ActualTests.Cisco.640-802.Exam.Q.and.A.08.15.08-DDU (1130589), страница 37
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The largest mask that can be used is the single IP host mask, which is /32. It is notpossible to use a /38 mask, unless of course IPv6 is being used.C, D, E. These masks will provide for a larger number of host addresses, and since only 2IP addresses are needed for a point to point link, these extra addresses are wasted.F: No available host addresses with a /32 maskQUESTION 296:What is the maximum number of IP addresses that can be assigned to hosts on aCertkiller subnet that uses the 255.255.255.224 subnet mask?A. 14B.
15C. 16D. 30E. 31F. 32Answer: DExplanation:The subnet mask 255.255.255.224 means that there are 27 network bits. The remaining 5bits are the host bits. The maximum possible combinations with 5 bits are 25 = 32. As allzero's and all one's hosts are not allowed so, maximum number of valid hosts with themask 255.255.255.224 are 25 -2 =32-2 = 30 HostsQUESTION 297:In a Certkiller network that supports VLSM, which network mask should be usedfor point-to-point WAN links in order to reduce waste of IP addresses?A. /24B. /30C.
/27D. /26E. /32F. None of the aboveAnswer: BExplanation:A 30-bit mask is used to create subnets with two valid host addresses. This is the exactnumber needed for a point-to-point connection.Actualtests.com - The Power of Knowing640-802QUESTION 298:The network 172.25.0.0 has been divided into eight equal subnets. Which of thefollowing IP addresses can be assigned to hosts in the third subnet if the ipsubnet-zero command is configured on the router? (Choose three)A.
172.25.78.243B. 172.25.98.16C. 172.25.72.0D. 172.25.94.255E. 172.25.96.17F. 172.25.100.16Answer: A, C, DExplanation:If we divide the address 172.25.0.0 in 8 subnets, the resulting subnets will be1. 172.25.0.02. 172.25.32.03. 172.25.64.0 This is the third subnet4. 172.25.96.05. 172.25.128.06. 172.25.160.07. 172.25.192.08. 172.25.224.0Addresses that fall in the 3rd subnet will be from 172.25.64.0 ---- 172.25.95.255Choices A, C and D lie in this network range.QUESTION 299:The Certkiller network administrator has designed the IP scheme as shown in thediagram below:Based on the information shown above, what effect will this addressing scheme haveon the network?Actualtests.com - The Power of Knowing640-802A. IP traffic between subnet A and B will be prevented.B.
Routing information will not be exchanged.C. The addressing scheme will allow all IP traffic between the LANs.D. IP traffic between all the LANs will be prevented.E. None of the aboveAnswer: CExplanation:This scheme will allow for communication between all networks, and uses all IPaddresses in the 192.168.1.0/24 IP network with no overlap. Note that RIPv2 is beingused instead of RIPv1. RIPv2 carries subnet mask information allowing for VLSMnetworks like the one shown here.QUESTION 300:The network with the IP address 172.31.0.0/19 is to be configured on the Certkillerrouter with the partial configuration shown in the graphic.
Which of the followingstatements describes the number of available subnets and hosts that will result fromthis configuration?Exhibit:A. There are 7 usable subnets, with 2046 usable host addresses.B. There are 8 usable subnets, with 30 usable host addresses.C. There are 7 usable subnets, with 30 usable host addresses.D.
There are 8 usable subnets, with 2046 usable host addresses.E. There are 7 usable subnets, with 8190 usable host addresses.F. There are 8 usable subnets, with 8190 usable host addresses.Actualtests.com - The Power of Knowing640-802Answer: FExplanation:The 172.31.0.0/19 will have 3 bits in the network portion, and 13 bits in the host portion.This will allow for 2^3 = 8 networks and 2^13 = 8192 hosts available for each network(8190 usable). Since the IP subnet-zero command is used the first network is available,making choice F correct.QUESTION 301:A portion of he Certkiller network is shown in the diagram below:Consider the 192.1.1.0/24 network in this exhibit.
This network uses RIP v2.Which combination of subnetwork assignments will satisfy the requirements fornetworks A, B, and C of this design? (Select three)A. Network A = 192.1.1.128/25B. Network A = 192.1.1.0/25C. Network B = 192.1.1.252/30D. Network B = 192.1.1.4/30E. Network C = 192.1.1.64/26F. Network C = 192.1.1.224/27Answer: A, D, EExplanation:To properly answer this question, it is best to start from the end, which is network C.Since network C requires at least 55 host addresses, a /26 network must be used.
Anetwork mask of /26 will provide for 62 usable IP addresses while a /27 network willonly provide for 30 so we must choose E. With choice E taken, hosts within the range of192.1.1.65-192.1.1.126 will be used.For network A, both choices A and B are using the correct subnet mask, but we are onlylimited to choice A since many of the hosts in choice B are already being used in networkC. Finally, for network B we are left with choice D since hosts in choice C are alreadybeing used by network A.Actualtests.com - The Power of Knowing640-802QUESTION 302:If an ethernet port on router CK1 was assigned an IP address of 172.16.112.1/20,what is the maximum number of hosts allowed on this LAN subnet?A.
