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Cisco 200-301 Practice Test 5

Question 40

A router learns identical `198.51.100.0/24` prefixes through OSPF with administrative distance 110 and a manually configured static route with administrative distance 5. Both next hops are reachable and no more-specific route exists. Which route is installed?

  1. The OSPF route because its metric is numerically smaller than the static route's distance.
  2. Both routes necessarily load-balance because they have the same prefix length.
  3. The router always installs OSPF over static routes regardless of AD values.
  4. The static route with AD 5, because it has the lower administrative distance for the same prefix.

Explanation

Correct Answer

D:

When multiple route sources offer the same destination prefix, IOS generally prefers the source with lower administrative distance for installation, provided the route is valid. The /24 matches in both cases, so the source trust ranking is relevant; OSPF cost is not used to compare against a static AD 5 candidate.

Incorrect Answers

A:

OSPF metric and administrative distance serve different decisions. Metrics compare eligible paths within a routing protocol, while AD compares route sources to an identical prefix. A reachable static route with AD 5 is preferred to the OSPF source AD 110.

B:

Same-prefix candidates are not necessarily equal-cost installed next hops. The significant AD difference normally causes one source to win rather than mixing static and OSPF forwarding by default. ECMP requires multiple eligible next hops under applicable selection rules.

C:

Default AD values may make connected or static entries preferable to OSPF, but administrators can override the static distance as this scenario does. IOS uses the configured AD for comparing the same prefix, not an unchangeable preference for dynamic protocols.

Question 41

A file transfer application must deliver large accounting records intact and in order, recovering lost segments without requiring the application to build its own reliability layer. Which transport suits the stated contract?

  1. TCP, using connection-oriented sequencing, acknowledgment and retransmission.
  2. UDP alone, because it automatically retransmits missing datagrams in sequence.
  3. ICMP echo messages as a substitute for a reliable file-transfer transport.
  4. Ethernet source MAC learning as the end-to-end retransmission mechanism.

Explanation

Correct Answer

A:

TCP presents applications with an ordered reliable byte stream and handles loss recovery, flow control and connection state. It is appropriate when correctness and ordered delivery matter more than minimizing transport overhead. The network still needs application-level integrity and session security where required.

Incorrect Answers

B:

UDP provides datagrams without built-in acknowledgment, ordered delivery or retransmission guarantees. Applications can implement reliability above UDP, but the prompt explicitly wants transport-layer loss recovery. That behavior is supplied by TCP rather than raw UDP.

C:

ICMP echo assists diagnostics and basic reachability checks, but it does not provide an ordered application byte stream, congestion/flow control and data retransmission. A successful ping is not proof a complete accounting file was transferred.

D:

Local MAC learning chooses how switches forward frames on a VLAN; it does not acknowledge application bytes across routed networks. A damaged or missing segment may require retransmission at a transport or application layer. TCP directly addresses the stated file-transfer requirement.

Question 42

A WLAN controller defines the SSID `Guests-2026` and WPA2 settings, but clients joining this WLAN are placed in the internal staff subnet instead of guest VLAN 140. What is the most relevant WLC GUI configuration to correct?

  1. Change the WLAN's WPA2 key so staff clients stop authenticating.
  2. Bind the guest WLAN/SSID profile to the intended client VLAN or interface mapping 140.
  3. Increase the 5 GHz radio channel width while leaving the existing VLAN mapping unchanged.
  4. Rename the SSID to `Guests-2027` without changing its mapped interface.

Explanation

Correct Answer

B:

Authentication and SSID settings determine how clients join, while the WLAN-to-interface/VLAN mapping determines the wired subnet where their traffic is placed. Correct the guest profile's mapping, verify the wired VLAN path, and confirm a test client receives the expected guest DHCP lease and policy.

Incorrect Answers

A:

The passphrase authenticates clients but does not determine which subnet authorized guests receive. The reported fault is guest traffic landing in the wrong VLAN, not a failed security handshake. Update and verify interface/VLAN mapping while maintaining an appropriate authentication policy.

