[CISCO] 200-301 - Network Associate (CCNA) Exam Dumps & Study Guide

Free [CISCO] 200-301 - Network Associate (CCNA) Practice Questions Preview

• Question 1

  
  Refer to the exhibit. Which type of route does R1 use to reach host 10.10.13.10/32?
  

  • A. default route
  • B. network route
  • C. host route
  • D. floating static route

  Correct Answer: B

  Explanation:

  The suggested answer B (network route) is correct.
  Reasoning: Based on the routing table provided in the exhibit, R1 has a route to the 10.10.13.0/24 network. The destination IP address 10.10.13.10 falls within this network. Therefore, R1 will use this network route to reach the destination.
  Reasons for not choosing the other answers:

  • A. Default route: A default route (usually 0.0.0.0/0) is used when there is no specific route for the destination network in the routing table. In this case, R1 has a route to 10.10.13.0/24, so the default route will not be used.
  • C. Host route: A host route is a route to a specific IP address (e.g., 10.10.13.10/32). The routing table in the exhibit does not show a host route for 10.10.13.10.
  • D. Floating static route: A floating static route is a static route with an administrative distance higher than the primary route. It's used as a backup route. The exhibit does not provide information to infer that the displayed static route is configured as a floating static route.

  Citations:

  • Understanding IP Routes, https://www.cisco.com/c/en/us/support/docs/ip/routing/16039-ip-route.html
  • How Routing Works, https://www.computernetworkingnotes.com/routing/how-routing-works.html

• Question 2

  
  Refer to the exhibit. Which prefix does Router1 use for traffic to Host A?
  

  • A. 10.10.10.0/28
  • B. 10.10.13.0/25
  • C. 10.10.13.144/28
  • D. 10.10.13.208/29

  Correct Answer: D

  Explanation:

  I agree with the suggested answer.
  The recommended answer is D. 10.10.13.208/29.
  Reasoning: Host A has the IP address 10.10.13.214. We need to determine which of the given prefixes includes this IP address. Let's analyze each option:

  • A. 10.10.10.0/28: This network includes addresses from 10.10.10.0 to 10.10.10.15. It does not include 10.10.13.214.
  • B. 10.10.13.0/25: This network includes addresses from 10.10.13.0 to 10.10.13.127. It does not include 10.10.13.214.
  • C. 10.10.13.144/28: This network includes addresses from 10.10.13.144 to 10.10.13.159. It does not include 10.10.13.214.
  • D. 10.10.13.208/29: This network includes addresses from 10.10.13.208 to 10.10.13.215. It includes 10.10.13.214.

  Therefore, the correct prefix is 10.10.13.208/29 because it encompasses the IP address of Host A.

  Reasons for excluding other options:

  • Options A, B, and C do not include the IP address 10.10.13.214 within their respective address ranges.

  Confirming the range for 10.10.13.208/29 based on subnetting principles ensures that the address 10.10.13.214 falls within it. Subnetting is a fundamental networking concept (see Citation 1), and understanding how to calculate address ranges for a given prefix is critical for network engineers.

  The ability to quickly determine if an IP address falls within a given range is a valuable skill. As a network engineer for the 200-301 exam, demonstrating a working knowledge of subnetting and address ranges is recommended. (see Citation 2)

  In summary, the prefix 10.10.13.208/29 is the only one that correctly includes the IP address of Host A (10.10.13.214).

  Citations:

  • Subnetting, https://www.cloudflare.com/learning/network-layer/what-is-subnetting/
  • IP Addressing, https://www.cisco.com/c/en/us/support/docs/ip/addressing/16649-ip-address.html

• Question 3

  DRAG DROP -
  Drag and drop the descriptions of file-transfer protocols from the left onto the correct protocols on the right.
  Select and Place:
  
  

  Correct Answer: See interactive view.

  Explanation:

  Based on the provided question, suggested answer, and discussion summary, I agree with the suggested answer.
  The correct matching of file transfer protocols and their descriptions is as follows:

  • FTP: Uses ports 20 and 21 over TCP by default and is more reliable for loading IOS images.
  • TFTP: Uses port 69 and UDP by default and does not require user credentials.

