root password
AnswerB

SUID (Set User ID) is a special permission bit on executable files that allows them to run with the permissions of the file's owner, rather than the user executing it. If a binary owned by root, like `find` or `nmap`, has SUID set and can be manipulated to execute arbitrary commands or spawn a shell, a low-privileged user can exploit this to gain root privileges. This is a highly effective and common method for local privilege escalation on Linux/Unix systems.

Why this answer

SUID (Set User ID) binaries execute with the privileges of the file owner, typically root. On a misconfigured Linux system, a low-privileged user can run a SUID-root binary (e.g., `find`, `vim`, `nmap`) to spawn a shell with root privileges, directly achieving privilege escalation without needing credentials or additional exploits.

Exam trap

The trap here is that candidates confuse establishing a reverse shell (which maintains the current privilege level) with privilege escalation, or they assume password cracking is feasible without first obtaining the hashed password file.

How to eliminate wrong answers

Option A is wrong because `netcat` is a network utility for establishing reverse shells or listening for connections; it does not escalate privileges—it only provides a remote shell at the current privilege level. Option C is wrong because John the Ripper cracking the shadow file requires read access to `/etc/shadow`, which a low-privileged shell typically does not have (shadow file is readable only by root or the shadow group). Option D is wrong because brute-forcing the root password is impractical: it requires network or console access, risks account lockout, and is noisy; moreover, the goal is to exploit a misconfiguration, not guess credentials.

104
MCQhard

An analyst reviews the following HTTP response: HTTP/1.1 200 OK Set-Cookie: sessionid=abc123; SameSite=None; Secure ... <html><body><p>Welcome back!</p></body></html>. What possible vulnerability exists if the application does not use CSRF tokens?

A.Cross-site request forgery (CSRF)
B.Clickjacking
C.Cross-site scripting (XSS)
D.Session fixation
AnswerA

Cross-site request forgery (CSRF) is a vulnerability where an attacker tricks an authenticated user into submitting an unintended request to a web application. If the HTTP response implies that session cookies are sent on cross-site requests (e.g., via SameSite=None without Secure or HttpOnly flags) and the application lacks anti-CSRF tokens, the application becomes susceptible. An attacker can craft a malicious page that, when visited by the victim, forces their browser to send a request to the vulnerable site, leveraging the victim's active session.

Why this answer

SameSite=None allows cross-site requests to include cookies, making CSRF possible if no CSRF tokens are used. SameSite=Lax or Strict would block some CSRF attacks.

105
Multi-Selectmedium

Which TWO of the following Nmap flags are used for evasion of IDS/IPS? (Choose two.)

Select 2 answers
A.-sV
B.-O
C.-D
D.-f
E.-sT
AnswersC, D

The -D flag enables decoy scanning, an effective evasion technique where Nmap sends scan packets from multiple spoofed source IP addresses in addition to the attacker's real IP. By interspersing legitimate scan packets with numerous fake ones, this method aims to confuse Intrusion Detection Systems (IDS) and obscure the true origin of the scan. This makes it significantly harder for security analysts to pinpoint the actual attacker's machine amidst a flood of seemingly disparate scan attempts, thus providing a layer of anonymity.

Why this answer

(-D) is correct because the Nmap decoy scan flag allows you to spoof multiple source IP addresses, making it difficult for IDS/IPS to distinguish the real scanning host from decoys. Option D (-f) is correct because fragmenting packets (e.g., using -f to split TCP headers into 8-byte fragments) evades signature-based detection by bypassing pattern-matching rules that expect complete packet headers.

Exam trap

EC-Council often tests the misconception that -sV or -O are evasion techniques because they are 'stealthy' in some contexts, but the CEH exam specifically requires knowing that decoys (-D) and fragmentation (-f) are the standard Nmap evasion flags.

106
MCQmedium

During a penetration test, you run the command `enum4linux -a 192.168.1.10` and receive output containing user account names, group memberships, and share listings. Which protocol is primarily being enumerated?

A.NFS
B.SMB
C.SNMP
D.SMTP
AnswerB

Server Message Block (SMB) is the correct protocol targeted by enum4linux. This tool is specifically engineered to perform comprehensive enumeration against Windows systems, leveraging the SMB protocol to extract critical security-relevant information. It can identify user accounts, shared resources, group memberships, operating system versions, and even password policies, providing valuable insights for penetration testers. This makes SMB enumeration a foundational step in assessing Windows network security.

Why this answer

enum4linux is a tool specifically designed to enumerate information from Windows and Samba systems via the SMB (Server Message Block) protocol. The command `enum4linux -a 192.168.1.10` performs a comprehensive scan that retrieves user accounts, group memberships, and share listings, all of which are exposed through SMB's named pipe and RPC mechanisms. SMB is the correct protocol because it is the primary means for file and printer sharing in Windows networks, and enum4linux leverages SMB's IPC$ share and SAMR/LSA RPC services to extract this data.

Exam trap

The trap here is that candidates often confuse enum4linux with tools like 'showmount' for NFS or 'snmpwalk' for SNMP, but the key is that enum4linux is explicitly built for SMB enumeration, and the output of user accounts and shares is a hallmark of SMB, not NFS or SNMP.

How to eliminate wrong answers

Option A is wrong because NFS (Network File System) is a Unix/Linux-based file-sharing protocol that does not expose user account or group membership details via enum4linux; enum4linux is designed for SMB/CIFS environments, not NFS mounts. Option C is wrong because SNMP (Simple Network Management Protocol) is used for network device monitoring and management, not for enumerating user accounts or shares; enum4linux does not interact with SNMP agents. Option D is wrong because SMTP (Simple Mail Transfer Protocol) is an email delivery protocol and has no mechanism for listing user accounts, group memberships, or file shares; enum4linux targets SMB services, not mail servers.

107
MCQmedium

Which of the following is a cryptographic attack that exploits collisions in hash functions?

A.Dictionary attack
B.Downgrade attack
C.Birthday attack
D.Replay attack
AnswerC

The birthday attack is a cryptographic attack that exploits the mathematics behind the birthday paradox to find collisions in hash functions more efficiently than a brute-force search. It works by generating a large number of distinct inputs and their corresponding hash outputs, then searching for two inputs that produce the same hash value. This probabilistic method significantly reduces the computational effort required to find a hash collision, making it a direct threat to the collision resistance property of hash functions.

Why this answer

A birthday attack exploits the birthday paradox to find two different inputs that produce the same hash output (collision).

108
MCQmedium

An analyst executes 'nmap -sU -p 161,162 10.0.0.1'. What is the primary purpose of this scan?

A.Detect TCP services on the target
B.Enumerate all open ports on the target
C.Discover SNMP services running on the target
D.Perform a SYN flood attack
AnswerC

The command `nmap -sU -p 161,162` specifically instructs Nmap to perform a UDP scan on ports 161 and 162. UDP port 161 is the well-known port for the Simple Network Management Protocol (SNMP) agent, used for management queries, while UDP port 162 is designated for SNMP trap messages, which are asynchronous notifications. Therefore, this scan is precisely configured to identify the presence and responsiveness of SNMP services and their associated trap listeners on the target.

Why this answer

The `-sU` flag instructs Nmap to perform a UDP scan, and the `-p 161,162` targets the default SNMP ports (UDP 161 for SNMP queries, UDP 162 for SNMP traps). This combination is specifically designed to discover SNMP services running on the target host, as SNMP operates exclusively over UDP. Option C is correct because the command's primary purpose is to probe for SNMP services.

Exam trap

The trap here is that candidates often confuse `-sU` with TCP scans or assume the command scans all ports, but CEH specifically tests the understanding that `-sU` with `-p 161,162` targets SNMP over UDP, not general port enumeration or attacks.

How to eliminate wrong answers

Option A is wrong because `-sU` specifies a UDP scan, not a TCP scan; TCP services are detected using `-sT` or `-sS`, not `-sU`. Option B is wrong because the command only scans ports 161 and 162, not all ports; enumerating all open ports would require a broader port range (e.g., `-p-`) or a different scan type. Option D is wrong because a SYN flood attack is a denial-of-service technique using TCP SYN packets, whereas this is a reconnaissance scan using UDP probes; Nmap does not perform attacks by default.

109
MCQeasy

A security analyst runs `nbtstat -A 192.168.1.10` and receives a response with the computer name, logged-in user, and domain. Which protocol is being queried?

