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Cisco 200-301 Practice Test 28

Question 270

A port-security review finds a switch interface carrying uncontrolled lab equipment into a production VLAN. The team isolates the port while it confirms an approved device inventory. What is this containment step?

  1. A mitigation that reduces the attack surface
  2. An example of data exfiltration
  3. The source of a known root bridge election
  4. A transport-layer retransmission mechanism

Explanation

Correct Answer

A:

Temporarily isolating an uncontrolled equipment port reduces the number of paths through which unapproved devices can reach the production VLAN. That is a mitigation or containment measure because it lowers exposure while the team checks the authorized asset inventory. It does not automatically eliminate all existing vulnerabilities or prove that an attacker was present. Restore connectivity only after validating endpoint identity, VLAN placement and approved operational requirements under the organization's change process.

Incorrect Answers

B:

Data exfiltration means unauthorized disclosure or transfer of sensitive information out of a protected environment. Disconnecting a suspicious lab port is a defensive action intended to prevent possible misuse, not evidence that data was sent elsewhere. The team has not observed an outbound transfer in the question. A security investigation may later check for exfiltration, but the described immediate control is isolation that reduces the potential attack surface.

C:

Spanning-tree root selection depends on bridge priorities, MAC identifiers and protocol BPDUs in an applicable Layer 2 topology. Isolating an unauthorized lab endpoint is not itself a description of a root bridge election, even if disabling a port could incidentally affect topology. The scenario's purpose is to contain an unapproved production VLAN attachment during inventory review. Labeling that as a root election misses the access-security objective.

D:

Transport-layer retransmission allows protocols such as TCP to recover missing segments when appropriate acknowledgments and sequence numbers are used. A switchport isolation action changes network access rather than retransmitting application data. The incident concerns uncontrolled equipment connected to a production VLAN and the team's deliberate risk reduction. Although isolated traffic may be lost, packet retransmission behavior is a separate consequence and not the security-control category being asked about.

Question 271

A backup route has a better administrative distance than a dynamic route to an identical prefix, but its configured next hop cannot be resolved to an active connected path. What is the key limitation?

  1. The router must always prefer a static route regardless of reachability
  2. The route must switch to IPv6 automatically
  3. The next hop must be reachable or recursively resolvable for the route to be installed and used
  4. The first-hop neighbor’s hostname must exactly match the route prefix

Explanation

Correct Answer

C:

A static route is eligible for active forwarding only when its next hop can be resolved through the current routing topology, or when an appropriate outgoing interface is known according to the configuration. Lower administrative distance influences preference among eligible sources for the same prefix, but cannot create missing neighbor reachability. The unresolved gateway therefore prevents the supposed backup from being used. Check recursion, interface state, ARP or link-local resolution and actual connected prefixes before comparing distances.

Incorrect Answers

A:

A static route can be assigned an attractive administrative distance, but IOS still requires its forwarding next hop to resolve through a usable path before the route can be installed and used. If that gateway is unreachable, the route cannot serve as an actual egress merely because its AD is lower than a competing dynamic route. The router may retain the static command in configuration without selecting it for forwarding. Reachability and preference are separate prerequisites.

B:

An unreachable IPv4 static next hop does not cause IOS to convert the route into IPv6 automatically. The two protocol families have different address representations, forwarding tables and static route commands. A gateway problem must be fixed in the applicable IPv4 routing context, such as by restoring a connected path or correcting the next-hop address. Changing address family would not repair the intended destination reachability or make the configured static route usable.

D:

Router next-hop resolution uses IP addresses, installed prefixes and link-layer reachability, not whether a neighboring device's hostname exactly matches the destination subnet text. Hostnames help operators label nodes but do not substitute for a route to the gateway. An otherwise correctly named next hop still fails if the path is down. The question explicitly says the static gateway cannot be resolved, so fixing names alone would not make the route installable.

Question 272

An access interface connected to a single workstation should enter STP forwarding promptly at link-up, without waiting for normal topology convergence. Which feature is appropriate when the port is truly an edge port?

  1. Root guard
  2. PortFast
  3. Loop guard
  4. EtherChannel LACP

Explanation

Correct Answer

B:

PortFast allows a properly designated STP edge port to transition promptly into forwarding instead of waiting through legacy convergence stages meant for inter-switch links. This improves the experience of workstations that need immediate DHCP and network access after link-up. Use it only where no switching loop is expected, often together with BPDU Guard for protection against an unexpected switch attachment. The feature does not disable STP across the network; it changes edge-port startup behavior.

