Oct 12, 2025 Reliable Study Materials for CNSP Exam Success For Sure
100% Latest Most updated CNSP Questions and Answers
NEW QUESTION # 23
According to the screenshot below, which of the following statements are correct?
- A. The application is running on port 443 and the HTTPS protocol.
- B. The application is running on port 80 and the HTTP protocol.
- C. The credentials have been submitted over the HTTP protocol.
- D. The credentials have been submitted over the HTTPS protocol.
Answer: A
Explanation:
The screenshot is from Wireshark, a network protocol analyzer, displaying captured network traffic. The relevant columns include the source and destination IP addresses, ports, protocol, and additional information about the packets. Let's break down the details:
Destination Port Analysis: The screenshot shows multiple packets with a destination port of 443 (e.g., in the "Destination" column, entries like "172.72.61.9:443"). Port 443 is the default port for HTTPS (HTTP Secure), which is HTTP traffic encrypted using SSL/TLS. This indicates that the application is communicating over HTTPS.
Protocol Analysis: The "Protocol" column lists "TLSv1.2" for most packets (e.g., frame numbers 2000084, 2000086). TLS (Transport Layer Security) is the cryptographic protocol used by HTTPS to secure HTTP communications. This confirms that the traffic is HTTPS, not plain HTTP.
Packet Details: The "Info" column provides additional context, such as "Application Data" for TLS packets, indicating encrypted application-layer data (typical of HTTPS). There are also HTTP packets (e.g., frame 2000088), but these are likely part of the HTTPS session (e.g., HTTP/2 over TLS, as noted by "HTTP2").
Now, let's evaluate the options:
Option A: "The application is running on port 443 and the HTTPS protocol." This is correct. The destination port 443 and the use of TLSv1.2 confirm that the application is using HTTPS. HTTPS is the standard protocol for secure web communication, and port 443 is its designated port. CNSP documentation emphasizes that HTTPS traffic on port 443 indicates a secure application-layer protocol, often used for web applications handling sensitive data.
Option B: "The credentials have been submitted over the HTTP protocol." This is incorrect. HTTP typically uses port 80, but the screenshot shows traffic on port 443 with TLS, indicating HTTPS. Credentials submitted over this connection would be encrypted via HTTPS, not sent in plaintext over HTTP. CNSP highlights the security risks of HTTP for credential submission due to lack of encryption, which isn't the case here.
Option C: "The credentials have been submitted over the HTTPS protocol." While this statement could be true (since HTTPS is in use, any credentials would likely be submitted securely), the question asks for the "correct" statement based on the screenshot. The screenshot doesn't explicitly show credential submission (e.g., a POST request with form data); it only shows the protocol and port. Option A is more directly supported by the screenshot as it focuses on the application's protocol and port, not the specific action of credential submission. CNSP notes that HTTPS ensures confidentiality, but this option requires more specific evidence of credentials.
Option D: "The application is running on port 80 and the HTTP protocol." This is incorrect. Port 80 is the default for HTTP, but the screenshot clearly shows port 443 and TLS, indicating HTTPS. CNSP documentation contrasts HTTP (port 80, unencrypted) with HTTPS (port 443, encrypted), making this option invalid.
Conclusion: Option A is the most accurate and comprehensive statement directly supported by the screenshot, confirming the application's use of port 443 and HTTPS. While Option C might be true in a broader context, it's less definitive without explicit evidence of credential submission in the captured packets.
NEW QUESTION # 24
Which is the correct command to change the MAC address for an Ethernet adapter in a Unix-based system?
- A. ifconfig eth0 hdwr ether AA:BB:CC:DD:EE:FF
- B. ifconfig eth0 hw ether AA:BB:CC:DD:EE:FF
- C. ifconfig eth0 hdw ether AA:BB:CC:DD:EE:FF
- D. ifconfig eth0 hwr ether AA:BB:CC:DD:EE:FF
Answer: B
Explanation:
In Unix-based systems (e.g., Linux), the ifconfig command is historically used to configure network interfaces, including changing the Media Access Control (MAC) address of an Ethernet adapter. The correct syntax to set a new MAC address for an interface like eth0 is ifconfig eth0 hw ether AA:BB:CC:DD:EE:FF, where hw specifies the hardware address type (ether for Ethernet), followed by the new MAC address in colon-separated hexadecimal format.