A.NetBIOS
B.SNMP
C.LDAP
D.SMTP
AnswerA

The `nbtstat` command is a dedicated utility for querying NetBIOS over TCP/IP (NetBT) information, which provides name resolution and session services for legacy Windows networking components. When an analyst runs `nbtstat -a <IP_address>`, it specifically attempts to retrieve the NetBIOS name table from the specified remote host. This command directly interacts with the NetBIOS protocol to gather details like registered names and MAC addresses, making NetBIOS the correct answer.

Why this answer

The `nbtstat -A` command performs a NetBIOS name service query (NBNS) against the target IP address using UDP port 137. It retrieves the NetBIOS name table, which includes the computer name, logged-in user, and domain membership, directly from the NetBIOS over TCP/IP (NetBT) protocol stack.

Exam trap

The trap here is that candidates confuse `nbtstat -A` (which queries NetBIOS over TCP/IP) with `nbtstat -a` (which queries by name) or assume it uses a different protocol like SMB, but the command specifically targets the NetBIOS name service on UDP 137.

How to eliminate wrong answers

Option B is wrong because SNMP (Simple Network Management Protocol) uses UDP ports 161/162 and is queried with tools like `snmpget` or `snmwalk`, not `nbtstat`. Option C is wrong because LDAP (Lightweight Directory Access Protocol) operates over TCP port 389 and is used to query directory services like Active Directory, not to retrieve NetBIOS names. Option D is wrong because SMTP (Simple Mail Transfer Protocol) uses TCP port 25 for email transfer and has no role in NetBIOS name resolution or enumeration.

110
MCQeasy

During a security assessment, a tester uses `nmap -sU 192.168.1.1`. What type of scan does this command perform?

A.UDP scan
B.TCP SYN scan
C.Ping sweep
D.OS fingerprinting
AnswerA

The `nmap -sU` command explicitly instructs Nmap to perform a UDP scan, which is the correct interpretation of the provided syntax. This method sends UDP packets to target ports and analyzes the responses (or lack thereof) to determine if a port is open, closed, or filtered. Unlike TCP, UDP is connectionless, making port state determination more challenging and often slower, as open ports may not send a response, while closed ports typically return an ICMP Port Unreachable message.

Why this answer

The `-sU` flag in Nmap explicitly instructs the tool to perform a UDP scan. This sends UDP packets to the target ports and analyzes responses (or lack thereof) to determine if a UDP port is open, closed, or filtered. Unlike TCP, UDP is connectionless, so the scan relies on ICMP unreachable messages or lack of response to infer port status.

Exam trap

The trap here is that candidates confuse the `-sU` flag with a TCP SYN scan (`-sS`) or assume it performs a general host discovery, but the question specifically tests knowledge of Nmap's scan type flags.

How to eliminate wrong answers

Option B is wrong because TCP SYN scan uses the `-sS` flag, not `-sU`, and relies on the TCP three-way handshake (sending a SYN packet) to determine port states. Option C is wrong because a ping sweep typically uses ICMP echo requests (or TCP/UDP probes to multiple hosts) to discover live hosts, not a single target with UDP probes; the command `nmap -sn` is used for ping sweeps. Option D is wrong because OS fingerprinting is performed with options like `-O` or `-A`, which analyze TCP/IP stack behavior, not a simple UDP scan.

111
MCQhard

A security analyst executes the command 'msfvenom -p windows/meterpreter/reverse_tcp LHOST=192.168.1.10 LPORT=4444 -f exe -o shell.exe' and transfers the file to a target. Which technique is being used?

A.Generating a Trojan
B.Creating a virus
C.Deploying a worm
D.Initiating a DoS attack
AnswerA

The msfvenom command, particularly when generating a Windows payload like a reverse shell or Meterpreter, is designed to create a malicious program that provides remote access and control over a compromised system. This type of payload establishes a persistent backdoor, allowing an attacker to execute commands and exfiltrate data. Such a program, which often relies on social engineering or embedding within legitimate software to trick users into execution, perfectly aligns with the definition and functionality of a Trojan.

Why this answer

The command uses msfvenom to generate a Windows executable payload that establishes a reverse TCP connection back to the attacker's IP (192.168.1.10) on port 4444. This creates a trojan horse—malicious code disguised as a legitimate file (shell.exe)—which, when executed on the target, provides unauthorized remote access. The technique is specifically trojanization, not virus or worm propagation.

Exam trap

EC-CEH often tests the distinction between trojans (user-executed, non-replicating) and viruses (self-replicating by infecting files), so candidates mistakenly choose 'virus' because they associate malicious executables with infection, ignoring the lack of replication mechanism.

How to eliminate wrong answers

Option B is wrong because a virus requires self-replication and insertion into other files or system areas, whereas this payload is a standalone executable that does not replicate. Option C is wrong because a worm autonomously spreads across networks without user interaction, but this file must be manually transferred and executed by the victim. Option D is wrong because a DoS attack aims to disrupt service availability, while this payload establishes a reverse shell for persistent remote control.

112
Multi-Selectmedium

A security analyst notices that a web application's search functionality returns database error messages in the response. The analyst suspects SQL injection. Which TWO techniques should the analyst use to confirm and exploit this vulnerability? (Choose TWO.)

Select 2 answers
A.Use out-of-band SQL injection with DNS exfiltration
B.Use a time-based blind SQL injection with SLEEP() function
C.Leverage error-based SQL injection with CONVERT() or double query
D.Use SQLMap with --union-col and --union-from flags
E.Implement parameterized queries in the application code
AnswersC, D

Leveraging error-based SQL injection with functions like CONVERT() or by employing double query techniques (e.g., using `EXTRACTVALUE` or `UPDATEXML` in MySQL) is highly effective when an application displays verbose database error messages. These methods intentionally trigger type conversion errors or XML parsing errors, forcing the database to include the results of an injected subquery within the error message itself. This allows the attacker to directly extract data, such as database version, table names, or user credentials, from the application's response.

Why this answer

Union-based SQL injection uses UNION SELECT to retrieve data from other tables. Error-based SQL injection leverages database error messages to extract information. Both are common in-band techniques suitable when errors are displayed.

113
MCQhard

While analyzing web server logs, an analyst finds the following entry: GET /../../../../etc/passwd HTTP/1.1 with a 200 OK response. Which vulnerability is indicated, and what is the MOST likely impact?

A.Command injection; remote shell access
B.Directory traversal; reading sensitive files
C.SQL injection; data exfiltration
D.Remote File Inclusion (RFI); arbitrary code execution
AnswerB

The presence of `../` sequences in the URL path is a definitive indicator of a directory traversal (also known as path traversal) attack. This technique exploits vulnerabilities in file handling routines to access files and directories stored outside the intended web root directory by manipulating relative paths. The goal is often to read sensitive system files, such as `/etc/passwd`, or configuration files, leading directly to unauthorized information disclosure.

Why this answer

The path contains '../' sequences indicating directory traversal. A 200 response suggests the attacker successfully read the /etc/passwd file, leading to disclosure of system user accounts. This can aid further attacks like password cracking.

114
MCQmedium

An incident responder analyzes logs and finds repeated failed zone transfer attempts from an external IP. The zone transfer requests are targeting the domain example.com. Which DNS record type, if misconfigured, would allow this attack to succeed?

A.NS records
B.AXFR
C.MX records
D.SOA records
AnswerB

AXFR is the DNS query type for zone transfers. Allowing AXFR from unauthorized hosts is a misconfiguration.

Why this answer

B is correct because AXFR (Asynchronous Full Transfer) is the DNS zone transfer protocol that, if misconfigured (i.e., allowing unrestricted AXFR queries from any IP), permits an external attacker to request and receive the entire DNS zone file for example.com. The repeated failed attempts indicate the attacker is probing for an open AXFR service, which would succeed if the DNS server is configured to allow zone transfers to any host without restriction.

Exam trap

The trap here is that candidates confuse the DNS record type (e.g., NS, SOA) with the protocol or query type (AXFR) used to perform the zone transfer, leading them to select a record type instead of recognizing AXFR as the specific misconfiguration that enables the attack.

How to eliminate wrong answers

Option A is wrong because NS records specify the authoritative name servers for a domain, not the mechanism for transferring zone data; misconfigured NS records could lead to delegation issues but do not directly allow zone transfer attacks. Option C is wrong because MX records define mail exchange servers for email routing and have no role in DNS zone transfers; they are irrelevant to the attack described. Option D is wrong because SOA records contain administrative metadata about the zone (e.g., serial number, refresh interval) but do not control or enable zone transfer requests; misconfigured SOA records might affect zone replication timing but not allow external AXFR queries.

