Answer Key

1. D and E

2. D

3. B

4. A and D

5. A and F

6. C

7. B

8. C

9. E

10. D and E

11. C

12. C

13. A

14. C

15. B, E, and F

16. A and B

17. B

18. D

19. C

20. D

21. A and B

22. B

23. A

24. D

25. B

26. A

27. A

28. A

29. E

30. D

31. B

32. D

33. B

34. C

35. C and D

36. A

37. D

38. A and B

39. B and G

40. B

41. A and D

42. D and E

43. C

44. A, D, and E

45. C and D

46. B

47. B

48. B, C, and D

49. A, C, and D

50. D

51. C

52. A & B

53. C

54. A, B, and E

55. D

A1:	Answers D and E are correct. D is correct because it allows for company growth providing 60 ports for use on the network. Once answer D is chosen , the only one of the second parts of the solution that mentions a 48-port switch is answer E. The router will mediate all traffic going to networks other than that hosted in the building. Answer A is incorrect, as not enough ports will be available for the company or its growth projection. Answer B is incorrect, as not enough ports will be available given the company's projected growth. Answer C is incorrect, as not enough ports will be available given the company's projected growth. Answer F is wrong, as it can only be true if answer B is true. Answer G is wrong, as it depends on answer A being true.
A2:	Answer D is correct. It allocates more than the correct number of ports to each building. It also mentions utilizing repeaters to bridge the distance between each building. Answer A is incorrect because it does not provide enough ports for Building C (two 12-port hubs = 24 hosts; Building C has 40 computers). Answer B is incorrect because the distance between buildings exceeds that supported for UTP. Repeaters are needed. Answer C is incorrect because Building B, which has 50 hosts , will not be able to connect all of them to the network. It also suffers from the problem of not having any repeaters on the cables that run between buildings .
A3:	Answer B is correct. In the subnetting scheme in answer B, 126 hosts are allocated for the first network, 62 are allocated to the second, 30 are allocated to the third, and 30 to the fourth. A quick formula for calculating the number of hosts on a network using its CIDR (Classless Inter Domain Routing) number (the one after the /) is (2(32-CIDR))-2. Hence number of hosts when CIDR = 25 is (2(32-25))-2 = 126. Answer A is incorrect because it only allocates 14 addresses for the final network and it must support at least 24. Answer C is incorrect because only 62 hosts are allocated for the first network and the first network must support 90 hosts. Answer D is incorrect because only 14 hosts are allocated to the third and 14 hosts allocated to the fourth network. The third network must support 30 hosts and the fourth network must support 24 hosts.
A4:	Answers A and D are correct because these two networks can be supernetted together. The secret to figuring this out is that the start of the network must be a multiple of four. See the main text for an overview of subnetting. Answer B is incorrect because these two networks cannot be supernetted together; 10.10.21.0 needs to be paired with 10.10.20.0 in a /23 network and 10.10.22.0 with 10.10.23.0. Answer C is incorrect; 10.10.130.0 and 10.10.131.0 would be part of the 10.10.128.0 /23 network. 10.10.132.0 and 10.10.133.0 would be part of the 10.10.132.0 /23 network.
A5:	Answers A and F are correct. IGRP and EIGRP are protocols designed and implemented by Cisco. They are not "open" standards, but proprietary and unlikely to be used on non-Cisco equipment. Answer B is incorrect. RIP is an open protocol and is supported by all router manufacturers. Answer C is incorrect. RIPv2 is an open protocol and is supported by all router manufacturers. Answer D is incorrect. OSPF is an open protocol and is supported by all router manufacturers. Answer E is incorrect. BGP is an open protocol and is supported by all router manufacturers. BGP is not used on internal internetworks but is used for routers on the larger internet.
A6:	Answer C is correct. This question asks which protocols could not map this network. RIPv2 can map routes up to 15 hops in diameter. Routes 16 hops away are considered an infinite distance away. RIPv2 would not be able to map this network. Answer A is incorrect because OSPF could map this network. Answer B is incorrect because IGRP could map this network. Answer D is incorrect because EIGRP could map this network.
