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Cisco 200-301 Practice Test 6

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.

Question 51

Two OSPFv2 routers on the same broadcast Ethernet /24 use area 0 and compatible timers. They each display router ID `10.1.1.1` due to copied configuration. What should the operator correct first?

  1. Force both OSPF routers to become DR simultaneously, assuming duplicate DR ownership is a prerequisite to FULL neighbor state.
  2. Increase OSPF interface cost on both routers to force their IDs to differ.
  3. Give each OSPF router a unique router ID and verify neighbor establishment again.
  4. Change one router to OSPF area 2 while retaining both router IDs.

Explanation

Correct Answer

C:

An OSPF router ID uniquely identifies the router in the link-state domain. Cloning that ID between neighbors can prevent reliable adjacency and create LSA identity conflicts. Correct one ID and restart or refresh the affected process carefully, then check show ip ospf neighbor and the route database.

Incorrect Answers

A:

A multiaccess adjacency uses a consistent DR/BDR election; forcing duplicate roles is not a repair for mismatched Hello or area parameters.

B:

OSPF cost determines route preference, not router identity. Cost adjustments do not change the router ID embedded in LSAs and neighbor relationships. Each router needs a distinct configured ID or an otherwise unique eligible address selected by the process.

D:

Routers sharing a link must agree on OSPF area to form the intended neighbor relationship. Changing one end to another area would add a mismatch. The duplicate router ID is the actual stated problem and should be resolved directly.

Question 52

A service desk reviews IPv4 interfaces on three lab PCs: 10.7.5.20, 172.16.90.10 and 192.168.4.50. An apprentice says only the last one is private. What is accurate about their RFC 1918 classification?

  1. Only 192.168.4.50 is private because 10.0.0.0/8 is public ISP space.
  2. The 172.16.90.10 address is always public because 172 is outside 192.168.
  3. All three are globally reachable on the Internet without translation because they are IPv4.
  4. All three are private IPv4 addresses, including the entire 10.0.0.0/8 range.

Explanation

Correct Answer

D:

RFC 1918 reserves 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16 for private addressing. These addresses are not meant to be globally routed across the public Internet without an appropriate gateway or translation design. The first octet alone does not make 172.16 addresses public.

Incorrect Answers

A:

The complete 10/8 block is private space, widely used inside enterprises and labs. Treating 10.7.5.20 as public may lead to routing or NAT mistakes. The 172.16.90.10 address also falls inside the 172.16–172.31 private range.

B:

RFC 1918 includes 172.16.0.0 through 172.31.255.255, which contains 172.16.90.10. Private addressing is defined by prefixes, not a single familiar 192.168 pattern. Check masks and documented ranges precisely.

C:

Private address blocks are not normally routed across the public Internet. An enterprise may connect them using VPNs or translate traffic at its boundary. Their IPv4 syntax does not make them public addresses, and using them as Internet endpoints requires additional design.

Question 53

A tagged Ethernet frame arrives on an interswitch 802.1Q trunk for VLAN 55. Both switches agree that VLAN 99 is native. How should VLAN 55 normally be represented on the trunk wire?

  1. With an 802.1Q VLAN 55 tag, because VLAN 55 is not the native VLAN.
  2. Untagged, because all VLANs on a trunk use the native VLAN's untagged format.
  3. With an IP subnet mask rather than an Ethernet VLAN tag.
  4. Describe the trunk native VLAN as the same as its advertised tag for VLAN 55, even though the frame contains an explicit VLAN 55 ID.

Explanation

Correct Answer

A:

In ordinary 802.1Q operation, frames for a nonnative VLAN are VLAN-tagged so the neighbor can place them in the correct broadcast domain. The agreed native VLAN is generally sent untagged by default, subject to platform-specific native-tagging settings. Verify both ends agree on native and allowed VLANs.

Incorrect Answers

B:

The trunk must distinguish multiple VLANs over one physical link. If every VLAN were untagged there would be no VLAN identity to separate them. Non-native VLAN 55 traffic normally carries the 802.1Q tag; only the designated native VLAN is untagged under default behavior.

C:

The VLAN identifier is a Layer 2 encapsulation field carried in the Ethernet frame header. Subnet masks are IP addressing constructs and cannot mark Ethernet frames for membership in VLAN 55. Routing and switching use different header information.

D:

A tagged frame is classified by the 802.1Q VLAN ID, not by the native VLAN reserved for normally untagged traffic.

Question 54

A new guest VLAN 55 contains DHCP clients but the DHCP server is on 192.0.2.20 in a different subnet. The routed SVI for VLAN 55 has no DHCP relay configured. Which interface-level change enables the server to receive client requests?

