2021年度思科认证考试题库.docx
CCNA640-802 V13题库试题分析题库解说:吴教师(艾迪飞CCIE实验室首发网站:1. What are two reasons that a network administrator would use access lists?(Choose two.)A. to control vty access into a routerB. to control broadcast traffic through a routerC. to filter traffic as it passes through a routerD. to filter traffic that originates from the routerE. to replace passwords as a line of defense against security incursionsAnswer： AC解释一下:在VTY线路下应用ACL,可以控制从VTY线路进来telnet流量。也可以过滤穿越一台路由器流量。2. A default Frame Relay WAN is classified as what type of physical network?A. point-to-pointB. broadcast multi-accessC. nonbroadcast multi-accessD. nonbroadcast multipointE. broadcast point-to-multipointAnswer： C解释一下:在默认状况,帧中继为非广播多路访问链路。但是也可以通过子接口来修改她 网络类型。3. Refer to the exhibit. How many broadcast domains exist in the exhibited topology?A. oneB. twoC. threeD. fourE. fiveF. sixAnswer： C解释一下:广播域问题,在默认状况下,每个互换机是不能隔离广播域,因此在同一种区域 所有互换机都在同一种广播域中,但是为了减少广播危害,将广播限制在种更小范畴,有 /VLAN概念，VLAN表达是种虚拟局域网，而她作用就是隔离广播。因此被VLAN隔离 了每个区域都表达种单独广播域，这样种VLAN中广播流量是不能传到其她区域，因此 在上题中就有3个广播域J。4. A single 802.11 g access point has been configured and installed in the center of a square office. A few wireless users are experiencing slow performance and drops while most users are operating at peak efficiency. What are three likely causes of this problem?(Choose three.)A. mismatched TKIP encryptionB. null SSIDC. cordless phonesD. mismatched SSIDE. metal file cabinetsF. antenna type or directionAnswer： CEF6. The command frame-relay map ip 10.121.16.8 102 broadcast was entered on the router. Which of the following statements is true concerning this command?A. This command should be executed from the global configuration mode.B. The IP address 10.121.16.8 is the local router port used to forward data.C. 102 is the remote DLCI that will receive the information.D. This command is required for all Frame Relay configurations.E. The broadcast option allows packets, such as RIP updates, to be forwarded across the PVC.Answer： E解释一下：关于命令frame-relay map ip 10.121.16.8 102 broadcast,这个命令用于手工静态添 加一条映射,到达10.121.16.8流量封装种DLCI号为102,并且这条PVC是支持广播流量, 例如RIP更新包。由于在默认状况下,帧中继网络为非广播,而R1P在其上是无法发包。8. Which of the following are associated with the application layer of the OS I model ?(Choose two.)1. ping8. TelnetC. FTPD. TCPE. IPAnswer： BC解释一下:在OSI 7层模型中位于应用层应用有lelnel和ftp这两种应用。9. For security reasons, the network administrator needs to prevent pings into the corporate networks from hosts outside the internetwork. Which protocol should be blocked with access control lists?A. IPB. ICMPC.TCPD. UDPAnswer： B解释一下:PING命令 运用ICMP合同echo,和echoreplay两个报文来检测链路与否连通。因 此如果要制止PING流量到网络,就只要过滤掉ICMP应用就可以了。10. Refer to the exhibit. The network administrator has created a new VLAN on Switch 1 and added host C and host D. The administrator has properly configured switch interfaces FastEthemetO/13 through FastEthemetO/24 to be members of the new VLAN. However, after the network administrator completed the configuration, host A could communicate with host B, but host A could not communicate with host C or host D. Which commands are required to resolve this problem?Router It show o roitfe192.168.2.15/24 192.168.2.15;24192.168.3.15124 192.168.3.16124A. Router(config)# interface fastethernet 0/1.3Router(config-if)# encapsulation dotlq 3Router(config-if)# ip address 192.168.3.1 255.255.255.0B. Router(config)# router ripRouter(con fig-router)# network 192.168.1.0Router(con fig-router)# network 192.168.2.0Router(config-router)# network 192.168.3.0C. Switch 1# vlan databaseSwitch l(vlan)# vtp v2-modeSwitch 1 (vlan)# vtp domain ciscoSwitch l(vlan)# vtp serverD. Switch 1 (config)# interface fastethernet 0/1Switch 1 (config-if)# switchport mode trunkSwitch 1 (config-if)# switchport trunk encapsulation islAnswer： A解释一下:这是种多VLAN间通讯问题,虽然都同在一台互换机上,但是由于处在不同 VLAN中,而导致了不同VLAN中主机是不能通讯。