Cisco 200-301 Practice Test 34
Question 330
A router configuration contains both an enable password and an enable secret. Under normal Cisco IOS behavior, which one is used when entering privileged EXEC mode?
- The console line password always wins
- The configured
enable secret takes precedence - Both strings must be entered in alphabetical order
- Neither is checked if SSH was used to log in
Explanation
Correct Answer
B:Cisco IOS normally prefers enable secret over the legacy enable password when both are configured for entry into privileged EXEC mode. The secret uses a more secure stored representation than the older enable password, depending on the configured algorithm and platform. A login over console or SSH does not by itself remove the privileged-mode check. Strong unique secrets and appropriate AAA privilege controls are still needed in a production configuration.
Incorrect Answers
A:A console line password controls entry through that management line under its configured login method and is distinct from the privileged EXEC credential. Once authenticated on the console, a user entering enable is governed by the device's enable secret when it exists. Having a line password does not make that credential outrank the separately configured privileged-mode secret.
C:Cisco IOS does not require entering both configured strings in alphabetical order. The effective privileged EXEC credential is selected by the IOS precedence rules, with enable secret normally taking priority. Treating the two configurations as a two-step ordered password sequence misinterprets their relationship and may lead to unnecessary password exposure during troubleshooting.
D:SSH encrypts the remote-management transport and authenticates a login according to VTY/AAA settings. It does not automatically bypass the later authorization or secret required for privileged EXEC mode. Operators may arrive with an already privileged account under some AAA models, but that is a separate authorization decision, not a general exemption created by choosing SSH.
Question 331
A router shows C 10.200.5.0/24 is directly connected, Gi0/3 and L 10.200.5.254/32 is directly connected, Gi0/3. A packet arrives with destination 10.200.5.254. Which entry specifically marks that destination as an address owned by the router?
- The 10.200.5.0/24 C entry, because it means every host is a local router address
- Any OSPF route, because routing-protocol preference overrides local interface ownership
- The 10.200.5.254/32 L entry
- The gateway of last resort, because a default route supersedes all /32 matches
Explanation
Correct Answer
C:The L code denotes a local route for an address owned by one of the router's interfaces. In this output, 10.200.5.254/32 marks the interface's exact IPv4 address, so traffic to that destination is handled as local delivery rather than forwarded to an adjacent host. The broader C 10.200.5.0/24 entry describes on-link neighbor reachability. The /32 is the more-specific match and unambiguously identifies local ownership.
Incorrect Answers
A:A connected C route records that the entire 10.200.5.0/24 network is attached to Gi0/3. It includes many possible neighbor destinations and does not mark every host address as belonging to the router itself. The host-specific L /32 entry is what identifies the interface's own address, so the broad connected prefix is not the best interpretation of ownership.
B:OSPF learns network destinations from neighbors and can compete with other routing sources for matching prefixes, subject to longest-prefix and administrative-distance rules. It does not override a router's own local interface address merely by being a dynamic routing protocol. The displayed L host route is direct evidence that 10.200.5.254 is configured locally, not learned externally.
D:A gateway of last resort is a default prefix such as 0.0.0.0/0 and is used only when no more-specific installed match exists. A local /32 for 10.200.5.254 is maximally specific, so a default route cannot supersede it. Confusing fallback forwarding with locally addressed packets would send control or management traffic in the wrong direction.
Question 332
show etherchannel summary displays Po5(SU) and member Gi1/0/6(P) on a Catalyst Layer 2 switch. Which interpretation correctly combines those flags?
- Po5 is a Layer 3 routed port-channel and the member is suspended
- Po5 is down and Gi1/0/6 is operating independently
- Po5 is a Layer 2 channel but the member is waiting to aggregate
- Po5 is an in-use Layer 2 EtherChannel and Gi1/0/6 participates in it
Explanation
Correct Answer
D:On commonly used Catalyst show etherchannel summary output, Po5(SU) means the logical channel is Layer 2 (S) and operationally in use (U). The member Gi1/0/6(P) is bundled in the port-channel, meaning its traffic contributes to the aggregated logical link. This is the expected healthy combination for the scenario. Port-channel numbering alone does not reveal whether LACP or PAgP formed it; inspect the group configuration separately if protocol identity matters.
