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- Question
Which of the following are valid ways to refer only to host 172.16.30.55 in an IP access list?
- 172.16.30.55 0.0.0.255
- 172.16.30.55 0.0.0.0
- any 172.16.30.55
- host 172.16.30.55
- 0.0.0.0 172.16.30.55
- ip any 172.16.30.55
Options- A. 1 and 4
- B. 2 and 4
- C. 1, 4 and 6
- D. 3 and 5
- Correct Answer
- 2 and 4
ExplanationThe wildcard 0.0.0.0 tells the router to match all four octets. This wildcard format alone can be replaced with the host command. - 1. You want to create a standard access list that denies the subnet of the following host: 172.16.50.172/20. Which of the following would you start your list with?
Options- A. access-list 10 deny 172.16.48.0 255.255.240.0
- B. access-list 10 deny 172.16.0.0 0.0.255.255
- C. access-list 10 deny 172.16.64.0 0.0.31.255
- D. access-list 10 deny 172.16.48.0 0.0.15.255 Discuss
- 2. You are working on a router that has established privilege levels that restrict access to certain functions. You discover that you are not able to execute the command show running-configuration. How can you view and confirm the access lists that have been applied to the Ethernet 0 interface on your router?
Options- A. show access-lists
- B. show interface Ethernet 0
- C. show ip access-lists
- D. show ip interface Ethernet 0 Discuss
- 3. Which of the following series of commands will restrict Telnet access to the router?
Options- A. Lab_A(config)#access-list 10 permit 172.16.1.1
Lab_A(config)#line con 0
Lab_A(config-line)#ip access-group 10 in - B. Lab_A(config)#access-list 10 permit 172.16.1.1
Lab_A(config)#line vty 0 4
Lab_A(config-line)#access-class 10 out - C. Lab_A(config)#access-list 10 permit 172.16.1.1
Lab_A(config)#line vty 0 4
Lab_A(config-line)#access-class 10 in - D. Lab_A(config)#access-list 10 permit 172.16.1.1
Lab_A(config)#line vty 0 4
Lab_A(config-line)#ip access-group 10 in
Discuss
- 4. You want to create a standard access list that denies the subnet of the following host: 172.16.198.94/19. Which of the following would you start your list with?
Options- A. access-list 10 deny 172.16.192.0 0.0.31.255
- B. access-list 10 deny 172.16.0.0 0.0.255.255
- C. access-list 10 deny 172.16.172.0 0.0.31.255
- D. access-list 10 deny 172.16.188.0 0.0.15.255 Discuss
- 5. You need to create an access list that will prevent hosts in the network range of 192.168.160.0 to 192.168.191.0. Which of the following lists will you use?
Options- A. access-list 10 deny 192.168.160.0 255.255.224.0
- B. access-list 10 deny 192.168.160.0 0.0.191.255
- C. access-list 10 deny 192.168.160.0 0.0.31.255
- D. access-list 10 deny 192.168.0.0 0.0.31.255 Discuss
- 6. How many non-overlapping channels are available with 802.11h?
Options- A. 3
- B. 12
- C. 23
- D. 40 Discuss
- 7. What is the maximum distance running the lowest data rate for 802.11b?
Options- A. About 100 feet
- B. About 175 feet
- C. About 300 feet
- D. About 350 feet Discuss
- 8. How many non-overlapping channels are available with 802.11b?
Options- A. 3
- B. 12
- C. 23
- D. 40 Discuss
- 9. What is the maximum data rate for the 802.11g standard?
Options- A. 6Mbps
- B. 11Mbps
- C. 22Mbps
- D. 54Mbps Discuss
- 10. In Cisco's Unified Wireless Solution, what is the split-MAC architecture?
Options- A. The split-MAC architecture uses MAC addresses to create a forward/filter table and break up collision domains.
- B. The split-MAC architecture allows the splitting of 802.11 protocol packets between the AP and the controller to allow processing by both devices.
- C. The split-MAC architecture uses MAC addresses on the wireless network and IP addresses on the wired network.
- D. The split-MAC architecture uses MAC addresses to create a forward/filter table and break up broadcast domains. Discuss
Security problems
Search Results
Correct Answer: access-list 10 deny 172.16.48.0 0.0.15.255
Explanation:
First, you must know that a /20 is 255.255.240.0, which is a block size of 16 in the third octet. Counting by 16s, this makes our subnet 48 in the third octet, and the wildcard for the third octet would be 15 since the wildcard is always one less than the block size.
Correct Answer: show ip interface Ethernet 0
Explanation:
The only command that shows which access lists have been applied to an interface is
show ip interface Ethernet 0. The command
show access-lists displays all configured access lists, and
show ip access-lists displays all configured IP access lists, but neither command indicates whether the displayed access lists have been applied to an interface.
Correct Answer: Lab_A(config)#access-list 10 permit 172.16.1.1
Lab_A(config)#line vty 0 4
Lab_A(config-line)#access-class 10 in
Explanation:
Telnet access to the router is restricted by using either a standard or extended IP access list inbound on the VTY lines of the router. The command
access-class is used to apply the access list to the VTY lines.
Correct Answer: access-list 10 deny 172.16.192.0 0.0.31.255
Explanation:
First, you must know that a /19 is 255.255.224.0, which is a block size of 32 in the third octet. Counting by 32, this makes our subnet 192 in the third octet, and the wildcard for the third octet would be 31 since the wildcard is always one less than the block size.
Correct Answer: access-list 10 deny 192.168.160.0 0.0.31.255
Explanation:
The range of 192.168.160.0 to 192.168.191.0 is a block size of 32. The network address is 192.168.160.0 and the mask would be 255.255.224.0, which for an access list must be a wildcard format of 0.0.31.255. The 31 is used for a block size of 32. The wildcard is always one less than the block size.
Correct Answer: 23
Explanation:
The IEEE 802.11h standard provides an addition 11 channels to the 802.11a standard's 12 non-overlapping channel for a total of 23 non-overlapping channels.
Correct Answer: About 350 feet
Explanation:
The IEEE 802.11b standard provides the lowest data rate at 1Mbps, but it also has the longest distance, which is about 350 feet.
Correct Answer: 3
Explanation:
The IEEE 802.11b standard provides 3 non-overlapping channels.
Correct Answer: 54Mbps
Explanation:
The IEEE 802.11g standard provides a maximum data rate of up to 54Mbps.
Correct Answer: The split-MAC architecture allows the splitting of 802.11 protocol packets between the AP and the controller to allow processing by both devices.
Explanation:
The split-MAC architecture allows the splitting of 802.11 protocol packets between the Cisco LWAPP-based AP, which handles real-time portions of the protocol, and the WLAN controller, which handles those items that are not time sensitive.
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