A7:	Answer B is correct. At least one router per branch office is required. Seven branch offices equals seven routers. Even if there are two subnets, only one router is required at the branch office. Answer A is incorrect. At least one router per branch office is required. Even if there are two subnets, only one router is required at the branch office. Answer C is incorrect. At least one router per branch office is required. Even if there are two subnets, only one router is required at the branch office. Answer D is incorrect. At least one router per branch office is required. Even if there are two subnets, only one router is required at the branch office.
A8:	Answer C is correct. Six lines will be required. The first requirement is one from each branch office to the head office. Each branch office can then be paired with a partner for the sake of redundancy. This creates three pairs and another three lines, for a total of nine lines. Answer A is incorrect. It provides no redundancy if a single ISDN line fails. Answers B and D are incorrect. Six lines, one from each branch office to the head office, compose the first requirement. Each branch office can be then paired with a partner for the sake of redundancy. This creates three pairs and another three lines for a total of nine lines. Tolerance of two lines failing is not required.
A9:	Answer E is correct. This will allow access from 10.10.x.x, which is the entire range that should be granted access. Answer A is incorrect because it permits hosts from networks x.x.x.0 (where x can be anything from 1 to 255). Answer B is incorrect. Access lists in the 200 range do not deal with TCP/IP, but protocol type codes. Answer C is incorrect because it allows access from networks x.x.0.0 (where x can be anything from 1 to 255). Answers D and F are incorrect because access lists in the 200 range do not deal with TCP/IP, but Ethernet type codes.
A10:	Answer D is correct. Traffic will be denied if the first 25 bits of the 32-bit binary host address match 00001010.1100100.00101101.0xxxxxxx, as 0.0.0.127 in binary is equal to 00000000.0000000.00000000.01111111 The last 7 digits of the final octet are wild. 25 in binary is 00011001. You can see from the decimal that the other first three octets already match, and the first binary digit is the same as well ”the list applies and the traffic is denied. Answer E is correct. Traffic will be denied if the first 25 bits of the 32-bit binary host address match 00001010 .1100100.00101101.0xxxxxxx, as 0.0.0.127 in binary is equal to 00000000.0000000.00000000.01111111 The last 7 digits of the final octet are wild. 118 in binary is 01110110. You can see from the decimal that the other first three octets already match, the first binary digit is the same as well ”the list applies and the traffic is denied. Answers A and B are incorrect. Traffic from these hosts is not stopped by the first line of access list 1 and is permitted by the second. Answer C is incorrect. The first line of access list 1 does not stop traffic from this host, only traffic from hosts on the 10.100.45.0 network with the last quad's value less than 128 (a one must be the first number in the final binary octet for it to not be a match). The second line of the access list permits this traffic.
A11:	Answer C is correct. Frame relay is relatively inexpensive; can provide the option of bursting, which increases data transfer rate over the agreed upon CIR; and allows new sites to be added to the network quickly. Answer A is incorrect. Leased lines tend to be a lot more expensive than packet-switched networks. It is also often a more involved process to add new sites to such a WAN network than it is with a technology like frame relay. Answer B is incorrect. Leased lines tend to be a lot more expensive than packet-switched networks. It is also often a more involved process to add new sites to such a WAN network than it is with a technology like frame relay. Answer D is incorrect. Although ATM, like frame relay, allows new sites to be added quickly, it is generally more expensive than frame relay and is less likely to offer extra bandwidth above the agreed upon amount.
A12:	Answer C is correct. HDLC is the default protocol used by Cisco routers. Answer A is incorrect because SLIP is not the default protocol used by Cisco routers. Answer B is incorrect. Although it is pervasive, PPP is not the default protocol used by Cisco routers. Answer D is incorrect. LAPB provides error recovery, but is not the default leased-line protocol used by Cisco routers.
A13:	Answer A is correct. This is the sequence of IOS commands that you would use to ensure that RIPv1, rather than RIPv1 and RIPv2, is being used on the network. Answer B is incorrect. You do not use the interface configuration mode to set up the specific RIP version that the router should use. Answer C is incorrect because the version of RIP must be specified after the router rip command has been entered, not on the same line. Answer D is incorrect because the syntax in this answer is incorrect.