Incorrect Answers
A:The S flag in Po5(SU) identifies a Layer 2 port-channel rather than a routed Layer 3 interface. The member's (P) means it is participating in the bundle, not suspended; a suspended state has different indicator text. The listed flags must be interpreted together instead of assuming every channel is a routed logical port.
B:The U flag indicates that the port-channel is in use, so Po5(SU) does not mean the logical channel is down. Likewise, Gi1/0/6(P) shows a bundled member rather than an independent standalone interface. A broken bundle would show different summary indicators and should be investigated against actual interfaces and protocol state.
C:The description correctly recognizes S as a Layer 2 channel, but it wrongly treats (P) as pending aggregation. On Catalyst EtherChannel summaries, (P) indicates a port in the port-channel bundle, not a waiting state. A member blocked from joining could be identified by other codes. Misreading this flag could cause needless troubleshooting of a healthy channel.
Question 333
A campus switch copies selected traffic to a monitoring appliance using a SPAN destination. Security wants alerts on suspicious traffic, but the appliance must not sit in the production forwarding path or drop packets. Which role is appropriate?
- A passive network intrusion detection sensor that analyzes mirrored traffic and reports events
- An inline intrusion prevention system configured to discard all matching packets
- A Layer 3 router that replaces the existing switch gateway for all client subnets
- A wireless LAN controller that negotiates SSID security for roaming laptops
Explanation
Correct Answer
A:A switch SPAN destination receives a copy of selected network frames while production switching continues independently. A passive intrusion detection sensor can examine that mirrored traffic, apply detection signatures or anomaly rules, and send security alerts without becoming the forwarding decision point. This exactly meets the requirement that the appliance neither sit inline nor discard production packets. Verify the SPAN session mirrors the intended VLANs and has sufficient monitoring capacity so important evidence is not missed.
Incorrect Answers
B:An inline IPS can actively block or reset suspicious flows because production packets traverse its enforcement position. That capability is useful when the organization wants prevention and accepts the change to the forwarding path, but it violates the request for observation only. Attaching a prevention appliance to a SPAN port does not make its drop decisions control the real forwarded packet stream.
C:A Layer 3 gateway routes packets between IP networks and may enforce ACLs or firewall policy, but it is not an out-of-band mirrored traffic observer. Replacing the current switch gateway would alter the operational topology and forwarding path instead of providing passive security alerts. The SPAN copy should go to a dedicated monitoring sensor without requiring router replacement.
D:A wireless LAN controller coordinates supported access points, SSIDs and client roaming policies; it is not by definition a network intrusion detector on a wired SPAN session. A controller may export telemetry or integrate with security tools, yet that is a distinct role. The question asks which appliance should analyze copied switch traffic for threats while remaining outside packet forwarding.
Question 334
R1 learns the exact same IPv4 10.40.0.0/16 prefix from two valid RIP neighbors. The first neighbor advertises hop count 2; the second advertises hop count 5. Their administrative distance is the same. Which route does ordinary RIP metric selection prefer?
- The five-hop path, because higher metrics always have higher routing priority
- Both routes must be rejected because equal administrative distance is invalid
- The destination IP address alone picks a neighbor without considering RIP metrics
- The two-hop path, because lower RIP hop count is the better metric for this same prefix
Explanation
Correct Answer
D:RIP uses hop count as its routing metric, with a smaller number representing a shorter path and therefore a preferable route when comparing the same prefix from equally preferred RIP neighbors. The first neighbor reports two hops, while the second reports five, so the two-hop candidate normally becomes the selected RIP path. This comparison is independent of the IPv4 host address inside 10.40.0.0/16. Equal administrative distance simply means protocol preference does not decide between these RIP entries.