115
Multi-Selectmedium

Which TWO of the following are effective mitigations against Cross-Site Request Forgery (CSRF)?

Select 2 answers
A.Using SameSite cookies
B.Input validation
C.Using anti-CSRF tokens
D.Using HTTPOnly cookies
E.Using CSRF tokens
AnswersA, E

SameSite cookies are an effective mitigation because they instruct the browser to restrict when cookies are sent with cross-origin requests. By setting `SameSite=Lax` or `SameSite=Strict`, the browser will not attach session cookies to requests initiated from a different site, thereby preventing an attacker's forged request from carrying the necessary authentication credentials to execute an unauthorized action.

Why this answer

SameSite cookies (option A) are effective against CSRF because they restrict the browser from sending cookies on cross-origin requests, preventing forged requests from carrying authentication credentials. CSRF tokens (option E) are also effective because they require a unique token in each request that the attacker cannot predict. Option C (anti-CSRF tokens) is essentially the same technique as option E, so it is not considered a distinct mitigation; thus, only options A and E are correct.

Exam trap

EC-Council often tests the distinction between CSRF and XSS mitigations, and the trap here is that candidates confuse HTTPOnly cookies (which protect against XSS) as a CSRF defense, or they think that input validation or redundant token naming (anti-CSRF vs CSRF) are separate valid options.

116
MCQmedium

You are a security analyst for a medium-sized company. The company uses a custom web application for internal project management. The application uses AES-256-CBC for encrypting sensitive data stored in the database. Recently, the company experienced a data breach where an attacker exfiltrated the entire database. Although the data was encrypted, the attacker was able to decrypt some records. Investigation reveals that the encryption key is stored in a configuration file on the same server, and the initialization vector (IV) is hardcoded in the application code. Additionally, the application uses the same key for all records. Which of the following is the most effective remediation to prevent future decryption of stolen encrypted data?

A.Change the encryption mode from CBC to GCM to provide authentication
B.Store the encryption key in a hardware security module (HSM) and use the same key
C.Rotate the encryption key every 24 hours
D.Implement per-record encryption keys derived from a master key combined with a unique salt
AnswerD

Implementing per-record encryption keys derived from a master key combined with a unique salt is the most robust solution. This method ensures that each individual record is encrypted with a distinct, cryptographically unique key generated using a Key Derivation Function (KDF). Consequently, if one derived record key is ever compromised, only that specific record is affected, leaving all other records protected by their respective unique keys and the secure master key. This significantly limits the impact of any single key compromise, enhancing overall data security.

Why this answer

Using per-record encryption keys derived from a master key combined with a unique salt ensures that even if an attacker exfiltrates the entire database, each encrypted record requires a separate key derivation operation. Without the unique salt per record, the attacker cannot decrypt all records even if they compromise the master key. This approach mitigates the risk of a single key compromise leading to bulk decryption, which is the core vulnerability in the current setup where the same AES-256-CBC key and hardcoded IV are reused across all records.

Exam trap

The trap here is that candidates often focus on key storage or rotation (options B and C) as the primary solution, overlooking that the real vulnerability is the reuse of a single key across all records, which allows an attacker to decrypt the entire dataset with a single key compromise.

How to eliminate wrong answers

Option A is wrong because changing the encryption mode from CBC to GCM adds authentication (integrity) but does not address the fundamental issue of a single static key and IV being reused; an attacker who steals the database and the key can still decrypt all records regardless of the mode. Option B is wrong because storing the encryption key in an HSM while still using the same key for all records does not prevent an attacker from decrypting all stolen data if they compromise the application at runtime or obtain the key from the HSM via authorized access; the HSM protects the key at rest but does not mitigate the single-key reuse vulnerability. Option C is wrong because rotating the encryption key every 24 hours only limits the window of exposure for future records; it does not protect already exfiltrated encrypted data that was encrypted with the old key, and the attacker can still decrypt all records encrypted before the rotation if they have the old key.

117
Multi-Selectmedium

Which THREE of the following are characteristics of asymmetric encryption?

Select 3 answers
A.Uses a single shared key for both encryption and decryption
B.Supports digital signatures
C.Provides key exchange without prior shared secret
D.Involves a public key and a private key
E.Typically faster than symmetric encryption
AnswersB, C, D

Asymmetric encryption is foundational for digital signatures, providing non-repudiation, integrity, and authenticity. The sender uses their unique private key to encrypt a hash of the message, creating the digital signature. Recipients then use the sender's publicly available corresponding public key to decrypt the signature and verify the message's integrity and the sender's identity. This process ensures the message originated from the claimed sender and has not been tampered with.

Why this answer

Asymmetric encryption uses two keys (public/private), provides key exchange, and supports digital signatures.

118
MCQhard

A penetration tester runs the following command against a Linux server: `smbclient -L //192.168.1.10 -N`. The output lists shares including 'IPC$', 'ADMIN$', and 'data'. Which of the following is the BEST next step to enumerate the 'data' share?

A.Run `nmap --script smb-enum-shares -p 445 192.168.1.10`
B.Run `enum4linux -a 192.168.1.10` to gather more information
C.Use `rpcclient -U '' 192.168.1.10` to enumerate users
D.Use `smbclient //192.168.1.10/data -N` to attempt a null session connection
AnswerD

The `smbclient` utility is the standard command-line tool for interacting with SMB/CIFS shares, making it the most appropriate choice for this scenario. The syntax `//192.168.1.10/data` correctly specifies the target server and the known share name. The `-N` flag is crucial as it instructs `smbclient` to attempt a null session connection, meaning it tries to connect anonymously without requiring a username or password. This is the most direct and efficient method to test if the 'data' share is anonymously accessible, which is a common misconfiguration.

Why this answer

The command `smbclient -L //192.168.1.10 -N` performs a null session (no password) listing of SMB shares. The output shows that the 'data' share exists and is accessible without authentication (since the -N flag succeeded). The best next step is to attempt a null session connection to that specific share using `smbclient //192.168.1.10/data -N`, which will mount the share and allow file enumeration.

This directly leverages the null session already confirmed by the initial scan.

Exam trap

The trap here is that candidates often choose a broad enumeration tool like enum4linux or an nmap script, thinking they need more information first, when the direct connection to the already-discovered share is the logical and efficient next step in a penetration test.

How to eliminate wrong answers

Option A is wrong because `nmap --script smb-enum-shares -p 445` would re-enumerate shares, which is redundant after already discovering the 'data' share via smbclient. Option B is wrong because `enum4linux -a` is a comprehensive enumeration tool that gathers users, groups, shares, and policies, but it is a broader, slower step that does not directly access the 'data' share; the immediate goal is to connect to the share, not gather more metadata. Option C is wrong because `rpcclient -U ''` is used for RPC-based enumeration (e.g., users, SIDs) via the IPC$ share, not for accessing a file share like 'data'; it would not list or retrieve files from the 'data' share.

119
Multi-Selectmedium

A security analyst observes a sudden increase in network traffic from many external IPs targeting the company's web server with multiple HTTP GET requests to the same page (/index.php?page=home). The requests appear legitimate but are coming at a very high rate. Which TWO types of attack is the analyst most likely witnessing?

Select 2 answers
A.Smurf attack
B.Volumetric attack
C.Application-layer (Layer 7) attack
D.SYN flood attack
E.Distributed denial-of-service (DDoS) attack
AnswersC, E

An application-layer (Layer 7) attack specifically targets the application layer of the OSI model, exploiting vulnerabilities or resource limitations within the application itself. The observation of a sudden increase in HTTP GET requests directed at a specific web page perfectly aligns with this definition. These requests consume server resources like CPU, memory, and database connections, ultimately leading to service degradation or denial for legitimate users without necessarily saturating network bandwidth.

Why this answer

The attack targets the application layer (Layer 7) by sending numerous HTTP GET requests to a specific page (/index.php?page=home). This type of attack aims to exhaust server resources like CPU, memory, or database connections, as each request appears legitimate but collectively overwhelms the web server's ability to process them. It is a classic example of an HTTP flood, which is a Layer 7 attack.

Exam trap

The trap here is that candidates might confuse a high-rate HTTP GET flood with a volumetric attack (Option B) or a SYN flood (Option D), but the key distinction is that this attack specifically targets the application layer by exhausting server resources through legitimate-looking HTTP requests, not by saturating bandwidth or exploiting TCP handshake mechanics.