  1. Increase the DHCP lease duration on the server without changing routing or relay.
  2. Configure `ip helper-address 192.0.2.20` on VLAN 55's routed SVI.
  3. Configure `ip helper-address` only on the remote DHCP server's Ethernet port.
  4. Create a local DHCP pool on the access switch without routing or relaying client broadcasts to the actual remote server.

Explanation

Correct Answer

B:

A DHCP client initially sends discovery broadcasts within its subnet. A router does not ordinarily forward those broadcasts across networks. An ip helper-address on the client-facing routed interface relays DHCP requests to the specified server, preserving client-subnet information for selecting the proper pool.

Incorrect Answers

A:

A longer lease only affects clients that have already received an address. New guests still cannot deliver a broadcast to a different subnet when the helper is missing. Relay configuration is the prerequisite for initial lease discovery across the routed boundary.

C:

The relay point needs to observe the client's local DHCP broadcast. The server is already on another network and cannot hear the guest VLAN's broadcast by default. Configure the helper on the client-facing VLAN 55 routed interface.

D:

A pool existing elsewhere does not automatically cross a routed subnet boundary; configure a suitable SVI relay or DHCP service path.

Question 55

An operations group must standardize NTP, DNS and interface descriptions on 120 IOS devices. It wants repeatable configuration tasks using version-controlled YAML and SSH/transport modules without a resident agent on every switch. Which tool is the appropriate fit?

  1. Use only SNMP GET polling to read current device settings, assuming read operations push approved configuration to all 120 devices.
  2. Copying a configuration screenshot by email to each operator without a shared source file.
  3. Ansible playbooks with suitable Cisco device modules and reviewed inventory.
  4. Terraform state alone, used as an SSH session script that manually types configuration commands on each switch.

Explanation

Correct Answer

C:

Ansible supports task-based configuration management using inventory, playbooks and modules, often connecting over SSH or supported network APIs without placing a separate agent on each device. Test idempotence and backups, protect credentials and compare effective device state after applying a change.

Incorrect Answers

A:

SNMP polling gathers state but is not a repeatable device configuration deployment or desired-state reconciliation system.

B:

Screenshots are not executable, versioned task definitions and leave each engineer to interpret and type device changes. That preserves manual drift and makes rollback difficult. Use reviewed playbooks with a documented inventory and result verification.

D:

Terraform is best known for declarative provisioning of infrastructure resources with providers and state. It can manage some device configuration through providers, but the specifically described playbook-style task execution and agentless device command workflow maps more directly to Ansible. Using state as a typing script would miss that distinction.

Question 56

A remote monitoring host at 198.51.100.74 must be reached through a special WAN next hop 203.0.113.6; all other addresses in 198.51.100.0/24 must keep following an existing route. What static destination mask selects only the monitoring host?

  1. A /24 static route for `198.51.100.0 255.255.255.0` through 203.0.113.6.
  2. A default static route 0.0.0.0/0 through 203.0.113.6.
  3. A /16 route for 198.51.0.0/16 through 203.0.113.6.
  4. A host route for `198.51.100.74 255.255.255.255` through 203.0.113.6.

Explanation

Correct Answer

D:

A /32 IPv4 static route matches the one destination host and is more specific than a /24 route, so it affects the monitoring address while preserving the existing network route for its neighbors. Verify the WAN next-hop reachability and return path before considering this steering complete.

Incorrect Answers

A:

This would match every host in the /24 and could redirect unrelated application traffic through the special WAN. The request is restricted to one specific monitoring IP, so the static route should use a /32 host prefix rather than the whole subnet.

B:

A default route applies to destinations without any more-specific match and cannot express a one-host exception by itself. It might also change general Internet forwarding. Use a specific /32 route so the exception affects only the named monitoring host.

C:

A /16 includes a much larger address block than the specified /24, potentially capturing many additional networks. It is less specific than the existing /24 for ordinary addresses and still does not precisely document the single-host exception. A /32 is required.

Question 57

A network team grants access to a router's privileged configuration mode only after validating an administrator's username and secret. This determines whether the person is who they claim to be, before separate command permissions are checked. Which AAA function is shown?

  1. Authentication, which establishes the administrator's identity at login.
  2. Authorization, because checking identity is the same as permitting every configuration command.
  3. Accounting, because login secrets are used mainly to log packet totals.
  4. Network address translation, because usernames are mapped to public identities.

Explanation

Correct Answer

A:

AAA distinguishes identity verification from the actions an authenticated person may perform and from records of activity. Checking a username, credential, certificate or other factor establishes authentication. Authorization follows to grant role or command access; accounting records what was done.