这时咱们就需要借助与trunk和三层路由 功能了，在互换机和路由器之间封装TRUNK,这样可以容许互换机间二层通讯,但是由于 两个VLAN是划分到不同网段中，因而需要借助路由器路由功能来实现三层可达，可以将 VLAN中主机网关指定为路由器与该VLAN相连子接口地址,这样VLAN中数据包就都会发 往网关,而由网关来进行进一步转发。在这个题中,题目给出了路由器子接口网段，而又给出了VLAN2与路由器相连接口IP地址, 因此剩余种网段就是给VLAN 3了,因此要在路由器上将与一种子接口划分到VLAN 3, 并给其分派另种网段中IP地址。这样就可以了。11. What are two recommended ways of protecting network device configuration files from outside network security threats?(Choose two.)A. Allow unrestricted access to the console or VTY ports.B. Use a firewall to restrict access from the outside to the network devices.C. Always use Telnet to access the device command line because its data is automatically encrypted.D. Use SSH or another encrypted and authenticated transport to access device configurations.E. Prevent the loss of passwords by disabling password encryption.Answer： BD解释一下:要保证外部安全站点可以访问我网络,这就涉及到了安全问题了,咱们可以 使用防火墙来限制外网中来设备;也可以通过SSH或加密和认证来控制。12. Refer to the exhibit. The access list has been configured on the S0/0 interface of router RTB in the outbound direction. Which two packets, if routed to the interface, will be denied?(Choosetwo.) access-list 101 deny tcp 192.168.15.32 0.0.0.15 any eq telnetaccess-list 101 permit ip any anyTelnelServerA. source ip address： 192.168.15.5； destination port： 21B. source ip address:» 192.168.15.37 destination port： 21C. source ip address:, 192.168.15.41 destination port： 21D. source ip address:, 192.168.15.36 destination port： 23E. source ip address： 192.168.15.46； destination port： 23F. source ip address:, 192.168.15.49 destination port： 23Answer： DE解释一下:这个访问列表定义两个语句:access-list 101 deny tcp 192.168.15.32 0.0.0.15 any eq telnet access-list 101 permit ip any any 在访问列表中匹配顺序是从上到下,如果匹配了某一句,就退出访问列表,如果没有就始终 往下匹配，在访问列表中有一句隐含回绝所有。因此不论怎么样均有一句是能被匹配。在上 题中，她定义第一句是回绝到从192.168.15.32-192.168.15.47发出任何telnet流量，然后第二 句定义就是容许所有IP流量。并且要明确telnet流量使用是端口23,因此这个题答案就很明确 了。13. Refer to the exhibit. Switch! has just been restarted and has passed the POST routine. Host A sends its initial frame to Host C. What is the first thing the switch will do as regards populating the switching table?Host BMAC - 000A.8A47.E612MAC - 000B.DB95.2EE9IP-192.168.23.4導IP -192.168.23.12HostDA. Switch 1 will add 192.168.23.4 to the switching table.B. Switch 1 will add 192.168.23.12 to the switching table.C. Switch 1 will add 000A.8A47.E612 to the switching table.D. Switch 1 will add 000B.DB95.2EE9 to the switching table.Answer： C解释一下:互换机重新启动了，这个时候互换机MAC地址表是空,当主机A发送数据给主机 C而通过互换机时,互换机依照她工作原理她要进行原MAC地址学习，而由于对于这个目 MAC地址无记录，而将这个流量从除收到这个接口外所有接口泛洪出去。因此在最开始步中，互换机是记录下主机AMAC地址000A.8A47.E612到她MAC地址表中。14. he user of Hostl wants to ping the DSL modem/router at 192.168.1.254. Based on the HostlARP table that is shown in the exhibit, what will Hostl do?A. send a unicast ARP packet to the DSL modem/routerB. send unicast ICMP packets to the DSL modem/routerC. send Layer 3 broadcast packets to which the DSL modem/router respondsD. send a Layer 2 broadcast that is received by Host2, the switch, and the DSL modem/routerAnswer： B解释下:在下面表中咱们可以看到ARP表中关于于192.168.1.254ARP条目,因此在这主机 都只需要发送单播ICMP包到DSL modem/router即可15. Refer to the exhibit. What is the most efficient summarization that R1 can use to advertise its networks to R2?172.1.5.0/24172.1.6.0/24172.1.4.128/25172.1.7.0/24A. 172.1.0.0/22B. 172.1.0.0/21C. 172.1.4.0/22D. 172.1.4.0/24172.1.5.0/24172.1.6.0/24172.1.7.0/24E. 172.1.4.0/25172.1.4.128/25172.1.5.0/24172.1.6.0/24172.1.7.0/24Answer： C解释一下:这还是种关于汇总问题。