127 free GIAC Certified Intrusion Analyst GCIA practice questions from the CertStash pack of 507, each with the correct answer and a full explanation. The rule is the same for every exam: ten free on the product page, 127 free here, all 507 in the pack, and each step includes the one before it.
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Question 1
An analyst runs tcpdump -n -S on a client's uplink and captures the start of a connection. The third segment is shown with its acknowledgment number removed.
10.1.1.20.51514 > 203.0.113.80.443: Flags [S], seq 2944051730, win 64240, options [mss 1460,sackOK,TS val 1803321 ecr 0,nop,wscale 7], length 0
203.0.113.80.443 > 10.1.1.20.51514: Flags [S.], seq 1180366102, ack 2944051731, win 65160, options [mss 1460,sackOK,TS val 99120 ecr 1803321,nop,wscale 7], length 0
10.1.1.20.51514 > 203.0.113.80.443: Flags [.], ack ??????????, win 502, length 0
Which acknowledgment number must the client's third segment carry to complete the three-way handshake?
Show answer and explanation
Correct answer: B. 1180366103
The SYN/ACK carries the server's initial sequence number, 1180366102. A SYN occupies one sequence number even though it carries no data, so the next sequence number the client expects from the server is 1180366102 + 1 = 1180366103, and that is the value its ACK must carry. The server applied the same rule in its own ACK: 2944051730 + 1 = 2944051731. A third segment that acknowledged the bare ISN, or any other value, would not complete a normal handshake.
Why the other options are wrong
- A. 2944051731 is the client's own next sequence number, which the server acknowledged, not a value the client acknowledges.
- C. 1180366102 is the server's ISN itself, and acknowledging it ignores the sequence number that the SYN consumes.
- D. 2944051732 adds two to the client's own ISN and has nothing to do with the server's sequence space.
Question 2
Sasha wants to add an entry to your DNS database for your mail server.
Which of the following types of resource records will she use to accomplish this?
Show answer and explanation
Correct answer: C. MX
MX (Mail Exchange) records are used in DNS to specify the mail servers responsible for receiving email for a domain. They are essential for directing email traffic to the correct mail server. ANAME is not a standard DNS record type, SOA defines zone parameters, and CNAME creates aliases for hostnames.
Why the other options are wrong
- A. ANAME is not a standard DNS resource record type.
- B. SOA records define Start of Authority parameters for a DNS zone, not mail server entries.
- D. CNAME records create canonical name aliases but are not used specifically for mail server entries.
Question 3
In Wireshark 4.x an analyst applies the display filter below, intending to hide all traffic to or from 10.0.0.5.
ip.addr !== 10.0.0.5
Which packets remain displayed?
Show answer and explanation
Correct answer: B. Any packet with an address other than 10.0.0.5
ip.addr stands for both the source and the destination address, so comparisons on it test two values. The !== operator (any_ne) is true when any occurrence differs from the value, so a packet from 10.0.0.5 to 10.0.0.9 still matches because its destination differs. Only a packet with 10.0.0.5 in both fields would be hidden. To hide the host's traffic, use ip.addr != 10.0.0.5, which since Wireshark 3.6 means all not equal, or the explicit !(ip.addr == 10.0.0.5).
Why the other options are wrong
- A. That result needs all-not-equal semantics, which is ip.addr != 10.0.0.5, not !==.
- C. !== is a valid C-like operator, the alias of any_ne.
- D. ip.addr represents both the source and destination addresses, and !== tests whether any of them differs.
Question 4
Which of the following proxy servers is also referred to as transparent proxies or forced proxies?
Show answer and explanation
Correct answer: D. Intercepting proxy server
Intercepting proxy servers, also called transparent proxies or forced proxies, intercept client requests without requiring explicit client configuration. They transparently handle traffic by intercepting it at the network level, forcing all traffic through the proxy without the user's knowledge or manual configuration. The other proxy types require explicit configuration or serve different purposes.
Why the other options are wrong
- A. Tunneling proxy servers establish encrypted tunnels for secure communication but are not referred to as transparent proxies.
- B. Reverse proxy servers handle requests from clients on behalf of backend servers and are not transparent proxies.
- C. Anonymous proxy servers hide the client's identity but require explicit configuration and are not the same as transparent proxies.
Question 5
Which of the following statements about a host-based Intrusion Detection System (HIDS) are true? Each correct answer represents a complete solution. Choose two.
Show answer and explanation
Correct answer: B, C. It can handle encrypted and unencrypted traffic equally; It cannot detect events scattered over the network.
Option B is correct because Intrusion Detection Systems can handle both encrypted and unencrypted traffic equally at the network layer by analyzing packet patterns and metadata. Option C is correct because an IDS cannot detect events scattered across the entire network without proper sensor placement and network architecture, it can only see traffic on the network segments where it is positioned. These two statements accurately describe the limitations and capabilities of IDS technology.
Why the other options are wrong
- A. IDS can detect events only on the network segments where sensors are deployed, not automatically across scattered network locations.
- D. This describes network load balancing or IP address sharing, which is unrelated to IDS/IPS detection capabilities.
Question 6
Victor works as a network administrator for DataSecu Inc. He uses a dual firewall Demilitarized Zone (DMZ) to insulate the rest of the network from the portions that is available to the Internet.
Which of the following security threats may occur if DMZ protocol attacks are performed?
Each correct answer represents a complete solution.
Choose all that apply.
Show answer and explanation
Correct answer: A, B, D. Attacker can perform Zero Day attack by delivering a malicious payload that is not a part of the intrusion detection/prevention systems guarding the network; Attacker can gain access to the Web server in a DMZ and exploit the database; Attacker can exploit any protocol used to go into the internal network or intranet of the com pany
Option A is correct because attackers can deliver zero-day exploits through the DMZ that bypass intrusion detection systems. Option B is correct because compromising a web server in the DMZ can provide a foothold to exploit backend databases. Option D is correct because attackers can exploit protocols used to communicate between the DMZ and internal network. Option C is incorrect because if an attacker breaks the first firewall, the second firewall would still protect the internal network if properly configured with different rules.
Why the other options are wrong
- C. A properly configured dual firewall DMZ means the second firewall would still defend the internal network even if the first is breached, assuming the firewalls have independent rule sets.
Question 7
tcpdump, run without -S, prints the following segment from the middle of an HTTP download.
203.0.113.80.80 > 10.1.1.20.51514: Flags [.], seq 1449:2897, ack 1, win 509, length 1448
Which interpretation of the seq field is correct?
Show answer and explanation
Correct answer: D. It carries relative bytes 1449 to 2896 inclusive, so the next ACK will be 2897.
tcpdump prints the sequence range as first:last, meaning sequence numbers from first up to but not including last. Here 2897 - 1449 = 1448, which matches length 1448, so the segment holds relative bytes 1449 through 2896. The receiver acknowledges the next byte it expects, 2897. Without -S (or --absolute-tcp-sequence-numbers), tcpdump prints numbers relative to the initial sequence number of the conversation, so the first data byte in each direction is 1.
Why the other options are wrong
- A. The segment length is 1448 bytes, not 2,897; 2897 is the end of the range.
- B. The last value in the range is excluded, so byte 2897 is not part of this segment.
- C. Without -S tcpdump prints relative numbers, so 1449 is an offset from the ISN, not the ISN.
Question 8
Ryan, a malicious hacker submits Cross-Site Scripting (XSS) exploit code to the Website of Internet forum for online discussion.
When a user visits the infected Web page, code gets automatically executed and Ryan can easily perform acts like account hijacking, history theft etc.
Which of the following types of Cross-Site Scripting attack Ryan intends to do?
Show answer and explanation
Correct answer: D. Persistent
The exploit code is submitted to a forum and stored on the site, so it is served back to every visitor of the infected page and executes automatically without the victim clicking a crafted link. Injecting script that is saved in the application's data store (message boards, comments, profiles) and replayed to other users is the defining characteristic of a persistent, or stored, XSS attack, and it enables session hijacking and history theft as described.
Why the other options are wrong
- A. DOM-based XSS occurs entirely in client-side script manipulating the document object model; the payload is not stored on the server and delivered to all forum visitors.
- B. Non-persistent (reflected) XSS requires the victim to follow a specially crafted request; the payload is not saved on the site as it is here.
- C. SAX is an event-driven XML parsing API, not a category of cross-site scripting.
Question 9
Hunting for scripted clients, an analyst wants every HTTP request whose User-Agent contains the word curl in any letter case, such as curl/8.5.0 or CURL-agent. Which display filter does this?
Show answer and explanation
Correct answer: D. http.user_agent matches "curl"
The matches operator applies a Perl-compatible regular expression to a string field, and Wireshark makes these matches case-insensitive by default, so the pattern curl finds curl, CURL or Curl anywhere in the header. The contains operator compares exact characters, so a case-insensitive contains test needs lower() or upper() on the field. Adding (?-i) at the start of a pattern switches matches back to case-sensitive.
Why the other options are wrong
- A. contains compares exact characters, so it would miss CURL-agent unless the field is passed through lower() first.
- B. (?-i) forces a case-sensitive match, so CURL-agent would be missed.
- C. == requires the whole header to equal curl exactly, so curl/8.5.0 would not match.
Question 10
You are the Network Administrator for a large corporate network. You want to monitor all network traffic on your local network for suspicious activities and receive a notification when a possible attack is in process.
Which of the following actions will you take for this?
Show answer and explanation
Correct answer: B. Install a network-based IDS
A network-based Intrusion Detection System (IDS) is specifically designed to monitor all traffic on a network segment in real-time and generate alerts when suspicious patterns or known attacks are detected. It operates at the network level, analyzing traffic across multiple hosts, making it ideal for monitoring an entire local network. Host-based IDS monitors individual systems, firewalls control access rather than detect attacks, and DMZ firewalls are architectural components, not monitoring tools.
Why the other options are wrong
- A. Verbose firewall logging records events but does not actively detect and alert on attacks in real-time.
- C. A DMZ firewall is a network architecture component used for segmentation, not for monitoring suspicious activities.
- D. Host-based IDS monitors individual systems only, not all network traffic across the local network.
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Question 11
During an exfiltration hunt, an analyst wants outbound flow records that each carried at least 50 MB (50,000,000 bytes). Which rwfilter switch selects them?
Show answer and explanation
Correct answer: C. --bytes=50000000-
Volume switches take a range of non-negative integers. When the upper limit is omitted, as in 50000000-, the record's byte count must be at least that value, which is exactly the at-least test needed. A single value with no hyphen matches only flows of exactly that size. Combined with --type=out, this finds large single transfers; slow exfiltration spread over many small flows needs aggregation with rwuniq or rwstats instead.
Why the other options are wrong
- A. A single value requires the byte count to match exactly, so almost no flow passes.
- B. A range of 0 to 50,000,000 passes flows of at most 50 MB, the opposite of the goal.
- D. --bytes-per-packet tests the average bytes per packet, not the flow's total bytes.
Question 12
A newly booted workstation broadcasts a request to dynamically obtain its IP address configuration from a server on the local subnet; to which destination port is that traffic sent?
Show answer and explanation
Correct answer: D. UDP 67
DHCP (Dynamic Host Configuration Protocol) operates on UDP port 67 for the server side. When a workstation boots and broadcasts a DHCP Discover message to obtain an IP address configuration, it sends the request to UDP port 67. TCP 161 is for SNMP, UDP 69 is for TFTP, and TCP 21 is for FTP, none are used for dynamic IP address assignment.
Why the other options are wrong
- A. TCP 161 is used for SNMP (Simple Network Management Protocol), not DHCP.
- B. UDP 69 is used for TFTP (Trivial File Transfer Protocol), not DHCP.
- C. TCP 21 is used for FTP (File Transfer Protocol), not DHCP.
Question 13
An analyst reviews a colleague's lab script that sends IP(dst=target)/TCP(dport=80) with no flags argument. Which TCP control flags are set in the packet that goes out?
Show answer and explanation
Correct answer: A. Only SYN
The TCP layer defaults its flags field to SYN, shown in Scapy as flags set to the S bit. So TCP() with no flags argument produces a lone SYN, the segment that normally opens a connection, which is why Scapy is convenient for connection probes. An analyst reading such a script should expect single SYN segments on the wire unless the flags argument overrides the default.
Why the other options are wrong
- B. ACK alone is not the default; the default sets the SYN bit.
- C. A SYN/ACK is a server response and is not what the client default produces.
- D. A segment with no flags set would require explicitly clearing the default SYN.
Question 14
You work as a Network Administrator for Tech Perfect Inc. Your company has a Windows 2000-based network. You want to verify the connectivity of a host in the network.
Which of the following utilities will you use?
Show answer and explanation
Correct answer: A. PING
PING is the standard utility used to verify connectivity to a host on a network. It sends ICMP Echo Request packets and waits for Echo Reply responses, confirming that a host is reachable and responding. TELNET is used to establish remote sessions, NETSTAT displays network statistics and connections, and TRACERT traces the path to a host, none of which are primary connectivity verification tools.
Why the other options are wrong
- B. TELNET establishes remote terminal sessions, not simply verify connectivity.
- C. NETSTAT displays network statistics, connection states, and routing information, not basic connectivity.
- D. TRACERT traces the route to a destination but is more complex than simple connectivity verification.
Question 15
An analyst has read a capture into a Scapy packet list and wants one short line per packet to scan the traffic quickly. Which call produces that?
Show answer and explanation
Correct answer: D. summary() on the list
For a packet list, summary() displays a list of one-line summaries, one per packet, which is ideal for scanning many packets at a glance. show2() assembles and expands a single packet, hexdump() dumps raw bytes for every packet, and ls() lists fields rather than a per-packet line. When triaging a crafted or captured set, the per-packet summary gives the fastest overview before drilling into individual packets.
Why the other options are wrong
- A. show2() expands and assembles a packet in full rather than giving one line each.
- B. hexdump() outputs raw bytes for each packet, not a concise line.
- C. ls() lists fields of a layer, not a per-packet summary.
Question 16
John works as a professional Ethical Hacker. He has been assigned a project for testing the security of www.we-are-secure.com. He wants to corrupt an IDS signature database so that performing attacks on the server is made easy and he can observe the flaws in the We-are-secure server. To perform his task, he first of all sends a virus that continuously changes its signature to avoid detection from IDS. Since the new signature of the virus does not match the old signature, which is entered in the IDS signature database, IDS becomes unable to point out the malicious virus.