Incorrect Answers
A:RIP hop count is a path-cost metric in which fewer router hops are generally preferred for the same destination. A five-hop route is worse than a two-hop route under ordinary RIP metric comparison. Higher values do not imply greater preference; a metric of 16 conventionally represents unreachable in RIP. The router therefore selects the lower-hop candidate when the sources and prefix lengths are otherwise comparable.
B:Routes learned by the same routing protocol can have identical administrative distances without being invalid. Administrative distance determines preference between sources of routing information; the protocol's internal metric compares eligible paths learned by that source. Since both are RIP routes to the same /16 prefix, normal RIP hop-count selection is the relevant next step, not rejecting both routes.
C:Destination prefix matching determines which network entry applies to the packet, but this case provides two RIP candidates for the exact same 10.40.0.0/16 destination. Once prefix length and source preference are equal, RIP compares its hop metrics to select the best path. The destination address alone does not distinguish between the two advertised next hops.
Question 335
A configuration management system has a playbook that sets the SSH-only VTY policy, and a separate infrastructure-as-code definition that creates a virtual network. Which mapping is reasonable?
- Use an NTP client to create the virtual network and SNMP to set VTY authentication
- Use Terraform to collect only logs while Ansible functions as a DNS resolver
- Use only a Layer 2 MAC table to represent both tasks
- Use Ansible for device configuration and Terraform for declarative infrastructure resources
Explanation
Correct Answer
D:Ansible playbooks and inventories can apply repeatable configuration to network devices, making them a reasonable choice for enforcing SSH-only VTY login settings. Terraform's declarative providers can create and reconcile infrastructure resources such as supported virtual networks. Both tools require correct provider/module support, credentials, state management and review of planned changes. The division here is device configuration orchestration versus lifecycle provisioning of managed infrastructure, not a claim that either tool is exclusive to one domain.
Incorrect Answers
A:NTP synchronizes device clocks and does not create virtual-network infrastructure, while SNMP primarily monitors or manages instrumented device objects rather than authoring SSH-only VTY policy in a task playbook. Neither pairing describes the tools in the question. Confusing management protocols with configuration and provisioning automation would miss the distinct operational responsibilities.
B:Terraform is an infrastructure-as-code tool for managing declarative resources, not merely a log collector; Ansible can configure network devices and is not a DNS resolver. The proposed roles invert both technologies' usual purpose. Logs and DNS can support operations, but neither represents the requested SSH line policy or creation of a managed virtual network.
C:A Layer 2 MAC table maps destination Ethernet addresses to switch ports, which is forwarding data rather than an infrastructure definition. It cannot create a cloud virtual network or set router VTY authentication policy. The scenario calls for two automation workflows that express desired configuration and infrastructure state through supported management interfaces.
Question 336
An inbound provider contract limits a flow to 10 Mb/s and requires packets beyond an allowance to be dropped or remarked immediately. Which TWO statements correctly distinguish this policy from shaping? Choose TWO.
- Policing must retain excess packets in a queue until the next second
- Shaping cannot be configured on an egress interface
- A policer can drop or remark traffic measured in excess of its configured rate or burst allowance
- DSCP values automatically increase the physical link capacity
- A shaper commonly delays excess packets, whereas the described policer can act immediately
Explanation
Correct Answers
C:A policer measures offered traffic against a configured committed rate and burst allowance, often using token-bucket logic. Packets that exceed the allowance can be dropped or remarked immediately according to policy, which matches the provider's enforcement requirement. Policing does not need to buffer all excess traffic for later emission. Exact behavior depends on the chosen action, meter interval, burst sizing, and platform support; 10 Mb/s alone does not specify every token-bucket parameter.