规定R1将所有网段用汇总条目发送给R2,由于这些 条目网络位是相似都为172.1,因此在这需要汇总只是第3个八位,将4, 4, 5, 6, 7这些写成二进制形式,然后找出相似位数,则有相似位数字节就是她们掩码位数,而最小有相似位 最小数字就是她们基数位,因此R1告示出去汇总条目为172.240/22。16. Refer to the exhibit. Assume that all router interfaces are operational and correctly configured.In addition, assume that OSPF has been correctly configured on router R2. How will the defaultroute configured on RI affect the operation of R2?A. Any packet destined fbr a network that is not directly connected to router RI will be dropped.B. Any packet destined for a network that is not directly connected to router R2 will be dropped immediately.C. Any packet destined for a network that is not directly connected to router R2 will be dropped immediately because of the lack of a gateway on RI.D. The networks directly connected to router R2 will not be able to communicate with the172.16.100.0, 172.16.100.128, and 172.16.100.64 subnetworks.E. Any packet destined for a network that is not referenced in the routing table of router R2 will be directed to R1. R1 will then send that packet back to R2 and a routing loop will occur.Answer： E解释一下:在RI上产生了一种OSPF缺省路由,出接口指定为S0/0,这条缺省路由以5类LSA 形式告示给J'R2,于是R2上也有了一条标记为E2 0.0.0.0/0出接口为SerialO/O路由。因此R2收到任何路由表中没有目网段时,就将指定给R1,而RI依照缺省路山出接口又将数据包 发往R2,这样就形成了一种路由环路。17. A network interface port has collision detection and carrier sensing enabled on a shared twisted pair network. From this statement, what is known about the network interface port?A. This is a 10 Mb/s switch port.B. This is a 100 Mb/s switch port.C.rPhis is an Ethernet port operating at half duplex.D. This is an Ethernet port operating at full duplex.E. This is a port on a network interface card in a PC.Answer： C解释一下:一种接口有冲突检测和载波侦听,并且是使用双绞线网络,那么对于这个接口咱 们可以推测出她是以太接口,并且是工作在半双工模式下。10.10.0.020. Refer to the topology and router configuration shown in the graphic. A host on the LAN is accessing an FTP server across the Internet. Which of the following addresses could appear as a source address for the packets forwarded by the router to the destination server?interface Striall«1ip address 200 2 2.18 256266 256 2622Ip nat outside t< InterfaceFastEthemetOSI： 200.2.2.18/30ip address 10.10.0.1 25555255.0fjQ 10 I 4開煙Testinside nat pool test 199 99 9.40 199593.62 netmMk 256 255255224 nat Inside source list 1 pool testip rout« 。 0 0X) 0 20022.17 (acc”粕 11 permit 10.1000 0.0.0 255A. 10.10.0.1B. 10.10.0.2C. 199.99.9.33D. 199.99.9.57E. 200.2.2.17F. 200.2.2.18Answer： D解释一下:这是个NAT地址转换题目,在这fi)/0接口连接下为私有地址,这些地址是不能同 外网进行通讯,这时就借助NAT,将内网私有地址转换为可以在公网上通讯地址,咱们看到 NAT POOL中定义转换后公有地址为199.99.9.40到199.99.9.62,则表达这段地址是我转换后 内网全局地址，因此HOST想要穿过INTERNET访问FTP服务器,则需要转换为公有地址 199.99.9.40到199.99.9.62之内地址，在上面答案中只有地址199.99.9.57满足条件，因此答案 就是D 了。21. A company is installing IP phones. The phones and office computers connect to the same device. To ensure maximum throughput for the phone data, the company needs to make sure that the phone traffic is on a different network from that of the office computer data traffic. What is the best network device to which to directly connect the phones and computers, and what technology should be implemented on this device?(Choose two.)A. hubB. routerC. switchD. STPE. subinterfacesF. VLANAnswer： CF解释一下:公司语音设备和办公设备都连在相似设备上,还要保证语音数据流在不同与公司 办公数据流量,最佳网络设备固然是互换机了,然后运用VLAN技术就完全可以满足所有规 定了。22. Refer to the exhibit. Which statement describes DLCI 17?A. DLCI 17 describes the ISDN circuit between R2 and R3.B. DLCI 17 describes a PVC on R2. It cannot be used on R3 or R1.C. DLCI 17 is the Layer 2 address used by R2 to describe a PVC to R3.D. DLCI 17 describes the dial-up circuit from R2 and R3 to the service provider.Answer： C解释下：DLCI是在Frame-relay中描述二层信息地址,她地位等同于以太网中MAC地址。