A14:	Answer C is correct. This answer correctly configures the EIGRP AS (Autonomous System) number and the correct network (10.10.50.0). Answer A is incorrect. This answer attempts to (incorrectly) configure RIP rather than EIGRP. The network is also incorrect. Answer B is incorrect. This answer configures the incorrect network. The network stated in the question is 10.10.50.0, not 10.50.10.0. Answer D is incorrect. This answer configures the wrong AS (Autonomous System) number and the wrong network.
A15:	Answers B, E, and F are correct because they specify the correct subnet mask and the correct IP address for the interface. Answers A, C, and D are incorrect because they specify an incorrect subnet mask for the network interface.
A16:	Answer A is correct. 10.10.10.31 is an addressable host if the subnet mask 255.255.255.0 is used. Answer B is correct. 10.10.10.31 is an addressable host if the subnet mask 255.255.255.192 is used. Answer C is incorrect. 10.10.10.31 is a broadcast address if the subnet mask is 255.255.255.224. Answer D is incorrect. 10.10.10.31 is a broadcast address if the subnet mask is 255.255.255.240. Answer E is incorrect. 10.10.10.31 is a broadcast address if the subnet mask is 255.255.255.248.
A17:	Answer B is correct. TACACS+ is used to support a router that runs as a Terminal Server. Cisco recommends this protocol over RADIUS. TACACS+ is proprietary to Cisco. Answer A is incorrect. Although RADIUS can be used on a terminal server router to manage other routers, Cisco recommends against this because it does not provide the level of administrative granularity that TACACS+ does. Answers C and D are incorrect. IGRP and OSPF are routing protocols, so they cannot be directly used to manage other routers from a router that is configured as a terminal server.
A18:	Answer D is correct. The Cisco Catalyst 1900 switch supports 64 VLANs with a separate spanning tree. Answer A is incorrect. The Cisco Catalyst 1900 switch supports 64 VLANs with a separate spanning tree. Answer B is incorrect. The Cisco Catalyst 1900 switch supports 64 VLANs with a separate spanning tree. Answer C is incorrect. The Cisco Catalyst 1900 switch supports 64 VLANs with a separate spanning tree. Answer E is incorrect. The Cisco Catalyst 1900 switch supports 64 VLANs with a separate spanning tree.
A19:	Answer C is correct. Show VLAN-membership will display a list of ports and their corresponding VLAN-membership. Answer A is incorrect. The IOS command to display the VLAN membership is show VLAN-membership . List ports is not a valid IOS command. Answer B is incorrect. The IOS command to display the VLAN membership is show VLAN-membership . List VLAN is not a valid IOS command. Answer D is incorrect. VLAN-membership is used to assign a port to a VLAN. It will not display current port membership status.
A20:	Answer D is correct. There are adequate ports in the administration area, and all of the cabling on the factory floor is shielded . Answer A is incorrect. The cable running from the administration area switch to the factory floor switch is unshielded. STP should be used rather than UTP because of the documented electromagnetic interference. Answer B is incorrect. The cable running from the factory floor switch to the hosts on the factory floor is unshielded . STP should be used rather than UTP because of the documented electromagnetic interference. Answer C is incorrect. The two switches that are allocated to the administration area will not have enough ports to handle the 50 workstations and three servers.
A21:	Answer A is correct because a bridge can segment a network in two, thereby reducing the amount of unicast traffic on a segment. Answer B is correct because a switch reduces the size of a segment to an individual host. Answer C is incorrect. The school asked not to have the network resubnetted, which would be necessary if a router was implemented. Answer D is incorrect because repeater does not reduce the size of a network segment.
A22:	Answer B is correct. Store-and-forward switching copies the entire frame into a buffer, performs a CRC (cyclic redundancy) check, and then forwards error-free frames to their destination. Answer A is incorrect. Fragment-free switching only examines the first 64 bytes of a frame before forwarding it. Answer C is incorrect because cut-through switching performs no error correction whatsoever on a frame. Answer D is incorrect because basic switching is the same as cut through switching. No error correction occurs using this method.