Incorrect Answers

B:

Authorization decides what a verified principal is allowed to do, such as a permitted command set or interface scope. The scenario describes checking who the person is before those privileges are assigned. A successful credential check alone should not imply unrestricted configuration rights.

C:

Accounting produces records of access sessions and actions for audit or billing. While a login event can be logged, the credential validation itself is the authentication step. Confusing these terms can lead to a design with audit records but no actual identity verification.

D:

NAT maps IP addresses and possibly ports at a network boundary; it does not validate a human administrator's username or secret. Router management security requires an AAA identity mechanism followed by separate authorization and accounting as needed.

Question 58

A device is assigned IPv6 address `2001:db8:0:0:0:0:0:42`. The engineer wants to use valid zero compression without losing hextet positions. Which rendering represents the same address?

  1. `2001:db8:0:0:0:0:0:0:42`
  2. `2001:db8::42`
  3. `2001:db8:42`
  4. `2001::db8::42`

Explanation

Correct Answer

B:

An IPv6 double-colon can replace one contiguous sequence of all-zero hextets, once in an address. In this value the five consecutive all-zero hextets between db8 and 42 can be compressed to ::. The resulting address preserves the same 128-bit quantity.

Incorrect Answers

A:

This rendering has nine hextets rather than eight, exceeding the fixed 128-bit IPv6 address width. The original has 2001, db8, five zeros and 42. Correct compression makes the value shorter without adding an additional group.

C:

Without a double-colon abbreviation the value contains too few hextets and lacks a legal way to infer the omitted zero groups. IPv6 text needs eight hextets after expansion; only :: provides the documented contiguous-zero compression.

D:

Using the double-colon twice makes expansion ambiguous because the address no longer states how many zero hextets belong to each omission. IPv6 allows at most one :: run. Compress the one contiguous zero sequence between db8 and 42.

Question 59

Switch A is root bridge for VLAN 20. Switch B has two independent uplinks to A: Gi1/0/1 has a path cost of 4; Gi1/0/2 has a path cost of 19. No other cost or priority tie-breaker applies. Which port is Switch B's root port?

  1. Both links must become root ports because they reach the same root bridge.
  2. Neither port can be a root port because root-port roles exist only on the root switch.
  3. Gi1/0/1, because it offers the lowest total path cost toward the root bridge.
  4. Gi1/0/2, because higher path cost represents a faster path.

Explanation

Correct Answer

C:

Every non-root bridge selects a root port representing its best path toward the VLAN's root bridge, using STP tie-breakers when necessary. With costs 4 and 19 and no offsetting conditions, Gi1/0/1 wins. The other link may be an alternate to avoid a Layer 2 loop.

Incorrect Answers

A:

STP normally selects one root port on each non-root bridge for a given VLAN and prevents loops by putting alternate redundant paths into a nonforwarding role. Having two forwarding root ports without the appropriate topology mechanism would defeat loop prevention.

B:

The root bridge has designated ports on its active segments; non-root bridges select root ports to reach it. Switch B is explicitly not the root and therefore needs a best upstream root port. The lower-cost Gi1/0/1 fulfills that role.

D:

STP path cost is a cost to minimize when selecting the root port. A lower cost generally reflects a better path given the configured speeds and costs. Cost 19 loses to cost 4 with all other factors equal; higher is not preferred.

Question 60

Two Cisco routers share an Ethernet /24 and attempt OSPFv2 adjacency. Both use matching authentication and hello intervals, but the first advertises area 0 and the second area 3 on that interface. The OSPF neighbor list remains empty. What is the primary correction?

  1. Set different hello timers on the interfaces to force a new OSPF negotiation.
  2. Increase the OSPF cost on the area 3 interface so its route wins the election.
  3. Treat the Ethernet 802.1Q trunk as a replacement for matching OSPF area and Hello timer settings between the two routers.
  4. Configure both directly connected OSPF interfaces in the same intended area.

Explanation

Correct Answer

D:

OSPF neighbor formation requires compatible area identity on the shared link. A router expecting area 0 does not form an adjacency with a peer sending area 3 Hello packets. Align the area configuration, then verify neighbor state; changing router IDs or route costs cannot substitute for matching area IDs.

Incorrect Answers

A:

Hello/dead timers are additional parameters that normally must be compatible. Creating a timer mismatch would introduce another obstacle while leaving the area IDs different. Fix the identified area mismatch and keep the already matching timers.

B:

OSPF interface cost influences route selection after adjacencies and LSDB exchange. It does not permit neighbors in different areas on the same link to establish adjacency. No route metric can correct a mismatch in the area field of neighbor Hellos.

C:

Layer 2 connectivity can carry protocol packets but cannot make OSPF neighbors form when their required control-plane parameters are incompatible.


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