Which of the following IDS evasion attacks is John performing?
Show answer and explanation
Correct answer: D. Polymorphic shell code attack
Polymorphic shell code attack is the correct answer because it describes malware that continuously changes its signature to evade detection. The virus mutates its code signature while maintaining the same functionality, making it unrecognizable to signature-based IDS systems that rely on static signature matching. This is the defining characteristic of polymorphic malware.
Why the other options are wrong
- A. Insertion attacks add redundant data to confuse IDS processing, not change malware signatures.
- B. Session splicing attacks split attacks across multiple packets to evade reassembly-based detection.
- C. Evasion attack is too generic and describes the overall category rather than this specific technique of signature mutation.
Question 17
An analyst sends a single crafted stimulus with sr1() in a lab and expects to study the reply. What does sr1() return?
Show answer and explanation
Correct answer: B. Only the one packet that answered the stimulus
sr1() is a variant of the send and receive function that returns only the single packet that answered the one sent. The full sr() returns two lists, the answered pairs and the unanswered stimuli, but sr1() is a convenience for the common case of one probe and one reply. Because Scapy hands back the whole decoded answer rather than an interpretation, an analyst can inspect every field of the response instead of a summary verdict.
Why the other options are wrong
- A. Returning paired lists is the behavior of sr(), not sr1().
- C. The unanswered list is part of the sr() return, not what sr1() provides.
- D. sr1() returns a packet, not a count.
Question 18
When no anomaly is present in an Intrusion Detection, but an alarm is generated, the response is known as __________.
Show answer and explanation
Correct answer: D. False positive
A false positive occurs when an IDS generates an alarm when no actual attack (anomaly) is present. This is a false alert, the system incorrectly identified normal activity as malicious, which is the opposite of the true state.
Why the other options are wrong
- A. True negative means no alarm is generated and no attack is present, which is correct behavior.
- B. True positive means an alarm is generated and an actual attack is present, which is correct detection.
- C. False negative means no alarm is generated but an attack is actually present, missing the attack.
Question 19
Which of the following IPv6 address types is a single address that can be assigned to multiple interfaces?
Show answer and explanation
Correct answer: B. Anycast
Anycast is an IPv6 address type that can be assigned to multiple interfaces, where packets are delivered to the nearest interface (in routing terms). Unicast addresses are assigned to a single interface, loopback is a special unicast address, and multicast is designed for one-to-many communication but is fundamentally different from anycast in its delivery model.
Why the other options are wrong
- A. Unicast addresses are assigned to and identify a single interface, not multiple.
- C. Loopback is a special type of unicast address used for testing, not for multiple interfaces.
- D. Multicast addresses are for one-to-many communication but use a different delivery mechanism than anycast.
Question 20
In which format is data encapsulated as it is handed off by the Data Link layer of the OSI model?
Show answer and explanation
Correct answer: D. Data frames
The Data Link layer (Layer 2) of the OSI model encapsulates data as frames. A frame is the protocol data unit (PDU) at the Data Link layer and includes the network layer packet plus header and trailer information (including MAC addresses).
Why the other options are wrong
- A. Protocols are rules/standards, not a format for data encapsulation.
- B. Raw bits are the physical layer representation, not the Data Link layer format.
- C. Data packets are the PDU at the Network layer (Layer 3), not the Data Link layer.
- E. Data segments are the PDU at the Transport layer (Layer 4), not the Data Link layer.
Question 21
Every packet from one Linux server carries IPv4 Identification 0x0000, and bytes 6 and 7 of each header are 0x4000. A new IDS rule flags the repeated ID as spoofing. How should the analyst assess the alerts?
Show answer and explanation
Correct answer: B. False positive: these are atomic datagrams, so the ID is unused
0x4000 = 0100 0000 0000 0000: the reserved bit is clear, DF is set, MF is clear and the fragment offset is 0. Such packets are atomic datagrams: not fragmented and barred from further fragmentation, so the Identification field has no reassembly role. Originators may set it to any value, and a constant zero is legal. Devices that examine IPv4 headers must ignore the ID of atomic datagrams, so the rule produces false positives.
Why the other options are wrong
- A. Uniqueness is not required for atomic datagrams, which never need reassembly.
- C. No Identification value is reserved, and zero is a legal value for an atomic datagram.
- D. A first fragment has MF set, while 0x4000 has DF set and MF clear, so the datagram is unfragmented.
Question 22
Which of the following is the default port for Hypertext Transfer Protocol (HTTP)?
Show answer and explanation
Correct answer: C. 80
Port 80 is the default port for Hypertext Transfer Protocol (HTTP). This is the standard, well-established port used by web browsers and HTTP servers for unencrypted web communication. Port 23 is for Telnet, port 21 is for FTP, and port 25 is for SMTP.
Why the other options are wrong
- A. Port 23 is the default port for Telnet, a remote access protocol.
- B. Port 21 is the default port for FTP (File Transfer Protocol).
- D. Port 25 is the default port for SMTP (Simple Mail Transfer Protocol).
Question 23
A log pipeline rewrites every IPv6 address into the RFC 5952 recommended text form so that searches match across tools. A firewall log records the source address below.
2001:0DB8:0000:0000:00A5:0000:0000:0F0F
Which string does the pipeline store?
Show answer and explanation
Correct answer: A. 2001:db8::a5:0:0:f0f
RFC 5952 requires leading zeros in each field to be dropped, the hex letters a to f to be lowercase, and "::" to replace the longest run of zero fields. This address has two runs of two zero fields (fields 3 and 4, and fields 6 and 7); when runs are equal, the first one must be shortened. The result is 2001:db8::a5:0:0:f0f. Storing one canonical form matters for analysts because the same host can otherwise appear under several strings in different logs.
Why the other options are wrong
- B. It is a valid address, but it compresses the second run of zeros, while the tie rule requires the first run to be shortened.
- C. The "::" symbol may appear only once in an address, so this string is not a valid IPv6 address.
- D. RFC 5952 requires the hexadecimal letters a through f to be written in lowercase.
Question 24
Which of the following IPv4 to IPv6 transition methods uses encapsulation of IPv6 packets to traverse IPv4 networks?
Show answer and explanation
Correct answer: C. Tunneling
Tunneling is the IPv4-to-IPv6 transition method that encapsulates IPv6 packets within IPv4 packets so they can traverse existing IPv4 networks. This allows IPv6 traffic to pass through IPv4 infrastructure as if the IPv6 packets are the payload of IPv4 packets. Dual-stack runs both protocols simultaneously, translation converts between protocols, and 'Stack' is not a standard transition method.
Why the other options are wrong
- A. Dual-stack involves running both IPv4 and IPv6 simultaneously, not encapsulation through IPv4 networks.
- B. Translation converts packets between IPv4 and IPv6 formats, not by encapsulation.
- D. Stack is not a recognized IPv4-to-IPv6 transition method.
Question 25
Internal hosts use 10.0.0.0/8. An analyst wants only packets sent from an internal host to an address outside that range, to review outbound traffic. Which display filter does this?
Show answer and explanation
Correct answer: A. ip.src == 10.0.0.0/8 && !(ip.dst == 10.0.0.0/8)
IPv4 address fields accept CIDR notation in display filters, so ip.src == 10.0.0.0/8 is true for any source in the 10.x.x.x block. Negating the same test on ip.dst keeps only packets whose destination lies outside the block, which together describe outbound traffic. Using ip.src and ip.dst rather than ip.addr matters because ip.addr tests both addresses at once, and leaving out the /8 compares against the single address 10.0.0.0.
Why the other options are wrong
- B. ip.addr covers both addresses, so no packet can both have and lack an address in the block; nothing is shown.
- C. Without a prefix length the value is the single address 10.0.0.0, so almost no packet matches.
- D. With || any internal source matches even when the destination is internal, and any external destination matches regardless of source.
Question 26
An analyst captures the same IPv4 packet on both sides of a router that performs no NAT, does not fragment the packet and finds no IP options in it. Which header fields differ between the two captures?
Show answer and explanation
Correct answer: B. Time to Live and header checksum
A router forwarding a datagram must decrease its TTL by at least one. Because the header checksum covers the header only and the TTL is part of that header, the checksum must then be recomputed. Identification, Total Length and the addresses stay unchanged when no fragmentation or translation occurs. An Identification or length change on such a hop would point to a middlebox rewriting packets.
Why the other options are wrong
- A. The Identification is set by the source and is not changed by a forwarding router.
- C. Total Length changes only when the router fragments the datagram, which it does not do here.
- D. The Identification stays the same, and the checksum must change along with the TTL.
Question 27
A sensor records these packets to a Linux server (tcpdump -n -S). No other replies appear.
203.0.113.9.40210 > 10.8.0.5.22: Flags [none], seq 3110429876, win 1024, length 0
203.0.113.9.40210 > 10.8.0.5.23: Flags [none], seq 3110429876, win 1024, length 0
10.8.0.5.23 > 203.0.113.9.40210: Flags [R.], seq 0, ack 3110429876, win 0, length 0
What do the packets show about ports 22 and 23?
Show answer and explanation
Correct answer: A. A null scan: port 22 is open or filtered and port 23 is closed
The probes carry no flags at all, which is an nmap null scan. Under standard TCP behavior, which Linux follows, a closed port answers any segment that lacks RST with a reset, while a listening port silently drops a segment without SYN, RST or ACK. Port 23 returned RST/ACK with seq 0 and ack 3110429876 + 0, the form used when the arriving segment has no ACK, so it is closed. The silence from port 22 means open or filtered.
Why the other options are wrong
- B. Reversed: a listening port drops the flagless probe, and only a closed port answers with RST.
- C. The probes show Flags [none], not F, so they are null probes, and silence from 22 is the expected open-port result.
- D. RST/ACK is the closed-port reply, so port 23 cannot be open.
Question 28
A Suricata sensor runs inline in IPS mode. Rule sid 9000200 alerts on SYN packets to the SSH server, and threshold.config contains this entry:
rate_filter gen_id 1, sig_id 9000200, track by_src, count 20, seconds 60, new_action drop, timeout 600
What does the entry do?
Show answer and explanation
Correct answer: A. Once a source reaches 20 hits in 60 seconds, its matching traffic is dropped for 600 seconds
A rate_filter changes a rule's action once a rate is reached instead of only thinning alerts. Here hits are tracked per source address; when one source reaches 20 hits on sid 9000200 within 60 seconds, the filter activates and new_action drop is applied to traffic matching that rule, so further SYNs from that source to the SSH server are dropped. The changed action stays in force for the 600-second timeout, while sources below the rate keep the rule's normal alert action.
Why the other options are wrong
- B. Capping alert counts is the job of threshold type limit; a rate_filter replaces the rule action with new_action.
- C. new_action applies only to traffic that matches the rule, not to every packet the source sends to any host.
- D. The new action starts after the count is reached within the period, not for the first matches, and it lasts for the timeout.
Question 29
An analyst examines the following IPv4 header from an ICMP packet.
45 00 00 54 3e 9d 40 00 40 01 0f c5 c0 a8 01 64 c6 33 64 07
How many bytes of ICMP message, header plus data, does the datagram carry?
Show answer and explanation
Correct answer: A. 64 bytes
Total Length, in bytes 2 and 3, counts the whole datagram, header and data. 0x0054 = 5 x 16 + 4 = 84 bytes. The first byte 0x45 gives IHL 5, a 20-byte header with no options, so the ICMP message occupies 84 - 20 = 64 bytes. The protocol byte 0x01 confirms ICMP. Comparing Total Length with the header length this way is how an analyst spots oversized, truncated or padded payloads.
Why the other options are wrong
- B. Eighty-four bytes is the Total Length, which includes the 20-byte IP header.
- C. Fifty-six bytes also removes an 8-byte ICMP header, but the message length includes that header.
- D. Seventy-six bytes removes only 8 bytes instead of the 20-byte IP header.
Question 30
You are using the ping command to verify the connectivity between two hosts. You want ping to send larger sized packets than the usual 32-byte ones. Which of the following commands will you use?
Show answer and explanation
Correct answer: D. ping -l
The ping -l command (lowercase L) is used to specify the packet size to be sent. On Windows and Linux systems, this flag allows you to set a custom packet size larger than the default 32 bytes. Option A (-a) resolves addresses to hostnames, B (-4) forces IPv4, and C (-t) sets continuous ping mode.
Why the other options are wrong
- A. The -a flag is used for address resolution, not packet size specification.
- B. The -4 flag forces IPv4 protocol selection, not packet size control.
- C. The -t flag enables continuous ping mode on Windows, not packet size adjustment.
Question 31
For a host to have successful Internet communication, which of the following network protocols are required? You should assume that the users will not manually configure the computer in any way and that the measure of success will be whether the user can access Web sites after powering the computer and logging on.
Each correct answer represents a complete solution.
Choose all that apply.
Show answer and explanation
Correct answer: A, B, C. DNS; HTTP/HTTPS; DHCP
With no manual configuration, the host must obtain its IP address, subnet mask, default gateway, and DNS server settings automatically, which requires DHCP. Once addressed, the host must translate the human-readable Web site names typed by the user into IP addresses, which requires DNS. Finally, the actual retrieval of Web pages is performed by HTTP or its secure form HTTPS. Together, DHCP, DNS, and HTTP/HTTPS are the protocols that must be functioning for the user to browse Web sites successfully after logging on.
Why the other options are wrong
- D. NTP keeps the system clock synchronized, but a host does not need it to obtain an address, resolve names or retrieve Web pages.
Question 32
In which of the following IKE phases the IPsec endpoints establish parameters for a secure ISAKMP session?
Show answer and explanation
Correct answer: C. IKE Phase 1
IKE Phase 1 is where IPsec endpoints establish parameters for a secure ISAKMP (Internet Security Association and Key Management Protocol) session. During Phase 1, peers authenticate each other and establish a secure, authenticated channel. Phase 2 uses this established channel to negotiate IPsec security associations. Phases 1.5 and 2.5 do not exist in the IKE standard.
Why the other options are wrong
- A. IKE Phase 2.5 does not exist in the standard IKE specification.
- B. IKE Phase 2 negotiates IPsec SAs after Phase 1 establishes the ISAKMP session.
- D. IKE Phase 1.5 does not exist in the standard IKE specification.