C:

Radio width affects RF capacity and interference but does not select a client subnet. Guests are already joining the SSID; the incorrect outcome occurs after association when traffic enters the wrong wired VLAN. Fix the WLAN mapping, not RF bandwidth.

D:

The SSID is the network's advertised name and is not a VLAN assignment. Renaming it could confuse clients but would not correct the staff-subnet placement. The controller must associate the WLAN with the intended guest VLAN and secure downstream connectivity.

Question 43

R1 connects to 172.20.10.0/24 locally and must reach a remote branch LAN 172.20.40.0/24 via next hop 192.0.2.2. The /30 transit interface is already up. Which IOS static network route is correct?

  1. `ip route 172.20.40.0 255.255.0.0 192.0.2.2`
  2. `ip route 172.20.40.0 255.255.255.0 192.0.2.1`
  3. `ip route 172.20.40.0 255.255.255.0 192.0.2.2`
  4. `ip route 172.20.10.0 255.255.255.0 192.0.2.2`

Explanation

Correct Answer

C:

A static network route includes the remote destination prefix, its subnet mask and a next hop reachable through a connected interface. The remote LAN is /24, so its mask is 255.255.255.0. Verify the next hop and reverse routing from the branch so traffic can return successfully.

Incorrect Answers

A:

The /16 mask describes 172.20.0.0/16, a much broader range than the specified /24. It could redirect traffic for other 172.20 subnets incorrectly. A route to the precise remote branch network requires mask 255.255.255.0.

B:

The proposed next hop is not the remote neighbor identified by the scenario; it is the local transit-side address in the illustrative topology. Static routes must direct packets to an appropriate reachable next hop toward the destination. Using oneself cannot provide the intended branch forwarding path.

D:

This targets R1's already local LAN rather than the missing branch destination. It could create confusing alternative route behavior without installing the required prefix. Use the actual remote LAN 172.20.40.0/24 as the static destination.

Question 44

A Cisco log collector filters messages to retain severity 3 and more urgent events. A technician claims severity 7 is the most urgent because the number is highest. What is the correct interpretation?

  1. Severity 7 represents emergencies and severity 0 represents routine debugging.
  2. Syslog severity is a MAC address value and cannot be used for event filtering.
  3. Facility code alone determines urgency; the severity field is only decorative.
  4. Severity 0 is most urgent (emergencies), while severity 7 is debugging.

Explanation

Correct Answer

D:

Cisco syslog severity numbers run from 0 emergencies through 7 debugging. A filter admitting severity 0–3 includes emergency, alert, critical and error categories, not debug noise. Facilities indicate message origin while severity indicates urgency; operators should preserve the configured logging threshold semantics.

Incorrect Answers

A:

This reverses the syslog priority numbering. It would cause administrators to filter the wrong events and potentially discard critical device failures. Interpret the numeric levels using the defined severity names rather than assuming larger numbers mean greater urgency.

B:

Syslog messages do have numeric severity categories, and operators can use them to control recording or forwarding volume. MAC addresses identify Layer 2 interfaces, which are unrelated to the syslog priority field. The collector's threshold must use the standard severity ordering.

C:

Facility commonly categorizes the source or subsystem of a syslog message, whereas severity carries the urgency level. Collectors can use both for routing and filtering. Ignoring severity would prevent the stated critical-only retention behavior.

Question 45

A lab router should allow management connections sourced from 192.0.2.0/24 and deny SSH connections from all other IPv4 source subnets. The requirement names only the source network, not a particular destination port. Which simple ACL type can match the authorized source?

  1. A standard IPv4 ACL matching source network 192.0.2.0 with wildcard 0.0.0.255.
  2. An ACL matching destination TCP port 443 but not source IP addresses.
  3. A standard ACL matching destination 192.0.2.0/24, regardless of client source.
  4. A port-security MAC limit that accepts only one Ethernet station on the router uplink.