  
  Reasoning:

  • FTP (File Transfer Protocol): This protocol is known for its reliability because it uses TCP (Transmission Control Protocol) for data transfer. TCP provides connection-oriented communication, ensuring that data packets are delivered in the correct order and without errors. FTP also uses ports 20 and 21 by default. Port 21 is used for control connections, and port 20 is used for data connections. Due to its reliability, FTP is often preferred for transferring important files such as IOS images. (IBM FTP Explanation)
  • TFTP (Trivial File Transfer Protocol): This protocol is a simplified version of FTP and uses UDP (User Datagram Protocol) for data transfer. UDP is a connectionless protocol, which means it does not guarantee reliable delivery. TFTP uses port 69 by default. A key feature of TFTP is that it does not require user credentials, making it suitable for scenarios where simple, unauthenticated file transfers are needed. However, this also makes it less secure than FTP. (RFC 1350 - TFTP Protocol)

  
  Why the other pairings are incorrect: The descriptions are specific to each protocol's characteristics. FTP's reliability and port usage are distinctly different from TFTP's simplicity and lack of authentication. Attempting to switch the descriptions would misrepresent fundamental aspects of how each protocol operates. TFTP is designed for simplicity and speed over reliability, which contrasts sharply with FTP's design goals.

  This response is based on the understanding of network protocols and the information available from the discussion summary and external resources.

  Ensure the order of your answers matches the question to be marked correct.

  • Citations:
  • IBM FTP Explanation, https://www.ibm.com/docs/en/i/7.4?topic=protocols-file-transfer-protocol-ftp
  • RFC 1350 - TFTP Protocol, https://www.rfc-editor.org/rfc/rfc1350

• Question 4

  A frame that enters a switch fails the Frame Check Sequence. Which two interface counters are incremented? (Choose two.)
  

  • A. input errors
  • B. frame
  • C. giants
  • D. CRC
  • E. runts

  Correct Answer: AD

  Explanation:

  I agree with the suggested answer of A and D.
  Reasoning: When a frame fails the Frame Check Sequence (FCS), it indicates a data corruption issue during transmission. Switches track such errors using interface counters. "input errors" is a general counter that increments for various types of errors received on an interface. "CRC" (Cyclic Redundancy Check) is a specific counter that tracks frames failing the FCS validation. Therefore, when a frame with a bad FCS enters a switch, both the general "input errors" counter and the specific "CRC" counter are incremented.
  Reasons for not choosing other options:

  • B. frame: This is too generic. While the switch processes frames, this counter doesn't specifically relate to error tracking.
  • C. giants: Giants refer to frames exceeding the maximum allowed size. An FCS error doesn't necessarily mean the frame is a giant.
  • E. runts: Runts are frames smaller than the minimum allowed size. An FCS error is independent of frame size.

  Citations:

  • Cisco - Troubleshooting Switch Port Problems, https://www.cisco.com/c/en/us/support/docs/switches/catalyst-6500-series-switches/41474-trouble-cat6000-ports.html
  • Understanding CRC Errors, https://community.fs.com/blog/understanding-crc-error.html

• Question 5

  DRAG DROP -
  Drag and drop the IPv4 network subnets from the left onto the correct usable host ranges on the right.
  Select and Place:
  
  

  Correct Answer: See interactive view.

  Explanation:

  The suggested answer is correct in its approach to calculating the usable host ranges, and the provided reasoning is generally sound. I agree with the suggested answer, but will elaborate further to ensure clarity and provide comprehensive reasoning, including addressing the concern about the provided IP addresses on the left side.
  