E:A shaper typically queues eligible excess traffic and schedules it later to achieve an average configured rate, assuming buffer space exists. A policer instead meters traffic and can immediately drop or change markings for out-of-profile packets. That difference directly matches the provider's requirement to enforce a 10-Mb/s ceiling without delaying every excess packet for later delivery. Queue overflow can still cause drops under shaping, so these are typical rather than absolute behavior descriptions.
Incorrect Answers
A:Policing generally measures traffic against rate and burst allowances and can discard, remark or otherwise act on excess packets without holding each excess packet until the next scheduling interval. Queueing excess traffic for later transmission instead describes a shaping mechanism. The provider's requirement for immediate handling on inbound traffic therefore aligns with a policer, not mandatory buffering.
B:Traffic shaping is commonly an egress queueing operation, so the claim that shaping cannot be applied on an egress interface reverses typical practice. Shaping can smooth bursty output by deferring transmission while the queue has capacity. The contractual demand in the stem is not for delayed delivery; it is for immediate enforcement when traffic exceeds 10 Mb/s.
D:DSCP carries packet classification information used by QoS policies, but its numeric marking cannot increase the physical interface's line speed. Remarking excess traffic may change how downstream devices queue it, not the underlying capacity. The contract remains governed by the policer's rate and burst controls; changing DSCP cannot exempt a flow from a 10-Mb/s cap.
Question 337
A standard ACL can match sources only. To limit access from a few clients to one particular server subnet without accidentally restricting their other destinations, where is a standard ACL generally placed?
- As close to the source networks as possible, regardless of topology
- Only on the management VTY line of every router
- Near the destination network being protected
- At a random intermediate switch port because the wildcard determines destination
Explanation
Correct Answer
C:Because a standard ACL matches only source addresses, placement near the protected server subnet limits collateral blocking of the same clients' traffic headed to other destinations. Applied in the appropriate direction on an interface close to that network, the filter can stop unauthorized sources just before they reach the servers. This is the classic standard-ACL placement guideline, subject to the actual topology and other policy layers. An extended ACL would allow source-plus-destination matching where a more precise placement is desirable.
Incorrect Answers
A:A standard ACL examines source IPv4 addresses but cannot express the intended server destination subnet. Applying it near the source would suppress that client's traffic to many unrelated destinations as soon as the source matches. The question requires restricting access to one protected network while preserving other connectivity, so place the source-only rule closer to where traffic reaches that destination.
B:An access-class on VTY lines can limit which source addresses may initiate device management sessions, but it is not the general routed traffic policy for users accessing a server subnet. The intended restriction concerns application packets destined to servers, not interactive management of router lines. A standard ACL must be attached to a suitable transit interface if used for this filtering objective.
D:A wildcard mask on a standard IPv4 ACL identifies which bits of the SOURCE address are relevant to a match; it does not specify a destination network. Placing the ACL at an arbitrary switchport cannot ensure only access to the designated server subnet is limited. The interface and direction must be chosen based on the traffic path and the destination boundary.
Question 338
A non-designated spanning-tree port is receiving BPDUs from its neighbor. An unidirectional failure makes those BPDUs disappear, raising a risk that the port might move into forwarding incorrectly. Which feature addresses this scenario?
- Root Guard
- PortFast
- BPDU Filter
- Loop Guard
Explanation
Correct Answer
D:Loop Guard prevents an STP non-designated port from transitioning to forwarding merely because expected BPDUs stop arriving. It can place the port in a loop-inconsistent state during a one-way or otherwise abnormal BPDU loss, preserving loop avoidance until messages return. This is appropriate for the stated unidirectional-failure scenario. It does not replace Root Guard on designated ports or BPDU Guard on end-host access ports; each protection addresses a different STP failure mode.