Question 33
Which of the following commands will you use with the tcpdump command to capture the traffic from a filter stored in a file?
Show answer and explanation
Correct answer: D. tcpdump -F file_name
The tcpdump command uses the -F flag to read a filter expression from a specified file. This allows users to apply complex, pre-defined packet capture filters stored in a file rather than typing them directly on the command line. The -A flag displays ASCII data, -D lists available interfaces, and -X displays hex and ASCII output, none of these load filters from a file.
Why the other options are wrong
- A. The -A flag displays packet payload in ASCII format, not used for loading filter files.
- B. The -D flag lists available network interfaces for capture, not for loading filters.
- C. The -X flag displays packet data in hexadecimal and ASCII formats, not for loading filter files.
Question 34
An analyst reviews this Zeek conn.log excerpt. The same pattern repeats for about 2,400 more lines within two minutes.
ts id.orig_h id.orig_p id.resp_h id.resp_p proto conn_state history
1759310401.120 10.20.4.31 50112 10.20.8.11 445 tcp S0 S
1759310401.121 10.20.4.31 50113 10.20.8.12 445 tcp S0 S
1759310401.123 10.20.4.31 50114 10.20.8.13 445 tcp S0 S
1759310401.124 10.20.4.31 50115 10.20.8.14 445 tcp S0 S
What does this activity most likely represent?
Show answer and explanation
Correct answer: A. A sweep for SMB listeners: SYNs to many hosts that never answered
Zeek sets conn_state S0 when it sees a connection attempt with no reply, and the history S records a single SYN without ACK from the originator. One source walking sequential addresses on TCP 445 about a millisecond apart, each attempt unanswered, is a horizontal scan for SMB listeners. Hosts that are absent or behind a silent filter produce exactly this S0 pattern. Established sessions would be logged as SF or S1 with handshake letters, refused attempts as REJ, and mid-stream traffic as OTH.
Why the other options are wrong
- B. Sessions that were established and closed are logged as SF with handshake history letters, not S0 with a lone S.
- C. An attempt answered with a RST is logged as REJ with an r in the history, not as S0.
- D. Traffic first seen mid-stream is logged as OTH, and these records show only an initial SYN.
Question 35
Which of the following is a reason to implement security logging on a DNS server?
Show answer and explanation
Correct answer: D. For monitoring unauthorized zone transfer
Security logging on a DNS server is primarily implemented to monitor unauthorized zone transfers. Zone transfers are sensitive operations that should only occur between authorized primary and secondary DNS servers, and logging allows detection of unauthorized attempts to obtain zone data.
Why the other options are wrong
- A. Recording the number of queries resolved is a performance metric, not a security reason for logging.
- B. Preventing malware attacks is not the primary function of DNS security logging; other mechanisms handle malware prevention.
- C. Measuring DNS server performance is a performance consideration, not a security logging reason.
Question 36
Which of the following tools is used to analyze the files produced by several popular packetcapture programs such as tcpdump, WinDump, Wireshark, and EtherPeek?
Show answer and explanation
Correct answer: D. tcptrace
Tcptrace is a tool specifically designed to analyze and process packet capture files from various capture programs including tcpdump, WinDump, Wireshark, and EtherPeek. It provides detailed traffic analysis and statistics from these packet capture formats. Fpipe is for piping data between sockets, tcptraceroute is for tracing network paths, and Sniffer is a packet capture tool itself rather than an analyzer.
Why the other options are wrong
- A. Fpipe is used to create a pipe between two sockets for network redirection, not for analyzing capture files.
- B. Tcptraceroute is used to trace network routing paths, not to analyze packet capture files.
- C. Sniffer is a packet capture tool used to collect network traffic, not to analyze existing capture files.
Question 37
Two IPv4 fragments share the same ID, addresses and protocol, and each has a 20-byte header.
Frag A: offset field 0, MF=1, total length 52
Frag B: offset field 2, MF=0, total length 44
How many bytes of data do the two fragments overlap?
Show answer and explanation
Correct answer: B. 16 bytes
Fragment A carries 52 - 20 = 32 data bytes covering bytes 0 to 31. Fragment B's offset field of 2 means it starts at 2 x 8 = 16, and it carries 44 - 20 = 24 data bytes covering bytes 16 to 39. The shared range is bytes 16 to 31, which is 16 bytes. Normal fragmentation never overlaps, so the receiving stack's reassembly behavior decides which copy of those bytes survives, which is how overlapping fragments can show a sensor different data than the target keeps.
Why the other options are wrong
- A. An 8-byte overlap would need fragment B to start at byte 24 (offset field 3), not at byte 16.
- C. 24 bytes is fragment B's whole data length, but bytes 32 to 39 of fragment B lie beyond the end of fragment A.
- D. 32 bytes is all of fragment A's data, but fragment B starts at byte 16, not at byte 0.
Question 38
An analyst crafts IP(dst='198.51.100.7')/UDP()/Raw(load=b'test') in Scapy without naming any ports. Which source and destination UDP ports does the packet use?
Show answer and explanation
Correct answer: B. Source 53 and destination 53
The UDP layer defaults both its source and destination ports to 53. So an unconfigured UDP() is sent from port 53 to port 53, which resembles DNS server to server traffic. An analyst who forgets this can misread lab captures as DNS activity, and a crafting script that leaves the ports unset will produce 53 to 53 flows that stand out from normal client behavior, where the source port is usually ephemeral.
Why the other options are wrong
- A. 20 and 80 are the TCP layer defaults, not the UDP ones.
- C. Port 0 is reserved and is not the Scapy UDP default.
- D. 67 and 68 are the DHCP ports and are not filled in automatically.
Question 39
Which of the following activities will you use to retrieve user names, and info on groups, shares, and services of networked computers?
Show answer and explanation
Correct answer: D. Network enumerating
Network enumeration is the systematic process of discovering and gathering detailed information about network resources, including usernames, group memberships, shares, and services running on networked computers. It is the active probing technique that extracts specific network configuration and resource information.
Why the other options are wrong
- A. Network tap is a passive monitoring technique that captures traffic for analysis, not actively retrieves user and service information.
- B. Packet crafting involves creating custom network packets for testing, not specifically retrieving user and service information.
- C. Network mapping identifies network topology and connected systems but does not specifically retrieve usernames, groups, shares, and services details.
Question 40
Which of the following TCP ports are used by the FTP service? Each correct answer represents a complete solution. Choose two.
Show answer and explanation
Correct answer: B, C. 21; 20
FTP (File Transfer Protocol) uses two ports: Port 21 for the control/command channel and Port 20 for the data channel. Port 80 is used for HTTP web traffic, and Port 443 is used for HTTPS encrypted web traffic. Both ports 20 and 21 are essential for complete FTP service functionality.
Why the other options are wrong
- A. Port 80 is used for HTTP, not FTP service.
- D. Port 443 is used for HTTPS, not FTP service.
Question 41
An analyst rebuilds the IPv4 header below for a replay test and must fill in the header checksum, shown as ?? ??.
45 00 00 34 6b 2e 40 00 40 06 ?? ?? c0 a8 0a 15 cb 00 71 32
What value belongs in the checksum field?
Show answer and explanation
Correct answer: C. 0xc8a5
The checksum is the one's complement of the one's complement sum of the header's 16-bit words, with the checksum field taken as zero. 4500 + 0034 + 6b2e + 4000 + 4006 + 0000 + c0a8 + 0a15 + cb00 + 7132 = 0x33757. Folding the carry gives 0x3757 + 0x3 = 0x375a, and its complement is 0xffff - 0x375a = 0xc8a5. A receiver summing all ten words including 0xc8a5 gets 0xffff, so the header verifies.
Why the other options are wrong
- A. 0x375a is the folded sum before the final one's complement is taken.
- B. 0xa5c8 is the correct value with its two bytes swapped out of network byte order.
- D. 0xc8a8 complements the sum without first adding the end-around carry of 3.
Question 42
Flow records from a yaf sensor carry initial and session TCP flags. An analyst wants flows whose very first packet had both SYN and ACK set, which can indicate backscatter or a collection that began mid-handshake. Which rwfilter switch fits?
Show answer and explanation
Correct answer: D. --flags-initial=SA/SA
Some flow generators, such as yaf, export the flags of the first packet separately from the OR of the remaining packets. --flags-initial applies the HIGH/MASK test only to that first packet, so SA/SA passes flows that started with a SYN/ACK. --flags-all looks at the OR across all packets, so an ordinary client flow whose SYN was followed by ACKs would also match, and --flags-session ignores the first packet entirely.
Why the other options are wrong
- A. --flags-all tests the OR of all packets, so ordinary client flows with a SYN and later ACKs also pass.
- B. --flags-session ignores the first packet and tests only the second through final packets.
- C. The single-flag switches test whether the flag was set at all in the flow, not on the first packet.
Question 43
You work as a Network Administrator for a bank. For securing the bank's network, you configure a firewall and an IDS. In spite of these security measures, intruders are able to attack the network.
After a close investigation, you find that your IDS is not configured properly and hence is unable to generate alarms when needed.
What type of response is the IDS giving?
Show answer and explanation
Correct answer: C. False Negative
A False Negative occurs when an IDS fails to detect and alarm on actual intrusion attempts. In this scenario, intruders are successfully attacking the network while the IDS remains silent, meaning the system is failing to generate alerts for real attacks. This is the definition of a false negative response.
Why the other options are wrong
- A. False Positive is when the IDS generates an alarm for traffic that is not actually malicious.
- B. True Negative is when the IDS correctly does not alert because there is no actual attack occurring.
- D. True Positive is when the IDS correctly detects and alerts on actual malicious activity.
Question 44
An analyst reviewing a capture sees an IPv4 packet whose header begins with the bytes below.
46 00 00 3c
How many bytes of IP options follow the fixed portion of the header?
Show answer and explanation
Correct answer: B. 4 bytes
The low nibble of the first byte is the Internet Header Length, counted in 32-bit words. 0x46 splits into version 4 and IHL 6, so the header is 6 x 4 = 24 bytes. A header without options is 5 words, or 20 bytes, so 24 - 20 = 4 bytes of options, including any padding, sit between the fixed header and the transport header. Any IHL above 5 signals options, which are rare in normal traffic and worth a closer look.
Why the other options are wrong
- A. Six is the IHL value in 32-bit words, not a count of option bytes.
- C. Twenty-four bytes is the length of the whole header, not of the options alone.
- D. Twenty bytes is the size of the fixed header that carries no options.
Question 45
Which TCP port is used by default for the Post Office Protocol version 3 (POP3) service when clients retrieve email from a mail server?
Show answer and explanation
Correct answer: B. 110
Port 110 is the default TCP port for Post Office Protocol version 3 (POP3), which is used by email clients to retrieve messages from a mail server.
Why the other options are wrong
- A. Port 21 is used for File Transfer Protocol (FTP), not POP3.
- C. Port 80 is used for HTTP (Hypertext Transfer Protocol), not POP3.
- D. Port 25 is used for SMTP (Simple Mail Transfer Protocol) for sending mail, not POP3 for retrieving mail.
Question 46
You are concerned about outside attackers penetrating your network via your company Web server. You wish to place your Web server between two firewalls.
One firewall between the Web server and the outside world. The other between the Web server and your network.
What is this called?
Show answer and explanation
Correct answer: A. DMZ
A DMZ (Demilitarized Zone) is a network segment that sits between two firewalls, one facing the external network and one facing the internal network. This architecture isolates the web server from direct access to internal resources while protecting it from the internet. The web server in a DMZ can be accessed from the outside while remaining separated from the internal network by a second firewall.
Why the other options are wrong
- B. SPI (Stateful Packet Inspection) firewalls are a type of firewall technology, not a network architecture design with two firewalls.
- C. IDS (Intrusion Detection System) is a monitoring tool that detects attacks, not a network architectural arrangement.
- D. Application Gateway firewalls operate at the application layer but do not specifically describe the dual-firewall architecture described in the question.
Question 47
An analyst runs the command below, intending to stop at 06:00 on 15 September.
rwfilter --start-date=2026/09/14 --end-date=2026/09/15:06 --type=all --proto=6 --pass=tcp.rw
Which hours of repository data does rwfilter process?
Show answer and explanation
Correct answer: B. Every hour of both 14 September and 15 September
When --start-date is given to day precision, rwfilter ignores any hour in --end-date and processes files from midnight on the start date through 23:59 on the end date. Here the start date has no hour, so the :06 is dropped and both full days are read. To stop at the 06:00 hour, the start must also carry an hour, for example --start-date=2026/09/14:00, so that both dates are hour precision and every hour in that range is processed.
Why the other options are wrong
- A. That range needs hour precision on both dates; with a day-precision start the end hour is ignored.
- C. Hour precision selects one continuous range of hours, never the same hour on each day.
- D. The end hour is ignored here, and even with hour precision the 06:00 hour itself would be included.
Question 48
Which TCP port does an FTP server use by default for its control connection, over which client commands are issued?
Show answer and explanation
Correct answer: A. 21
FTP (File Transfer Protocol) uses port 21 by default for control connections, where client commands are issued. Port 25 is for SMTP (email), port 23 is for Telnet, and port 80 is for HTTP web traffic. The control connection on port 21 remains open throughout the FTP session and carries all user commands and server responses.
Why the other options are wrong
- B. Port 25 is the default port for SMTP (Simple Mail Transfer Protocol), used for email transmission.
- C. Port 23 is the default port for Telnet, a remote terminal protocol.
- D. Port 80 is the default port for HTTP (Hypertext Transfer Protocol), used for web browsing.
Question 49
A sensor on an access switch captures the following IPv6 packet: a 40-byte IPv6 header followed by an ICMPv6 message.
0000 60 00 00 00 00 10 3a 40 fe 80 00 00 00 00 00 00
0010 02 a0 c9 ff fe 14 7b 31 ff 02 00 00 00 00 00 00
0020 00 00 00 00 00 00 00 01 86 00 ef 41 40 00 07 08
0030 00 00 00 00 00 00 00 00
How should a receiving host process this message?
Show answer and explanation
Correct answer: C. Discard it, because a Router Advertisement must arrive with Hop Limit 255
Next Header 0x3a = 58 is ICMPv6, and the first ICMPv6 byte 0x86 = 134 makes this a Router Advertisement from fe80::2a0:c9ff:fe14:7b31 to all-nodes ff02::1. Byte 7 of the IPv6 header is the Hop Limit, 0x40 = 64. Hosts must silently discard any Router Advertisement whose Hop Limit is not 255, because only 255 proves the packet was not forwarded by a router. A value of 64 points to a crafted or off-link advertisement and is worth an alert, even though the source and destination look normal.