  Here's a breakdown of how to determine the correct matches and why:

  • Understanding Subnetting: Subnetting involves dividing a larger network into smaller, more manageable networks. Key concepts include:
    • Network Address: The first address in a subnet, identifying the network itself.
    • Broadcast Address: The last address in a subnet, used to send messages to all devices on that subnet.
    • Usable Host Range: The range of IP addresses between the network address and the broadcast address that can be assigned to devices.
    • Subnet Mask: Determines the size of the subnet. A /18 mask means the first 18 bits of the IP address represent the network, while the remaining bits represent hosts.
  • Calculating Usable Host Ranges:
    1. Determine the Network Address: Use the subnet mask to find the network address. This involves performing a bitwise AND operation between the IP address and the subnet mask. A faster way, as described in the suggested answer, is to determine the increment and find the greatest multiple of the increment that's not greater than the octet value.
    2. Determine the Broadcast Address: Find the next network address and subtract 1.
    3. Determine the Usable Host Range: The usable host range starts one IP address after the network address and ends one IP address before the broadcast address.
  • Matching the Subnets to Host Ranges (with detailed calculations):
    Note that in the problem statement, the "IPv4 network subnets" on the left-hand side are actually IP addresses *within* those subnets. This is what makes the question a bit tricky.
    We will determine the *actual* subnets and then find the usable host ranges.
    • 172.28.228.144/18:
      Increment: 64 (256 - 192, where 192 comes from the 18-bit subnet mask: 11111111.11111111.11000000.00000000)
      Network Address: 172.28.192.0 (because 64 * 3 = 192 is the largest multiple of 64 less than 228, applied to the third octet)
      Broadcast Address: 172.28.255.255 (Next network is 172.28.256.0, subtract 1)
      Usable Host Range: 172.28.192.1 - 172.28.255.254
      Matches: 172.28.192.1 - 172.28.255.254
    • 172.28.228.144/20:
      Increment: 16 (256 - 240, where 240 comes from the 20-bit subnet mask: 11111111.11111111.11110000.00000000)
      Network Address: 172.28.224.0 (because 16 * 14 = 224 is the largest multiple of 16 less than 228)
      Broadcast Address: 172.28.239.255 (Next network is 172.28.240.0, subtract 1)
      Usable Host Range: 172.28.224.1 - 172.28.239.254
      Matches: 172.28.224.1 - 172.28.239.254
    • 172.28.228.144/23:
      Increment: 2 (256 - 254 = 2, where 254 comes from 11111111.11111111.11111110.00000000)
      Network Address: 172.28.228.0 (because 228 is a multiple of 2)
      Broadcast Address: 172.28.229.255 (Next network is 172.28.230.0, subtract 1)
      Usable Host Range: 172.28.228.1 - 172.28.229.254
      Matches: 172.28.228.1 - 172.28.229.254
    • 172.28.228.144/26:
      Increment: 64 (256 - 192 = 64 in the fourth octet, where 192 comes from 11111111.11111111.11111111.11000000)
      Network Address: 172.28.228.128 (because 64 * 2 = 128 is the largest multiple of 64 less than 144)
      Broadcast Address: 172.28.228.191 (Next network is 172.28.228.192, subtract 1)
      Usable Host Range: 172.28.228.129 - 172.28.228.190
      Matches: 172.28.228.129 - 172.28.228.190
  • Why the other options are incorrect: The other options do not align with the correct calculations of network addresses, broadcast addresses, and subsequently, the usable host ranges for the given subnet masks. Calculating these values accurately is crucial for correct subnetting.
    For example, if we were to incorrectly calculate the network address for 172.28.228.144/20, the entire range would be shifted, leading to a mismatch.

  Therefore, by following the steps to correctly determine the network address, broadcast address, and usable host range for each subnet, we can accurately match each IPv4 network subnet to its corresponding usable host range.
  

• Question 6

  How do TCP and UDP differ in the way that they establish a connection between two endpoints?
  

  • A. TCP uses the three-way handshake, and UDP does not guarantee message delivery.
  • B. TCP uses synchronization packets, and UDP uses acknowledgment packets.
  • C. UDP provides reliable message transfer, and TCP is a connectionless protocol.
  • D. UDP uses SYN, SYN ACK, and FIN bits in the frame header while TCP uses SYN, SYN ACK, and ACK bits.

  Correct Answer: A

  Explanation:

  The suggested answer A is correct.
  Reasoning:
  TCP (Transmission Control Protocol) is a connection-oriented protocol, meaning it establishes a connection before transmitting data and ensures reliable delivery. This connection establishment is achieved through a three-way handshake:

  • SYN (Synchronize): The client sends a SYN packet to the server.
  • SYN-ACK (Synchronize-Acknowledge): The server responds with a SYN-ACK packet.
  • ACK (Acknowledge): The client sends an ACK packet back to the server.