Why A is correct: The hw ether argument is the standard and correct syntax recognized by ifconfig to modify the MAC address. This command temporarily changes the MAC address until the system reboots or the interface is reset, assuming the user has sufficient privileges (e.g., root). CNSP documentation on network configuration and spoofing techniques validates this syntax for testing network security controls.
Why other options are incorrect:
B: hdw is not a valid argument; it's a typographical error and unrecognized by ifconfig.
C: hdwr is similarly invalid; no such shorthand exists in the command structure.
D: hwr is incorrect; the full keyword hw followed by ether is required for proper parsing.
NEW QUESTION # 25
Which of the following techniques can be used to bypass network segmentation during infrastructure penetration testing?
- A. All of the above
- B. VLAN hopping
- C. DNS tunneling
- D. Covert channels
Answer: A
Explanation:
Network segmentation isolates network zones for security, but certain techniques can circumvent these controls, a focus of CNSP penetration testing.
Why D is correct:
A: DNS tunneling encodes data in DNS queries, bypassing segmentation via legitimate DNS traffic.
B: VLAN hopping exploits switch misconfigurations (e.g., double tagging) to access other VLANs.
C: Covert channels use hidden communication paths (e.g., timing channels) to evade segmentation.
All are valid techniques per CNSP for testing segmentation controls.
Why other options are incomplete: A, B, or C alone exclude other viable methods, making D the comprehensive answer.
NEW QUESTION # 26
How many octets are there in an IPv6 address?
- A. 0
- B. 1
- C. 2
- D. 3
Answer: B
Explanation:
An IPv6 address, defined in RFC 4291, is a 128-bit address designed to replace IPv4's 32-bit scheme, vastly expanding address space (2^128 vs. 2^32). An octet is 8 bits (1 byte). To calculate octets in IPv6:
128 bits ÷ 8 bits/octet = 16 octets.
Representation:
IPv6 is written as eight 16-bit hexadecimal blocks (e.g., 2001:0db8:85a3:0000:0000:8a2e:0370:7334), separated by colons.
Each block is 16 bits (2 bytes), so 8 blocks = 16 octets.
Contrast with IPv4 (e.g., 192.168.0.1), which has 4 octets (32 bits).
Technical Note: Your original input flagged this question's phrasing as potentially misleading, suggesting "octets" is an IPv4 term, while IPv6 uses "16-bit groups" or "hextets." While technically accurate (RFC 4291 uses "16-bit blocks"), "octets" remains a common, if informal, term in security contexts for byte-wise analysis (e.g., packet crafting). CNSP might use "octets" to test byte-level understanding, though "groups" is more precise for IPv6. Here, 16 octets (128 bits) is correct either way.
Security Implications: IPv6's larger address space complicates scanning (e.g., Nmap struggles with 2^128 possibilities) but introduces risks like misconfigured Neighbor Discovery Protocol (NDP). Understanding its structure aids in firewall rules and IDS signatures.
Why other options are incorrect:
B . 32: Implies 256 bits (32 × 8), far exceeding IPv6's 128-bit design.
C . 64: Suggests 512 bits (64 × 8), unrelated to IPv6 or any IP standard.
D . 128: Misinterprets octets as bits; 128 bits = 16 octets, not 128 octets.
Real-World Context: IPv6 packet analysis (e.g., Wireshark) breaks addresses into 16 octets for raw data inspection.
NEW QUESTION # 27
Which of the following is a valid DNS record type?
- A. All of the above
- B. SRV record
- C. NAPTR record
- D. TXT record
Answer: A
Explanation:
DNS (Domain Name System) records define how domain names are mapped to various types of data, each serving a specific purpose in network operations. The question asks for valid DNS record types, and all listed options are recognized.
Why D is correct:
A . NAPTR record: The Naming Authority Pointer (NAPTR) record is used for service discovery and mapping domain names to services, protocols, and ports (e.g., in SIP or ENUM systems).
B . SRV record: The Service (SRV) record specifies the hostname and port for specific services (e.g., LDAP, XMPP), aiding in service location.