A23:	Answer A is correct. Fragment free switching checks the first 64 bytes of a frame for errors. If the frame is error free, it is forwarded to its destination. Because only the first 64 bytes are checked, latency for this method of switching does not increase with the size of the transmitted frame. Answer B is incorrect. Store and forward switching checks the entire frame. This means that latency increases as frame size increases. The larger the frame, the longer it takes to check. Answers C and D are incorrect because basic switching and cut through switching provide no error correction.
A24:	Answer D is correct. You do not have to enter the IP address of the TFTP server until after this command is issued (it will prompt for one). Answer A is incorrect because backup is not an IOS command. Answer B is incorrect because backup is not an IOS command. Answer C is incorrect; this will copy a startup-config file stored on the TFTP server to the switch.
A25:	Answer B is correct. This shorthand is commonly used to copy the startup configuration to the running configuration of a router. Answer A is incorrect. This will copy the running configuration to the startup configuration. Answers C and D are incorrect because overwrite is not a command in IOS.
A26:	Answer A is correct. The hostname command is used to set the hostname of a router. Answer B is incorrect. The set name command does not exist within IOS. Answer C is incorrect. The configure name command does not exist within IOS. Answer D is incorrect. The apply name command does not exist within IOS.
A27:	Answer A is correct. Enable password CAPRICORNUS will set the enable password to CAPRICORNUS. Line con 0 will put the router into the mode where the console password can be set. Password AQUILA will set the appropriate password. Answers B and D are incorrect because enable password CAPRICORNUS will set the enable password to CAPRICORNUS. Answer C is incorrect. Line vty 0 4 will configure the VTY rather than CONSOLE password.
A28:	Answer A is correct. CDP is enabled by default on Cisco Catalyst 1900 switches. Answer B is incorrect. By default, there is no console password on a Cisco Catalyst 1900 switch. Answer C is incorrect. By default, the Spanning Tree protocol is enabled on a Cisco Catalyst 1900 switch. Answer D is incorrect. By default, the switching mode is set to FragmentFree. Answer E is incorrect. By default, the 10BaseT port is set to half duplex.
A29:	Answer E is correct. It assigns the correct IP, default gateway, and the correct subnet mask to the switch. Answer A is incorrect because it assigns the wrong subnet mask and default gateway to the switch. Answer B is incorrect because it assigns the wrong default gateway to the switch. Answer C is incorrect because it assigns the wrong IP address, default gateway, and subnet mask to the switch. Answer D is incorrect because it assigns the wrong subnet mask to the switch.
A30:	Answer D is correct. The first goal is to stop traffic from a particular internal network reaching the Internet. Traffic to the Internet from the internal network travels inbound (into the router) across interface S1 and outbound (out of the router) across interface E0. Access list 4 applied inbound on s1 will meet the first goal, as it blocks the traffic from the specified subnet. To meet the second goal, Access list 2 should be applied to e0 inbound (that is, from the Internet to the router) to restrict traffic from the Internet to that which is specified. Access list 2 could be applied to s1 out and access list 4 to e0 out to achieve the same result; however, this would require more work from the router. It was also not presented as an option in this question. Answer A is incorrect. Traffic from the Internet is inbound on interface e0 and outbound on interface s1. These access lists also use the incorrect wildcard mask (Access list 1 permits traffic from x.x.x.0 where X is anywhere from 1 to 255). Answer B is incorrect. These access lists use the incorrect wildcard mask (Access list 1 permits traffic from x.x.x.0 where X is anywhere from 1 to 255). Answer C is incorrect. Traffic from the Internet is inbound on interface e0 and outbound on interface s1.
A31:	Answer B is correct. Although this access list should be applied to incoming traffic on interface e0, applying it to outgoing traffic on interface s1 will work as well. Answer A is incorrect because the wildcard mask specifies the first three quads rather than the final quad of 192.168.20.0. Answer C is incorrect. Applying this access list to outgoing traffic on interface e0 will not work because all traffic from that particular network originates on the network that is connected to the interface and will therefore be incoming on e0. Answer D is incorrect because the access list mask specifies the first three quads rather than the final quad of 192.168.20.0. Answer E is incorrect; because answer B will work, there is a correct answer.