120
MCQmedium

Refer to the exhibit. A penetration tester runs hashcat to crack NTLM hashes. Which hash mode (-m) would be correct for NTLM?

A.1100
B.1000
C.3000
D.5500
AnswerB

Hashcat mode 1000 is the correct and standard choice for cracking NTLM (NT LAN Manager) hashes, which are the cryptographic one-way functions of a user's password used for authentication in Windows environments. These hashes are commonly extracted from the Security Account Manager (SAM) database, Active Directory's NTDS.DIT file, or captured via various network protocols. This mode directly targets the NTLM hash format, making it ideal for direct password recovery from stored hashes.

Why this answer

NTLM hash mode is 1000. The exhibit shows -m 1000, which is correct for NTLM.

121
MCQmedium

During a penetration test, a tester captures a WPA2 4-way handshake. Which of the following is the NEXT step to attempt to recover the Wi-Fi passphrase?

A.Use aircrack-ng to crack the WEP key from the handshake
B.Run a dictionary attack using aircrack-ng with a wordlist
C.Brute-force the WPS PIN using Reaver
D.De-authenticate the client from the network again to capture another handshake
AnswerB

Running a dictionary attack using aircrack-ng with a wordlist is the correct approach because the captured WPA2 4-way handshake contains the necessary cryptographic elements, such as the ANonce, SNonce, and the Message Integrity Code (MIC). Aircrack-ng can iterate through a wordlist, calculate the Pairwise Master Key (PMK) and subsequent MIC for each potential Pre-Shared Key (PSK), and compare it to the MIC within the captured handshake. A match indicates the correct PSK has been found, allowing the attacker to decrypt network traffic.

Why this answer

After capturing the handshake, the tester must perform a dictionary attack against the handshake file. Tools like aircrack-ng or hashcat can compare the handshake against a wordlist of potential passphrases.

122
MCQmedium

A penetration tester uses the following Google dork: site:example.com filetype:pdf inurl:confidential. What is the MOST likely goal of this search?

A.Retrieve all PDF files from example.com regardless of content
B.Identify all web pages on example.com that link to PDF files
C.Find PDF files on example.com that have 'confidential' in their filename or path
D.Discover PDF files that contain the word 'confidential' on example.com
AnswerC

This option accurately describes the dork's function. The 'inurl:confidential' operator specifically targets URLs that contain the string "confidential," which encompasses both the directory path and the filename components of a URL. When combined with 'filetype:pdf' and 'site:example.com', the dork precisely identifies PDF documents on the specified domain where "confidential" appears within their web address.

Why this answer

The Google dork `site:example.com filetype:pdf inurl:confidential` combines the `site` operator to restrict results to example.com, `filetype:pdf` to filter for PDF files, and `inurl:confidential` to require that the URL or path contains the word 'confidential'. This targets PDF files whose filename or directory path includes 'confidential', making option C correct. The `inurl` operator matches the URL string, not the file content, so it does not search within the PDF text.

Exam trap

The trap here is confusing `inurl` (which searches the URL string) with `intext` or content-based search, leading candidates to incorrectly assume the dork finds PDFs containing the word 'confidential' inside the document.

How to eliminate wrong answers

Option A is wrong because the dork includes `inurl:confidential`, which narrows results to PDFs with 'confidential' in the URL, not all PDFs. Option B is wrong because the dork retrieves PDF files directly, not web pages that link to PDFs; `filetype:pdf` returns the PDF file itself. Option D is wrong because `inurl` searches the URL string, not the content of the PDF; to search within file content, one would use `intext` or `filetype:pdf` combined with a content search term like `"confidential"` without `inurl`.

123
MCQmedium

A security analyst receives an alert about a workstation repeatedly sending large volumes of ICMP echo request packets to a broadcast address. Which type of attack is this indicative of?

A.Smurf attack
B.Ping of Death
C.SYN flood
D.Slowloris
AnswerA

A Smurf attack is a distributed denial-of-service (DDoS) attack where an attacker sends a large number of Internet Control Message Protocol (ICMP) echo requests to a network's broadcast address. The crucial element is that the source IP address of these requests is spoofed to be the victim's workstation. Consequently, all hosts on the broadcast network respond to the victim's IP address with ICMP echo replies, overwhelming the workstation with traffic and causing a denial of service.

Why this answer

A Smurf attack is a distributed denial-of-service (DDoS) attack that exploits ICMP echo request packets sent to a network broadcast address. The source IP is spoofed to be the victim's address, causing all hosts on the broadcast network to reply to the victim, overwhelming it with ICMP echo replies. This matches the alert description of large volumes of ICMP echo requests to a broadcast address.

Exam trap

EC-CEH often tests the distinction between Smurf and Ping of Death, where candidates confuse the volume-based amplification of Smurf with the oversized-packet exploit of Ping of Death.

How to eliminate wrong answers

Option B (Ping of Death) is wrong because it involves sending a malformed ICMP packet larger than the maximum allowed size (65535 bytes), causing a buffer overflow, not repeated large volumes of echo requests to a broadcast address. Option C (SYN flood) is wrong because it exploits the TCP three-way handshake by sending many SYN packets with spoofed source IPs, leaving half-open connections, and does not use ICMP or broadcast addresses. Option D (Slowloris) is wrong because it is an application-layer DDoS attack that holds many HTTP connections open by sending partial requests, targeting web servers, not ICMP or network-layer broadcast traffic.

124
MCQeasy

Which of the following Google dorks would an attacker MOST likely use to find login pages of web applications that are publicly accessible?

A.intitle:login
B.inurl:robots.txt
C.filetype:pdf
D.cache:example.com
AnswerA

The `intitle:login` Google dork is a highly effective reconnaissance tool for attackers, specifically designed to locate web pages where the HTML `<title>` tag contains the word "login". This operator is crucial for identifying potential authentication interfaces, administrative panels, or other sensitive entry points on target systems. By focusing the search on page titles, it efficiently filters out irrelevant results, allowing attackers to quickly pinpoint web pages that likely require user credentials. This precision makes it an invaluable technique for initial information gathering and identifying potential attack vectors.

Why this answer

The Google dork 'intitle:login' is most effective for finding login pages because it searches for the word 'login' in the HTML title tag of web pages. Attackers use this to quickly identify publicly accessible authentication portals, which are common entry points for brute-force or credential-stuffing attacks. This dork directly targets the page title, a standard HTML element that often contains the word 'login' on authentication pages.

Exam trap

EC-Council often tests the distinction between operators that find specific page content (like 'intitle:') versus those that find file types or cached data, leading candidates to confuse 'inurl:robots.txt' (which finds a specific file) with finding login pages.

How to eliminate wrong answers

Option B is wrong because 'inurl:robots.txt' is used to find the robots.txt file, which discloses directories that the site owner wants to hide from search engines, not login pages. Option C is wrong because 'filetype:pdf' restricts results to PDF files, which are unlikely to be login pages (login pages are typically HTML). Option D is wrong because 'cache:example.com' shows the cached version of a specific domain, not a search for login pages across multiple sites.

125
MCQmedium

Which of the following tools would be BEST to use for identifying all live hosts in a large IP range (e.g., 10.0.0.0/8) quickly?

A.Masscan
B.OpenVAS
C.Nmap with -sL flag
D.hping3
AnswerA

Masscan is exceptionally well-suited for rapid host discovery across vast IP ranges due to its asynchronous, custom-built TCP SYN scanner. It can transmit millions of packets per second, bypassing the operating system's network stack by using raw sockets to achieve unparalleled scanning speed. This makes it the optimal choice for identifying live hosts quickly over large networks or the entire internet.

Why this answer

Masscan is the best choice because it is designed for high-speed scanning across large IP ranges, capable of transmitting packets at rates exceeding 10 million packets per second. It uses asynchronous transmission and raw sockets to quickly identify live hosts by sending SYN probes and analyzing responses, making it ideal for scanning a /8 subnet (16.7 million addresses) in minutes.

Exam trap

EC-Council often tests the distinction between scanning speed and functionality, where candidates mistakenly choose Nmap (a versatile tool) for large-range host discovery without recognizing that its default scanning modes are too slow for a /8 subnet, whereas Masscan is purpose-built for speed.