C . TXT record: The Text (TXT) record stores arbitrary text data, often for SPF, DKIM, or domain verification.
All are valid DNS record types per RFC standards and CNSP documentation, making "All of the above" the correct answer.
Why other options are incomplete: A, B, or C alone exclude other valid types listed, so D is the most comprehensive response.
NEW QUESTION # 28
The Active Directory database file stores the data and schema information for the Active Directory database on domain controllers in Microsoft Windows operating systems. Which of the following file is the Active Directory database file?
- A. NTDS.MDB
- B. MSAD.MDB
- C. NTDS.DAT
- D. NTDS.DIT
Answer: D
Explanation:
The Active Directory (AD) database on Windows domain controllers contains critical directory information, stored in a specific file format.
Why D is correct: The NTDS.DIT file (NT Directory Services Directory Information Tree) is the Active Directory database file, located in C:\Windows\NTDS\ on domain controllers. It stores all AD objects (users, groups, computers) and schema data in a hierarchical structure. CNSP identifies NTDS.DIT as the key file for AD data extraction in security audits.
Why other options are incorrect:
A . NTDS.DAT: Not a valid AD database file; may be a confusion with other system files.
B . NTDS.MDB: Refers to an older Microsoft Access database format, not used for AD.
C . MSAD.MDB: Not a recognized file for AD; likely a misnomer.
NEW QUESTION # 29
If a hash begins with $2a$, what hashing algorithm has been used?
- A. SHA512
- B. SHA256
- C. MD5
- D. Blowfish
Answer: D
Explanation:
The prefix $2a$ identifies the bcrypt hashing algorithm, which is based on the Blowfish symmetric encryption cipher (developed by Bruce Schneier). Bcrypt is purpose-built for password hashing, incorporating:
Salt: A random string (e.g., 22 Base64 characters) to thwart rainbow table attacks.
Work Factor: A cost parameter (e.g., $2a$10$ means 2^10 iterations), making it computationally expensive to brute-force.
Format: $2a$[cost]$[salt][hash]
Example: $2a$10$N9qo8uLOickgx2ZMRZoMyeIjZAgcfl7p92ldGxad68LJZdL17lhWy
$2a$: Bcrypt variant (original is $2$; $2a$ fixes a minor bug).
$10$: 1024 iterations.
Next 22 characters: Salt.
Remaining: Hashed password.
Used in /etc/shadow on Linux, bcrypt's adaptive nature ensures it remains secure as hardware improves. CNSP likely includes it in cryptography modules for its strength over older algorithms like MD5.
Why other options are incorrect:
B . SHA256: Part of the SHA-2 family, outputs a 64-character hexadecimal string (e.g., e3b0c442...), no $ prefix. It's faster, less suited for passwords.
C . MD5: Produces a 32-character hex string (e.g., d41d8cd9...), no prefix. It's cryptographically broken (collisions found).
D . SHA512: SHA-2 variant, 128-character hex (e.g., cf83e135...), no $ prefix, not salted by default.
Real-World Context: Bcrypt protects SSH keys and web app passwords (e.g., in PHP's password_hash()).
NEW QUESTION # 30
Which Kerberos ticket is required to generate a Silver Ticket?
- A. Ticket-Granting Ticket
- B. There is no specific ticket required for generating a Silver Ticket
- C. Session Ticket
- D. Service Account Ticket
Answer: D
Explanation:
A Silver Ticket is a forged Kerberos Service Ticket (TGS - Ticket Granting Service) in Active Directory, granting access to a specific service (e.g., MSSQL, CIFS) without KDC interaction. Unlike a Golden Ticket (TGT forgery), it requires:
Service Account's NTLM Hash: The target service's account (e.g., MSSQLSvc) hash, not a ticket.
Forgery: Tools like Mimikatz craft the TGS (e.g., kerberos::golden /service:<spn> /user:<user> /ntlm:<hash>).
Kerberos Flow (RFC 4120):
TGT (Ticket-Granting Ticket): Obtained via AS (Authentication Service) with user creds.
TGS: Requested from TGS (Ticket Granting Service) using TGT for service access.
Silver Ticket Process:
No TGT needed; the attacker mimics the TGS step using the service account's stolen hash (e.g., from a compromised host).