Why the other options are wrong
- A. A link-local source is required but not sufficient, and this advertisement fails the Hop Limit 255 check.
- B. The all-nodes address ff02::1 is the normal destination of an unsolicited Router Advertisement.
- D. Router Lifetime is counted in seconds, and the message must be discarded anyway because its Hop Limit is 64.
Question 50
Which notation is used to represent IPv6 addresses?
Show answer and explanation
Correct answer: B. Colon-hexadecimal
IPv6 addresses are 128 bits long and are written in colon-hexadecimal notation: eight groups of four hexadecimal digits separated by colons, for example 2001:0db8:0000:0000:0000:ff00:0042:8329, with leading zeros omitted and one run of zero groups optionally compressed to a double colon.
Why the other options are wrong
- A. Colon-dot is not a defined address format for IPv6.
- C. Hexadecimal-dot notation is not used; hexadecimal groups in IPv6 are separated by colons, not dots.
- D. Dot (dotted decimal) notation is the four-octet format used for IPv4 addresses.
Question 51
Which of the following can be monitored by using the host intrusion detection system (HIDS)?
Each correct answer represents a complete solution.
Choose two.
Show answer and explanation
Correct answer: B, D. File system integrity; System files
A Host Intrusion Detection System (HIDS) monitors file system integrity by detecting unauthorized changes to files and system files by tracking modifications, deletions, and additions. System files are critical assets that HIDS specifically protects. Computer performance monitoring and storage space monitoring are system administration functions not part of HIDS's intrusion detection mission.
Why the other options are wrong
- A. Computer performance monitoring is a system management function, not an intrusion detection function.
- C. Storage space monitoring is capacity planning, not intrusion detection.
Question 52
Which of the following terms describes an attempt to transfer DNS zone data?
Show answer and explanation
Correct answer: A. Reconnaissance
Attempting to transfer DNS zone data is a reconnaissance technique called a DNS zone transfer. Reconnaissance is the gathering of information about a target system before launching an attack, and DNS zone transfers are a common method to enumerate DNS records and discover network infrastructure.
Why the other options are wrong
- B. Encapsulation refers to wrapping data with protocol headers, not DNS information gathering.
- C. Dumpster diving involves searching through physical trash for sensitive information, not attempting DNS zone transfers.
- D. Spam is unsolicited bulk messaging, not related to DNS zone data extraction.
Question 53
A router must forward a 4,000-byte IPv4 datagram (20-byte header, no options, DF clear) onto a link with a 1,500-byte MTU, and it builds the largest fragments the link allows. What are the Fragment Offset field value and the Total Length of the third fragment?
Show answer and explanation
Correct answer: B. Offset 370, Total Length 1040
The datagram carries 4000 - 20 = 3980 data bytes. Each fragment can hold 1500 - 20 = 1480 data bytes, which is 185 blocks of 8, so non-final fragments stay on 8-octet boundaries. The fragments carry 1480 bytes at offset 0, 1480 bytes at offset 185, and the remaining 3980 - 2960 = 1020 bytes at offset 2960 / 8 = 370. The third fragment's Total Length is 1020 + 20 = 1040 bytes, and its MF bit is clear because it ends the datagram.
Why the other options are wrong
- A. 2960 is the byte position of the data, but the Fragment Offset field counts 8-octet units, so it holds 370.
- C. 1020 bytes is the data alone, while Total Length also counts the 20-byte header of the fragment.
- D. 375 equals 3000 / 8, which wrongly counts each earlier fragment's 20-byte header as part of the data.
Question 54
You are analyzing a possibly compromised Windows host and need to identify the sessions it currently has established with remote systems. Which command-line utility displays the local host's active TCP and UDP connections?
Show answer and explanation
Correct answer: C. NETSTAT
NETSTAT is the command-line utility that displays all active TCP and UDP connections on a local host, along with listening ports and associated process information. This is essential for analyzing suspicious network activity on a potentially compromised system.
Why the other options are wrong
- A. PING is used to test connectivity to a remote host by sending ICMP echo requests.
- B. TRACERT traces the route packets take to a destination but does not show active connections.
- D. NSLOOKUP is used for DNS queries and name resolution, not for displaying active connections.
Question 55
A sensor records an IPv4 fragment with IHL 5, Total Length 120, MF clear and a Fragment Offset field of 0x1ffd. What is anomalous about this fragment?
Show answer and explanation
Correct answer: A. Reassembly would produce a 65,632-byte datagram, above the 65,535 maximum
0x1ffd = 8189 units, and 8189 x 8 = 65,512 bytes. The fragment carries 120 - 20 = 100 data bytes, so the data would end at byte 65,612, and adding a 20-byte header gives 65,632 bytes. Total Length is a 16-bit field, so no datagram may exceed 65,535 octets. IPv6 applies the same rule explicitly: a fragment whose length and offset would push the Payload Length of the reassembled packet past 65,535 octets must be discarded. Such fragments are a classic denial-of-service signature.
Why the other options are wrong
- B. The field is already counted in 8-octet units, so every offset value is aligned by definition.
- C. A last fragment has MF clear and normally a nonzero offset; MF clear with offset 0 is an unfragmented datagram.
- D. Only non-final fragments must carry multiples of 8 bytes, and this fragment is the last one.
Question 56
A firewall drops inbound TCP segments that have SYN set and ACK clear. A capture shows the first fragment of an inbound TCP datagram with IHL 5, Total Length 28, MF set and an offset of 0. Which TCP header field is missing from this fragment, letting it pass the rule?
Show answer and explanation
Correct answer: C. Control flags
Total Length 28 minus the 20-byte IP header leaves only 8 bytes of TCP header in this fragment: source port (2 bytes), destination port (2 bytes) and the sequence number (4 bytes). The flags sit in byte 13 of the TCP header, so they arrive in the next fragment. A filter that tests SYN and ACK cannot evaluate them in the first fragment and usually ignores later fragments, which is the tiny fragment attack. Filters defend by dropping fragments too short to hold the full transport header.
Why the other options are wrong
- A. The source port occupies TCP bytes 0 and 1, which are inside the 8 data bytes of this fragment.
- B. The destination port occupies TCP bytes 2 and 3, which are inside the 8 data bytes of this fragment.
- D. The sequence number occupies TCP bytes 4 to 7, the last four of the 8 data bytes in this fragment.
Question 57
Session splicing is an IDS evasion technique in which an attacker delivers data in multiple smallsized packets to the target computer. Hence, it becomes very difficult for an IDS to detect the attack signatures of such attacks.
Which of the following tools can be used to perform session splicing attacks?
Each correct answer represents a complete solution.
Choose all that apply.
Show answer and explanation
Correct answer: C, D. Whisker; Fragroute
Whisker and Fragroute are tools specifically designed to perform session splicing and packet fragmentation attacks to evade IDS detection. Whisker is an HTTP scanning tool capable of IDS evasion, and Fragroute is specifically designed to intercept, modify, and rewrite egress traffic for the purpose of evading network intrusion detection systems by fragmenting packets.
Why the other options are wrong
- A. Nessus is a vulnerability scanner tool used for identifying security weaknesses, not for performing session splicing IDS evasion attacks.
- B. Y.A.T. does not appear to be a recognized tool for session splicing or IDS evasion in standard security literature.
Question 58
Which of the following work as traffic monitoring tools in the Linux operating system?
Each correct answer represents a complete solution.
Choose all that apply.
Show answer and explanation
Correct answer: A, C, D. MRTG; IPTraf; Ntop
MRTG (Multi Router Traffic Grapher), IPTraf, and Ntop are all legitimate traffic monitoring and network analysis tools for Linux. MRTG creates graphs of network traffic, IPTraf provides real-time network statistics, and Ntop monitors network traffic by protocol. John the Ripper is a password cracking tool, not a traffic monitoring tool.
Why the other options are wrong
- B. John the Ripper is a password cracking utility, not a traffic monitoring tool.
Question 59
Which of the following tools can be used for passive OS fingerprinting?
Show answer and explanation
Correct answer: D. tcpdump
TCPDUMP is a packet capture utility that allows passive observation and analysis of network traffic without sending probes or actively querying systems. By analyzing captured packets, an analyst can identify operating systems through characteristics like TTL values, TCP window sizes, and other passive indicators. This makes it ideal for passive OS fingerprinting. DIG is a DNS query tool, NMAP is an active scanner that sends probes, and PING actively sends ICMP echo requests, all three require active network interaction rather than passive observation.
Why the other options are wrong
- A. DIG is used for DNS lookups, not OS fingerprinting.
- B. NMAP performs active scanning by sending probes, not passive fingerprinting.
- C. PING is an active probe that sends ICMP echo requests, not a passive analysis tool.
Question 60
An analyst building a lab packet with Scapy types the expression below and then serializes it to bytes.
raw(IP(dst='10.0.0.5')/TCP(dport=443))
What value does Scapy place in the IPv4 Protocol field of the resulting packet?
Show answer and explanation
Correct answer: C. 6, selecting TCP as the next header
The / operator composes layers, and a lower layer has default fields overloaded by the layer stacked above it. The IP layer alone defaults its proto field to 0, but placing TCP above it overloads that field so the serialized packet carries the IANA protocol number for TCP, which is 6. This is why Scapy prints proto=TCP for IP()/TCP(). The overload happens only because a recognized upper layer is bound to IP.
Why the other options are wrong
- A. The bare IP default of 0 is replaced once TCP is stacked above it.
- B. 1 is the number for ICMP, which is not the stacked layer here.
- D. 17 is the number for UDP, but the upper layer is TCP.
Question 61
An analyst must decrypt a browser session that negotiates TLS 1.3 with an ECDHE key exchange and will capture it again while reproducing the activity. The web team offers the server's RSA private key, and the workstation team can start the browser with the SSLKEYLOGFILE environment variable set. Which approach will decrypt the traffic?
Show answer and explanation
Correct answer: C. Point the TLS key log preference at the browser's key log file
A key log file holds per-session secrets that browsers write, through libraries such as NSS, OpenSSL or BoringSSL, when SSLKEYLOGFILE is set, and it enables decryption even when a Diffie-Hellman exchange is used. Wireshark reads it through the TLS preference (Pre)-Master-Secret log filename (tls.keylog_file). An RSA private key only helps when the cipher suite does not use (EC)DHE and the version is TLS 1.2 or earlier, so it cannot decrypt this session. The secrets must come from the same sessions that are captured.
Why the other options are wrong
- A. An RSA key works only for non-(EC)DHE cipher suites in SSLv3 and TLS 1.0 to 1.2, so it cannot decrypt TLS 1.3.
- B. Decode As changes which dissector parses the bytes; it cannot remove the encryption.
- D. Following a TLS stream shows decrypted data only after decryption secrets have been provided.
Question 62
You work as a Network Administrator for McRobert Inc. Your company has a TCP/IP-based network. You want to get the protocol statistics and the active TCP/IP network connections of your computer.
Which of the following will you use?
Show answer and explanation
Correct answer: E. NETSTAT
NETSTAT is the correct command-line utility for displaying protocol statistics and active TCP/IP network connections on a computer. It shows listening ports, established connections, and network activity in real-time, making it essential for network administrators to monitor network status and troubleshoot connectivity issues.
Why the other options are wrong
- A. IPSTAT is not a standard Windows or Linux command for displaying network statistics.
- B. SNMP is a network management protocol used for monitoring and managing network devices remotely, not for displaying local connection statistics.
- C. ARP is used to resolve IP addresses to MAC addresses, not to display protocol statistics or active connections.
- D. NBTSTAT displays NetBIOS over TCP/IP statistics and is used for NetBIOS name resolution, not general TCP/IP connection information.
Question 63
Every network device contains a unique built in Media Access Control (MAC) address, which is used to identify the authentic device to limit the network access.
Which of the following addresses is a valid MAC address?
Show answer and explanation
Correct answer: A. A3-07-B9-E3-BC-F9
A valid MAC address consists of 48 bits represented as six pairs of hexadecimal digits separated by hyphens or colons (A3-07-B9-E3-BC-F9 is correct). Option B is an IP address format, C uses an incorrect notation with periods and longer segments, and D uses binary representation with incorrect separators for MAC addresses.
Why the other options are wrong
- B. This is an IPv4 address format with decimal octets, not a MAC address.
- C. This format uses periods and groups of four hexadecimal characters, which is not valid MAC address notation.
- D. This appears to be binary representation with incorrect separators; MAC addresses use hexadecimal with hyphens or colons.
Question 64
You work as a Network Administrator for TechPerfect Inc. The company has a corporate intranet setup. A router is configured on your network to connect outside hosts to the internetworking. For security, you want to prevent outside hosts from pinging to the hosts on the internetwork.
Which of the following steps will you take to accomplish the task?
Show answer and explanation
Correct answer: A. Block the ICMP protocol through ACL.
Ping uses the ICMP (Internet Control Message Protocol) to send echo requests and receive echo replies. To prevent outside hosts from pinging internal hosts, you must block ICMP protocol through an Access Control List (ACL) on the router. TCP and UDP are transport layer protocols used for other services, while IPv6 is a network layer protocol for addressing.
Why the other options are wrong
- B. IPv6 is a network protocol for addressing; blocking it would affect all IPv6 traffic, not just ping.
- C. UDP is a transport protocol; blocking it would affect many services beyond ping.
- D. TCP is a transport protocol; blocking it would affect many services unrelated to ping.
Question 65
Suricata alerts on overlapping IPv4 fragments with conflicting data sent to a Linux server at 10.20.30.40. The host-os-policy section of suricata.yaml is shown.
host-os-policy:
windows: [0.0.0.0/0]
bsd: []
linux: []
What should the analyst change so Suricata reassembles that traffic the way the server will?
Show answer and explanation
Correct answer: D. Add 10.20.30.40 to the linux list so its overlap handling is mimicked
Operating systems resolve overlapping fragment data in different ways. If a sensor keeps different bytes than the target, it inspects a datagram the target never sees, and an attack can slip past. Suricata's host-os-policy maps addresses to operating systems so the defrag and stream engines handle anomalies the way each host does. Here 0.0.0.0/0 maps every address to windows, so the Linux server's address belongs in the linux list.