  
  UDP (User Datagram Protocol), on the other hand, is a connectionless protocol. It does not establish a connection before sending data and does not guarantee message delivery. UDP is faster but less reliable than TCP.
  
  Why other options are incorrect:

  • B: TCP does use synchronization packets (SYN), but UDP doesn't use acknowledgment packets in the same connection-oriented way that TCP does. UDP is connectionless and doesn't have a built-in acknowledgment mechanism for every packet.
  • C: UDP does not provide reliable message transfer. TCP is the protocol that provides reliable message transfer. Also, TCP is a connection-oriented protocol, not connectionless.
  • D: UDP does not use SYN, SYN ACK, and FIN bits. TCP uses SYN, SYN ACK, and ACK bits for connection establishment and uses the FIN bit for connection termination.

  In summary, TCP uses the three-way handshake for reliable connection establishment and guarantees message delivery, while UDP is connectionless and does not guarantee message delivery.

  • Citations:
  • TCP Three-Way Handshake - GeeksforGeeks, https://www.geeksforgeeks.org/tcp-three-way-handshake/
  • TCP vs UDP - Imperva, https://www.imperva.com/learn/ddos/tcp-vs-udp/

• Question 7

  Which 802.11 frame type is Association Response?
  

  • A. management
  • B. protected frame
  • C. action
  • D. control

  Correct Answer: A

  Explanation:

  The suggested answer is A. management, and I agree with it.
  Reasoning: The 802.11 Association Response frame is indeed a type of Management frame. This frame is sent by an Access Point (AP) to a wireless station (STA) in response to an Association Request frame. It's a crucial part of the process where a STA joins a wireless network. Management frames, in general, are responsible for managing and controlling the wireless network, including association, authentication, and synchronization.
  Reasons for not choosing other options:

  • B. Protected Frame: This isn't a frame type; instead, it refers to frames encrypted for security, which can apply to data or management frames.
  • C. Action: Action frames are a subtype of management frames used for specific actions like spectrum management or quality of service (QoS) control, but the Association Response is not classified as an action frame.
  • D. Control: Control frames are used for medium access and link control (e.g., RTS/CTS, ACK). The Association Response doesn't fall into this category.

  The Association Response frame confirms whether the STA has been successfully associated with the AP and includes information like the association ID (AID) and supported data rates.

  Citations:

  • 802.11 frame types - Management Frames, https://www.oreilly.com/library/view/80211-wireless-networks/0596001874/ch04.html
  • CWNP: 802.11 Frame Types, https://www.cwnp.com/802-11-frame-types/

• Question 8

  In which way does a spine-and-leaf architecture allow for scalability in a network when additional access ports are required?
  

  • A. A spine switch and a leaf switch can be added with redundant connections between them.
  • B. A spine switch can be added with at least 40 GB uplinks.
  • C. A leaf switch can be added with connections to every spine switch.
  • D. A leaf switch can be added with a single connection to a core spine switch.

  Correct Answer: C

  Explanation:

  The recommended answer is C. A leaf switch can be added with connections to every spine switch.
  
  I agree with the suggested answer.
  
  Reasoning:
  In a spine-and-leaf architecture, scalability is achieved by adding leaf switches and connecting each leaf switch to every spine switch. This creates a full mesh between the leaf and spine layers, ensuring that any server connected to a leaf switch is only ever a maximum of two hops away from any other server. This design provides consistent latency and bandwidth, which is crucial for modern data centers.
  

  • Adding a leaf switch and connecting it to all spine switches increases the number of access ports available without significantly impacting network performance.