C . Service Account Ticket: Misnomer-it's the hash of the service account (e.g., MSSQLSvc) that enables forgery, not a pre-existing ticket. CNSP's phrasing likely tests this nuance.
Security Implications: Silver Tickets are stealthier than Golden Tickets (service-specific, shorter-lived). CNSP likely stresses hash protection (e.g., LAPS) and Kerberos monitoring.
Why other options are incorrect:
A . Session Ticket: Not a Kerberos term; confuses session keys.
B . TGT: Used for Golden Tickets, not Silver.
D: Incorrect; the service account's hash (implied by "ticket") is essential.
Real-World Context: Silver Tickets exploited in APT29 attacks (2020 SolarWinds) for lateral movement.
NEW QUESTION # 31
On a Microsoft Windows Operating System, what does the following command do?
net localgroup administrators
- A. List domain admin users for the current domain
- B. Displays the local administrators group on the computer
Answer: B
Explanation:
The net command in Windows is a legacy tool for managing users, groups, and network resources. The subcommand net localgroup <groupname> displays information about a specified local group on the machine where it's run. Specifically:
net localgroup administrators lists all members (users and groups) of the local Administrators group on the current computer.
The local Administrators group grants elevated privileges (e.g., installing software, modifying system files) on that machine only, not domain-wide.
Output Example:
Alias name administrators
Comment Administrators have complete and unrestricted access to the computer Members
------------------------------------------------------------------------------- Administrator Domain Admins The command completed successfully.
Technical Details:
Local groups are stored in the Security Accounts Manager (SAM) database (e.g., C:\Windows\System32\config\SAM).
This differs from domain groups (e.g., Domain Admins), managed via Active Directory.
Security Implications: Enumerating local admins is a reconnaissance step in penetration testing (e.g., to escalate privileges). CNSP likely covers this command for auditing and securing Windows systems.
Why other options are incorrect:
A . List domain admin users for the current domain: This requires net group "Domain Admins" /domain, which queries the domain controller, not the local SAM. net localgroup is strictly local.
Real-World Context: Attackers use this command post-compromise (e.g., via PsExec) to identify privilege escalation targets.
NEW QUESTION # 32
What user account is required to create a Golden Ticket in Active Directory?
- A. Domain User account
- B. Service account
- C. Local User account
- D. KRBTGT account
Answer: D
Explanation:
A Golden Ticket is a forged Kerberos Ticket-Granting Ticket (TGT) in Active Directory (AD), granting an attacker unrestricted access to domain resources by impersonating any user (e.g., with Domain Admin privileges). Kerberos, per RFC 4120, relies on the KRBTGT account-a built-in service account on every domain controller-to encrypt and sign TGTs. To forge a Golden Ticket, an attacker needs:
The KRBTGT password hash (NTLM or Kerberos key), typically extracted from a domain controller's memory using tools like Mimikatz.
Additional domain details (e.g., SID, domain name).
Process:
Compromise a domain controller (e.g., via privilege escalation).
Extract the KRBTGT hash (e.g., lsadump::dcsync /user:krbtgt).
Forge a TGT with arbitrary privileges using the hash (e.g., Mimikatz's kerberos::golden command).
The KRBTGT account itself isn't "used" to create the ticket; its hash is the key ingredient. Unlike legitimate TGTs issued by the KDC, a Golden Ticket bypasses authentication checks, persisting until the KRBTGT password is reset (a rare event in most environments). CNSP likely highlights this as a high-severity AD attack vector.
Why other options are incorrect:
A . Local User account: Local accounts are machine-specific, lack domain privileges, and can't access the KRBTGT hash stored on domain controllers.
B . Domain User account: A standard user has no inherent access to domain controller credentials or the KRBTGT hash without escalation.
C . Service account: While service accounts may have elevated privileges, they don't automatically provide the KRBTGT hash unless compromised to domain admin level-still insufficient without targeting KRBTGT specifically.
Real-World Context: The 2014 Sony Pictures hack leveraged Golden Tickets, emphasizing the need for KRBTGT hash rotation post-breach (a complex remediation step).
NEW QUESTION # 33
How many usable TCP/UDP ports are there?