Incorrect Answers

A:

Root Guard is an STP protection feature intended to prevent certain ports from accepting superior BPDU information that would move the root bridge unexpectedly. It is not the standard edge-port optimization for immediately moving an ordinary workstation access port into forwarding at link-up. The requirements describe one trusted endpoint-facing interface without switching redundancy, making PortFast the appropriate speed-up. Root Guard solves a different topology-control concern.

C:

Loop Guard helps protect against certain unexpected loss-of-BPDU conditions on appropriate non-designated spanning-tree ports, reducing risk that a blocked link incorrectly begins forwarding. It is not the ordinary control that accelerates a workstation access port at link-up. The question is about an explicitly true edge port with a single endpoint, not a redundant uplink receiving intermittent STP information. Use PortFast for the edge transition and separate loop-prevention features as needed.

D:

LACP negotiates aggregation membership among compatible Ethernet interfaces and can provide bandwidth distribution or resilience across multiple physical links. It does not instruct a single workstation-facing STP edge port to skip normal convergence delays. A lone access port connected to a PC generally is not an EtherChannel member. The desired capability concerns a spanning-tree port startup role, which is configured with PortFast rather than link aggregation.

Question 273

An IPv6 anycast service uses the same address on routers in three geographically separated sites. What behavior should clients expect from ordinary routing?

  1. Ordinary routing delivers each packet to every instance as if the address were multicast
  2. Routing always chooses the geographically closest server even when network metrics prefer another
  3. Routing delivers packets to a topologically nearest reachable anycast instance
  4. All clients are statically pinned to the first server that registered the address

Explanation

Correct Answer

C:

With anycast, several devices can announce the same service address from different locations and normal routing directs each packet toward an eligible instance selected by network topology and metrics. This is commonly described as the topologically nearest instance, not necessarily the shortest physical distance. Routing changes may move client traffic to another site during failure or policy changes. Applications should consider session persistence because ordinary anycast forwarding is not a permanent client-to-server binding.

Incorrect Answers

A:

Anycast assigns the same unicast-form destination address to multiple service instances, but ordinary routing chooses one reachable instance according to its best path; it does not clone every packet to all instances. Delivery to a listener group is multicast behavior, not anycast. The routers may advertise the anycast prefix from several sites, with forwarding selecting one based on routing information. Treating it as multicast would misrepresent how a single client session reaches a service instance.

B:

The anycast instance selected by routing is topologically preferred according to installed routes and metrics, not guaranteed to be the geographically nearest server. Network policy, provider peering and congestion can make a more distant physical site the preferred path. An engineer should measure routing outcomes rather than promise geographical proximity. The question states three distributed sites, but provides no basis for choosing one solely from map distance; the route table remains decisive.

D:

Anycast forwarding does not normally depend on which server first registered an address, nor does it permanently pin every client to the initial advertising instance. Routing control can change the preferred path as topology and policy evolve. A long-lived session may be disrupted if traffic shifts to a different instance without shared state. The scenario asks for ordinary routing behavior across advertised sites, which is path-based selection rather than immutable enrollment order.

Question 274

An administrator must route traffic for exactly one IPv4 monitoring host, 192.0.2.77, through next hop 198.51.100.2. Other addresses in 192.0.2.0/24 must follow their existing route. Which Cisco IOS static route accomplishes this?

  1. ip route 192.0.2.0 255.255.255.0 198.51.100.2
  2. ip route 192.0.2.77 255.255.255.255 198.51.100.2
  3. ip route 0.0.0.0 0.0.0.0 198.51.100.2
  4. ip route 192.0.2.77 255.255.0.0 198.51.100.2

Explanation

Correct Answer

B:

ip route 192.0.2.77 255.255.255.255 198.51.100.2 creates an IPv4 host route to exactly 192.0.2.77 through the specified next hop. Its /32 prefix normally wins over an existing /24 for that one destination while leaving other /24 hosts on their previous path. Verify that 198.51.100.2 is reachable and no policy overrides lookup. The host mask, rather than a broad network route, provides the required steering precision.

Incorrect Answers

A:

A static route to 192.0.2.0/24 covers the entire 256-address subnet, so it would also redirect other monitoring and application hosts whose addresses share that prefix. That contradicts the requirement that only 192.0.2.77 take the alternate next hop. A broader route can be useful for planned subnet migration, but it is too coarse for one host exception. Use a /32 destination to isolate the single address.

