4 04 2014
Latest 2014 Pass4sure and Lead2pass Cisco 642-902 Practice Tests (1-10)
QUESTION 1
Which statement is true about the implementation of IPv6 in an already existing IPv4 network?
A. IPv6 can be routed using the same routing protocol versions as IPv4
B. A router routing for IPv6 and IPv4 must convert IPv4 packets to IPv6 packets to route them.
C. IPv4 and IPv6 networks can be routed simultaneously.
D. Only OSPF version 3 can be utilized for routing IPv4 and IPv6.
Answer: C
QUESTION 2
Refer to the exhibit. The DHCP configuration that is shown is configured on a Cisco router. Which statement is true?
A. The router will distribute IP addresses from pool 1 until its addresses are exhausted. Then the router
will begin distributing addresses from pool 2.
B. The router will choose which pool to use based upon the interface the DHCP request was received on.
C. The configuration is invalid because the DHCP options are global configuration commands.
D. The configuration is incomplete until the DHCP pools are bound to the appropriate interface or interfaces.
Answer: B
QUESTION 3
Refer to the exhibit. On the basis of the information presented, which statement is true?
A. A default route is configured on the local router.
B. Network 6.0.0.0/8 was learned from an OSPF neighbor within the area.
C. OSPF router 5.0.0.2 is an ABR.
D. The default route is learned from an OSPF neighbor.
Answer: B
QUESTION 4
Drag and Drop question
Answer:
QUESTION 5
Which two reductions are the correct reductions of the IPv6 address 2001:0d02:0000:0000:0014:0000:0000:0095? (Choose two.)
A. 2001:d02::14::95
B. 2001:0d02:::0014:::0095
C. 2001:0d02:::0014:0:0:0095
D. 2001:d02::14:0:0:95
E. 2001:d02:0:0:14::95
F. FF::0014:0:0:0095
Answer: DE
We can’t use triple colons (:::) in IPv6 presentation. Also We can’t use double colons (::) twice. You can use it only once in any address because if two double colons are placed in the same address, there will be no way to identify the size of each block of 0s. Remember the following techniques to shorten an IPv6 address:
– Omit leading 0s in the address field, so :0000 can be compressed to just :0 and :0d02 can be com-pressed to :d02 (but :1d00 can not be compressed to :1d)
– Use double colons (::), but just once, to represent a contiguous block of 0s, so 2001:0d02:0000:0000:0014:0000:0000:0095 can be compressed to 2001:0d02::14:0:0:95 or 2001:0d02:0:0:14::95
QUESTION 6
Hotspot – EIGRP
Answer: AF
Explanation:
To understand the output of the “show ip eigrp topology all-links” command, let’s analyze an entry (we choose the second entry because it is better for demonstration than the first one)
The first box tells us there is only 1 successor for the path to 10.24.0.0/24 network but there are 2 boxes below. So we can deduce that one box is used for successor and the other is used for another route to that network. Each of these two boxes has 2 parameters: the first one (“156160″ or “157720″) is the Feasible Distance (FD) and the second (“128256″ or “155160″) is the Advertised Distance (AD) of that route.
The next thing we want to know is: if the route via 172.18.10.2 (157720/155160) would become the feasible successor for the 10.24.0.0/24 network. To figure out, we have to compare the Advertised Distance of that route with the Feasible Distance of the successor’s route, if AD < FD then it will become the feasible successor. In this case, because AD (155160) < FD (156160) so it will become the feasible successor. Therefore we can conclude the network 10.24.0.0/24 has 1 feasible successor.
After understanding the output, let’s have a look at the entire output:
Because the question asks about feasible successor so we just need to focus on entries which have more paths than the number of successor. In this case, we find 3 entries that are in green boxes because they have only 1 successor but has 2 paths, so the last path can be the feasible successor.
By comparing the value of AD (of that route) with the FD (of successor’s route) we figure out there are 2 entries will have the feasible successor: the first and the second entry. The third entry has AD = FD (30720) so we eliminate it.
Answer: B C D
Explanation:
First indicate the positions of these networks:
Network 172.18.3.128/25 has 2 successors, therefore the two paths below are both successors. Network 172.18.2.0/24 has only 1 successor, therefore the path lies right under it is the successor.
Answer: C E F
Explanation:
First, we should notice about the entry in the orange box, it shows that the network 172.18.10.0/24 is directly connected with this router and has a FD of 28160. So we can guess the networks that directly connected with router at 172.18.10.2 will be shown with an AD of 28160. From that, we find out 3 networks which are directly connected to the router at 172.18.10.2 (they are green underlined). The network172.18.10.0/24 is surely directly connected to the router at 172.18.10.2 (in fact it is the network that links the router at 172.18.10.2 with Switch1 router).
QUESTION 7
Refer to the exhibit. EIGRP is configured on all routers in the network. On the basis of the output provided, which statement is true?
A. Because the key chain names do not match, router R1 will not be able to ping routers R2 and R3 .
B. Because the key strings do not match, router R1 will not be able to ping routers R2 and R3.
C. Because authentication is misconfigured on interfaces Gi0/0 and Gi0/1 on router R2, router R1 will
not be able to ping routers R2 and R3.
D. Because autosummarization needs to be turned on for EIGRP on all routers, router R1 will not be able
to ping routers R2 and R3.
E. Router R1 will be able to ping routers R2 and R3.
Answer: E
Explanation:
Here we see that all of the routers have correctly included the proper networks in the EIGRP process, and the authentication is also correct so all networks would be reachable from R1. Even though the name of the authentication keys are different, the actual keys are identical so authentication will work.
QUESTION 8
A network administrator is troubleshooting a redistribution of OSPF routes into EIGRP. Given the exhibited commands, which statement is true?
A. Redistributed routes will have an external type of 1 and a metric of 1.
B. Redistributed routes will have an external type of 2 and a metric of 20.
C. Redistributed routes will maintain their original OSPF routing metric.
D. Redistributed routes will have a default metric of 0 and will be treated as reachable and advertised.
E. Redistributed routes will have a default metric of 0 but will be treated as unreachable and not advertised.
Answer: B
Explanation:
By default, all routes redistributed into OSPF will be tagged as external type 2 (E2) with a metric of 20, except for BGP routes (with a metric of 1).
Note: The cost of a type 2 route is always the external cost, irrespective of the interior cost to reach that route. A type 1 cost is the addition of the external cost and the internal cost used to reach that route.
QUESTION 9
Which two statements are true about EIGRP manual summarization? (Choose two.)
A. Manual summarization is configured on a per interface basis.
B. Manual summaries can be configured with the classful mask only.
C. When manual summarization is configured, autosummarization is automatically disabled by default.
D. The summary address is assigned an administrative distance of 10 by default.
E. The summary address is entered into the routing table and is shown to be sourced from the Null0
interface.
Answer: AE
QUESTION 10
Which is the correct command format to configure EIGRP summary route?
A. ip auto-summary as-number address mask
B. ip summary-address as-number address mask
C. ip auto-summary eigrp as-number address mask
D. ip summary-route eigrp as-number address mask
E. ip summary-address eigrp as-number address mask
Answer: E
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