A32:	Answer D is correct. The interface is correctly chosen, the encapsulation type is correct, and the keepalive setting is correctly set at 20 seconds. Answer A is incorrect because the encapsulation is set to ppp rather than frame-relay. Answer B is incorrect because the interface is set to s1 instead of s0. Answer C is incorrect because the encapsulation is set to ppp rather than frame-relay, and the keepalive is set to 5 rather than 20.
A33:	Answer B is correct, and answers A, C, and D are incorrect. Layer 1 deals with Bits. Layer 2 deals with Frames. Layer 3 deals with Packets. Layer 4 deals with Segments.
A34:	Answer C is correct. IP addresses are represented by Layer 3 of the OSI model. Answers A, B, and D are incorrect, as they represent IP addresses occurring at other layers of the OSI model.
A35:	Answers C and D are correct. Repeaters and hubs do nothing to stop or filter broadcast traffic; therefore, these devices could not be used to reduce this type of congestion on the LAN. Answer A is incorrect. The Catalyst 1900 switch can be configured with VLANs. VLANs do not forward broadcast traffic; therefore, segmenting the network into several VLANs might alleviate this problem. Answer B is incorrect. Routers do not forward broadcast traffic. Segmenting the network with a router may alleviate the broadcast traffic problem.
A36:	Answer A is correct because it provides the correct syntax and is also the only one that has the correct series of numbers . Answers B, C, and D are incorrect because the syntax and number series are incorrect.
A37:	Answer D is correct. To answer this question you must work out what the network address and broadcast address will be for each of the /29 networks that this /24 network has been divided into. You can only assign an address that is neither a network nor a broadcast address to a router interface. In answer D, this address is the first available host address on network 192.168.10.152 /29. Answer A is incorrect. 192.168.10.15 is the broadcast address of network 192.168.10.8 /29. Legitimate addresses would have been from .9 through to .14. Answer B is incorrect. 102.168.10.223 is the broadcast address of network 192.168.10.116 /29. Legitimate addresses would have been from .117 through to .222. Answer C is incorrect. 192.168.10.184 is the network address of network 192.168.10.184 /29. Legitimate addresses would have been from .185 through to .190.
A38:	Answers A and B are correct. To find which hosts are addressable and which hosts are not, you must find what the networks are. To do this divide the two /24 address spaces into a total of 16 equal-sized networks. This comes to 8 equal sized networks per /24 address space. As the networks are of equal size, the subnet mask of this new network is straightforward to calculate. A /24 network is 1 network of 254 hosts. A /24 network divided in two is 2 networks at /25 with 126 hosts. A /24 network divided in four is at /26 and has 62 hosts. A /24 network divided in eight is at /27 and has 30 hosts per network division. So in this question, taking two /24 networks and putting them into 16 equal- sized networks results in 16 networks of 30 possible hosts. The final octet network addresses of these networks will be .0, .32, .64, .96, .128, .160, .192, and .224. If any of the answers have these numbers in the final address, they cannot be assigned to hosts on the network. Similarly the broadcast addresses of .31., .63, .95, .127, .159, .191, .223, and .255 cannot be addressed. Any answer that has these numbers in the final octet cannot be assigned to a host. The question is looking for which addresses cannot be assigned. Answer A (192.168.100.31) cannot be assigned, as 192.168.100.31 is the broadcast address of network 192.168.100.0 /27. Therefore, answer A is a correct answer. Answer B is a correct answer. 192.168.100.63 is the broadcast address of network 192.168.100.32 /27. Answer C is incorrect because it is a host address on the 192.168.101.192 /27 network. This address can be assigned and hence is a wrong answer. Answer D is incorrect because it is a host address on the 192.168.100.224 /27 network. This address can be assigned and hence is a wrong answer. Answer E is incorrect because it is a host address on the 192.168.100.128 /27 network. This address can be assigned and hence is a wrong answer.