C:

A default route 0.0.0.0/0 is the least-specific IPv4 forwarding fallback and could influence many unrelated destinations that lack a longer route. It does not express a single monitoring-host exception. Existing /24 routes normally outrank a default for those addresses anyway, so installing only the proposed /0 might not even redirect the intended host. The desired policy should use an exact /32 destination prefix with a reachable next hop.

D:

The route 192.0.2.77 255.255.0.0 ... uses a /16 mask while naming a noncanonical host address in the destination field. Even if IOS normalized or rejected it depending on context, the stated mask would represent a much broader network than one host. The user needs only 192.0.2.77 to change path, not every address in 192.0.0.0/16. A /32 host mask is the correct level of specificity.

Question 275

An API payload contains {"site":null,"managed":false,"tags":["campus","lab"]}. Which THREE claims about the JSON values are accurate? Choose THREE.

  1. site is explicitly assigned the JSON null value
  2. managed is the string "false"
  3. tags is an array of two strings
  4. The document is invalid because objects cannot hold arrays
  5. The literal null means the numeric value zero
  6. managed is a boolean whose value is false

Explanation

Correct Answers

A:

The JSON literal null represents an explicit absence or null value, not a string containing the word null. In the payload, the property site exists and is set to that null value, which differs from omitting the field entirely. API semantics for null can vary—for example, clearing a stored property—but syntactically its type is clear. This interpretation is one of the three correct readings of the supplied object.

C:

The tags property contains a JSON array enclosed in square brackets with two elements, each a double-quoted string: campus and lab. Array order and element types are meaningful to an API according to its schema, even when the values are short labels. JSON objects can legitimately contain arrays, so this nested structure is valid. The correct statement identifies both the array type and the fact that its two entries are strings.

F:

The managed field is assigned the unquoted JSON primitive false, which has boolean type and indicates the negative truth value. It is not a text string and does not require further interpretation to become a boolean in a correctly parsing API client. The difference matters when software branches on management state or validates field types. The payload explicitly includes the property, so it is not absent or null either.

Incorrect Answers

B:

The field managed uses the unquoted literal false, which is a JSON boolean value rather than a quoted string containing false. Strings require enclosing double quotation marks, and the distinction can affect validation and application logic. A controller API expecting a boolean may reject a textual false because it has a different type. The payload's syntax intentionally demonstrates the boolean primitive, not a quoted string that happens to spell the same word.

D:

JSON supports nested values including arrays within objects and objects within arrays. Having tags refer to an array of strings is valid syntax, not a parse error. A particular API might impose its own schema constraints on allowed tags, but that would be an application validation decision, not a general JSON prohibition. The question is about understanding the literal payload's types and structure; nested arrays are one of JSON's basic supported forms.

E:

The JSON literal null indicates an explicit null or missing-value sentinel rather than the numeric value zero. Zero is represented by the unquoted number 0 and can be used in arithmetic, while null has separate semantic meaning. Treating them as interchangeable can cause configuration APIs to clear fields unexpectedly or confuse an unknown setting with a measured zero. The site field here is not a numerical count or index.

Question 276

A pool supplies only three global IPv4 addresses for dynamic NAT without overload. Five independent inside hosts start long-lived outbound sessions. What limitation should the engineer anticipate?

  1. All five sessions always share the first global IP using different TCP ports
  2. The router creates three public IP addresses for each local host
  3. The router forwards all unmatched sessions using inside-local addresses without translation
  4. A fourth unique inside-local binding cannot obtain a pool address until an address becomes available

Explanation

Correct Answer

D:

With only three global IPv4 addresses and dynamic one-to-one NAT without overload, at most three independent inside-local hosts can hold simultaneous unique mappings from that pool. A fourth new binding cannot obtain a global address while all three are occupied; it must wait for an available address or an expanded design. Long-lived sessions make exhaustion more likely. NAT overload could scale using port distinctions, but that mechanism is deliberately excluded by the question.

Incorrect Answers

A:

Dynamic NAT without overload normally creates one-to-one active bindings from inside-local addresses to available global pool addresses. Sharing one global IP among several simultaneous independent hosts by distinguishing transport ports is Port Address Translation or NAT overload, which the scenario explicitly excludes. A three-address pool does not automatically multiplex five unique hosts through the first address. The fourth binding can be blocked by pool exhaustion until a global address is released.