咱们以R2上DLCI 17来看,DLCI 17描述是:从这个接口出去目地为R3接口这条PVC二层地 址为17。23. Which routing protocol by default uses bandwidth and delay as metrics?A. RIPB. BGPC. OSPFD. EIGRPAnswer： D解释一下:在咱们路由合同中使用复合度量合同只有IGP和EIGPR,而她们在默认状况下是 使用带宽和延时来计算度量。25. In the implementation of VLSM techniques on a network using a single Class C IP address, which subnet mask is the most efficient for point-to-point serial links?A. 255.255.255.0B. 255.255.255.240C. 255.255.255.248D. 255.255.255.252E. 255.255.255.254Answer： D解释下:在点到点链路上由于只需要分派两个地址给两端就可以J,因此加上网络地址和 广播地址,这个网段也就只需要有4个地址了,因此网络位需要匹配30位,掩码就为 255.255.255.252.26. Refer to the exhibit. The networks connected to router R2 have been summarized as a 192.168.176.0/21 route and sent to RI. Which two packet destination addresses will RI forward toR2?(Choose two.)OtherNetworksA. 192.168.194.160B. 192.168.183.41C. 192.168.159.2D. 192.168.183.255E. 192.168.179.4F. 192.168.184.45Answer： BE解释一下:这个题其实就是考察汇总问题,她说意思是R2发送了一种汇总路由 192.168.176.0/21给R1,哪两个包文目地R1仍将转发给R2。这还是汇总问题种反向考察, 依照21位掩码位数可以推断在第3个八位字节前5位是相似,不同是背面3位，而将176写成二 进制形式为1011 0000,因此可以看出来明细路由可以是176-183,因此在上面答案中可以很 容易看到答案B和E是咱们明细路由。27. Refer to the exhibit. Switch-1 needs to send data to a host with a MAC address of 00b0,d056.efa4. What will Switch-1 do with this data?Switch-1# show mac address-tableDynamic Addresses Count3Secure Addresses (User-defined) Count0Static Addresses (User-defined) Count0System Self Addresses Count41Total Mad/却鱒芈s1 56 Non-static AtidrSssw CDestination Address Address Type VLAN Destination Port0010 OdeO e289Dynamic1FastEttiemetO/10010 7b00 1540Dynamic2FastEthernetO/30010 7boe1,1545Dynamic2FastEthemetO/2A. Switch-1 will drop the data because it does not have an entry for that MAC address.B. Switch-1 will flood the data out all of its ports except the port from which the data originated.C. Switch-1 will send an ARP request out all its ports except the port from which the data originated.D. Switch-1 will forward the data to its default gateway.Answer： B解释一下:一方面Switch 1需要发送种数据到MAC地址为00b0.d056.efa4主机,理解到目地 后,就查看她MAC地址表,然后发当前MAC地址表中没有这个MAC地址条目存在。互换 机在收到未知单播,组播和广播时,都采用是泛洪方式,往除收到数据这个接口外所有接口 都发送。因此在这儿,Switch 1也采用上泛洪方式。28. wo routers named Atlanta and Brevard are connected by their serial interfaces as shown in the exhibit, but there is no data connectivity between them. The Atlanta router is known to have a correct configuration. Given the partial configurations shown in the exhibit, what is the problem on the Brevard router that is causing the lack of connectivity?AtlantaBrevardAtlanla show intertaces MSerialO is upf line protocol is upHaidwdre is HD6457Q 0 Internet address is 192.168.10.1/24MTU 1500 bytes. BW 1544 Kbit reliablity 255.255Encapsulation HDLC, loopback not set Keepalive set (10 sec)Brevardfl show interfaces s1S»riatl is up. ine protocol is upHm 小ioRHM4570Internet address H 192.168.11.2/24MTU 1500 bytes, BW 56000 Kbit.reliability 255/255Enc4ip«iilation HDLC, loopback not set Keepalive set (10 sec)A. A loopback is not set.B. The IP address is incorrect.C. The subnet mask is incorrect.D. The serial line encapsulations are incompatible.E.rrhe maximum transmission unit (MTU) size is too large.F. The bandwidth setting is incompatible with the connected interface.Answer： B解释一下:很明显错误啊,两台路由器串行接口地址配备错误,不是在相似网段,从而导致 了不能通讯。29. Which two values are used by Spanning Tree Protocol to elect a root bridge?(Choose two.)A. amount of RAMB. bridge priorityC. IOS versionD. IP addressE. MAC addressF. speed of the linksAnswer： BE解释一下:生成树选举问题,根桥选举是通过比较B I D,而