Why the other options are wrong
- A. The memcap sets how much memory the defrag engine may use, not which copy of overlapping bytes is kept.
- B. The timeout controls how long incomplete fragments wait, not how overlapping data is resolved.
- C. fragbits:M; matches every fragment that has MF set, including normal ones, and does not change reassembly.
Question 66
At which layers of the OSI and TCP/IP models does IP addressing function?
Show answer and explanation
Correct answer: D. OSI Layer 3 and TCP/IP Internet Layer
IP addressing operates at OSI Layer 3 (Network Layer) and the TCP/IP Internet Layer, which are functionally equivalent layers in their respective models. Both layers handle logical addressing and routing of packets across networks using IP addresses.
Why the other options are wrong
- A. OSI Layer 5 is the Session Layer, not related to IP addressing; TCP/IP Transport Layer handles ports, not IP addresses.
- B. OSI Layer 2 is the Data Link Layer which handles MAC addresses, not IP addresses; this is incorrect.
- C. OSI Layer 4 is the Transport Layer for TCP/UDP; TCP/IP Application Layer does not handle IP addressing functions.
Question 67
An analyst runs Wireshark on a Linux server with TCP checksum validation turned on. Every packet the server sends shows an invalid TCP checksum, while every packet it receives validates. What is the most likely cause?
Show answer and explanation
Correct answer: D. Checksum offloading leaves the field for the NIC to fill
With checksum offloading the operating system hands the NIC a packet whose checksum field is empty or garbage, and the hardware computes the real value during transmission. A capture taken on the sending host sees the packet before the NIC fills the field, so Wireshark flags it as invalid even though the packet on the wire is correct. Received packets crossed real hardware and validate normally. Because offloading is so common, recent Wireshark releases disable checksum validation by default.
Why the other options are wrong
- A. Corruption on the wire happens after the capture point, which records outbound packets before they reach the cable.
- B. Failures limited to locally sent packets, with every received packet valid, match offloading at the capture point rather than tampering.
- C. A switch handles frames after the capture point on the sending host, so it cannot change what that capture recorded.
Question 68
John, a malicious hacker, forces a router to stop forwarding packets by flooding it with many open connections simultaneously so that all hosts behind it are effectively disabled.
Which of the following attacks is John performing?
Show answer and explanation
Correct answer: B. DoS attack
John is flooding the router with many open connections simultaneously to exhaust its resources and prevent it from forwarding packets to hosts behind it. This is a classic Denial of Service (DoS) attack, which aims to make systems or services unavailable by overwhelming them with traffic or requests.
Why the other options are wrong
- A. Rainbow attack is a type of password cracking attack using precomputed hash tables, not a network flooding attack.
- C. ARP spoofing involves forging ARP messages to associate an attacker's MAC address with a victim's IP address, not flooding connections.
- D. Replay attack involves capturing and resending valid data transmissions, not flooding a router with connections.
Question 69
Which of the following firewalls keeps track of the state of network connections traveling across the network?
Show answer and explanation
Correct answer: A. Stateful firewall
A stateful firewall keeps track of the state of network connections by maintaining a state table of active connections, allowing it to make intelligent decisions about whether to allow return traffic based on the established connection state.
Why the other options are wrong
- B. Application-level firewalls operate at Layer 7 and inspect application data, not specifically track connection state.
- C. Packet filtering firewalls examine individual packets in isolation without maintaining connection state information.
- D. Circuit-level firewalls work at the Session Layer but do not maintain detailed connection state tracking like stateful firewalls.
Question 70
An analyst wants packets that use any TCP port from 6660 to 6669 and writes the filter below. The result also shows ordinary web traffic, such as a client on port 54012 talking to a server on port 80.
tcp.port >= 6660 && tcp.port <= 6669
Which filter returns only packets with a port in that range?
Show answer and explanation
Correct answer: A. tcp.port in {6660..6669}
tcp.port holds both the source and the destination port, and each comparison is satisfied when any occurrence matches. In the web packet, 54012 satisfies tcp.port >= 6660 and 80 satisfies tcp.port <= 6669, so the combined test passes although neither port is in the range. The membership operator tests a single field occurrence against the whole range, so tcp.port in {6660..6669} matches only when one port value itself lies between 6660 and 6669.
Why the other options are wrong
- B. A range is written inside a membership set with the in operator; == does not accept a range.
- C. With || almost every packet matches, for example any source port at or above 6660.
- D. With || every port satisfies one side or the other, so all TCP traffic is shown.
Question 71
An analyst right-clicks a packet, chooses Follow, TCP Stream, reads the conversation and closes the dialog with the Close button. The packet list now shows only that one conversation. Why, and how could the earlier view have been kept?
Show answer and explanation
Correct answer: C. Follow applied a stream display filter; Back would have restored it
Following a stream applies a display filter that selects all packets of that stream, for TCP of the form tcp.stream eq N, and the dialog shows the reassembled data with client-to-server bytes in red and server-to-client bytes in blue. Closing with Close leaves the stream filter in effect, which is why only one conversation remains. Closing with Back restores the previous display filter, and Filter out this stream removes the stream from view instead.
Why the other options are wrong
- A. Marking is a separate feature; following a stream sets a display filter rather than marking packets.
- B. Follow shows the data in a dialog and can save it with Save as, but it does not open a new capture file.
- D. Ignoring packets is a separate feature; the other packets are hidden by the stream's display filter.
Question 72
Which of the following proxy servers is placed anonymously between the client and remote server and handles all of the traffic from the client?
Show answer and explanation
Correct answer: C. Forced proxy server
A forced proxy server is placed anonymously between the client and remote server, transparently intercepting and handling all client traffic without the client's explicit configuration. It operates transparently at the network level, forcing clients to route through it. Caching proxies focus on content storage, web proxies are client-configured, and open proxies allow any external access, none of these describe forced transparent interception.
Why the other options are wrong
- A. Caching proxy servers focus on storing and serving cached content, not transparent traffic interception.
- B. Web proxy servers require explicit client configuration and are not placed anonymously.
- D. Open proxy servers allow external access but do not specifically describe anonymous transparent interception.
Question 73
What does a firewall check to prevent certain ports and applications from getting the packets into an Enterprise?
Show answer and explanation
Correct answer: D. The transport layer port numbers and the application layer headers
Firewalls inspect both transport layer port numbers (TCP/UDP ports) and application layer headers to determine whether to allow or block packets. This dual-layer inspection enables firewalls to make granular decisions about which ports and applications can access the enterprise network. The presentation and session layers (options B and C) are not typically where port-level filtering occurs, and option A incorrectly combines network layer headers with transport layer port numbers.
Why the other options are wrong
- A. Network layer headers do not contain port information; ports are a transport layer concept.
- B. The presentation and session layers are not where firewall port filtering occurs.
- C. Network layer headers lack port numbers, and session layer is not involved in port-level packet filtering.
Question 74
Which of the following tools is described below?
It is a set of tools that are used for sniffing passwords, e-mail, and HTTP traffic. Some of its tools include arpredirect, macof, tcpkill, tcpnice, filesnarf, and mailsnarf. It is highly effective for sniffing both switched and shared networks. It uses the arpredirect and macof tools for switching across switched networks. It can also be used to capture authentication information for FTP, telnet, SMTP, HTTP, POP, NNTP, IMAP, etc.
Show answer and explanation
Correct answer: A. Dsniff
Dsniff is a collection of network auditing and penetration testing tools that captures passwords, e-mail, and HTTP traffic. Its component utilities include arpredirect, macof, tcpkill, tcpnice, filesnarf, and mailsnarf, and it uses arpredirect and macof to intercept traffic on switched networks as well as shared ones. It parses many application protocols to extract authentication data, including FTP, Telnet, SMTP, HTTP, POP, NNTP, and IMAP.
Why the other options are wrong
- B. Libnids is a programming library that reassembles TCP streams for network intrusion detection tools; it is not a password, e-mail and HTTP sniffing toolkit with utilities such as arpredirect and mailsnarf.
- C. Cain is a Windows password recovery and sniffing tool, but it does not contain arpredirect, macof, tcpkill, filesnarf, or mailsnarf.
- D. LIDS (Linux Intrusion Detection System) is a kernel patch that hardens Linux with access controls rather than a ready set of sniffing tools.
Question 75
A sensor shows one internal workstation sending SYNs to 1,000 ports on a server. For every open port it completes the three-way handshake and then closes the connection at once without sending data; closed ports answer with RST/ACK. The user has no administrator rights and no raw-socket tools. Which nmap scan type matches?
Show answer and explanation
Correct answer: B. TCP connect scan (-sT)
A connect scan asks the operating system to open each connection with the connect system call, the same call ordinary applications use. The OS therefore completes the full three-way handshake to every open port, and nmap then closes the connection without sending data. It needs no raw-packet privileges, which is why nmap falls back to it for unprivileged users, and why services may log the brief connections. The completed handshakes separate it from the half-open SYN scan.
Why the other options are wrong
- A. A SYN scan needs raw-packet privileges and resets each open port instead of completing the handshake.
- C. An ACK scan sends unsolicited ACKs and never completes handshakes; it maps filtering only.
- D. A null scan sends flagless raw probes, which need privileges and never form connections.
Question 76
An analyst crafts a TCP segment in Scapy and leaves the chksum field at its default. What is written into the checksum field when the packet is serialized for sending?
Show answer and explanation
Correct answer: C. A correct checksum computed over the packet
Both IP and TCP default the chksum field to None, and when the packet is built Scapy computes the correct checksum for the current contents. This is one of the sensible defaults that let a one-line expression produce a valid packet. The behavior matters for analysts: a captured packet whose checksum does not match its contents was hand-set to a wrong value, corrupted in transit, or captured on the sending host before checksum offloading filled it in.
Why the other options are wrong
- A. TCP has no way to omit the checksum, and Scapy does not write 0x0000 by default.
- B. The checksum is calculated, not randomized.
- D. No fixed placeholder is written; the field is computed.
Question 77
A sensor logs an IPv4 packet whose header begins with the bytes below.
43 00 00 28
What is the most accurate assessment of the header?
Show answer and explanation
Correct answer: D. Malformed IPv4 header, since IHL 3 is below 5
0x43 is version 4 with an IHL of 3. IHL counts 32-bit words, so the field describes a 3 x 4 = 12 byte header, yet the fixed IPv4 header alone is 20 bytes and the minimum valid IHL is 5. A receiver cannot locate the transport header correctly, so the packet is malformed. Normal stacks never emit such a value; it usually points to a packet crafting tool, fuzzing, or an attempt to confuse a parser or sensor.
Why the other options are wrong
- A. Options exist only when IHL is above 5, and an IHL of 3 cannot even hold the fixed header.
- B. 0x43 is two nibbles, version 4 and IHL 3, not a decimal length of 43.
- C. IPv4 defines no short header form, and 20 bytes is the smallest valid header.
Question 78
In a capture, a TCP ACK appears as a 60-byte Ethernet II frame (FCS not captured). The IPv4 Total Length is 0x0028 and the IHL is 5. The last 6 bytes of the frame are all 0x00. What are those 6 bytes?
Show answer and explanation
Correct answer: C. Ethernet padding added to reach the minimum data size
0x0028 is 40 bytes: a 20-byte IPv4 header plus a 20-byte TCP header with no data. With the 14-byte Ethernet header the packet uses 54 bytes, so 60 - 54 = 6 bytes remain. Ethernet requires a data field of at least 46 bytes; the 40-byte IP packet falls 6 bytes short, so the sender pads with zeros. The padding is not part of the IP packet and is not counted in Total Length. Non-zero or varying trailer bytes would deserve a closer look.
Why the other options are wrong
- A. Total Length shows the datagram ends at byte 40, and zero bytes that bring the frame exactly to the minimum size are padding.
- B. TCP options sit inside the IP datagram and would be counted in Total Length, which covers only two 20-byte headers here.
- D. An IHL of 5 means the IPv4 header carries no options.
Question 79
An analyst has a text file, bad.txt, with one threat-intelligence IP address per line and wants every flow in which either endpoint is on that list. Which approach works?
Show answer and explanation
Correct answer: A. rwsetbuild bad.txt bad.set, then rwfilter --anyset=bad.set
The --sipset, --dipset and --anyset switches need a binary IPset file. rwsetbuild creates one from text input, reading the first argument and writing the second. --anyset then passes a record if its source or its destination address is in the set, so both inbound and outbound contact with the listed hosts is found in one query. --sipset alone would test only the source address.
Why the other options are wrong
- B. rwset builds IPsets from flow records, not from text, and --sipset needs a binary IPset rather than a text file.
- C. --any-cidr takes a comma-separated list of addresses and CIDR blocks, not the name of a file.
- D. --sipset tests only the source address, so flows where a listed host is the destination are missed.
Question 80
Crafted scan traffic often uses a bare 20-byte TCP header, while real clients add options. An analyst wants a filter that matches TCP segments whose header is longer than 20 bytes, meaning options are present. Which filter does this?
Show answer and explanation
Correct answer: B. tcp[12] & 0xf0 > 0x50
The data offset is the high nibble of TCP byte 12 and counts 32-bit words. A header with no options is 5 words (20 bytes), which appears in byte 12 as 0x50 after masking with 0xf0. Any value above 0x50, such as 0x80 for a 32-byte header, means the header carries options. Reversing the test (equal to 0x50) together with a SYN check is a quick way to surface tool-generated SYNs that lack the options normal stacks send.
Why the other options are wrong
- A. The low nibble of byte 12 holds reserved bits, so it does not reflect the header length.
- C. Without masking, any normal byte 12 value such as 0x50 is already greater than 5, so every segment matches.
- D. Byte 13 is the flags byte, so masking it with 0xf0 tests CWR, ECE, URG and ACK, not the header length.
Question 81
While reviewing traffic captured from a network management segment, an analyst needs to identify polling requests sent from a management station to a device's SNMP agent. Which port do these requests target?
Show answer and explanation
Correct answer: D. UDP port 161
SNMP (Simple Network Management Protocol) agents listen on UDP port 161 for polling requests from management stations. This is the standard port for SNMP query operations. TCP port 110 is for POP3, port 25 is for SMTP, and port 80 is for HTTP, none of which are used for SNMP management.