2046B. 1024C. 4096D. 8190E. 4094F. None of the aboveAnswer: EExplanation:Since a /20 equates to 12 bits used for the subnet mask, 4094 hosts can be uniquelyaddressed.Number of Bits in the Host orSubnet FieldMaximum number of Hosts orSubnets (2n-2)102236414530662712682549510101022112046Actualtests.com - The Power of Knowing640-8021240941381901416,382QUESTION 303:Part of the Certkiller WAN is shown below:A new subnet with 12 hosts has been added to the Certkiller network shown above.Which subnet address should this network use to provide enough useable addresses,while wasting the fewest number of IP addresses?A. 192.168.10.80/29B. 192.168.10.80/28C. 192.168.10.96/28D.
192.168.10.96/29E. None of the aboveAnswer: CQUESTION 304:DRAG DROPCertkiller has three locations and has plans to redesign the network accordingly. Thenetworking team received 192.168.151.0 to use as the addressing for entire networkfrom the administrator. After subnetting the address, the team is ready to assign theaddress.The administrator plans to configure ip subnet-zero and use RIP v2 as the routingprotocol.
As a member of the networking team, you must address the network andActualtests.com - The Power of Knowing640-802at the same time converse unused addresses for future growth.Being mindful of these goals, drag the host addresses on the left to the correct routerinterface. One of the routers is partially configured.
Move the mouse over a routerto view its configuration (** This information is missing**). Not all of the hostaddresses on the left will be used.Actualtests.com - The Power of Knowing640-802Answer:QUESTION 305:Part of the Certkiller WAN is shown below:All of the Certkiller routers in this network segment are configured with the "ipsubnet-zero" IOS command. Because of this, which network addresses should beused for Link Certkiller and Network Certkiller shown above? (Choose two.)Actualtests.com - The Power of Knowing640-802A. Network Certkiller - 172.16.3.128/25B.
Link Certkiller - 172.16.3.40/30C. Network Certkiller - 172.16.3.192/26D. Link Certkiller - 172.16.3.112/30E. Link Certkiller - 172.16.3.0/30F. Network Certkiller - 172.16.3.48/26Answer: A, EQUESTION 306:DRAG DROPCertkiller has three locations and has plans to redesign the network accordingly. Thenetworking team received 192.168.151.0 to use as the addressing for entire networkfrom the administrator. After subnetting the address, the team is ready to assign theaddress.The administrator plans to configure ip subnet-zero and use RIP v2 as the routingprotocol.
As a member of the networking team, you must address the network andat the same time converse unused addresses for future growth.Being mindful of these goals, drag the host addresses on the left to the correct routerinterface. One of the routers is partially configured. Move the mouse over a routerto view its configuration (** This information is missing**). Not all of the hostaddresses on the left will be used.Actualtests.com - The Power of Knowing640-802Answer:QUESTION 307:A host on the Certkiller network has been configured with the IP address10.16.3.66/23. Which two statements describe this IP address? (Choose two)Actualtests.com - The Power of Knowing640-802A. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.B.
This network is not subnetted.C. The last valid host address in the subnet is 10.16.2.254 255.255.254.0D. The subnet address is 10.16.3.0 255.255.254.0.E. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.Answer: A, EExplanation:A subnet mask of /23 translates to 255.255.254.0 and will provide for up to 512 IPaddresses.If we take the 10.16.X.X network using the /23 subnet mask, the first network available is10.16.0.0/23, which will provide host address from 10.16.0.1 to 10.16.2.254, with10.16.2.255 being the broadcast address. The next available network in the 10.16.X.Xcovers our example in this question of 10.16.3.66.In this case, the first useable IP address is (10.16.2.1 choice E), and the broadcast addressis 10.16.3.255 (choice A).In closing, the partial reference table on IPv4 subnets:CIDR | Netmask | Addresses-----+-----------------+----------/18 | 255.255.192.0 | 16384/19 | 255.255.224.0 | 8192/20 | 255.255.240.0 | 4096/21 | 255.255.248.0 | 2048/22 | 255.255.252.0 | 1024/23 | 255.255.254.0 | 512/24 | 255.255.255.0 | 256/25 | 255.255.255.128 | 128/26 | 255.255.255.192 | 64/27 | 255.255.255.224 | 32/28 | 255.255.255.240 | 16QUESTION 308:Part of the Certkiller network is shown below:Actualtests.com - The Power of Knowing640-802In the Certkiller network shown above the IP address space of 128.107.7.0/24 hasbeen allocated for all devices.
All devices must use the same subnet mask and allsubnets are usable. Which subnet mask is required to apply the allocated addressspace to the configuration that is shown?A. 255.255.255.192B. 255.255.255.128C. 255.255.255.0D. 255.255.255.224E. 255.255.254.0F. None of the aboveAnswer: AExplanation:In this example the requirement is that the company needs 3 subnets and at least 58 hostsper subnet. Referring to the following formula we see that 6 bits of subnet masking isneeded.Number of Bits in the Host orSubnet FieldMaximum number of Hosts orSubnets (2n-2)102236414530Actualtests.com - The Power of Knowing640-80266271268254With 6 bits used for the subnet portion, we get will get 4 different subnets with 62 usableIP addresses in each. The subnet mask for this /28 network translates to 255.255.255.192.QUESTION 309:DRAG DROPCertkiller has three locations and has plans to redesign the network accordingly.
Thenetworking team received 192.168.151.0 to use as the addressing for entire networkfrom the administrator. After subnetting the address, the team is ready to assign theaddress.The administrator plans to configure ip subnet-zero and use RIP v2 as the routingprotocol. As a member of the networking team, you must address the network andat the same time converse unused addresses for future growth.Being mindful of these goals, drag the host addresses on the left to the correct routerinterface.
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