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Q1. - (Topic 5)
What are three features of the IPv6 protocol? (Choose three.)
A. optional IPsec
B. autoconfiguration
C. no broadcasts
D. complicated header
E. plug-and-play
F. checksums
Answer: B,C,E
Explanation:
An important feature of IPv6 is that it allows plug and play option to the network devices by allowing them to configure themselves independently. It is possible to plug a node into an IPv6 network without requiring any human intervention. This feature was critical to allow network connectivity to an increasing number of mobile devices. This is accomplished by autoconfiguration.
IPv6 does not implement traditional IP broadcast, i.e. the transmission of a packet to all hosts on the attached link using a special broadcast address, and therefore does not define broadcast addresses. In IPv6, the same result can be achieved by sending a packet to the link-local all nodes multicast group at address ff02::1, which is analogous to IPv4 multicast to address 224.0.0.1.
Q2. - (Topic 5)
On which options are standard access lists based?
A. destination address and wildcard mask
B. destination address and subnet mask
C. source address and subnet mask
D. source address and wildcard mask
Answer: D
Explanation:
Standard ACL’s only examine the source IP address/mask to determine if a match is made. Extended ACL’s examine the source and destination address, as well as port information.
Q3. - (Topic 3)
Before installing a new, upgraded version of the IOS, what should be checked on the router, and which command should be used to gather this information? (Choose two.)
A. the amount of available ROM
B. the amount of available flash and RAM memory
C. the version of the bootstrap software present on the router
D. show version
E. show processes
F. show running-config
Answer: B,D
Explanation:
When upgrading a new version of the IOS we need to copy the IOS to the Flash so first we have to check if the Flash has enough memory or not. Also running the new IOS may require more RAM than the older one so we should check the available RAM too. We can check both with the “show version” command.
Q4. - (Topic 5)
Which statement is correct regarding the operation of DHCP?
A. A DHCP client uses a ping to detect address conflicts.
B. A DHCP server uses a gratuitous ARP to detect DHCP clients.
C. A DHCP client uses a gratuitous ARP to detect a DHCP server.
D. If an address conflict is detected, the address is removed from the pool and an administrator must resolve the conflict.
E. If an address conflict is detected, the address is removed from the pool for an amount of time configurable by the administrator.
F. If an address conflict is detected, the address is removed from the pool and will not be reused until the server is rebooted.
Answer: D
Explanation:
An address conflict occurs when two hosts use the same IP address. During address assignment, DHCP checks for conflicts using ping and gratuitous ARP. If a conflict is detected, the address is removed from the pool. The address will not be assigned until the administrator resolves the conflict.
(Reference: http://www.cisco.com/en/US/docs/ios/12_1/iproute/configuration/guide/1cddhcp.html)
Q5. - (Topic 8)
Which component of the routing table ranks routing protocols according to their preferences?
A. administrative distance
B. next hop
C. metric
D. routing protocol code
Answer: A
Explanation:
Administrative distance - This is the measure of trustworthiness of the source of the route. If a router learns about a destination from more than one routing protocol, administrative distance is compared and the preference is given to the routes with lower administrative distance. In other words, it is the believability of the source of the route.
Q6. - (Topic 7)
Which protocol can cause overload on a CPU of a managed device?
A. Netflow
B. WCCP
C. IP SLA
D. SNMP
Answer: D
Explanation:
Sometimes, messages like this might appear in the router console:
%SNMP-3-CPUHOG: Processing [chars] of [chars]
They mean that the SNMP agent on the device has taken too much time to process a request.
You can determine the cause of high CPU use in a router by using the output of the show process cpu command.
Note: A managed device is a part of the network that requires some form of monitoring and
management (routers, switches, servers, workstations, printers…).
Q7. - (Topic 5)
What are three benefits of GLBP? (Choose three.)
A. GLBP supports up to eight virtual forwarders per GLBP group.
B. GLBP supports clear text and MD5 password authentication between GLBP group members.
C. GLBP is an open source standardized protocol that can be used with multiple vendors.
D. GLBP supports up to 1024 virtual routers.
E. GLBP can load share traffic across a maximum of four routers.
F. GLBP elects two AVGs and two standby AVGs for redundancy.
Answer: B,D,E
Q8. - (Topic 5)
Refer to the exhibit.
Which subnet mask will place all hosts on Network B in the same subnet with the least amount of wasted addresses?
A. 255.255.255.0
B. 255.255.254.0
C. 255.255.252.0
D. 255.255.248.0
Answer: B
Explanation:
310 hosts < 512 = 29 -> We need a subnet mask of 9 bits 0 -> 1111 1111.1111 1111.1111
1110.0000 0000 -> 255.255.254.0
Q9. - (Topic 4)
Which Layer 2 protocol encapsulation type supports synchronous and asynchronous circuits and has built-in security mechanisms?