B:

A dynamic NAT pool with three global addresses has a total of three distinct usable public mapping identities available for its specified bindings, not three new addresses generated per inside host. NAT does not manufacture globally assigned addresses on demand. If five hosts request independent long-lived one-to-one translations, the pool can become exhausted. The design would need more global addresses or a suitable overload policy to support more simultaneous independent inside hosts.

C:

An unmatched inside packet does not automatically become valid Internet-facing traffic just because the NAT pool is exhausted. Depending on configuration and routing, untranslated private-source traffic may be dropped or fail beyond the provider edge, and the stated NAT requirement is not satisfied. The question asks about pool allocation capacity for dynamic translations. Assuming automatic bypass into the public network would conceal the exhaustion condition and could undermine policy controls.

Question 277

During onboarding, junior engineers learn they must report a lost console cable or misplaced configuration USB drive promptly. What is the primary purpose of that security program instruction?

  1. Replace switch port-security rules with an honor system
  2. Establish expected handling and timely incident reporting behavior
  3. Make all exported configuration files public information
  4. Guarantee every lost storage device has been remotely wiped

Explanation

Correct Answer

B:

Security awareness guidance establishes expected handling of sensitive physical items and tells employees how promptly to report loss or suspected exposure. A missing console cable or configuration USB drive may carry operational or credential risk, so timely notification allows access revocation, investigation and containment. Training does not guarantee no incident occurs, but it improves the likelihood of early response. The question asks for the purpose of the instruction rather than a technical enforcement guarantee.

Incorrect Answers

A:

Switch port-security rules can limit MAC behavior on access ports, but they do not educate personnel about handling lost equipment or reporting suspected compromise. A lost configuration USB drive may expose credentials or network topology even if all switch ports enforce strict MAC limits. Training and technical controls complement one another; one does not replace the other. The stated onboarding instruction is about human reporting behavior, not removing device-level protections.

C:

Network configuration files can contain addressing plans, access controls, device identities, certificates or secrets and should be handled under appropriate classification and protection rules. A lost USB drive does not make such content public information. Publishing configurations would often increase the attack surface by revealing administrative details. The onboarding requirement instead encourages secure custody and fast incident reporting when a potentially sensitive item is missing.

D:

Prompt reporting of a lost device can trigger investigation and possible remote revocation or wipe if the platform and management capabilities support those actions. It does not guarantee that every storage device is enrolled for remote management or reachable after loss. A passive USB drive may have no remote-wipe functionality at all. The primary benefit is getting the security team informed quickly so it can assess actual exposure and respond with available controls.

Question 278

In a supported Cisco outdoor wireless mesh, one AP has a wired Ethernet path into the distribution LAN while several other mesh APs connect by wireless backhaul. Which role should the wired mesh gateway normally perform?

  1. An LLDP relay AP
  2. A Local-mode radio monitor
  3. A nonroot MAP with no wired infrastructure access
  4. A Root Access Point (RAP)

Explanation

Correct Answer

D:

A Root Access Point, or RAP, is the supported mesh role with an Ethernet connection toward the wired distribution infrastructure. Remote Mesh Access Points can use wireless backhaul to reach that root and carry client or mesh traffic through it. The scenario explicitly identifies one wired AP and several wireless-only backhaul nodes, which matches this RAP/MAP relationship. Verify controller and AP support, mesh security and radio planning before deployment.

Incorrect Answers

A:

LLDP advertises link-neighbor information such as chassis and port identity on directly connected Layer 2 interfaces. It does not designate a wireless mesh node as the wired gateway for radio backhaul traffic. An access point could use LLDP on its Ethernet port while also participating in a mesh, but those are separate functions. The scenario's distinguishing feature is one AP with a wired distribution connection serving remote wireless mesh nodes.

B:

Local mode and monitor mode are different Cisco lightweight AP roles; monitor mode is associated with RF scanning rather than normal mesh traffic forwarding. Neither label directly describes the wired root that anchors a supported outdoor wireless mesh. The described AP must bridge traffic between its Ethernet uplink and remote APs over backhaul radios. That role is a root access point, not a controller Local-mode radio merely collecting measurements.