How to eliminate wrong answers

Option B (OpenVAS) is wrong because it is a vulnerability scanner that performs in-depth analysis on identified hosts, not a tool for rapid host discovery across large ranges; its scanning speed is too slow for a /8 subnet. Option C (Nmap with -sL flag) is wrong because the -sL flag performs a list scan that only resolves DNS names without sending any packets, so it cannot identify live hosts. Option D (hping3) is wrong because it is a packet crafting tool used for targeted testing and firewall auditing, not designed for high-speed scanning of massive IP ranges; its sequential packet transmission makes it impractical for a /8 subnet.

126
MCQmedium

During a penetration test, an ethical hacker runs the following command: aireplay-ng -0 5 -a 00:11:22:33:44:55 -c 66:77:88:99:AA:BB wlan0mon. What is the immediate effect of this command?

A.It performs a WEP injection attack to generate traffic
B.It cracks the pre-shared key using a dictionary
C.It forces the client to disconnect and reconnect, capturing the WPA handshake
D.It initiates a brute force attack on the WPS PIN
AnswerC

The `aireplay-ng -0` command executes a deauthentication attack by sending specially crafted deauthentication frames to a target client or broadcast to all clients associated with an access point. This action forcibly disconnects the client from the Wi-Fi network. When the client automatically attempts to re-establish its connection, it performs the crucial WPA/WPA2 4-way handshake with the access point, which can then be captured by a monitoring tool like `airodump-ng` for subsequent offline cracking attempts.

Why this answer

The -0 flag sends deauthentication packets to force a client to reconnect, enabling capture of the WPA handshake.

127
MCQmedium

A security analyst suspects that an attacker is scanning their network. They notice a large number of TCP SYN packets being sent to various ports on a single host, but no SYN-ACK responses are returned. Which type of scan is most likely being used?

A.TCP connect scan
B.UDP scan
C.SYN scan
D.FIN scan
AnswerC

SYN scan sends SYN packets; lack of SYN-ACK indicates filtered/closed ports.

Why this answer

C is correct because a SYN scan (also known as a half-open scan) sends TCP SYN packets to target ports and does not complete the three-way handshake. If no SYN-ACK is returned, it indicates the port is filtered or the host is not responding, which matches the scenario where the attacker receives no SYN-ACK responses. This scan is stealthier than a full TCP connect scan because it never establishes a full connection.

Exam trap

The trap here is that candidates often confuse SYN scan with TCP connect scan, thinking that any TCP scan must complete the handshake, but the key distinction is that SYN scan never sends the final ACK, making it half-open and stealthier.

How to eliminate wrong answers

Option A is wrong because a TCP connect scan completes the full three-way handshake (SYN, SYN-ACK, ACK) and would result in SYN-ACK responses for open ports, not the absence of them. Option B is wrong because a UDP scan sends UDP packets, not TCP SYN packets, and relies on ICMP unreachable messages or lack of response, not TCP SYN-ACK behavior. Option D is wrong because a FIN scan sends TCP packets with the FIN flag set, not SYN packets, and expects RST responses for closed ports, not SYN-ACKs.

128
MCQmedium

A penetration tester calls an employee claiming to be from the IT help desk and asks for their password to perform a 'security update'. The employee provides the password. Which social engineering technique is being used?

A.Pretexting
B.Tailgating
C.Quid pro quo
D.Phishing
AnswerA

Pretexting uses a fabricated scenario to obtain information.

Why this answer

The attacker is fabricating a scenario (IT help desk performing a security update) to manipulate the target into revealing sensitive information. This is the essence of pretexting, where the attacker creates a false identity or situation to gain trust and extract data. Unlike phishing, which typically uses malicious links or attachments, this attack relies purely on verbal impersonation and social manipulation.

Exam trap

The trap here is that candidates confuse pretexting with phishing because both involve deception, but phishing specifically uses electronic channels (email, fake login pages) while pretexting can occur over voice or in person without any technical payload.

How to eliminate wrong answers

Option B is wrong because tailgating involves physically following an authorized person into a restricted area without their consent, not deceiving someone over the phone. Option C is wrong because quid pro quo involves offering a service or benefit in exchange for information (e.g., 'I'll fix your computer if you give me your password'), whereas here the attacker simply demands the password under a false pretense. Option D is wrong because phishing typically uses electronic communication (email, SMS, fake websites) to trick victims into clicking a link or downloading malware, not a direct phone call asking for credentials.

129
MCQeasy

During a penetration test, the tester wants to discover all subdomains of a target domain using an OSINT technique. Which tool is specifically designed for subdomain enumeration via search engines and public records?

A.theHarvester
B.Maltego
C.Shodan
D.dnsrecon
AnswerA

theHarvester is designed to gather emails, subdomains, and other information from public sources.

Why this answer

theHarvester is specifically designed to perform OSINT-based subdomain enumeration by querying search engines (e.g., Google, Bing) and public data sources (e.g., PGP key servers, DNSDumpster). It collects email addresses, subdomains, IPs, and virtual hosts without direct interaction with the target's infrastructure, making it ideal for passive reconnaissance.

Exam trap

EC-Council often tests the distinction between passive OSINT tools (theHarvester) and active reconnaissance tools (dnsrecon), so candidates mistakenly choose dnsrecon because it is a DNS tool, but the question explicitly requires an OSINT technique using search engines and public records.

How to eliminate wrong answers

Option B (Maltego) is wrong because it is a general-purpose OSINT and link-analysis platform that requires transforms (some of which are paid) and is not solely focused on subdomain enumeration via search engines; it is overkill for this specific task. Option C (Shodan) is wrong because it is a search engine for internet-connected devices and services (e.g., IoT, servers), not for enumerating subdomains of a target domain via search engines or public records. Option D (dnsrecon) is wrong because it performs active DNS reconnaissance (e.g., zone transfers, brute-force subdomain discovery) and is not an OSINT technique that relies on search engines and public records.

130
Multi-Selecteasy

Which TWO of the following are types of malware that specifically aim to demand payment from victims?

Select 2 answers
A.Keylogger
B.Spyware
C.Scareware
D.Ransomware
E.Adware
AnswersC, D

Scareware displays fake alerts to trick users into paying for removal.

Why this answer

Scareware is a type of malware that displays fake security alerts or warnings to trick users into believing their system is infected, then demands payment to remove the nonexistent threat. Ransomware encrypts the victim's files or locks the system and demands a ransom payment for decryption or restoration. Both specifically aim to extort money from victims.

Exam trap

EC-CEH often tests the distinction between malware that demands payment (scareware and ransomware) versus malware that simply annoys or spies (adware, spyware, keyloggers), so candidates mistakenly classify scareware as a form of adware or spyware instead of recognizing its extortion-based goal.

131
MCQhard

During a penetration test, you discover an LDAP server on port 389 that allows anonymous binds. Which of the following enumeration techniques would provide the MOST comprehensive information about the directory structure?

A.Run nmap with the smb-enum-shares script
B.Perform a DNS zone transfer
C.Use ldapsearch to query the directory for all attributes
D.Use net view to list domain resources
AnswerC

ldapsearch can retrieve all objects and attributes from an LDAP directory, especially with anonymous bind.

Why this answer

`ldapsearch` with anonymous bind allows querying the LDAP directory for all attributes and entries, providing comprehensive information about the directory structure, including user accounts, groups, organizational units, and other objects. LDAP servers on port 389 often expose the entire directory tree when anonymous binds are permitted, making `ldapsearch` the most effective enumeration technique.

Exam trap

The trap here is that candidates confuse LDAP enumeration with SMB or DNS enumeration, assuming any network discovery tool will work, but only LDAP-specific queries (like `ldapsearch`) can extract directory structure from an LDAP server.

How to eliminate wrong answers

Option A is wrong because `nmap` with the `smb-enum-shares` script targets SMB (port 445) and enumerates Windows file shares, not LDAP directory structure on port 389. Option B is wrong because DNS zone transfer (using `dig` or `nslookup`) retrieves DNS records (A, MX, CNAME, etc.) from a DNS server, not LDAP directory attributes or objects. Option D is wrong because `net view` is a Windows command that lists SMB shared resources on a network, not LDAP directory entries.

132
MCQeasy

Which of the following best describes the attack where an attacker uses a valid session token to impersonate a user without needing to authenticate?

A.Cross-site scripting
B.Session hijacking
C.Phishing
D.Brute-force attack
AnswerB

Session hijacking is a sophisticated attack where an attacker successfully obtains a legitimate user's valid session ID or token and then uses it to impersonate that user, gaining unauthorized access to their active session. By presenting the stolen, yet valid, session token to the web server, the attacker effectively bypasses the initial authentication process and can perform actions as if they were the legitimate user. This allows them to take over an already established and authenticated session without needing the user's credentials.