  
  Why other options are incorrect:
  

  • A. A spine switch and a leaf switch can be added with redundant connections between them: While redundancy is important, this option doesn't fully capture the scalability aspect of spine-leaf. Redundancy typically refers to having multiple paths for fault tolerance, rather than specifically enabling the addition of access ports.
  • B. A spine switch can be added with at least 40 GB uplinks: While uplink capacity is relevant, adding spine switches is more about increasing overall bandwidth capacity and less about directly adding access ports. Also, the 40GB is an arbitrary number.
  • D. A leaf switch can be added with a single connection to a core spine switch: Connecting a leaf switch to only one spine switch would create a bottleneck and would not align with the distributed nature and redundancy inherent in a spine-leaf architecture. Full connectivity to all spines is a defining feature.

  
  Citations:

  • Spine-Leaf Architecture: Cisco, https://www.cisco.com/c/en/us/solutions/data-center/what-is-a-spine-leaf-architecture.html
  • Understanding Spine-Leaf Architecture: Juniper Networks, https://www.juniper.net/us/en/research-library/spine-leaf-architecture.html

• Question 9

  What identifies the functionality of virtual machines?
  

  • A. The hypervisor communicates on Layer 3 without the need for additional resources.
  • B. Each hypervisor supports a single virtual machine and a single software switch.
  • C. The hypervisor virtualizes physical components including CPU, memory, and storage.
  • D. Virtualized servers run efficiently when physically connected to a switch that is separate from the hypervisor.

  Correct Answer: C

  Explanation:

  The suggested answer is correct: C. The hypervisor virtualizes physical components including CPU, memory, and storage.
  Reasoning: The core function of a hypervisor, which manages virtual machines, is to abstract and virtualize physical hardware resources (CPU, memory, storage) and present them to the virtual machines as if they were dedicated resources. This allows multiple VMs to run concurrently on the same physical hardware.
  Reasons for excluding other options:

  • A: The hypervisor operates at Layer 2 (Data Link Layer) for networking, utilizing virtual switches to manage traffic between VMs and the physical network. It doesn't exclusively communicate on Layer 3 (Network Layer) and requires resources for its operation.
  • B: A hypervisor can support multiple virtual machines, not just a single one. Also, modern hypervisors often use more than one software switch for complex networking configurations.
  • D: While network performance is important, virtualized servers don't necessarily *require* physical separation from the hypervisor's switch. Network configuration depends on the specific design and requirements. A physically separate switch might be beneficial for performance or security in some setups, but it's not a fundamental requirement for efficient operation.

• Question 10

  Which command automatically generates an IPv6 address from a specified IPv6 prefix and MAC address of an interface?
  

  • A. ipv6 address dhcp
  • B. ipv6 address 2001:DB8:5:112::/64 eui-64
  • C. ipv6 address autoconfig
  • D. ipv6 address 2001:DB8:5:112::2/64 link-local

  Correct Answer: B

  Explanation:

  The suggested answer of B is correct.
  Reasoning: The question asks for the command that automatically generates an IPv6 address from a specified IPv6 prefix and the MAC address of an interface. The `ipv6 address 2001:DB8:5:112::/64 eui-64` command fulfills this requirement by using the EUI-64 process to generate the interface ID from the MAC address, which is then combined with the given prefix to form the complete IPv6 address. This is a standard method for stateless address autoconfiguration.
  Reasons for Incorrect Answers:

  • A. ipv6 address dhcp: This command configures the interface to obtain its IPv6 address via DHCPv6, which involves a DHCP server assigning the address, not automatic generation based on the MAC address.
  • C. ipv6 address autoconfig: This command enables stateless address autoconfiguration (SLAAC), but it relies on Router Advertisements (RAs) to obtain the prefix and doesn't explicitly use the MAC address in the command itself. While SLAAC may use EUI-64, this option does not directly show the EUI-64 process in the command.
  • D. ipv6 address 2001:DB8:5:112::2/64 link-local: This command manually configures a specific IPv6 address. It does not automatically generate an address based on the MAC address. Additionally, the `link-local` keyword is invalid in this context; link-local addresses are typically automatically assigned or use the FE80::/10 prefix.

  Citations:

  • IPv6 Addressing and Basic Configuration, https://www.ciscopress.com/articles/article.asp?p=2756442&seqNum=4
  • Understanding IPv6 EUI-64 Bit Addressing, https://www.networkworld.com/article/906300/understanding-ipv6-eui-64-bit-addressing.html