Why the other options are wrong
- A. TCP port 110 is used for POP3 email protocol, not SNMP.
- B. TCP port 25 is used for SMTP email protocol, not SNMP.
- C. TCP port 80 is used for HTTP web protocol, not SNMP.
Question 82
Which of the following Web attacks is performed by manipulating codes of programming languages such as SQL, Perl, Java present in the Web pages?
Show answer and explanation
Correct answer: B. Code injection attack
Code injection attacks involve manipulating programming language code (such as SQL, Perl, Java, or Python) embedded in web pages to execute malicious instructions. This includes SQL injection, command injection, and other variants where attacker-supplied input is interpreted as executable code. While command injection is a specific type of code injection focusing on OS commands, the broader category that encompasses manipulation of multiple programming languages is code injection attack.
Why the other options are wrong
- A. Command injection is a subset of code injection focusing specifically on operating system commands.
- C. Cross-Site Request Forgery (CSRF) tricks users into performing unwanted actions, not manipulating code.
- D. Cross-Site Scripting (XSS) injects scripts into web pages but is not primarily about manipulating existing code languages like SQL or Perl.
Question 83
An analyst reviews the following Snort 3 rule.
alert tcp $HOME_NET any -> any 22 (msg:"Client SSH version is not 2.0"; flow:established,to_server; content:"SSH-",depth 4; content:!"SSH-2.0",depth 7; sid:1000005; rev:1;)
Which client data makes this rule alert?
Show answer and explanation
Correct answer: B. An identification line that begins with SSH-1.5-PuTTY_0.58
Every option must pass. The first content requires SSH- in the first four payload bytes, and the negated content passes only when SSH-2.0 is absent from the first seven bytes. A client string beginning SSH-1.5 satisfies both, so the rule flags clients offering an old protocol version. A normal SSH-2.0 client fails the negated check, and data not starting with SSH- fails the first check. Negated content is a common way to flag the unusual variant of an otherwise expected value.
Why the other options are wrong
- A. SSH-2.0 is present in the first seven bytes, so the negated content check fails.
- C. The first content match requires SSH- in the first four bytes, so such packets never alert.
- D. The header and flow:to_server limit the rule to client data sent to port 22, so server lines are not inspected.
Question 84
You work as a Network Administrator for Net Perfect Inc. The company has a Windows Server 2008-based network. You have created a test domain for testing IPv6 addressing.
Which of the following types of addresses are supported by IPv6?
Each correct answer represents a complete solution.
Choose all that apply.
Show answer and explanation
Correct answer: A, B, D. Unicast; Multicast; Anycast
IPv6 supports three types of addresses: unicast (one-to-one communication), multicast (one-to-many communication), and anycast (one-to-nearest communication). Broadcast addressing, which was used in IPv4, is not supported in IPv6; multicast is used instead for similar purposes. The three supported types enable flexible communication patterns for modern networking requirements.
Why the other options are wrong
- C. Broadcast addressing is not supported in IPv6; multicast addresses replace this functionality.
Question 85
A local Suricata rule with gid 1 and sid 9000101 alerts on connections to a file-sharing service. Every alert involving the backup server 10.20.0.15 is a known false positive, whether that server is the source or the destination, and the rule must keep alerting for all other hosts. Which threshold.config entry achieves this?
Show answer and explanation
Correct answer: D. suppress gen_id 1, sig_id 9000101, track by_either, ip 10.20.0.15
A suppress entry in threshold.config silences one rule for a host or network while leaving it active for everything else. The track option decides which address is compared: by_src checks the source, by_dst the destination, and by_either suppresses the alert when either the source or the destination matches. Because the backup server appears on both sides of the alerting traffic, by_either with ip 10.20.0.15 removes exactly the known false positives. Suppression stops only the alert; state changes such as flowbits set by the rule still happen.
Why the other options are wrong
- A. Without track and ip, the entry turns the rule into noalert for every host, hiding true positives as well.
- B. track by_src suppresses only alerts where 10.20.0.15 is the source, so alerts with the server as destination would remain.
- C. A limit threshold only caps how many alerts each source produces per period; it still alerts on the backup server.
Question 86
An attacker makes an attempt against a Web server. The result is that the attack takes the form of URLs. These URLs search for a certain string that identifies an attack against the Web server.
Which IDS/IPS detection method do the URLs use to detect and prevent an attack?
Show answer and explanation
Correct answer: D. Signature-based detection
Signature-based detection identifies attacks by looking for known patterns or signatures within network traffic and URLs. The question describes searching for specific strings that identify attacks, which is the definition of signature-based detection. Anomaly-based detection looks for deviations from normal behavior, policy-based detection enforces security policies, and honeypot detection uses decoy systems, none match the described string-matching approach.
Why the other options are wrong
- A. Anomaly-based detection identifies deviations from normal behavior, not specific attack signatures.
- B. Policy-based detection enforces predefined security policies, not signature pattern matching.
- C. Honeypot detection uses decoy systems to trap attackers, not URL string analysis.
Question 87
An authorized nmap FIN scan (-sF) of a Windows file server reports every port closed, although the server is known to offer SMB on TCP 445. The capture shows RST/ACK returned for every FIN probe, including the one to port 445. What explains the result?
Show answer and explanation
Correct answer: A. The stack sends RST to these probes whether the port is open or not
FIN, null and Xmas scans rely on the TCP rule that a closed port answers a segment without SYN, RST or ACK with a reset, while an open port drops it. Not every stack follows that rule: Microsoft Windows, many Cisco devices, BSDI and IBM OS/400 send RST whether the port is open or not, so nmap labels every port closed. A defender reviewing such a scan should not read the RSTs as proof that services are down; a SYN scan separates the states reliably.
Why the other options are wrong
- B. A rewritten SYN would draw SYN/ACK from the open SMB port, not RST/ACK.
- C. A bare FIN from a new source port matches no existing session on the server.
- D. Under the standard TCP behavior a closed port sends RST and an open port stays silent, not the reverse.
Question 88
Which of the following technologies is used to detect unauthorized attempts to access and manipulate computer systems locally or through the Internet or an intranet?
Show answer and explanation
Correct answer: B. Intrusion detection system (IDS)
An Intrusion Detection System (IDS) is specifically designed to detect unauthorized attempts to access and manipulate computer systems, whether locally or remotely via Internet or intranet. It monitors network traffic and system activity for suspicious patterns that indicate potential attacks. DMZ is a network architecture concept, firewalls are preventive barriers, and packet filtering is a basic access control method, none of these detect intrusions the way an IDS does.
Why the other options are wrong
- A. A DMZ is a network zone between internal and external networks, not a detection mechanism.
- C. A firewall blocks traffic based on rules but does not detect intrusion attempts in the way an IDS does.
- D. Packet filtering is a basic access control technique that examines packet headers, not a detection system.
Question 89
An analyst extracts the ARP payload (the bytes after the 14-byte Ethernet header) from a frame captured on a user VLAN:
0000 00 01 08 00 06 04 00 02 3c 52 82 1a 7f 44 c0 a8
0010 0a 01 a4 bb 6d 10 22 9e c0 a8 0a 25
What does this ARP message assert?
Show answer and explanation
Correct answer: B. 192.168.10.1 is at 3c:52:82:1a:7f:44, in a reply sent to 192.168.10.37
ARP fields follow a fixed order: hardware type 0x0001 (Ethernet), protocol type 0x0800 (IPv4), hardware length 6, protocol length 4, then the 2-byte opcode, sender MAC, sender IP, target MAC and target IP. Opcode 0x0002 is a reply. The sender MAC is 3c:52:82:1a:7f:44 and the sender IP is c0 a8 0a 01 = 192.168.10.1; the target is a4:bb:6d:10:22:9e at c0 a8 0a 25 = 192.168.10.37. If 192.168.10.1 is the gateway and its known MAC differs, the reply deserves a closer look.
Why the other options are wrong
- A. Opcode 0x0002 is a reply, and a4:bb:6d:10:22:9e with 192.168.10.37 sits in the target fields, not the sender fields.
- C. A request carries opcode 0x0001, while this message carries 0x0002.
- D. The hardware address paired with 192.168.10.1 in the sender fields is 3c:52:82:1a:7f:44; a4:bb:6d:10:22:9e is the target's address.
Question 90
A single 9-hour SSH session appears in SiLK as several flow records. rwcut shows the attributes field as T on the first record, CT on the middle records and C on the last. What explains this?
Show answer and explanation
Correct answer: D. The flow generator's active timeout split a long-lived session
Flow generators close the record for a long-lived connection when it reaches the active timeout and start a new record for the traffic that follows. SiLK marks the first record T (it hit the timeout), the middle records CT (a continuation that also timed out) and the last record C (a continuation). To measure the real session, an analyst combines the records, for example summing bytes per 5-tuple with rwuniq, instead of counting each as a separate connection.
Why the other options are wrong
- A. C only marks a record created as a continuation after a timeout and says nothing about who sent the packets.
- B. Equal packet sizes are marked with S, not T.
- C. The T and C markers record active-timeout splits, not packet loss at the sensor.
Question 91
Which of the following types of firewall functions by creating two different communications, one between the client and the firewall, and the other between the firewall and the end server?
Show answer and explanation
Correct answer: B. Proxy-based firewall
A proxy-based firewall operates by acting as an intermediary between clients and servers, creating two separate communication channels: one between the client and the firewall (proxy), and another between the firewall and the destination server. This breaks the direct connection and allows the firewall to inspect, filter, and control both directions of traffic while hiding internal systems from external view.
Why the other options are wrong
- A. Stateful firewalls track connection states but do not necessarily create two separate communications channels.
- C. Packet filter firewalls examine individual packets but do not create intermediary connections.
- D. Endian is a firewall vendor/product, not a type of firewall architecture.
Question 92
You are responsible for security at a company that specializes in e-commerce. You realize that given the high volume of Web traffic, there is a significant chance of someone being able to breach your perimeter. You want to make sure that should this occur, you can redirect the attacker away from sensitive data.
How would you best accomplish this?
Show answer and explanation
Correct answer: B. Implement a honey pot.
A honeypot is a decoy system designed to attract and redirect attackers away from real, sensitive assets by appearing as an attractive target. This directly accomplishes the stated goal of redirecting attackers away from sensitive data. An IDS passively monitors or actively blocks threats but does not redirect attackers. A stateful firewall filters traffic but does not provide deception or redirection capabilities.
Why the other options are wrong
- A. A passive IDS only monitors and logs traffic without actively redirecting attackers away from sensitive data.
- C. A stateful firewall filters and inspects packets but does not provide deception or active redirection of attackers.
- D. A network-based IDS detects attacks but does not inherently redirect attackers away from sensitive systems.
Question 93
An analyst pulls a day of data with the command below and finds no flows leaving the network, although outbound DNS is known to occur.
rwfilter --start-date=2026/09/14 --proto=17 --pass=udp.rw
What is the most likely reason?
Show answer and explanation
Correct answer: A. The class's default types often cover only incoming traffic
Without --type or --flowtypes, rwfilter uses the default type list of the selected class, and those defaults often include only incoming traffic. Types usually encode the direction of the flow, so outbound records sit under other types. Adding --type=out, or --type=all, brings in the missing direction, and rwsiteinfo --fields=class,type,default-type shows the types and defaults defined at the site.
Why the other options are wrong
- B. Without --sensors, rwfilter uses every sensor valid for the class, so sensors are not the cause.
- C. Types typically describe the direction of the flow, not its protocol, so UDP is not stored separately.
- D. A day-precision start date with no end date processes the complete day.
Question 94
Which of the following is used as a default port by the TELNET utility?
Show answer and explanation
Correct answer: C. 23
TELNET uses port 23 as its default port for remote terminal access. This is the standard port assigned to the TELNET protocol for establishing unencrypted remote connections to servers.
Why the other options are wrong
- A. Port 21 is used by FTP (File Transfer Protocol), not TELNET.
- B. Port 80 is used by HTTP (HyperText Transfer Protocol) for web browsing, not TELNET.
- D. Port 20 is used by FTP for data transfer, not TELNET.
Question 95
Which of the following protocols is used by voice over IP (VoIP) applications?
Show answer and explanation
Correct answer: A. UDP
UDP (User Datagram Protocol) is the primary transport protocol for VoIP applications because it offers low latency and minimal overhead, which are essential for real-time voice communication. While some VoIP signaling may use TCP, the actual voice data transmission relies on UDP for speed. TCP is connection-oriented and has higher overhead, making it less suitable for real-time voice. ICMP is used for diagnostic purposes like ping and traceroute, not for VoIP. IPv6 is a network layer protocol that does not define transport mechanisms for voice.
Why the other options are wrong
- B. TCP is connection-oriented with acknowledgment requirements, causing latency unsuitable for real-time voice communication.
- C. ICMP is an Internet Control Message Protocol used for diagnostics, not for data transport.
- D. IPv6 is a network layer protocol addressing scheme, not a transport protocol for VoIP applications.
Question 96
An IDS logs the first 14 bytes of a TCP header from an inbound packet.
c3 51 01 bb 5e 2a 91 07 00 00 00 00 50 29
What kind of probe is it?
Show answer and explanation
Correct answer: D. An Xmas scan probe that has FIN, PSH and URG set
Byte 12 is 0x50: data offset 5, a 20-byte header, with the reserved bits clear. Byte 13 holds the flags: 0x29 = 0010 1001, which sets bit 5 (URG, 0x20), bit 3 (PSH, 0x08) and bit 0 (FIN, 0x01), while ACK (0x10), RST (0x04) and SYN (0x02) are clear. FIN, PSH and URG together is the nmap Xmas scan. The zero acknowledgment field fits a probe without ACK, and destination port 0x01bb is 443.
Why the other options are wrong
- A. The SYN bit, 0x02, is clear in 0x29, so it is not a SYN probe.
- B. The ACK bit, 0x10, is clear in 0x29, so FIN/ACK is not set.
- C. The flags byte is nonzero and byte 12's reserved bits are clear, so this is not a null probe.
Question 97
Which of the following are not functions of the SNORT application?
Each correct answer represents a complete solution.
Choose two.
Show answer and explanation
Correct answer: B, C. Virus detection; Hard disk drive scanning
SNORT is a network intrusion detection and prevention system. It performs packet sniffing, packet logging, and intrusion detection, but it is not designed for virus detection or hard disk drive scanning, those are antivirus and storage diagnostics functions respectively, outside SNORT's scope as a network-based tool.