Why this answer

Session hijacking involves stealing or using a valid session token to impersonate a user, bypassing authentication.

133
MCQhard

An IoT device uses the MQTT protocol without TLS. An attacker on the same network captures messages and publishes a fake temperature reading. Which attack is being executed?

A.Replay attack
B.Firmware reversing attack
C.Man-in-the-middle attack
D.Denial of service attack
AnswerC

A Man-in-the-Middle (MITM) attack is precisely what occurs when an attacker intercepts communications between two parties, in this case, an MQTT client and broker, without either party being aware. Since MQTT is used without TLS, the communication channel is unencrypted, allowing the attacker to easily intercept, read, modify, or inject arbitrary fake messages into the cleartext data stream. This direct manipulation of active network traffic, including the injection of new, crafted messages, is the hallmark of a successful MITM attack.

Why this answer

MQTT over plain TCP allows message interception and injection (man-in-the-middle) because no encryption or authentication is enforced.

134
MCQeasy

Which type of malware spreads by replicating itself across a network without requiring a host file to attach to?

A.Trojan
B.Ransomware
C.Worm
D.Virus
AnswerC

Worms self-propagate across networks independently.

Why this answer

A worm is a standalone malware that replicates itself across a network without needing a host file, exploiting vulnerabilities or using network protocols like SMB, RDP, or email to spread autonomously. Unlike viruses, worms do not attach to a host program; they self-propagate via network connections, often consuming bandwidth and system resources.

Exam trap

A common trap on the CEH exam is confusing 'self-replication' with 'requires a host file'—a worm replicates independently across a network, while a virus must attach to a host file to propagate.

How to eliminate wrong answers

Option A is wrong because a Trojan disguises itself as legitimate software but does not self-replicate; it relies on user execution for installation. Option B is wrong because ransomware encrypts files or locks systems for ransom but typically spreads via attachments or exploits, not by autonomous network replication without a host. Option D is wrong because a virus requires a host file or program to attach to and depends on user action (e.g., opening a file) to execute and spread, unlike a worm which is self-contained and network-propagating.

135
MCQeasy

Which of the following tools is commonly used for dynamic malware analysis by executing the malware in an isolated environment and monitoring system changes?

A.Strings
B.PEiD
C.VirusTotal
D.Cuckoo Sandbox
AnswerD

Cuckoo Sandbox is an open-source automated malware analysis system specifically designed for dynamic analysis. It executes suspicious files within an isolated virtual environment, commonly referred to as a sandbox, and meticulously monitors their runtime behavior. This includes observing API calls, network traffic, file system changes, and process interactions, providing a comprehensive report on the malware's actions and intent, which is the core objective of dynamic analysis.

Why this answer

Cuckoo Sandbox is the correct answer because it is an open-source automated malware analysis system specifically designed for dynamic analysis. It executes suspicious files in an isolated, virtualized environment (e.g., VirtualBox, KVM) and monitors system changes such as file system modifications, registry changes, network connections, and process behavior in real time, providing a detailed report of the malware's runtime activities.

Exam trap

EC-CEH often tests the distinction between static and dynamic analysis tools, and the trap here is that candidates may confuse VirusTotal's file scanning (which is static and signature-based) with true dynamic sandbox execution, or assume that Strings or PEiD can perform runtime monitoring when they are purely static analysis utilities.

How to eliminate wrong answers

Option A is wrong because Strings is a static analysis tool that extracts readable ASCII and Unicode strings from a binary file, not a dynamic analysis tool that executes malware. Option B is wrong because PEiD is a static analysis tool used to detect packers, compilers, and cryptors in PE files by signature matching, not for executing malware or monitoring runtime behavior. Option C is wrong because VirusTotal is a multi-engine file scanning service that aggregates static detection results from numerous antivirus engines, but it does not execute malware in an isolated sandbox for dynamic behavioral monitoring.

136
MCQeasy

Which of the following tools is commonly used to automate the detection and exploitation of SQL injection vulnerabilities?

A.SQLMap
B.Metasploit
C.Nmap
D.Burp Suite
AnswerA

SQLMap is an open-source penetration testing tool specifically designed to automate the detection and exploitation of SQL injection flaws and database server takeovers. It supports a wide array of SQL injection techniques, including boolean-based blind, time-based blind, error-based, UNION query-based, stacked queries, and out-of-band methods. Its primary function is to identify vulnerable parameters, extract data, and even access the underlying file system or execute commands on the compromised database server, making it the definitive choice for automated SQLi.

Why this answer

SQLMap is a dedicated, open-source penetration testing tool that automates the process of detecting and exploiting SQL injection flaws. It supports a wide range of database management systems (e.g., MySQL, Oracle, PostgreSQL) and injection techniques (e.g., boolean-based blind, time-based blind, UNION query, stacked queries), making it the standard choice for this specific task.

Exam trap

The trap here is that candidates often confuse Metasploit's broad exploitation capabilities with the specialized automation of SQL injection detection, leading them to choose Metasploit when SQLMap is the precise tool for this specific vulnerability class.

How to eliminate wrong answers

Option B (Metasploit) is wrong because it is a general exploitation framework used for developing and executing exploit code against a variety of vulnerabilities, not specifically designed or optimized for automating SQL injection detection and exploitation. Option C (Nmap) is wrong because it is a network scanning tool used for host discovery, port scanning, and service enumeration, lacking any built-in capability to detect or exploit SQL injection vulnerabilities. Option D (Burp Suite) is wrong because it is an intercepting proxy and web application security testing platform that requires manual configuration and extension (e.g., using SQLiPy or custom plugins) to perform automated SQL injection; it is not a dedicated automation tool for SQL injection like SQLMap.

137
MCQhard

A penetration tester runs `nmap -sI 192.168.1.10 -p 80 10.0.0.1` and receives output indicating port 80 is open. The scan uses a zombie host. Which type of scan is this?

A.Idle scan
B.SYN scan
C.Decoy scan
D.Fragmentation scan
AnswerA

The Nmap idle scan (-sI) is a highly stealthy technique that allows a penetration tester to scan a target without sending any packets directly from their own IP address. Instead, it leverages a "zombie" host's IP ID sequence to infer open ports on the target. By observing changes in the zombie's IP ID, the scanner can determine if the target responded to a forged packet sent from the zombie, thus achieving a truly blind scan, matching the 'si' flag in the command.

Why this answer

The `-sI` flag in Nmap specifies an idle scan, which uses a zombie host (192.168.1.10) to probe the target (10.0.0.1). By observing changes in the zombie's IP ID sequence, the attacker can infer whether a port on the target is open or closed without revealing their own IP address. The output indicating port 80 is open confirms the scan type as an idle scan.

Exam trap

The trap here is that candidates confuse the `-sI` flag with decoy scans (`-D`) because both involve spoofing, but idle scans uniquely require a zombie host and IP ID analysis, not just multiple decoy IPs.

How to eliminate wrong answers

Option B is wrong because a SYN scan uses the `-sS` flag, not `-sI`, and does not involve a zombie host; it sends raw SYN packets directly from the attacker's IP. Option C is wrong because a decoy scan uses the `-D` flag to spoof multiple source IPs to obscure the real scanner, but it does not rely on a zombie host's IP ID sequence. Option D is wrong because a fragmentation scan uses the `-f` flag to split packets into smaller fragments to evade detection, not to leverage a zombie host for stealth.

138
MCQeasy

What is the primary purpose of the 4-way handshake in WPA/WPA2-Personal?

A.To establish encryption keys without transmitting the pre-shared key
B.To authenticate the user with a username and password
C.To synchronize the beacon intervals between client and AP
D.To exchange digital certificates between client and AP
AnswerA

The WPA/WPA2 4-way handshake's primary function is to securely derive a unique set of session keys, including the Pairwise Transient Key (PTK) and Group Temporal Key (GTK), for encrypting subsequent data traffic. This critical process is achieved by exchanging nonces and cryptographic hashes, ensuring that the pre-shared key (PSK) itself is never transmitted over the air. By keeping the PSK off the network, the handshake effectively prevents eavesdroppers from capturing and compromising the long-term secret, thus establishing confidentiality and integrity for the wireless communication.

Why this answer

The 4-way handshake confirms that both client and AP possess the pre-shared key (PSK) without exposing it, and generates temporal keys for encryption.

