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The dumps for CNSP exam was last updated on Dec 12,2025 .

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Question#1

Which of the following services do not encrypt its traffic by default?

A. DNS

B. SSH

C. FTPS

D. All of these

Explanation:
Encryption ensures confidentiality and integrity of network traffic. Analyzing defaults:
A. DNS (Domain Name System):
Default: Unencrypted (UDP/TCP 53), per RFC 1035. Queries/responses (e.g., “google.com → 142.250.190.14”) are plaintext.
Modern Options: DNS over HTTPS (DoH, TCP 443) or DNS over TLS (DoT, TCP 853) encrypt, but aren’t default in most systems (e.g., pre-2020 Windows). B. SSH (Secure Shell):
Default: Encrypted (TCP 22), per RFC 4251. Uses asymmetric (e.g., RSA) and symmetric (e.g., AES)
crypto for all sessions.
C. FTPS (FTP Secure):
Default: Encrypted (TCP 21 control, dynamic data ports). Extends FTP with SSL/TLS (e.g., RFC 4217),
securing file transfers.
Technical Details:
DNS: Plaintext exposes queries to eavesdropping (e.g., ISP snooping) or spoofing (e.g., cache poisoning).
SSH/FTPS: Encryption is baked into their standards; disabling it requires explicit misconfiguration.
Security Implications: Unencrypted DNS risks privacy and integrity (e.g., Kaminsky attack). CNSP likely pushes DoH/DoT adoption.
Why other options are incorrect:
B, C: Encrypt by default.
D: False, as only DNS lacks default encryption.
Real-World Context: The 2013 Snowden leaks exposed DNS monitoring; DoH uptake (e.g., Cloudflare 1.1.1.1) counters this.
Reference: CNSP Official Study Guide (Protocol Security); RFC 1035 (DNS), RFC 4251 (SSH).

Question#2

How many octets are there in an IPv6 address?

A. 16

B. 32

C. 64

D. 128

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.
Reference: CNSP Official Documentation (IPv6 Networking); RFC 4291 (IP Version 6 Addressing Architecture).

Question#3

Which of the following represents a valid Windows Registry key?

A. HKEY_LOCAL_MACHINE

B. HKEY_INTERNAL_CONFIG

C. HKEY_ROOT_CLASSES

D. HKEY_LOCAL_USER

Explanation:
The Windows Registry is a hierarchical database storing system and application settings, organized into predefined root keys (hives). Only specific names are valid as top-level keys.
Why A is correct: HKEY_LOCAL_MACHINE (HKLM) is a standard root key containing hardware and system-wide configuration data. CNSP references it for security settings analysis (e.g., auditing policies).
Why other options are incorrect:
B: HKEY_INTERNAL_CONFIG is not a valid key; no such hive exists.
C: HKEY_ROOT_CLASSES is a misspelling; the correct key is HKEY_CLASSES_ROOT (HKCR).
D: HKEY_LOCAL_USER is incorrect; the valid key is HKEY_CURRENT_USER (HKCU).
Reference: CNSP "Windows Registry Security" (Section on Registry Structure) lists HKEY_LOCAL_MACHINE as a valid hive, detailing its role in system configuration.

Question#4

Which of the following is an example of a SUID program?

A. /bin/ls

B. /usr/bin/curl

C. /usr/bin/passwd

D. None of the above

Explanation:
In Linux/Unix, the SUID (Set User ID) bit allows a program to execute with the owner’s permissions, typically root, rather than the caller’s. It’s denoted by an s in the user execute field (e.g., -rwsr-xr-x). Common SUID programs perform privileged tasks requiring temporary elevation.
Analysis: C. /usr/bin/passwd:
Purpose: Updates user passwords in /etc/shadow (root-owned, 0600 perms).
Permissions: Typically -rwsr-xr-x, owned by root. The SUID bit lets non-root users modify shadow securely.
Command: ls -l /usr/bin/passwd confirms SUID (s in user execute).
A. /bin/ls:
Purpose: Lists directory contents, no privileged access needed.
Permissions: -rwxr-xr-x (no SUID). Runs as the calling user.
B. /usr/bin/curl:
Purpose: Transfers data over HTTP/FTP, no root privileges required by default.
Permissions: -rwxr-xr-x (no SUID).
Technical Details:
SUID Bit: Set via chmod u+s <file> or chmod 4755.
Security: SUID binaries are audited (e.g., find / -perm -u=s) due to escalation risks if writable or poorly coded (e.g., buffer overflows).
Security Implications: CNSP likely highlights SUID as an attack vector (e.g., CVE-1996-0095 exploited passwd flaws). Hardening removes unnecessary SUID bits.
Why other options are incorrect:
A, B: Lack SUID; no privileged operations.
D: Incorrect, as /usr/bin/passwd is a SUID example. Real-World Context: SUID on /bin/su or /usr/bin/sudo similarly enables privilege escalation, often targeted in exploits.
Reference: CNSP Official Documentation (Linux Security); Linux File Permissions Guide.

Question#5

Which of the following techniques can be used to bypass network segmentation during infrastructure penetration testing?

A. DNS tunneling

B. VLAN hopping

C. Covert channels

D. All of the above

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.
Reference: CNSP "Penetration Testing Techniques" (Section on Network Segmentation Bypass) lists DNS tunneling, VLAN hopping, and covert channels as effective methods.

Exam Code: CNSP Q & A: 60 Q&As Updated: Dec 12,2025

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Exam Code: CNSP
Q & A: 60 Q&As
Updated: Dec 12,2025

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