Why the other options are wrong
- A. Packet logging is a core function of SNORT.
- D. Packet sniffing is fundamental to SNORT's operation.
- E. Intrusion detection is the primary purpose of SNORT.
Question 98
An analyst wants tcpdump to catch nmap Null scan probes, which carry a TCP header with no flag bits set at all. Which filter matches them?
Show answer and explanation
Correct answer: B. tcp[tcpflags] == 0
tcp[tcpflags] reads byte 13 of the TCP header, the flags byte, and a Null scan sends that byte as 0x00. Comparing the whole byte with 0 therefore matches only segments with no flags at all. Legitimate TCP stacks always set at least one flag (SYN to open, ACK once established, RST or FIN to close), so a flagless segment is a strong sign of scanning or crafted traffic.
Why the other options are wrong
- A. It only requires FIN to be clear, which is true of most SYN and ACK segments.
- C. It matches every segment with at least one flag, which is the opposite of a Null probe.
- D. Byte 12 holds the data offset, which is at least 5 in any valid header, so it never equals 0 in normal traffic.
Question 99
Which of the following are the main components of an intrusion detection system? Each correct answer represents a complete solution. Choose two.
Show answer and explanation
Correct answer: A, E. Sensor; Console
Sensors and Consoles are key components of an Intrusion Detection System. Sensors are deployed throughout the network to monitor and collect traffic data for analysis, while the Console is the management interface where security analysts review alerts and manage the IDS. Gateways, Firewalls, and Modems are networking devices but are not intrinsic components of an IDS architecture.
Why the other options are wrong
- B. Gateway is a networking device for connecting networks but is not a dedicated IDS component.
- C. Firewall is a separate security tool used for access control, not a component of an IDS system.
- D. Modem is a communication device for establishing connections but is not related to IDS components.
Question 100
You work as a network administrator for BlueWell Inc. You have to convert your 48-bit host address (MAC address) to an IPv6 54-bit address.
Using the IEEE-EUI-64 conversion process, how do you convert the 48-bit host address (MAC address) to an IPv6 54-bit address?
Show answer and explanation
Correct answer: D. Add FF. FE between the third and fourth bytes
The IEEE EUI-64 process converts a 48-bit MAC address to a 64-bit interface identifier by inserting FFFE (not FFEE, FEEE, or EFFE) between the third and fourth octets of the MAC address. This is the standard method for creating IPv6 link-local addresses.
Why the other options are wrong
- A. EFFE is not the correct insertion value for EUI-64 conversion.
- B. FEEE is not the correct insertion value for EUI-64 conversion.
- C. FFEE is close but incorrect; the correct value is FFFE.
Question 101
An analyst hunting for a TCP covert channel wants segments that carry a nonzero urgent pointer even though the URG flag is clear. Which filter does this?
Show answer and explanation
Correct answer: C. tcp[18:2] != 0 and tcp[13] & 0x20 == 0
After the window (bytes 14-15) and checksum (16-17), the urgent pointer occupies bytes 18-19, so tcp[18:2] reads it. URG is 0x20 in the flags byte, and tcp[13] & 0x20 == 0 means it is clear. The urgent pointer is only interpreted when URG is set, so receivers ignore it otherwise, which makes a nonzero value with URG clear a place to hide data or a mark of crafted packets.
Why the other options are wrong
- A. Bytes 16 and 17 are the checksum, which is nonzero in nearly every segment.
- B. 0x10 is the ACK bit, so this looks for segments without ACK rather than without URG.
- D. Byte 12 holds the data offset and reserved bits, so 0x20 there is part of the header length, not URG.
Question 102
John works as a professional Ethical Hacker. He has been assigned the project of testing the security of www.we-are-secure.com. He wants to send malicious data packets in such a manner that one packet fragment overlaps data from a previous fragment so that he can perform IDS evasion on the We-are-secure server and execute malicious data.
Which of the following tools can he use to accomplish the task?
Show answer and explanation
Correct answer: C. Mendax
Mendax is a specialized tool designed to perform IDS evasion attacks through packet fragmentation overlap techniques, where fragments are deliberately crafted so that overlapping data from one fragment overwrites or obscures data from previous fragments, allowing malicious payloads to evade signature-based detection systems.
Why the other options are wrong
- A. Hunt is a network tool used for connection hijacking and session takeover, not for creating overlapping packet fragments for IDS evasion.
- B. Alchemy Remote Executor is a tool for remote code execution, not specifically designed for packet fragmentation-based IDS evasion.
- D. Ettercap is a network sniffer and man-in-the-middle tool, not specifically designed for overlapping fragment IDS evasion techniques.
Question 103
A Snort 3 rule that looks for the string PASS in client data sent to TCP port 21 keeps firing on single crafted packets sent without any handshake. Which flow option limits the rule to client data inside an established TCP connection?
Show answer and explanation
Correct answer: A. flow:established,to_server;
flow checks session properties tracked by Snort. established matches only packets that belong to an established TCP connection, so injected packets with no completed three-way handshake are ignored. to_server, the same as from_client, restricts the rule to traffic from the client toward the server, which is the direction an FTP PASS command travels. Combining state and direction is one of the simplest ways to cut false positives from crafted or scan traffic and to reduce the packets a rule must inspect.
Why the other options are wrong
- B. stateless matches regardless of stream state, so crafted packets outside a session still trigger the rule.
- C. to_client selects server responses, but the PASS command is sent by the client.
- D. not_established matches only when no TCP connection exists, which is exactly the crafted traffic to exclude.
Question 104
An analyst runs rwfilter with --pass=stdout directly in a terminal to look at the matching flows, and rwfilter refuses to write the output there. What should the analyst do?
Show answer and explanation
Correct answer: C. Pipe the pass stream into rwcut to print the records
rwfilter's pass, fail and all streams are always binary SiLK Flow records. They must go to a file or into another SiLK tool, and rwfilter complains when it detects that the stream is headed for a terminal. rwcut reads binary flow records and prints the chosen fields as bar-delimited text, so a pipeline of rwfilter with --pass=stdout feeding rwcut is the standard way to view results on screen.
Why the other options are wrong
- A. --print-statistics prints only summary counts, never the records themselves.
- B. rwsort writes binary records as well, so its output still needs rwcut to be read.
- D. The all destination is also binary, so it cannot be sent to a terminal either.
Question 105
Which of the following techniques allows probing firewall rule-sets and finding entry points into the targeted system or network?
Show answer and explanation
Correct answer: D. Packet crafting
Packet crafting is the technique that allows attackers to probe firewall rule-sets and find entry points by creating custom packets with specific characteristics to test how firewalls respond. By crafting packets with various combinations of flags, ports, and protocols, an attacker can identify which types of traffic the firewall allows or blocks. Network enumeration gathers information but doesn't directly probe firewall rules, packet collision is not a standard security technique, and DCC is an email anti-spam system.
Why the other options are wrong
- A. Network enumeration is used to discover network resources and systems but does not specifically probe firewall rule-sets.
- B. Packet collision is not a recognized technique for probing firewalls or finding network entry points.
- C. Distributed Checksum Clearinghouse (DCC) is a spam detection mechanism, not a firewall probing technique.
Question 106
An analyst sees these packets from an external host to an internal server behind a firewall (tcpdump -n -S). Port 82 never answers.
203.0.113.50.45001 > 10.10.0.8.80: Flags [.], ack 2875431109, win 1024, length 0
10.10.0.8.80 > 203.0.113.50.45001: Flags [R], seq 2875431109, win 0, length 0
203.0.113.50.45001 > 10.10.0.8.81: Flags [.], ack 2875431109, win 1024, length 0
10.10.0.8.81 > 203.0.113.50.45001: Flags [R], seq 2875431109, win 0, length 0
203.0.113.50.45001 > 10.10.0.8.82: Flags [.], ack 2875431109, win 1024, length 0
What is the external host learning?
Show answer and explanation
Correct answer: C. That 80 and 81 are unfiltered and 82 is filtered
Unsolicited ACKs with no prior handshake are an nmap ACK scan. A host that receives an ACK with no matching connection answers with a bare RST whose sequence number is copied from the probe's ACK field, here 2875431109, whether the port is open or closed. A RST therefore proves only that the probe got through (unfiltered). Silence or an ICMP unreachable means a filter dropped it, so port 82 is filtered. The scan maps firewall rules rather than open ports.
Why the other options are wrong
- A. Open and closed ports both answer an unsolicited ACK with RST, so RSTs cannot show which ports are open.
- B. Both RSTs carry window 0, so even the window scan's heuristic would not separate the two ports.
- D. No response to an ACK probe means filtered, not open, and the RSTs do not prove ports closed.
Question 107
An analyst runs the following rwfilter command against the SiLK repository.
rwfilter --start-date=2026/09/14 --type=in --proto=6 --dport=22 --pass=ssh.rw --fail=other.rw
Which records are written to other.rw?
Show answer and explanation
Correct answer: A. Every selected record that is not TCP to destination port 22
Partitioning switches are combined with AND: a record passes only if it satisfies every test, here protocol 6 and destination port 22, and it goes to ssh.rw. A record that fails any single test goes to the fail destination. other.rw therefore holds TCP flows to other ports, UDP and ICMP flows, and everything else selected for that day. Writing both streams in one pass is efficient because the repository is read only once.
Why the other options are wrong
- B. Non-TCP records also fail the protocol test, so they are written to other.rw as well.
- C. TCP records to other ports fail the port test and also land in other.rw.
- D. Failing any one predicate is enough, so a record does not need to fail both tests.
Question 108
An analyst extracts the following 20-byte IPv4 header, in hex, from a capture.
45 00 05 dc 1c 46 20 00 40 11 21 bb 0a 01 01 05 0a 02 02 09
What does the header show about fragmentation?
Show answer and explanation
Correct answer: B. The first fragment of a fragmented UDP datagram, with 1480 data bytes
Bytes 6 and 7 hold the flags and fragment offset: 0x2000 = 0010 0000 0000 0000, so the reserved bit is 0, DF is 0, MF is 1 and the 13-bit offset is 0. MF set with offset 0 marks the first fragment, the only one that carries the UDP header. Total Length 0x05dc = 1500 bytes minus the 20-byte header (IHL 5) leaves 1480 data bytes, which is 185 blocks of 8 as every non-final fragment requires. Protocol 0x11 = 17 identifies UDP.
Why the other options are wrong
- A. DF is the 0x4000 bit, which is clear here, and an unfragmented datagram has MF clear rather than set.
- C. The offset is only the low 13 bits (mask 0x1fff), which are all zero here, so the data starts at byte 0.
- D. A last fragment has MF clear and a nonzero offset, while this header has MF set and an offset of 0.
Question 109
Which of the following UDP ports are used by the Simple Network Management Protocol (SNMP) for agent queries and trap notifications? Each correct answer represents a complete solution. Choose two.
Show answer and explanation
Correct answer: B, D. UDP port 161; UDP port 162
UDP port 161 is used by SNMP (Simple Network Management Protocol) for network management and monitoring, and UDP port 162 is used by SNMP Trap for receiving notifications from managed devices. These are standard network management service ports.
Why the other options are wrong
- A. UDP port 69 is used by TFTP (Trivial File Transfer Protocol), a file transfer utility, not a primary network management service.
- C. UDP port 137 is used by NetBIOS Name Service for Windows name resolution, not a management service in the context of this question.
Question 110
The fragment summary below covers one second of traffic from one source to one destination (offsets shown in bytes).
10.4.4.4 > 10.9.9.9: id 4021, offset 0, flags [+], proto UDP (17), length 1500
10.4.4.4 > 10.9.9.9: id 4021, offset 0, flags [+], proto ICMP (1), length 1500
10.4.4.4 > 10.9.9.9: id 4021, offset 1480, flags [none], proto UDP (17), length 120
10.4.4.4 > 10.9.9.9: id 4021, offset 1480, flags [none], proto ICMP (1), length 120
How many original datagrams will the destination reassemble?
Show answer and explanation
Correct answer: D. Two, because the protocol field separates the UDP and ICMP fragment sets
A receiver groups fragments that match on all four fields: identification, source, destination and protocol. Here the ID and addresses match, but two fragments carry protocol 17 and two carry protocol 1, so they form two separate sets. Each set is complete: a first fragment with MF set holding 1480 data bytes, and a last fragment at byte 1480 holding 100 data bytes. A sender only has to keep an ID unique per source, destination and protocol, so this reuse is legal.
Why the other options are wrong
- A. Matching ID and addresses is not enough, because protocol is also part of the reassembly key.
- B. Fragments are combined, not treated as separate datagrams, whenever all four key fields match.
- C. The same offsets appear in two different datagrams, so no fragments overlap within any one datagram.
Question 111
Which of the following commands will you use with the tcpdump command to display the contents of the packets?
Show answer and explanation
Correct answer: D. tcpdump -A
The tcpdump -A flag displays the contents of packets in ASCII format, allowing analysts to view the actual payload data within captured packets. The -q flag produces quiet output with less detail, -v increases verbosity but focuses on header information, and -n prevents DNS name resolution, none of these display packet contents like -A does.
Why the other options are wrong
- A. The -q flag enables quiet mode, showing less information, not packet contents.
- B. The -v flag increases verbosity of header information, not payload content display.
- C. The -n flag disables DNS name resolution but does not display packet contents.
Question 112
A Snort 3 rule flags unusually large DNS queries with the option below. Each candidate packet is IPv4 with a 20-byte header (IHL 5) carrying UDP to port 53 and no IP options.
dsize:>512;
Which IP total length value belongs to a packet that matches?
Show answer and explanation
Correct answer: D. 548
dsize tests only the payload, so the 20-byte IPv4 header and the 8-byte UDP header are subtracted from the IP total length. 548 minus 28 leaves 520 bytes of DNS data, which is greater than 512. The other values leave 504, 512 and 508 bytes, and 512 does not pass a strict greater-than test. DNS over UDP was originally limited to messages of 512 octets or less, which is why oversized queries can indicate tunneling or abuse.
Why the other options are wrong
- A. 532 minus 28 header bytes leaves 504 bytes of payload, which is below 512.
- B. 540 minus 28 leaves exactly 512 bytes, and the > operator requires more than 512.
