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Q1. - (Topic 5)
Refer to the exhibit.
A network administrator has configured a Catalyst 2950 switch for remote management by pasting into the console the configuration commands that are shown in the exhibit. However, a Telnet session cannot be successfully established from a remote host. What should be done to fix this problem?
A. Change the first line to interface fastethernet 0/1.
B. Change the first line to interface vlan 0/1.
C. Change the fifth line to ip default-gateway 192.168.17.241.
D. Change the fifth line to ip route 0.0.0.0 0.0.0.0 192.168.17.1.
E. Change the sixth line to line con 0.
Answer: C
Explanation:
The default gateway must reside on the same IP subnet as the device. Here, the subnet mask for the VLAN interface is /228. Only choice C will fix this issue, as then the default gateway is on the same subnet as the 2950.
Q2. - (Topic 1)
Refer to the exhibit.
The output is from a router in a large enterprise. From the output, determine the role of the router.
A. A Core router.
B. The HQ Internet gateway router.
C. The WAN router at the central site.
D. Remote stub router at a remote site.
Answer: D
Explanation:
Since the routing table shows only a single default route using the single interface serial 0/0, we know that this is most likely a remote stub site with a single connection to the rest of the network. All the other answer options would mean that this router would have more connections, and would contain more routes.
Q3. - (Topic 2)
How many simultaneous Telnet sessions does a Cisco router support by default?
A. 1
B. 2
C. 3
D. 4
E. 5
F. 6
Answer: E
Explanation:
By default, Cisco routers support virtual terminal interfaces 0-4 (5 total) which are used for telnet sessions.
Q4. - (Topic 1)
Refer to the exhibit.
The host in Kiev sends a request for an HTML document to the server in Minsk. What will be the source IP address of the packet as it leaves the Kiev router?
A. 10.1.0.1
B. 10.1.0.5
C. 10.1.0.6
D. 10.1.0.14
E. 10.1.1.16
F. 10.1.2.8
Answer: E
Explanation:
Although the source and destination MAC address will change as a packet traverses a network, the source and destination IP address will not unless network address translation (NAT) is being done, which is not the case here.
Q5. - (Topic 5)
Refer to the exhibit.
The junior network support staff provided the diagram as a recommended configuration for the first phase of a four-phase network expansion project. The entire network expansion will have over 1000 users on 14 network segments and has been allocated this IP address space.
192.168.1.1 through 192.168.5.255
192.168.100.1 through 192.168.100.255
What are three problems with this design? (Choose three.)
A. The AREA 1 IP address space is inadequate for the number of users.
B. The AREA 3 IP address space is inadequate for the number of users.
C. AREA 2 could use a mask of /25 to conserve IP address space.
D. The network address space that is provided requires a single network-wide mask.
E. The router-to-router connection is wasting address space.
F. The broadcast domain in AREA 1 is too large for IP to function.
Answer: A,C,E
Explanation:
The given IP addresses of areas 1 and 3 along with network masks of /24 cannot accommodate 500 users so are inadequate, while the area 2 is having over capacity so its network mask can be reduced to /25 to accommodate the only 60 users it has.
Q6. - (Topic 3)
Refer to the exhibit.
The internetwork is using subnets of the address 192.168.1.0 with a subnet mask of
255.255.255.224. The routing protocol in use is RIP version 1. Which address could be assigned to the FastEthernet interface on RouterA?
A. 192.168.1.31
B. 192.168.1.64
C. 192.168.1.127
D. 192.168.1.190
E. 192.168.1.192
Answer: D
Explanation:
Subnet mask 255.255.255.224 with CIDR of /27 which results in 32 hosts per.
192.168.1.31 is the broadcast address for sunbet '0'
192.168.1.64 is the network address for subnet '2'
192.168.1.127 is the broadcast address for subnet '3'
192.168.1.192 is the network address for subnet '6'
Q7. - (Topic 1)
To what type of port would a cable with a DB-60 connector attach?
A. Serial port
B. Console port
C. Ethernet port
D. Fibre optic port
Answer: A
Explanation:
Serial Connection
cl_3_dte_male
cl_2_dce
The picture on the left shows a V.35 DTE cable with a male DB60 connector and a male standard 34-pin Winchester-type connector. The right picture shows a V.35 DCE serial cable with a male DB60 connector and a female 34-pin Winchester-type connector. As you probably guessed already, the male connector of the DTE cable is attached to the DCE cable's female connector, this is depicted in the picture below. This is known as a back-to-back connection, and 'simulates' a WAN link. In a real world setup, the DTE cable's male connector typically connects to a port on a CSU/DSU provided by a service provider (i.e. telco), which in turn connects to a CSU/DSU at another location, thru a T1 link for example.
The DB60 connector connects to a Serial interface on a router.
cl_4_malefemale Reference: http://www.techexams.net/techlabs/ccna/lab_hardware.shtml
Q8. - (Topic 7)
Which MTU size can cause a baby giant error?
A. 1500
B. 9216
C. 1600
D. 1518
Answer: C
Explanation: http://www.cisco.com/c/en/us/support/docs/switches/catalyst-4000-series-switches/29805-175.html
Q9. - (Topic 3)
R1 is configured with the default configuration of OSPF. From the following list of IP addresses configured on R1, which address will the OSPF process select as the router ID?
A. 192.168.0.1
B. 172.16.1.1
C. 172.16.2.1
D. 172.16.2.225
Answer: A
Explanation:
The Router ID (RID) is an IP address used to identify the router and is chosen using the following sequencE.
+
The highest IP address assigned to a loopback (logical) interface. + If a loopback interface is not defined, the highest IP address of all active router's physical interfaces will be chosen.
+
The router ID can be manually assigned In this case, because a loopback interface is not configured so the highest active IP address 192.168.0.1 is chosen as the router ID.
Q10. - (Topic 3)
What is the subnet address for the IP address 172.19.20.23/28?
A. 172.19.20.0
B. 172.19.20.15
C. 172.19.20.16
D. 172.19.20.20
E. 172.19.20.32
Answer: C
Explanation:
From the /28 we can get the following:
Increment: 16 (/28 = 11111111.11111111.11111111.11110000)
Network address: 172.19.20.16 (because 16 < 23)
Broadcast address: 172.16.20.31 (because 31 = 16 + 16 – 1)