Incorrect Answers
A:Root Guard protects a designated port from accepting a superior bridge advertisement that would cause an unexpected downstream switch to become root. In this question, the risk is loss of expected BPDUs on an already non-designated or alternate port because of a unidirectional fault. The concern is an unintended forwarding transition after BPDU reception stops, which is Loop Guard's problem space rather than a superior-BPDU root-election takeover.
B:PortFast speeds access-edge transition to forwarding for suitable end-host-facing ports; it does not preserve a blocked port's protection when BPDUs disappear from a network interswitch link. Enabling PortFast on an unsuitable interswitch port can increase loop risk, especially when coupled with an unseen bridging neighbor. The question describes a port already participating in STP, not endpoint startup timing.
C:BPDU Filter suppresses BPDU transmission or reception depending on configuration, potentially hiding a bridging connection from spanning-tree loop detection. The failure described is precisely about losing BPDUs unintentionally, so deliberately filtering them would aggravate rather than mitigate the hazard. BPDU Guard has a different edge-port protection role; neither is the same as Loop Guard on non-designated ports.
Question 339
A router interface has subnet mask 255.255.255.248 in dotted decimal. Which CIDR prefix length expresses the same mask?
- /27
- /29
- /28
- /30
Explanation
Correct Answer
B:The mask 255.255.255.248 has twenty-four one-bits in the first three octets and five leading one-bits in the last octet, because 248 is binary 11111000. Adding them gives a /29 network prefix. Such a subnet spans eight IPv4 addresses and normally offers six conventional host addresses after network and broadcast reservations. The calculation is a mask conversion, not an instruction to assume the interface has precisely six active endpoints.
Incorrect Answers
A:A /27 prefix has 27 one-bits in its mask, leaving three network bits in the final octet and producing 255.255.255.224. The given final octet is 248, which has five leading one-bits rather than three. Although /27 and /29 are both ordinary subnet masks, they create different block sizes and cannot be interchanged in route or interface configuration.
C:A /28 has only four high one-bits in the last mask octet, which equals 240 rather than 248. It would describe a block of sixteen addresses instead of the /29's eight. Counting octets without converting their binary values can make these two masks look deceptively similar; check the exact final-octet bit pattern in the scenario.
D:A /30 mask contains six one-bits in the fourth octet and is written 255.255.255.252. It is often used for conventional two-endpoint point-to-point links, but that application does not alter the binary mask given here. The last octet 248 contains only five one-bits, so /30 would incorrectly narrow the subnet by one additional bit.
Question 340
For an existing IPv6 static network route, an engineer needs to examine the router’s installed IPv6 routes and check its prefix/next hop. Which operational command is appropriate?
show ipv6 routeshow ip routeshow mac address-tableshow vlan brief
Explanation
Correct Answer
A:The operational show ipv6 route command displays installed IPv6 prefixes, route source codes, next-hop information, and egress context where supported. It therefore lets the engineer verify that a configured IPv6 static route is actually present in the router's forwarding information. Unlike a running-configuration listing, it shows the active route-table view; if an expected route is missing, next-hop resolution or competition with another route may require further investigation. It does not change configuration.
Incorrect Answers
B:show ip route is the ordinary IPv4 route-table command; the IPv6 question requires the ipv6 variant. A router can have valid IPv4 entries while lacking the corresponding IPv6 static prefix, so an IPv4 output cannot establish the requested route's presence or next hop. Verify the address family before interpreting a table or troubleshooting a seemingly missing destination.
C:show mac address-table lists learned Ethernet destination addresses and associated bridge ports for Layer 2 forwarding on capable switches. A static IPv6 route operates at Layer 3 and references an IPv6 destination prefix and next-hop resolution. MAC learning does not prove that the router installed the requested IPv6 network route.
D:show vlan brief summarizes VLAN database entries and access-port membership, not IPv6 routing prefixes or next-hop associations. Even if an IPv6 router interface is connected to a particular VLAN, the existence of that VLAN does not prove the static route was installed. The engineer needs the routing information base view for the correct address family.