- C. 536 minus 28 leaves 508 bytes, which is still under the threshold.
Question 113
Which of the following is a signature-based intrusion detection system (IDS) ?
Show answer and explanation
Correct answer: A. Snort
Snort is a widely-recognized open-source intrusion detection system that uses signature-based detection to identify known attack patterns and malicious traffic by comparing network traffic against a database of known attack signatures.
Why the other options are wrong
- B. StealthWatch is a behavioral and NetFlow-based monitoring tool, not signature-based.
- C. RealSecure is primarily an intrusion prevention system using multiple detection methods, not purely signature-based.
- D. Tripwire is a file integrity monitoring tool, not an intrusion detection system.
Question 114
You work as a technician for Tech Perfect Inc. You are troubleshooting an Internet name resolution issue. You ping your ISP's DNS server address and find that the server is down. You want to continuously ping the DNS address until you have stopped the command.
Which of the following commands will you use?
Show answer and explanation
Correct answer: C. ping -t
The ping -t command enables continuous ping mode, where the ping command will continue sending packets indefinitely until manually stopped by the user (typically via Ctrl+C). This is useful for monitoring network connectivity over time.
Why the other options are wrong
- A. ping -a is used to resolve addresses to hostnames, not for continuous pinging.
- B. ping -l specifies the packet size/buffer length, not continuous mode operation.
- D. ping -n specifies the number of echo requests to send, providing a finite count rather than continuous mode.
Question 115
An analyst filters a web server capture with the filter below to find receivers that stopped accepting data, but the results are dominated by RST segments, which also carry a window of 0.
tcp.window_size == 0
Which filter lists only true zero-window advertisements?
Show answer and explanation
Correct answer: D. tcp.analysis.zero_window
Wireshark's TCP analysis sets the ZeroWindow flag only when the receive window is zero and none of SYN, FIN or RST is set, so the field tcp.analysis.zero_window ignores resets that merely carry a zero window value. A true zero window tells the sender to pause because the receiver cannot accept more data, which usually points to a capacity problem on the receiving end. Window Full and Zero Window Probe describe the sender's side of such stalls rather than the advertisement itself.
Why the other options are wrong
- A. Excluding SYN leaves the RST segments in place, because they do not carry SYN either.
- B. Window Full marks a sender's segment that fills the other side's advertised window, not a zero-window advertisement.
- C. A Zero Window Probe is the one-byte segment a sender uses to test a zero window, not the advertisement itself.
Question 116
Which of the following statements about the File Transfer Protocol (FTP) is true?
Show answer and explanation
Correct answer: D. It transfers files between computers.
FTP (File Transfer Protocol) is a standard network protocol used to transfer files between computers over a network or the Internet. It operates on the application layer and provides a method for uploading and downloading files between a client and server. This is its primary function and the most accurate description of what FTP does.
Why the other options are wrong
- A. POP3 (Post Office Protocol 3) is used for retrieving email messages, not FTP, and FTP does not handle files transmitted through POP3.
- B. Network device management is handled by protocols like SNMP (Simple Network Management Protocol), not FTP.
- C. FTP connects computers for file transfer, not specifically file servers on the World Wide Web; that function is better served by HTTP/HTTPS.
- E. FTP typically requires authentication with a username and password; it is not a password-free protocol, though anonymous FTP exists in limited cases.
Question 117
A client sends the SYN below (tcpdump -n -S). No process listens on port 8443 and no firewall is in the path.
10.3.3.7.55210 > 192.0.2.60.8443: Flags [S], seq 4026531900, win 64240, options [mss 1460], length 0
Which reply should the analyst expect from 192.0.2.60?
Show answer and explanation
Correct answer: D. Flags [R.], seq 0, ack 4026531901, win 0, length 0
With no listener, the connection is in the CLOSED state and any segment that is not itself a RST draws a reset. Because the SYN has no ACK flag, the reset uses sequence number zero, sets ACK, and acknowledges SEG.SEQ + SEG.LEN. SEG.LEN counts the SYN as one, so the ACK is 4026531900 + 1 = 4026531901. On the wire that is a RST/ACK with seq 0 and ack 4026531901, which a scanner reads as a closed port.
Why the other options are wrong
- A. A reset without ACK is used only when the arriving segment had ACK set, and its seq comes from that ACK field.
- B. The reset's sequence number is zero, and its ACK must acknowledge the SYN at 4026531901.
- C. A SYN/ACK would mean the port is listening, which contradicts the setup.
Question 118
An analyst lists the fragments of one IPv4 datagram. All share the same ID, addresses and protocol, and each has a 20-byte header.
Frag 1: offset field 0, MF=1, total length 1500
Frag 2: offset field 185, MF=1, total length 1020
Frag 3: offset field 310, MF=1, total length 517
Frag 4: offset field 372, MF=0, total length 77
Which fragment breaks the fragmentation rules?
Show answer and explanation
Correct answer: C. Fragment 3, because it has MF set but carries 497 data bytes, not a multiple of 8
Fragmented data must be broken on 8-octet boundaries, and only the last fragment may carry a length that is not a multiple of 8. Fragment 3 has MF set, so it is not the last, yet it carries 517 - 20 = 497 data bytes, and 497 / 8 = 62.125. Its data starts at 310 x 8 = 2480 and would run to byte 2976, one byte into the range of fragment 4, which starts at 372 x 8 = 2976. Normal stacks never build such a fragment, so it points to crafted traffic.
Why the other options are wrong
- A. 1500 - 20 = 1480 = 185 x 8 data bytes is the normal maximum for a 1500-byte MTU, and no 1472-byte rule exists.
- B. 1480 / 8 = 185, so offset 185 starts exactly where the 1480 data bytes of fragment 1 end.
- D. The last fragment may carry any number of bytes, so its 57 data bytes are valid.
Question 119
Which of the following algorithms is used as a default algorithm for ESP extension header in IPv6?
Show answer and explanation
Correct answer: B. Cipher Block Chaining (CBC) Mode
Cipher Block Chaining (CBC) Mode is specified as the default encryption algorithm for the ESP (Encapsulating Security Payload) extension header in IPv6 according to IPsec standards. CBC mode provides strong encryption by making each ciphertext block dependent on the previous one, enhancing security against pattern analysis.
Why the other options are wrong
- A. Propagating Cipher Block Chaining (PCBC) Mode is not used as the standard for IPv6 ESP encryption.
- C. Electronic Codebook (ECB) Mode is considered insecure for most purposes due to pattern preservation and is not the IPv6 ESP default.
- D. Cipher Feedback (CFB) Mode is a stream cipher mode, not the standard block cipher mode used for IPv6 ESP encryption.
Question 120
An analyst wants TCP flow records in which the SYN flag was seen but the ACK flag never was, a common sign of scanning or unanswered connection attempts. Which rwfilter switch does this?
Show answer and explanation
Correct answer: C. --flags-all=S/SA
In a HIGH/MASK pair the flags left of the slash must be on and any other flags in the mask must be off. S/SA therefore requires SYN on and ACK off, with all other flags free. Because --flags-all examines the bit-wise OR of the flags across every packet in the flow, a pass means no packet in that record ever carried ACK, so the initiator never completed a handshake within it.
Why the other options are wrong
- A. SA/SA requires both SYN and ACK to be on, the opposite of the goal.
- B. A list of pairs passes when any pair matches, so S/S,A/A passes flows with either SYN or ACK.
- D. The high set must be a subset of the mask, so SA/S is not a valid pair.
Question 121
A perimeter sensor on the Internet-facing link records inbound TCP SYN packets with source address 127.0.0.1 and destination 198.51.100.20. How should the analyst classify them?
Show answer and explanation
Correct answer: A. Spoofed: loopback addresses must never appear outside a host
127.0.0.0/8 is the internal host loopback network, and addresses of this form must not appear outside a host. A packet carrying 127.0.0.1 as its source on a physical link was therefore forged, typically by a tool that randomizes sources or deliberately picks impossible ones to test whether a target stack or filter accepts them. Ingress filters should drop such packets, and an IDS should treat them as crafted.
Why the other options are wrong
- B. Loopback traffic never leaves the host, so it cannot cross the Internet-facing link.
- C. No network may number devices from 127.0.0.0/8, which is reserved for host loopback.
- D. The source address alone proves forgery, whatever value the TTL holds.
Question 122
A Snort 3 rule meant to catch percent-encoded directory traversal never fires, although captured requests contain /cgi-bin/%2e%2e%2f%2e%2e%2fetc/passwd. The rule uses these options.
http_uri; content:"%2e%2e%2f",nocase;
Which change lets the rule match the encoded sequence?
Show answer and explanation
Correct answer: A. Use the http_raw_uri sticky buffer instead of http_uri
The http_uri buffer holds the normalized URI: Snort percent-decodes values and removes directory traversals before placing the result there, so the literal text %2e%2e%2f no longer exists in that buffer. http_raw_uri keeps the URI exactly as the client sent it, so the encoded traversal can be matched there. The raw buffer is the usual choice when the evasion encoding itself is the indicator, while the normalized buffer suits matching the decoded resource name.
Why the other options are wrong
- B. http_header holds the request header fields, not the request URI.
- C. Normalization removes path traversals from http_uri, so ../ cannot be relied on in that buffer either.
- D. fast_pattern only changes which content drives the pattern matcher and cannot restore text the buffer no longer contains.
Question 123
A sensor records the following exchange (tcpdump -n -S). The same pattern continues across hundreds of ports.
198.51.100.23.61022 > 10.9.1.10.443: Flags [S], seq 1520147893, win 1024, options [mss 1460], length 0
10.9.1.10.443 > 198.51.100.23.61022: Flags [S.], seq 2210078812, ack 1520147894, win 64240, options [mss 1460], length 0
198.51.100.23.61022 > 10.9.1.10.443: Flags [R], seq 1520147894, win 0, length 0
198.51.100.23.61022 > 10.9.1.10.444: Flags [S], seq 1520147893, win 1024, options [mss 1460], length 0
10.9.1.10.444 > 198.51.100.23.61022: Flags [R.], seq 0, ack 1520147894, win 0, length 0
Which conclusion fits the traffic?
Show answer and explanation
Correct answer: B. A half-open SYN scan that found 443 open and 444 closed
The scanner sends a SYN, and port 443 answers with SYN/ACK, so it is listening. Instead of the ACK that would complete the handshake, the scanner returns a bare RST whose sequence number, 1520147894, is taken from the SYN/ACK's ACK field, which is how a stack with no matching connection rejects it. Port 444 answers the SYN with RST/ACK (seq 0, ack 1520147893 + 1), so it is closed. Never completing the handshake is the nmap half-open SYN scan.
Why the other options are wrong
- A. A connect scan completes the handshake to open ports, and here the third segment is a RST, not an ACK.
- C. SYN/ACK marks an open port and RST marks a closed one; filtered ports give no reply or ICMP errors.
- D. A flood source does not answer SYN/ACKs or move across hundreds of ports with one probe each.
Question 124
Distributed Checksum Clearinghouse (DCC) is a hash sharing method of spam email detection.
Which of the following protocols does the DCC use?
Show answer and explanation
Correct answer: D. UDP
Distributed Checksum Clearinghouse (DCC) uses UDP (User Datagram Protocol) for its hash sharing operations. UDP is preferred for DCC because it is lightweight, connectionless, and efficient for rapid hash lookups and sharing across distributed systems. TCP would be slower for this purpose, while TELNET and ICMP are not used for DCC hash sharing.
Why the other options are wrong
- A. TCP is connection-oriented and slower than needed for DCC's distributed hash-sharing operations.
- B. TELNET is a remote login protocol unrelated to spam detection or hash sharing.
- C. ICMP is used for network diagnostics (ping, traceroute), not for spam detection hash sharing.
Question 125
Which of the following techniques is used to identify attacks originating from a botnet?
Show answer and explanation
Correct answer: C. Passive OS fingerprinting
Passive OS fingerprinting is used to identify attacks originating from botnets by analyzing network traffic characteristics to determine the operating system and system configuration of attacking hosts without direct interaction, revealing patterns consistent with compromised machines in a botnet.
Why the other options are wrong
- A. IFilter is a text filtering utility used for content indexing and searching; it does not identify botnet attacks.
- B. BPF (Berkeley Packet Filter) is a packet filtering mechanism used for network packet capture and analysis, not specifically for botnet identification.
- D. Recipient filtering is an email security technique that blocks messages to invalid recipients; it does not identify botnet attacks.
Question 126
An analyst wants a tcpdump filter that captures every IPv4 fragment except the first one, meaning every fragment that carries no transport header. Which filter does this?
Show answer and explanation
Correct answer: A. ip[6:2] & 0x1fff != 0
The 16 bits at IPv4 header offset 6 hold three flag bits followed by the 13-bit fragment offset. Masking with 0x1fff keeps only the offset, which is nonzero for every fragment after the first, including the last fragment whose MF bit is clear. The opposite test, ip[6:2] & 0x1fff = 0, catches unfragmented datagrams and first fragments, so negating it leaves exactly the later fragments.
Why the other options are wrong
- B. Bit 0x20 of byte 6 is MF, which matches the first fragment and misses the last fragment, whose MF bit is clear.
- C. Mask 0x3fff also includes MF, so it matches the first fragment (MF set, offset 0) as well as the later ones.
- D. Bit 0x4000 is the Don't Fragment flag, which marks datagrams that must not be fragmented at all.
Question 127
John works as a professional Ethical Hacker for SecureEnet Inc. The company has a Windowsbased network. All client computers run on Windows XP. A project has been assigned to John to investigate about the open ports responsible for various malicious attacks on the network.
John wants to use the DOS command-line utility to find out the open ports.
Which of the following DOS commands will John use to accomplish the task?
Show answer and explanation
Correct answer: D. netstat
The netstat command is a DOS/Windows command-line utility that displays network statistics and active connections, including open ports and the services using them. This makes it ideal for identifying open ports that could be vulnerable to malicious attacks. Tracert and pathping are route tracing tools, nslookup is for DNS queries, and nbtstat is for NetBIOS name resolution, none of these directly reveal open ports.
Why the other options are wrong
- A. Tracert and pathping are used for tracing network routes, not identifying open ports.
- B. Nslookup is used for DNS name resolution queries, not port enumeration.
- C. Nbtstat is used for NetBIOS name resolution and cache information, not for identifying open ports.
That was 127 of 507.
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