A39:	Answer B and Answer G are correct. See Figure 16.1 for a diagram. Figure 16.1. Answer B is correct. As host 192.168.10.165 is hosted off switch three, this host would not be able to be pinged if switch three went down. Answer G is correct. Although host 192.168.10.165 is hosted off switch three and is unreachable, the break could have occurred on the repeater or the switch. Answer A is incorrect. Host 192.168.10.80 can ping host 192.168.10.200, thereby indicating that the router is functional. Answer C is incorrect. Host 192.168.10.200 is contactable. This means that switch four must be active. Answer D is incorrect. Switch one lies between host 192.168.10.70 and host 192.168.10.200, and these hosts can contact each other. Answer E is incorrect. Switch two hosts 192.168.10.70, which is able to communicate across the network. Answer F is incorrect. The repeater between switches one and two lies between host 192.168.10.70 and host 192.168.10.200, and these hosts can contact each other.
A40:	Answer B is correct. At no point do packets successfully traverse interface s1 of the router. Use the diagram with the question to trace each ping path . Of the options presented, this is the most likely from the information you are given in the question. Answer A is incorrect. Interface e0 is functional because packets can pass from host 10.10.10.14 to host 10.10.30.68. Answer C is incorrect. Interface s2 is functional because packets can pass from host 10.10.10.14 to host 10.10.30.68. Answer D is incorrect. Host 10.10.20.58 can be pinged; this would be impossible if the switch was down. Answer E is incorrect. Switch Omega is functional because packets can pass from host 10.10.10.14 to host 10.10.30.68. Answer F is incorrect. Switch Beta is functional because packets can pass from host 10.10.10.14 to host 10.10.30.68.
A41:	Answer A is correct. All other traffic bar 10.10.100.0 /24 and 10.10.150.0 /24 should be allowed. This line is irrelevant to our purpose. Answer D is correct. A router "reads" an access list from top to bottom. Once traffic matches, the router stops "reading." Access list 1 matches all traffic from the network 10.10.0.0 /16, hence the traffic we are interested in denying has already been permitted. Answers B and E are incorrect. These lines do what we want them to, but they do not have a chance to do so until line one is moved to the bottom of the list. Answer C is incorrect. Line three is irrelevant to our purpose and should be deleted.
A42:	Answer D is correct. This change will allow HTTP access to all hosts on the 192.168.0.0 /24 network; this was not specified in the goals. Answer E is correct. This will allow, rather than deny as specified, telnet access from target network to the destination network. Answer A is incorrect. This modification must be made to meet the goal of allowing all hosts access to the Web server on host 192.168.10.21. Answer B is incorrect. This modification must be made to achieve the goal of blocking telnet access to any host on the 192.168.10.0 /24 network from the 192.168.20.0 /24 network. Answer C is incorrect. This modification is required, or else hosts on 192.168.44.0 /24 will not be able to FTP to host 192.168.10.22.
A43:	Answer C is correct, and answers A, B, and D are incorrect. Bits represent Layer 1 communication. Frames represent Layer 2 communication. Packets represent Layer 3 communication. Segments represent Layer 4 communication and Datagrams represent Layer 5 (and above) communication.
A44:	Answer A is correct because Bits represent Layer 1 communication. Answer D is correct because Segments represent Layer 4 communication. Answer E is correct because Datagrams represent Layer 5 (and above) communication. Answer B is incorrect. Frames represent Layer 2 communication. Packets represent Layer 3 communication. Answer C is incorrect. Frames represent Layer 2 communication. Packets represent Layer 3 communication.
A45:	Answers C and D are correct because spanning tree is designed to ensure that there is only one path through the network. If multiple paths exist, data can loop through the network indefinitely until it expires . Answers A and B are incorrect because spanning tree is designed to ensure that there is only one path through the network.
A46:	Answer B is correct, and answers A, C, and D are incorrect. There will be an STP root bridge for each broadcast domain. Two TCP/IP subnets separated by a Cisco router equals two broadcast domains.
A47:	Answer B is correct, and answers A, C, and D are incorrect. Unicast is 1:1. Multicast is 1:Many. Broadcast is 1:All.
A48:	Answer B is correct. Switches do reduce the size of unicast collision domains. Answer C is correct. Routers do reduce the size of broadcast domains as they do not forward broadcast packets. Answer D is correct. VLANs do not forward broadcast packets to other VLANs. Implementing VLANs on a switch reduces the size of the broadcast domain. Answer A is incorrect. Switches, without VLANs, forward broadcast packets on all interfaces; therefore, they do not reduce the size of the broadcast domain.