139
MCQmedium

A web application uses a URL parameter to fetch a file from the server, e.g., 'download.php?file=report.pdf'. An attacker changes the parameter to '../../etc/passwd' and retrieves the password file. This attack is known as:

A.Command injection
B.Local File Inclusion (LFI)
C.Directory traversal
D.Server-Side Request Forgery (SSRF)
AnswerC

Directory traversal, also known as path traversal, is a web security vulnerability that allows an attacker to read arbitrary files on the server's file system. This is achieved by manipulating file paths in user-supplied input, typically using sequences like "../" (dot-dot-slash) to move up in the directory hierarchy, or its encoded forms. The objective is to access files and directories stored outside the intended web root directory, such as configuration files, source code, or system files like /etc/passwd.

Why this answer

Directory traversal (also known as path traversal) allows an attacker to access files outside the intended directory by using '../' sequences.

140
MCQhard

After gaining initial access, an attacker attempts to escalate privileges by exploiting a misconfigured service running as SYSTEM. They find that the service's binary path is writable by the Everyone group. Which privilege escalation technique is the attacker MOST likely using?

A.SUID/GUID abuse
B.Unquoted service path
C.Weak service permissions
D.Token impersonation
AnswerC

Weak service permissions refer to overly permissive access control lists (ACLs) on a Windows service executable or its configuration, allowing non-privileged users to modify or replace the service binary. If an attacker can write to the service executable, they can replace it with a malicious payload, which will then execute with the service's elevated privileges (often `SYSTEM`) upon the next service restart or system reboot. This direct manipulation of the service binary is a classic and highly effective privilege escalation method.

Why this answer

The attacker can exploit weak service permissions where the binary path of a service running as SYSTEM is writable by the Everyone group. This allows the attacker to replace the legitimate service binary with a malicious executable, and when the service restarts (or the system reboots), the malicious code executes with SYSTEM privileges, achieving privilege escalation.

Exam trap

The trap here is that candidates often confuse 'weak service permissions' (writable binary path) with 'unquoted service path' (missing quotes in the path), but the question explicitly states the binary path is writable, not that it contains spaces.

How to eliminate wrong answers

Option A is wrong because SUID/GUID abuse is a Linux/Unix privilege escalation technique involving setuid or setgid bits on executables, not applicable to Windows services. Option B is wrong because an unquoted service path exploits spaces in the service binary path to execute an attacker-controlled executable placed earlier in the path, not the writability of the binary itself. Option D is wrong because token impersonation involves stealing or duplicating access tokens (e.g., via SeImpersonatePrivilege) to assume another user's identity, not modifying a service binary.

141
Multi-Selectmedium

Which TWO of the following are common attack vectors against IoT devices? (Select TWO.)

Select 2 answers
A.Default credentials left unchanged
B.Regular firmware updates
C.Insecure protocols such as plaintext MQTT
D.Use of strong encryption protocols
E.Use of certificate-based authentication
AnswersA, C

Default credentials left unchanged represent a critical attack vector because many devices, especially in IoT, ship with easily guessable or publicly known usernames and passwords. Attackers can leverage automated scanning tools and credential stuffing techniques to gain unauthorized access, often leading to full device control, data exfiltration, or recruitment into botnets without requiring complex exploits.

Why this answer

IoT devices often have default credentials that are not changed, and they use insecure protocols like MQTT without encryption. These are common entry points for attackers.

142
MCQeasy

Which type of malware encrypts the victim's files and demands payment for the decryption key?

A.Keylogger
B.Spyware
C.Adware
D.Ransomware
AnswerD

Ransomware is a malicious software that encrypts a victim's files, rendering them inaccessible, and subsequently demands a payment, typically in cryptocurrency, in exchange for the decryption key. This type of attack is specifically designed for financial extortion by holding critical data hostage. It directly matches the description of encrypting files and demanding a ransom for their recovery.

Why this answer

Ransomware is the correct answer because it specifically encrypts the victim's files using symmetric encryption (e.g., AES) and then demands a ransom payment, typically in cryptocurrency, in exchange for the decryption key. Unlike other malware types, its primary purpose is data hostage for financial extortion, often leveraging asymmetric encryption (e.g., RSA) to secure the symmetric key.

Exam trap

The trap here is that candidates may confuse ransomware with scareware (which displays fake warnings but does not encrypt files) or mistakenly think spyware or adware could also demand payment, but only ransomware specifically encrypts data and demands a decryption key in return.

How to eliminate wrong answers

Option A is wrong because a keylogger is designed to capture keystrokes to steal credentials or sensitive data, not to encrypt files or demand payment. Option B is wrong because spyware covertly monitors user activity and collects information (e.g., browsing habits, login details) without encrypting files or issuing ransom demands. Option C is wrong because adware automatically displays or downloads advertisements, often for revenue generation, and lacks any file-encryption or extortion capabilities.

143
MCQmedium

A penetration tester is enumerating an SMTP server on port 25. They issue the command `VRFY root` and receive a 250 response, then `VRFY admin` also returns 250. What does this indicate about the SMTP server?

A.Both root and admin are valid email accounts on the server
B.The SMTP server supports password authentication
C.The command failed due to syntax errors
D.The SMTP server has disabled user verification
AnswerA

The 250 SMTP response code indicates that the requested mail action was successful and completed. In the context of the VRFY command, a 250 response explicitly confirms that the specified user, such as 'root' or 'admin', is a valid and existing account on the server. This direct confirmation of user existence is precisely what the penetration tester is looking for during enumeration, making both accounts valid.

Why this answer

The VRFY command is used to verify whether a specific user exists on an SMTP server. A 250 response code indicates that the user is valid and the mailbox exists. Since both `root` and `admin` returned 250, this confirms that both are valid email accounts on the server, making option A correct.

Exam trap

The trap here is that candidates may confuse the VRFY command's 250 success response with authentication success or syntax errors, when in fact it strictly indicates user existence on the server.

How to eliminate wrong answers

Option B is wrong because the VRFY command does not test password authentication; it only checks user existence, and SMTP authentication (AUTH) is a separate mechanism. Option C is wrong because a 250 response indicates success, not a syntax error; syntax errors would return a 5xx code (e.g., 500 or 501). Option D is wrong because a 250 response proves user verification is enabled and functional; if disabled, the server would return a 252 (cannot verify) or a 550 (no such user).

144
MCQmedium

During a penetration test, you discover a process named 'svch0st.exe' running on a Windows server with high CPU usage. The file is not digitally signed. Which type of malware is MOST likely present?

A.Polymorphic virus
B.Ransomware
C.Trojan
D.Worm
AnswerC

The process masquerades as a legitimate service (svchost.exe) to avoid detection, typical of a Trojan or backdoor.

Why this answer

The process name 'svch0st.exe' mimics the legitimate Windows service host 'svchost.exe' but uses a zero instead of 'o', a common masquerading technique. The lack of a digital signature and high CPU usage indicate malicious activity, and because it appears to be a standalone executable disguised as a system process, it fits the definition of a Trojan—malware that deceives users into running it by appearing legitimate.

Exam trap

The trap here is that candidates confuse a process name mimicking a legitimate service with a worm or virus, but the key differentiator is the lack of self-replication or code mutation, making it a Trojan that relies on user deception.

How to eliminate wrong answers

Option A is wrong because a polymorphic virus changes its code signature with each infection to evade detection, but the question describes a single suspicious process name and high CPU usage, not self-modifying code behavior. Option B is wrong because ransomware typically encrypts files or locks the system and demands payment, but no symptoms like file encryption, ransom notes, or system lockout are mentioned. Option D is wrong because a worm self-replicates across networks without user interaction, often exploiting vulnerabilities, whereas the scenario focuses on a single process on one server with no indication of network propagation.

145
MCQmedium

Which of the following tools is specifically designed to perform password cracking using rainbow tables?

A.John the Ripper
B.Ophcrack
C.RainbowCrack
D.Hashcat
AnswerC

RainbowCrack is a pioneering and highly specialized tool explicitly engineered for password cracking using the rainbow table technique. It not only performs efficient lookups against pre-computed rainbow tables but also includes functionalities for generating these large, memory-intensive tables for various hash algorithms. Its design is centered around the mathematical principles of rainbow tables, making it the definitive tool for both their creation and utilization in password recovery.