C:

A mesh access point without wired infrastructure connectivity may forward traffic wirelessly toward an upstream root through a mesh backhaul, but it is not the device described as physically connected to the distribution LAN. The scenario asks for the wired mesh gateway role. Calling the wired node a nonroot MAP would reverse the topology: the remote nodes rely on the wired root for upstream transport. Role assignment must reflect where Ethernet backhaul actually enters the mesh.

Question 279

Two adjacent 2.4-GHz APs are configured on the same channel and their coverage strongly overlaps. What RF issue increases when both cells transmit frequently?

  1. The APs use separate independent airtime pools because they advertise different SSIDs
  2. Co-channel contention makes the overlapping WLANs share airtime
  3. Carrier-sense contention disappears automatically when two APs share the same channel
  4. The physically stronger AP automatically disables the other AP’s transmitter

Explanation

Correct Answer

B:

Two nearby access points using the same 2.4-GHz channel in strongly overlapping coverage areas must share airtime under the applicable carrier-sense medium-access behavior. Frequent transmissions increase co-channel contention and can reduce throughput or increase delay for both cells. This is not necessarily a collision in the classic shared-Ethernet sense; clients often defer when energy or frames indicate the medium is busy. Channel planning and cell size should reduce harmful overlap.

Incorrect Answers

A:

Different SSID names identify different wireless services but do not create isolated radio spectrum when nearby APs use the same 2.4-GHz channel. Their transmissions still contend for the shared medium under 802.11 access rules. If coverage overlaps, one WLAN's traffic can consume airtime needed by another regardless of the logical network names. The question asks about RF capacity under common-channel overlap, not SSID identity or IP subnet segmentation.

C:

Carrier-sense and contention mechanisms remain active when multiple APs share a radio channel; that is precisely when competition for airtime matters most. The medium does not become contention-free merely because all transmitters agree on a channel number. Frequent activity can force clients to defer or retry, depending on RF conditions. A good design uses appropriate nonoverlapping channels and power planning rather than assuming shared-channel occupancy eliminates the need for coordination.

D:

Ordinary overlapping Wi-Fi access points do not automatically switch off the weaker transmitter because a stronger neighbor uses the same channel. Radio resource management may adjust channels or power under certain controller policies, but simple received signal strength alone does not mandate disabling a neighboring AP. Both cells can continue transmitting and contend for airtime. The problem's symptom is shared-channel competition, not an automatic active/standby radio election.

Question 280

R1 and R2 compete to become DR on a fresh OSPF broadcast segment. Their interface priorities are both 10 and their router IDs are 1.1.1.1 and 2.2.2.2. Which candidate wins the tie?

  1. R1 because 1.1.1.1 is the numerically smaller router ID
  2. Whichever router carries the numerically smaller IPv4 default gateway
  3. Both become DR simultaneously because equal priority cannot be resolved
  4. R2 because RID 2.2.2.2 is higher than 1.1.1.1

Explanation

Correct Answer

D:

Both candidates have OSPF interface priority 10, so the DR election compares their router IDs and favors the numerically higher 2.2.2.2 on R2. This is the normal tie-break for a fresh OSPF broadcast-network election. Router IDs must be unique in the routing domain, and an established DR is not automatically displaced solely by a later higher-priority arrival. The fresh-election assumption avoids that nonpreemptive behavior and makes R2 the correct candidate.

Incorrect Answers

A:

When OSPF interface priorities tie in a fresh DR election, the router ID is used as a tie-breaker, and the numerically higher router ID wins. R1's 1.1.1.1 is lower than R2's 2.2.2.2, so R1 is not favored under the stated equal priorities. Both routers remain eligible because their priorities are nonzero. This is separate from spanning-tree elections, which may prefer lower bridge identifiers.

B:

An IPv4 default gateway is a forwarding next hop for otherwise unmatched destinations; its numerical size does not determine OSPF DR selection. The election uses OSPF interface priorities and then router IDs when necessary. Routers may have different default routes and still form OSPF adjacencies on the same broadcast link. The question provides the tie-break fields directly, so a default-route address is not relevant to deciding the DR.

C:

Equal interface priorities do not force a permanent tie with two simultaneous OSPF designated routers. The protocol has a router-ID tie-break procedure for eligible candidates, assuming a fresh election and appropriate broadcast network type. Here 2.2.2.2 is higher than 1.1.1.1 and resolves the comparison. Normal OSPF DR elections are also nonpreemptive after establishment, but the stem expressly describes a new election rather than an incumbent DR.

Cisco Certified Network Associate (CCNA) (200-301) Practice Tests