Explanation

Correct Answer

A:

Standard ACLs filter IPv4 source addresses. A permit matching 192.0.2.0/24 with wildcard 0.0.0.255 can be applied at the appropriate VTY access-class management boundary. An implicit deny then handles other sources. Verify the effect without cutting off authorized administrative access.

Incorrect Answers

B:

HTTPS port 443 is unrelated to SSH and does not implement the requested source-network restriction. An extended ACL could filter by protocols and ports if needed, but the stated rule simply differentiates source networks for management access.

C:

Cisco standard IPv4 ACL entries examine source address, not the packet destination. Reinterpreting the source-based rule as a destination match could permit unintended clients or block the wrong traffic. The 192.0.2.0/24 value describes authorized management clients.

D:

Port security operates at the switch Layer 2 port/MAC level and does not specify which routed source subnets may open a VTY session. Remote SSH sources may traverse routers, so the correct control is a management-plane IP source restriction.

Question 46

A network inventory dashboard requests the current list of devices from a controller's documented REST endpoint. The operation must not change controller state. Which HTTP method best expresses the intended action?

  1. Use HTTP PATCH without a request body because PATCH is always a harmless query.
  2. Use HTTP GET to retrieve the collection, with authorized API credentials and a defined response format.
  3. Use HTTP DELETE so the controller returns the set it would remove.
  4. Use HTTP POST to recreate each device before reading the inventory.

Explanation

Correct Answer

B:

GET is designed to request a resource representation without intentionally creating, replacing or deleting managed objects. The client should authenticate, handle pagination and examine the response status and JSON fields. REST method semantics help make the inventory call repeatable and non-mutating.

Incorrect Answers

A:

PATCH expresses a partial modification of a resource and may be rejected or trigger changes depending on the API contract. An empty body does not turn it into a safe read. GET is the appropriate method for inspecting current inventory.

C:

DELETE represents removal of a target resource, not reading its inventory. Issuing it on a management API could be destructive if the credentials permit that action. The requirement explicitly prohibits changes, so a retrieval method is needed.

D:

POST frequently creates subordinate objects or starts operations. It does not provide the usual safe, read-only collection lookup and might change the controller's inventory. Use the published GET endpoint for the resource list rather than inventing objects.

Question 47

An HSRP group provides VLAN 10's virtual default gateway. Two routers participate: one is active and currently forwards traffic for the virtual IP, while the other is standby. What happens when the active router fails and the standby remains healthy?

  1. Both routers always forward equally for the same HSRP group without an active/standby distinction.
  2. Assume a change in HSRP role automatically rewrites the default-gateway DHCP option on every attached workstation.
  3. The standby can assume active forwarding responsibility for the shared virtual gateway.
  4. Every host must immediately reconfigure its default gateway to the standby router's physical SVI IP.

Explanation

Correct Answer

C:

HSRP coordinates ownership of a virtual IP and associated virtual MAC so clients can keep the same default-gateway setting while the active role transfers. The timing depends on timers and tracking. Verify the new active state and upstream routes after failover; HSRP alone cannot repair a failed WAN path beyond the gateway.

Incorrect Answers

A:

Standard HSRP group operation has active and standby roles and a single active gateway for that group, unlike designs that use multiple groups or other load-balancing approaches. Assuming simultaneous unrestricted forwarding ignores role coordination and can cause misdiagnosis.

B:

HSRP preserves a virtual IP across active/standby failover; no automatic per-client DHCP gateway reconfiguration is required.

D:

The purpose of a first-hop redundancy virtual address is to avoid requiring host default-gateway changes during member failover. A surviving participant assumes the virtual role. Clients should still use the same virtual IP rather than learn a new physical interface address.

Question 48

A host is configured as `192.168.42.77/26`. The network engineer must document the correct subnet network address and directed broadcast address. Which pair is correct?

  1. Network 192.168.42.0 and broadcast 192.168.42.255.
  2. Network 192.168.42.64 and broadcast 192.168.42.95.
  3. Network 192.168.42.77 and broadcast 192.168.42.126.
  4. Network 192.168.42.64 and broadcast 192.168.42.127.

