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NEW QUESTION: 1
CORRECT TEXT

- 10.10.10.0/24
- 190.200.250.32/27
- 190.200.250.64/27





Answer:
Explanation:
There are two ways to configure interVLAN routing in this case:
+ Use RouterC as a "router on a stick" and SwitchC as a pure Layer2 switch. Trunking must be established between RouterC and SwitchC.
+ Only use SwitchC for interVLAN routing without using RouterC, SwitchC should be configured as a Layer 3 switch (which supports ip routing function as a router). No trunking requires.
The question clearly states "No trunking has been configured on RouterC" so RouterC does not contribute to interVLAN routing of hosts H1 & H2 -> SwitchC must be configured as a Layer 3 switch with SVIs for interVLAN routing.
We should check the default gateways on H1 & H2. Click on H1 and H2 and type the "ipconfig" command to get their default gateways.
The Answer: \>ipconfig
We will get the default gateways as follows:
Host1:
+ Default gateway: 190.200.250.33
Host2:
+ Default gateway: 190.200.250.65
Now we have enough information to configure SwitchC (notice the EIGRP AS in this case is 650)
Note: VLAN2 and VLAN3 were created and gi0/10, gi0/11 interfaces were configured as access ports so we don't need to configure them in this sim.
SwitchC# configure terminal
SwitchC(config)# int gi0/1
SwitchC(config-if)#no switchport -> without using this command, the simulator does not let you assign IP address on Gi0/1 interface.
SwitchC(config-if)# ip address 10.10.10.2 255.255.255.0 ->RouterC has used IP
10.10.10.1 so this is the lowest usable IP address.
SwitchC(config-if)# no shutdown
SwitchC(config-if)# exit
SwitchC(config)# int vlan 2
SwitchC(config-if)# ip address 190.200.250.33 255.255.255.224
SwitchC(config-if)# no shutdown
SwitchC(config-if)# int vlan 3
SwitchC(config-if)# ip address 190.200.250.65 255.255.255.224
SwitchC(config-if)# no shutdown
SwitchC(config-if)#exit
SwitchC(config)# ip routing (Notice: MLS will not work without this command)
SwitchC(config)# router eigrp 65010
SwitchC(config-router)# network 10.10.10.0 0.0.0.255
SwitchC(config-router)# network 190.200.250.32 0.0.0.31
SwitchC(config-router)# network 190.200.250.64 0.0.0.31
NOTE: THE ROUTER IS CORRECTLY CONFIGURED, so you will not miss within it in the exam, also don't modify/delete any port just do the above configuration. Also some reports said the "no auto-summary" command can't be used in the simulator, in fact it is not necessary because the network 190.200.0.0/16 is not used anywhere else in this topology.
In order to complete the lab, you should expect the ping to SERVER to succeed from the
MLS, and from the PCs as well.
Also make sure you use the correct EIGRP AS number (in the configuration above it is 650 but it will change when you take the exam) but we are not allowed to access RouterC so the only way to find out the EIGRP AS is to look at the exhibit above. If you use wrong AS number, no neighbor relationship is formed between RouterC and SwitchC.
In fact, we are pretty sure instead of using two commands "network 190.200.250.32
0.0.0.31 and "network 190.200.250.64 0.0.0.31 we can use one simple command "network
190.200.0.0 because it is the nature of distance vector routing protocol like EIGRP: only major networks need to be advertised; even without "no auto-summary" command the network still works correctly. But in the exam the sim is just a flash based simulator so we should use two above commands, just for sure. But after finishing the configuration, we can use "show run" command to verify, only the summarized network 190.200.0.0 is shown.

NEW QUESTION: 2
A network device compares a computed value of its major software components against a stored value. While the network device is loading, the device halts and alerts the administrator that the values are different. Which of the following is the name of the value in use, and what does this process provide? (Choose two.)
A. Provides authentication
B. Provides availability
C. Provides confidentiality
D. A snapshot
E. Provides integrity
F. A hash
G. A key
H. A password service
Answer: E,F

NEW QUESTION: 3
Sunil, aged 33 years, is having a policy of Rs. 1 Lac sum assured and is paying premium of Rs. 1,800/- for the last 10 years. The cash surrender value of this policy is at the end of previous year was Rs. 20,000. It is estimated that by this year end, the cash surrender value of this policy would be Rs. 22,900.
There is another term insurance of sum assured of Rs. 80,000 costs Rs. 300/- per annum which is available to him . If rate of interest is 6%, then first calculate the CPT of existing and new policy respectively and then advise Sunil if it is better to continue this policy or to discontinue it?
A. 4.13 ,3.97.Disontinue this policy
B. 2.69 ,3.77.Continue this policy
C. 3.45,4.33.Continue this policy
D. 3.19 ,2.89.Discontinue this policy
Answer: B