- A. 0
- B. 1
- C. 2
- D. 3
Answer: A
Explanation:
TCP (Transmission Control Protocol) and UDP (User Datagram Protocol) port numbers are defined by a 16-bit field in their packet headers, as specified in RFC 793 (TCP) and RFC 768 (UDP). A 16-bit integer ranges from 0 to 65,535, yielding a total of 65,536 possible ports (2^16). However, port 0 is universally reserved across both protocols and is not considered "usable" for standard network communication. According to the Internet Assigned Numbers Authority (IANA), port 0 is designated for special purposes, such as indicating an invalid or dynamic port assignment in some systems (e.g., when a client requests an ephemeral port). In practice, operating systems and applications avoid binding to port 0 for listening services, and it's often used in error conditions or as a placeholder in protocol implementations (e.g., socket programming).
Thus, the usable port range spans from 1 to 65,535, totaling 65,535 ports. These ports are categorized by IANA into:
Well-Known Ports (0-1023): Reserved for system services (e.g., HTTP on 80/TCP). Note that 0 is still reserved within this range.
Registered Ports (1024-49151): Assigned to user applications.
Dynamic/Ephemeral Ports (49152-65535): Used temporarily by clients.
From a security perspective, understanding the usable port count is critical for firewall configuration, port scanning (e.g., with Nmap), and detecting anomalies (e.g., services binding to unexpected ports). Misconfiguring a system to use port 0 could lead to protocol errors or expose vulnerabilities, though it's rare. The CNSP curriculum likely emphasizes this distinction to ensure practitioners can accurately scope network security assessments.
Why other options are incorrect:
A . 65536: This reflects the total number of possible ports (0-65535), but it includes the reserved port 0, which isn't usable for typical TCP/UDP communication. In security contexts, including port 0 in a count could lead to misconfigured rules or scanning errors.
C . 63535: This is an arbitrary number with no basis in the 16-bit port structure. It might stem from a typo or misunderstanding (e.g., subtracting 2000 from 65535 incorrectly), but it's invalid.
D . 65335: Similarly, this lacks grounding in protocol standards. It could be a miscalculation (e.g., subtracting 200 from 65535), but it doesn't align with TCP/UDP specifications.
Real-World Context: In penetration testing, tools like Nmap scan ports 1-65535 by default, excluding 0 unless explicitly specified (e.g., -p0-65535), reinforcing that 65,535 is the practical usable count.
NEW QUESTION # 34
What ports can be queried to perform a DNS zone transfer?
- A. 53/TCP
- B. 53/UDP
- C. Both 1 and 2
- D. None of the above
Answer: A
Explanation:
A DNS zone transfer involves replicating the DNS zone data (e.g., all records for a domain) from a primary to a secondary DNS server, requiring a reliable transport mechanism.
Why A is correct: DNS zone transfers use TCP port 53 because TCP ensures reliable, ordered delivery of data, which is critical for transferring large zone files. CNSP notes that TCP is the standard protocol for zone transfers (e.g., AXFR requests), as specified in RFC 5936.
Why other options are incorrect:
B . 53/UDP: UDP port 53 is used for standard DNS queries and responses due to its speed and lower overhead, but it is not suitable for zone transfers, which require reliability over speed.
C . Both 1 and 2: This is incorrect because zone transfers are exclusively TCP-based, not UDP-based.
D . None of the above: Incorrect, as 53/TCP is the correct port for DNS zone transfers.
NEW QUESTION # 35
Which SMB (Server Message Block) network protocol version introduced support for encrypting SMB traffic?
- A. SMBv3
- B. SMBv1
- C. None of the above
- D. SMBv2
Answer: A
Explanation:
The SMB protocol, used for file and printer sharing, has evolved across versions, with significant security enhancements in later iterations.
Why C is correct: SMBv3, introduced with Windows 8 and Server 2012, added native support for encrypting SMB traffic. This feature uses AES-CCM encryption to protect data in transit, addressing vulnerabilities in earlier versions. CNSP notes SMBv3's encryption as a critical security improvement.
Why other options are incorrect:
A . SMBv1: Lacks encryption support and is considered insecure, often disabled due to vulnerabilities like WannaCry exploitation.