A. HDLC
B. PPP
C. X.25
D. Frame Relay
Answer: B
Explanation:
PPP: Provides router-to-router and host-to-network connections over synchronous and asynchronous circuits. PPP was designed to work with several network layer protocols, including IP. PPP also has built-in security mechanisms, such as Password Authentication Protocol (PAP) and Challenge Handshake Authentication Protocol (CHAP).
Q10. - (Topic 6)
Refer to the exhibit.
Statements A, B, C, and D of ACL 10 have been entered in the shown order and applied to interface E0 inbound, to prevent all hosts (except those whose addresses are the first and last IP of subnet 172.21.1.128/28) from accessing the network. But as is, the ACL does not restrict anyone from the network. How can the ACL statements be re-arranged so that the system works as intended?
A. ACDB
B. BADC
C. DBAC
D. CDBA
Answer: D
Explanation:
Routers go line by line through an access list until a match is found and then will not look any further, even if a more specific of better match is found later on in the access list. So, it it best to begin with the most specific entries first, in this cast the two hosts in line C and D. Then, include the subnet (B) and then finally the rest of the traffic (A).
Q11. - (Topic 7)
Refer to the exhibit.
HostA cannot ping HostB. Assuming routing is properly configured, what is the cause of this problem?
A. HostA is not on the same subnet as its default gateway.
B. The address of SwitchA is a subnet address.
C. The Fa0/0 interface on RouterA is on a subnet that can't be used.
D. The serial interfaces of the routers are not on the same subnet.
E. The Fa0/0 interface on RouterB is using a broadcast address.
Answer: D
Explanation:
Now let’s find out the range of the networks on serial link: For the network 192.168.1.62/27:
Increment: 32
Network address: 192.168.1.32
Broadcast address: 192.168.1.63 For the network 192.168.1.65/27: Increment: 32
Network address: 192.168.1.64
Broadcast address: 192.168.1.95
-> These two IP addresses don’t belong to the same network and they can’t see each other.
Q12. - (Topic 5)
The network administrator is asked to configure 113 point-to-point links. Which IP addressing scheme defines the address range and subnet mask that meet the requirement and waste the fewest subnet and host addresses?
A. 10.10.0.0/16 subnetted with mask 255.255.255.252
B. 10.10.0.0/18 subnetted with mask 255.255.255.252
C. 10.10.1.0/24 subnetted with mask 255.255.255.252
D. 10.10.0.0/23 subnetted with mask 255.255.255.252
E. 10.10.1.0/25 subnetted with mask 255.255.255.252
Answer: D
Explanation:
We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits (because 2^7 = 128).
The network used for point-to-point connection should be /30. So our initial network should be 30 – 7 = 23.
So 10.10.0.0/23 is the correct answer.
You can understand it more clearly when writing it in binary form:
/23 = 1111 1111.1111 1110.0000 0000
/30 = 1111 1111.1111 1111.1111 1100 (borrow 7 bits)
Q13. - (Topic 5)
Refer to the exhibit.
What is the most efficient summarization that R1 can use to advertise its networks to R2?
A. 172.1.0.0/22
B. 172.1.0.0/21
C. 172.1.4.0/22
D. 172.1.4.0/24
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24
E. 172.1.4.0/25
172.1.4.128/25
172.1.5.0/24
172.1.6.0/24
172.1.7.0/24
Answer: C
Explanation:
The 172.1.4.0/22 subnet encompasses all routes from the IP range 172.1.4.0 – 172.1.7.255.
Q14. - (Topic 3)
What is the effect of using the service password-encryption command?
A. Only the enable password will be encrypted.
B. Only the enable secret password will be encrypted.
C. Only passwords configured after the command has been entered will be encrypted.
D. It will encrypt the secret password and remove the enable secret password from the configuration.
E. It will encrypt all current and future passwords.
Answer: E
Explanation:
Enable vty, console, AUX passwords are configured on the Cisco device. Use the show run command to show most passwords in clear text. If the service password-encryption is used, all the passwords are encrypted. As a result, the security of device access is improved.
Q15. - (Topic 5)
How does a DHCP server dynamically assign IP addresses to hosts?
A. Addresses are permanently assigned so that the host uses the same address at all times.
B. Addresses are assigned for a fixed period of time. At the end of the period, a new request for an address must be made, and another address is then assigned.
C. Addresses are leased to hosts. A host will usually keep the same address by periodically contacting the DHCP server to renew the lease.
D. Addresses are allocated after a negotiation between the server and the host to determine the length of the agreement.
Answer: C
Explanation:
DHCP works in a client/server mode and operates like any other client/server relationship. When a PC connects to a DHCP server, the server assigns or leases an IP address to that PC. The PC connects to the network with that leased IP address until the lease expires. The host must contact the DHCP server periodically to extend the lease. This lease mechanism ensures that hosts that move or power off do not hold onto addresses that they do not need. The DHCP server returns these addresses to the address pool and reallocates them as necessary.