A49:	Answer A is correct. RIPv2 uses distance vector logic to determine routes. Answer C is correct. IGRP uses distance vector logic to determine routes. Answer D is correct. EIGRP is a hybrid protocol that makes use of both distance vector and link-state logic to determine routes. Answer B is incorrect. OSPF uses link-state logic to determine routes.
A50:	Answer D is correct. BGP is an exterior routing protocol. Answer A is incorrect. RIPv2 is an interior routing protocol. Answer B is incorrect. IGRP is an interior routing protocol. Answer C is incorrect. OSPF is an interior routing protocol.
A51:	Answer C is correct. TFTP is used for high-speed/low overhead file transfer. It is commonly used on Cisco systems for backing up and restoring config files and IOS images. Answer A is incorrect. HTTP is generally not used for bidirectional file transfer. It also has greater overhead than TFTP. Answer B is incorrect. Telnet is generally not used for bidirectional file transfer. Answer D is incorrect. Although FTP can be used to transfer files, TFTP is a better choice in this scenario because FTP has a greater overhead than TFTP.
A52:	Answer A is correct. A laptop running appropriate software can connect via a special cable to the console port for direct router configuration. Answer B is correct. A laptop running appropriate software can connect via a special cable to the auxiliary port for direct router configuration. Answer C is incorrect. A laptop cannot configure a router's initial setup via the UTP port. Later, when the router is configured, a Telnet session may be established via this port. Answer D is incorrect. A laptop cannot configure a router's initial setup via the serial port.
A53:	Answer C is correct. FLASH stores the IOS as well as additional configuration files and alternate IOS images (depending on the size of FLASH memory). Answer A is incorrect. RAM stores the running configuration file, working memory, and routing tables. Answer B is incorrect. The POST and Bootstrap are stored in ROM. Answer D is incorrect. NVRAM stores the startup config file and the config register, but does not store the IOS image.
A54:	Answer A is correct. According to line one of the access list, FTP traffic from 10.10.x.x to host 192.168.10.24 will be discarded. Answer B is correct. Although none of the access lists specifically mention this traffic, all access lists have an implicit deny statement at the end for any traffic that does not match. Answer E is correct. According to line 2 of the access list, these packets will be discarded. Answer C is incorrect. According to line 3 of the access list, these packets will be allowed to pass through. Answer D is incorrect. According to line 5 of the access list, these packets will be allowed to pass through.
A55:	Answer D is correct. This is a difficult question that requires you to understand access lists and their masks extremely well. The mask on the second line of the access list: ACCESS LIST 102 deny tcp 10.60.0.128 0.0.255.127 192.168.10.24 eq ftp stipulates that the first bit in the final octet is set but that other bits are wild (127 = 011111111). The first bit is set by the access list at 1 (as it is 128 in decimal, 10000000 in binary), meaning that this line of the list will catch all values in the final decimal quad over and including 128. In real life, you would not write an access list like this. The third quad is wild. The final decimal quad is 200; therefore, the packet is discarded. Answer A is incorrect. The fifth line of the access list will allow FTP packets from this host to traverse the interface when this list is applied. Once a packet matches a line in the access list, it stops being processed . Answer B is incorrect. The fifth line of the access list will allow FTP packets from this host to traverse the interface when this list is applied. Once a packet matches a line in the access list, it stops being processed. Answer C is incorrect. The mask on the second line of the access list stipulates that the first bit in the final octet is set but that other bits are wild (127 = 011111111). The first bit is set by the access list at 1, meaning that this line of the list will catch all values in the final decimal quad over and including 128. In real life, you would not write an access list like this. The third quad is wild. The final decimal quad is 57; therefore, the packet passes . If the final decimal quad had been over 128, like in answer D, then this packet would have been discarded. Answer E is incorrect. The fifth line of the access list will allow ftp packets from this host to traverse the interface when this list is applied. Once a packet matches a line in the access list, it stops being processed.