B . SMBv2: Introduces performance improvements but does not support encryption natively.
D . None of the above: Incorrect, as SMBv3 is the version that introduced encryption.
NEW QUESTION # 36
If you find the 111/TCP port open on a Unix system, what is the next logical step to take?
- A. None of the above.
- B. Run "rpcinfo -p <hostname>" to enumerate the RPC services.
- C. Telnet to the port, send "GET / HTTP/1.0" and gather information from the response.
- D. Telnet to the port to look for a banner.
Answer: B
Explanation:
Port 111/TCP is the default port for the RPC (Remote Procedure Call) portmapper service on Unix systems, which registers and manages RPC services.
Why A is correct: Running rpcinfo -p <hostname> queries the portmapper to list all registered RPC services, their programs, versions, and associated ports. This is a logical next step during a security audit or penetration test to identify potential vulnerabilities (e.g., NFS or NIS services). CNSP recommends this command for RPC enumeration.
Why other options are incorrect:
B . Telnet to the port to look for a banner: Telnet might connect, but RPC services don't typically provide a human-readable banner, making this less effective than rpcinfo.
C . Telnet to the port, send "GET / HTTP/1.0" and gather information from the response: Port 111 is not an HTTP service, so an HTTP request is irrelevant and will likely fail.
D . None of the above: Incorrect, as A is a valid and recommended step.
NEW QUESTION # 37
WannaCry, an attack, spread throughout the world in May 2017 using machines running on outdated Microsoft operating systems. What is WannaCry?
- A. Ransomware
- B. Malware
Answer: A
Explanation:
WannaCry is a ransomware attack that erupted in May 2017, infecting over 200,000 systems across 150 countries. It exploited the EternalBlue vulnerability (MS17-010) in Microsoft Windows SMBv1, targeting unpatched systems (e.g., Windows XP, Server 2003). Developed by the NSA and leaked by the Shadow Brokers, EternalBlue allowed remote code execution.
Ransomware Mechanics:
Encryption: WannaCry used RSA-2048 and AES-128 to encrypt files, appending extensions like .wcry.
Ransom Demand: Displayed a message demanding $300-$600 in Bitcoin, leveraging a hardcoded wallet.
Worm Propagation: Self-replicated via SMB, scanning internal and external networks, unlike typical ransomware requiring user interaction (e.g., phishing).
Malware Context: While WannaCry is malware (malicious software), "ransomware" is the precise subcategory, distinguishing it from viruses, trojans, or spyware. Malware is a broad term encompassing any harmful code; ransomware specifically encrypts data for extortion. CNSP likely classifies WannaCry as ransomware to focus on its payload and mitigation (e.g., patching, backups).
Why other options are incorrect:
B . Malware: Correct but overly generic. WannaCry's defining trait is ransomware behavior, not just maliciousness. Specificity matters in security taxonomy for threat response (e.g., NIST IR 8019).
Real-World Context: WannaCry crippled NHS hospitals, highlighting patch management's criticality. A kill switch (a domain sinkhole) halted it, but variants persist.
NEW QUESTION # 38
What is the response from a closed TCP port which is not behind a firewall?
- A. A RST and an ACK packet
- B. A SYN and an ACK packet
- C. A FIN and an ACK packet
- D. ICMP message showing Port Unreachable
Answer: A
Explanation:
TCP uses a structured handshake, and its response to a connection attempt on a closed port follows a specific protocol when unobstructed by a firewall.
Why C is correct: A closed TCP port responds with a RST (Reset) and ACK (Acknowledgment) packet to terminate the connection attempt immediately. CNSP highlights this as a key scanning indicator.
Why other options are incorrect:
A: ICMP Port Unreachable is for UDP, not TCP.
B: FIN/ACK is for closing active connections, not rejecting new ones.
D: SYN/ACK indicates an open port during the TCP handshake.
NEW QUESTION # 39
......
New The SecOps Group CNSP Dumps & Questions: https://www.passsureexam.com/CNSP-pass4sure-exam-dumps.html
Try with 100% Real Exam Questions and Answers: https://drive.google.com/open?id=1nqmdi_jN4rN8dDfmWTa_jevaEXGfe-Oa