Explanation

Correct Answer

D:

A /26 mask allocates blocks of 64 addresses in the final octet: 0–63, 64–127, 128–191, and 192–255. Host .77 lies in the 64–127 block. The network address is .64 and the broadcast is .127; usable ordinary hosts fall between them.

Incorrect Answers

A:

Those boundaries describe a /24, not the /26 configured on the host. A /26 creates four separate 64-address blocks within the /24. Treating all 256 addresses as one subnet would incorrectly include hosts from unrelated /26 segments.

B:

The pair would imply a 32-address /27 range, not a 64-address /26. The configured prefix covers .64 through .127, so .95 is an ordinary usable host address. Always derive the block size from the prefix length.

C:

The individual host IP is neither the subnet network nor the directed broadcast. The last usable host in this /26 is .126, while .127 is the broadcast. Network boundary selection is based on masked bits, not the observed client address.

Question 49

An IP phone with an attached computer connects to a Catalyst access port. Data clients belong to VLAN 15 while phone voice traffic belongs to VLAN 115. Which access-port configuration provides the intended separation?

  1. Use `switchport access vlan 15` and `switchport voice vlan 115` on the phone-facing access port.
  2. Use only `switchport access vlan 115` so both phone and PC share the voice subnet.
  3. Change the phone-facing port to dynamic desirable and rely on the PC to negotiate a trunk.
  4. Configure only the access VLAN for voice-tagged traffic and omit the configured voice VLAN for the phone.

Explanation

Correct Answer

A:

A voice VLAN allows supported phones to tag voice traffic separately while the attached PC normally sends untagged data traffic in the access VLAN. Verify the port's voice/data behavior and the required switch/phone capabilities. The switchport need not be changed into a general trunk for this common access design.

Incorrect Answers

B:

That would put untagged data traffic into the voice network rather than preserve the intended separation. The PC should remain in data VLAN 15 while phone media and signaling use the designated voice VLAN, subject to the site's QoS and security policies.

C:

A normal desktop is not an interswitch trunk peer and is not intended to negotiate multiple VLANs through DTP. The phone/access-port design uses a data VLAN and a configured voice VLAN instead of relying on dynamic trunk negotiation from an endpoint.

D:

An IP phone can tag frames for its voice VLAN, and the access port must explicitly support that separate voice mapping from the PC data VLAN.

Question 50

A security technician enables DHCP snooping on VLAN 35 after discovering an unauthorized laptop that offers bogus DHCP leases. What behavior should the feature enforce when the attacker sends a DHCP OFFER from an untrusted user-facing port?

  1. Disable MAC learning on every access port in VLAN 35 as the primary way to detect DHCP servers.
  2. Drop server-originated DHCP messages received on an untrusted port while learning legitimate bindings.
  3. Permit all DHCP server responses from every access port so clients always get a fast lease.
  4. Translate the DHCP OFFER's source address using NAT so it appears to come from the official server.

Explanation

Correct Answer

B:

DHCP snooping identifies trusted ports toward authorized servers and untrusted ports toward clients. A rogue OFFER received on an untrusted port is discarded; approved exchanges can populate a binding table used by other protections such as Dynamic ARP Inspection. Trust only legitimate uplinks rather than all access ports.

Incorrect Answers

A:

A switch still needs appropriate MAC learning for ordinary client forwarding. DHCP snooping examines DHCP message direction and ingress trust state; broadly disabling MAC learning would harm client connectivity without explicitly enforcing authorized DHCP server behavior.

C:

Allowing unsolicited server replies from untrusted endpoints defeats the primary anti-rogue-server purpose. Clients might receive a malicious router or DNS server. The feature should classify client-facing ports as untrusted and allow server-originated responses only from authorized sources.

D:

NAT changes IP address translation, not the Layer 2 trust decision required to stop rogue DHCP offers. It could even hide the origin of the unauthorized response. DHCP snooping is designed to discard the wrong message on an untrusted ingress port.


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