Why this answer

RainbowCrack is specifically designed to perform password cracking using precomputed rainbow tables, which are time-memory trade-off structures that allow hashes to be reversed quickly without brute-forcing each password. Unlike other tools that rely on brute force, dictionary attacks, or hybrid methods, RainbowCrack's core functionality is built around generating and using rainbow tables to crack LM, NTLM, MD5, SHA1, and other hash types.

Exam trap

EC-Council often tests the distinction between tools that use rainbow tables (RainbowCrack) versus tools that use brute-force or dictionary methods (John the Ripper, Hashcat), and candidates mistakenly associate Ophcrack with general rainbow table cracking when it is actually limited to LM hashes.

How to eliminate wrong answers

Option A is wrong because John the Ripper is a password cracking tool that primarily uses dictionary, brute-force, and incremental modes, not rainbow tables. Option B is wrong because Ophcrack is a Windows password cracker that uses LM hash rainbow tables but is limited to Windows LAN Manager hashes and is not a general-purpose rainbow table tool like RainbowCrack. Option D is wrong because Hashcat is a high-speed password cracker that uses GPU acceleration for brute-force, dictionary, and rule-based attacks, but it does not natively use rainbow tables for cracking.

146
Multi-Selectmedium

Which TWO of the following attacks can be prevented by properly validating and sanitizing user input? (Select 2)

Select 2 answers
A.Cross-Site Request Forgery (CSRF)
B.SQL injection
C.Man-in-the-Middle (MitM) attack
D.Clickjacking
E.Cross-Site Scripting (XSS)
AnswersB, E

SQL injection attacks occur when an attacker inserts malicious SQL code into user input fields, which is then executed by the database. Proper input validation and sanitization are highly effective against this threat. Techniques such as using parameterized queries (prepared statements) or escaping special characters ensure that user-supplied data is treated strictly as data literals, preventing it from being interpreted as executable SQL commands and neutralizing the injection attempt.

Why this answer

SQL injection and XSS are both injection attacks that can be prevented by input validation and sanitization. CSRF requires tokens, and clickjacking requires frame-busting headers.

147
MCQeasy

Which of the following commands is used to enumerate SNMP information from a network device using a specific community string?

A.ldapsearch -x -h 192.168.1.1 -b dc=domain,dc=com
B.enum4linux -a 192.168.1.1
C.snmpwalk -c public -v 2c 192.168.1.1
D.nbtstat -a 192.168.1.1
AnswerC

snmpwalk is the correct and standard command-line utility for enumerating SNMP (Simple Network Management Protocol) information from a target device. The `-c public` option specifies the community string, which acts as a password for accessing SNMP data, with "public" being a common default. The `-v 2c` flag designates SNMP version 2c, a widely supported version, while `192.168.1.1` is the target IP address from which to retrieve Management Information Base (MIB) data. This command systematically queries the entire MIB tree, revealing device configurations, network statistics, and system details.

Why this answer

`snmpwalk` is the standard command-line tool for enumerating SNMP (Simple Network Management Protocol) information from a network device. By specifying the community string (`-c public`) and SNMP version (`-v 2c`), it retrieves the entire Management Information Base (MIB) tree from the target IP address, allowing an attacker to discover system details, running processes, and network interfaces.

Exam trap

The trap here is that candidates confuse SNMP enumeration tools with other network enumeration tools (like LDAP, SMB, or NetBIOS), leading them to pick a command that targets a different protocol entirely.

How to eliminate wrong answers

Option A is wrong because `ldapsearch` is used for querying LDAP directories (port 389), not for SNMP enumeration; it requires a base DN and is unrelated to community strings. Option B is wrong because `enum4linux` is a tool for enumerating SMB/CIFS shares, users, and policies from Windows systems (port 445), not for SNMP queries. Option D is wrong because `nbtstat` is a Windows utility for NetBIOS over TCP/IP name resolution and cache management (port 137), not for SNMP enumeration.

148
MCQmedium

A penetration tester wants to crack Windows NTLM hashes using rainbow tables. Which tool is specifically designed for this purpose?

A.Hashcat
B.John the Ripper
C.RainbowCrack
D.Ophcrack
AnswerC

RainbowCrack is a specialized tool explicitly designed to utilize precomputed rainbow tables for efficient password recovery. It leverages the space-time trade-off inherent in rainbow table methodology, where extensive precomputation allows for exceptionally fast lookup during the cracking phase. By using these tables, RainbowCrack can quickly reverse NTLM hashes into their original plaintexts, provided the plaintext exists within the table's precomputed range, making it the most direct and effective tool for this specific task.

Why this answer

RainbowCrack is specifically designed for cracking hashes using precomputed rainbow tables. It works by performing a lookup in a rainbow table to reverse a hash into its plaintext, which is the core mechanism described in the question. While Hashcat and John the Ripper can also crack NTLM hashes, they are not specifically designed for rainbow table attacks; they primarily use brute-force or dictionary-based methods.

Exam trap

In the CEH exam, candidates often confuse Ophcrack (which is also a rainbow table tool but limited to LM/NTLM on older Windows) with RainbowCrack (the general-purpose tool), or they assume Hashcat or John the Ripper are the primary tools for rainbow table attacks when they are not specifically designed for that purpose.

How to eliminate wrong answers

Option A is wrong because Hashcat is a GPU-accelerated password recovery tool that uses brute-force, dictionary, and rule-based attacks, not rainbow tables. Option B is wrong because John the Ripper is a password cracking tool that primarily uses dictionary and brute-force attacks, and while it can use rainbow tables via external plugins, it is not specifically designed for that purpose. Option D is wrong because Ophcrack is a tool specifically designed for cracking LM and NTLM hashes using rainbow tables, but it is limited to Windows XP/Vista/7 and is not the general-purpose rainbow table tool for NTLM hashes; the question asks for a tool specifically designed for this purpose, and RainbowCrack is the correct answer.

149
MCQmedium

Which of the following commands would a tester use to enumerate NetBIOS names and their associated IP addresses on a local subnet?

A.nbtstat -n
B.nbtstat -c
C.nbtstat -a 192.168.1.10
D.nbtstat -A 192.168.1.10
AnswerD

The "nbtstat -A" command is specifically designed to perform a remote NetBIOS name table query against a target identified by its *IP address*. This command sends a NetBIOS Adapter Status Request to the specified IP, retrieving the NetBIOS names registered by that host, including workgroup/domain membership and services. This direct query capability makes "nbtstat -A 192.168.1.10" the correct choice for enumerating NetBIOS information from a remote system. It directly fulfills the requirement to enumerate a remote host using its IP.

Why this answer

The `nbtstat -A` command (with a capital 'A') performs a NetBIOS name table lookup against a remote IP address, listing the NetBIOS names registered by that host along with their associated IP addresses. This is the standard method for enumerating NetBIOS names on a specific target within a local subnet, as it queries the NetBIOS name service (UDP port 137) directly.

Exam trap

The trap here is that candidates often confuse the lowercase `-a` (which expects a hostname) with the uppercase `-A` (which expects an IP address), leading them to incorrectly select option C when the question specifies an IP address.

How to eliminate wrong answers

Option A is wrong because `nbtstat -n` displays only the local NetBIOS names registered on the tester's own machine, not names from other hosts on the subnet. Option B is wrong because `nbtstat -c` shows the local NetBIOS name cache, which contains recently resolved names and their IP addresses, but does not actively enumerate all hosts on the subnet. Option C is wrong because `nbtstat -a` (lowercase 'a') performs a NetBIOS name table lookup using a hostname, not an IP address, so it would fail or produce incorrect results when given an IP address.

150
MCQmedium

A security analyst observes repeated de-authentication packets targeting clients on a corporate Wi-Fi network. What is the MOST likely goal of the attacker?

A.To perform a denial-of-service attack and disrupt all wireless connectivity
B.To capture the WPA2 4-way handshake for offline password cracking
C.To install malware on the client devices
D.To exploit a vulnerability in the RADIUS server
AnswerB

The primary purpose of repeatedly sending deauthentication packets in a WPA2 environment is to force connected clients to disconnect from the access point and then reinitiate the authentication process. This forced reconnection allows an attacker, who is passively monitoring the wireless traffic with tools like airodump-ng, to capture the WPA2 4-way handshake. This handshake contains cryptographic material that can then be used for offline brute-force or dictionary attacks to recover the Pre-Shared Key (PSK).

Why this answer

De-authentication attacks force clients to reconnect, allowing the attacker to capture the 4-way handshake for offline cracking of the PSK.

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