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The SecOps Group Certified Network Security Practitioner Sample Questions (Q12-Q17):

NEW QUESTION # 12
If you find the 111/TCP port open on a Unix system, what is the next logical step to take?

• A. Telnet to the port, send "GET / HTTP/1.0" and gather information from the response.
• B. Telnet to the port to look for a banner.
• C. None of the above.
• D. Run "rpcinfo -p <hostname>" to enumerate the RPC services.

Answer: D

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 # 13
Which of the following protocols is not vulnerable to address spoofing attacks if implemented correctly?

• A. IP
• B. TCP
• C. ARP
• D. UDP

Answer: B

Explanation:
Address spoofing fakes a source address (e.g., IP, MAC) to impersonate or amplify attacks. Analyzing protocol resilience:
C . TCP (Transmission Control Protocol):
Mechanism: Three-way handshake (SYN, SYN-ACK, ACK) verifies both endpoints.
Client SYN (Seq=X), Server SYN-ACK (Seq=Y, Ack=X+1), Client ACK (Ack=Y+1).
Spoofing Resistance: Spoofer must predict the server's sequence number (randomized in modern stacks) and receive SYN-ACK, impractical without session hijacking or MITM.
Correct Implementation: RFC 793-compliant, with anti-spoofing (e.g., Linux tcp_syncookies).
A . UDP:
Connectionless (RFC 768), no handshake. Spoofed packets (e.g., source IP 1.2.3.4) are accepted if port is open, enabling reflection attacks (e.g., DNS amplification).
B . ARP (Address Resolution Protocol):
No authentication (RFC 826). Spoofed ARP replies (e.g., fake MAC for gateway IP) poison caches, enabling MITM (e.g., arpspoof).
D . IP:
No inherent validation at Layer 3 (RFC 791). Spoofed source IPs pass unless filtered (e.g., ingress filtering, RFC 2827).
Security Implications: TCP's handshake makes spoofing harder, though not impossible (e.g., blind spoofing with sequence prediction, mitigated since BSD 4.4). CNSP likely contrasts this with UDP/IP's vulnerabilities in DDoS contexts.
Why other options are incorrect:
A, B, D: Lack handshake or authentication, inherently spoofable.
Real-World Context: TCP spoofing was viable pre-1990s (e.g., Mitnick attack); modern randomization thwarts it.

NEW QUESTION # 14
On a Microsoft Windows operating system, what does the following command do?
net localgroup Sales Sales_domain /add

• A. Display the list of the users of a local group Sales
• B. Add a domain group to the local group Sales
• C. Add a new user to the local group Sales
• D. Add a local group Sales to the domain group

Answer: B

Explanation:
The net localgroup command manages local group memberships on Windows systems, with syntax dictating its action.
Why B is correct: net localgroup Sales Sales_domain /add adds the domain group Sales_domain to the local group Sales, granting its members local group privileges. CNSP covers this for privilege escalation testing.
Why other options are incorrect:
A: Displaying users requires net localgroup Sales without /add.
C: Adding a user requires a username, not a group name like Sales_domain.
D: The reverse (local to domain) uses net group, not net localgroup.

NEW QUESTION # 15
Which of the following represents a valid Windows Registry key?

• A. HKEY_ROOT_CLASSES
• B. HKEY_LOCAL_MACHINE
• C. HKEY_INTERNAL_CONFIG
• D. HKEY_LOCAL_USER

Answer: B

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).

NEW QUESTION # 16
What will be the subnet mask for 192.168.0.1/18?

• A. 255.225.192.0
• B. 255.255.192.0
• C. 255.225.225.0
• D. 255.255.255.0

Answer: B

Explanation:
An IP address with a /18 prefix (CIDR notation) indicates 18 network bits in the subnet mask, leaving 14 host bits (32 total bits - 18). For IPv4 (e.g., 192.168.0.1):
Binary Mask: First 18 bits are 1s, rest 0s.
1st octet: 11111111 (255)
2nd octet: 11111111 (255)
3rd octet: 11000000 (192)
4th octet: 00000000 (0)
Decimal: 255.255.192.0
Calculation:
Bits: /18 = 2

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