ActualTests.Cisco.640-802.Exam.Q.and.A.08.15.08-DDU (1130589), страница 35
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192.168.1.160/28D. 192.168.1.144/28E. 192.168.1.143/28Answer: BExplanation:The available subnets and IP ranges that are available using a /28 (255.255.255.240)subnet mask is shown below:Actualtests.com - The Power of Knowing640-802Based on this information, we need to choose an IP address within the 145-158 range,since the IP address of the Fa0/0 on the router is 192.168.1.158, leaving only answerchoice C as feasible.Reference: http://www.more.net/technical/netserv/tcpip/subnet.html#28QUESTION 275:You are a systems administrator and you are about to assign static IP addresses tovarious servers on your network. For the network 192.168.20.24/29 the router isActualtests.com - The Power of Knowing640-802assigned to the first usable host address, while the last usable host address goes toyour Sales server.
Which one of the following commands would you enter into theIP properties box of the sales server?A. IP address: 192.168.20.14 Subnet Mask: 255.255.255.248 Default Gateway:192.168.20.9B. IP address: 192.168.20.254 Subnet Mask: 255.255.255.0 Default Gateway:192.168.20.1C. IP address: 192.168.20.30 Subnet Mask 255.255.255.248 Default Gateway:192.168.20.25D. IP address: 192.168.20.30 Subnet Mask 255.255.255.240 Default Gateway:192.168.20.17E.
IP address: 192.168.20.30 Subnet Mask 255.255.255.240 Default Gateway:192.168.20.25Answer: CExplanationA subnet mask uses 29 bits. This means that it uses 5 bits in the 4th octet. This equates to255.255.255.248. This network has 3 bits for hosts. Using the 2n-2 formula (23-2) in thiscase), we are left with 6 (2*2*2 - 2 = 6) host addresses. 192.168.20.24 is the networkaddress. Therefore the next address (192.168.20.25) would be the first host address. Thisaddress must be assigned to the router, which serves as the gateway for the network. Thelast available host address would be 192.168.20.30 (192.168.20.24+6).
This address isassigned to the server. The broadcast address is 192.168.20.31.QUESTION 276:You've been assigned a single Class C address. From this, you need 8 subnets, andyour subnet mask is 255.255.255.224. Which one of the following configurationcommands would you have to use before you begin?A. Router(config)# ip classlessB. Router(config)# ip subnet-zeroC. Router(config)# ip version 6D. Router(config)# no ip classfulE. Router(config)# ip unnumberedF. Router(config)# ip all-netsAnswer: BExplanation: To get 8 subnets from a class C address, and a mask of 255.255.255.224use the reserved subnet space.
To do this, you need the command 'ip subnet-zero.'This will allow the router to use the very first subnet, which is normally reservedand unused as the network address.Prior to Cisco IOS(r) Software Release 12.0,Cisco routers, by default, did not allow an IP address belonging to subnet zero to beActualtests.com - The Power of Knowing640-802configured on an interface. However, if a network engineer working with a CiscoIOS software release older than 12.0 finds it safe to use subnet zero, the ipsubnet-zero command in the global configuration mode can be used to overcomethis restriction.
As of Cisco IOS Software Release 12.0, Cisco routers now have ipsubnet-zero enabled by default, but if the network engineer feels that it is unsafe touse subnet zero, the no ip subnet-zero command can be used to restrict the use ofsubnet zero addresses.In versions prior to Cisco IOS Software Release 8.3, the service subnet-zero commandwas used.It should be noted that even though it was discouraged, the entire address space includingsubnet zero and the all-ones subnet have always been usable.
The use of the all-onessubnet was explicitly allowed and the use of subnet zero is explicitly allowed since CiscoIOS Software Release 12.0. Even prior to Cisco IOS Software Release 12.0, subnet zerocould be used by entering the ip subnet-zero global configuration command.On the issue of using subnet zero and the all-ones subnet, RFC 1878 states, "This practice(of excluding all-zeros and all-ones subnets) is obsolete. Modern software will be able toutilize all definable networks." Today, the use of subnet zero and the all-ones subnet isgenerally accepted and most vendors support their use.
However, on certain networks,particularly the ones using legacy software, the use of subnet zero and the all-ones subnetcan lead to problems.QUESTION 277:Three Certkiller routers are connected as shown below:Taking the information shown above, which command line below would correctlyconfigure serial port0 on the Certkiller 2 router with the LAST usable host addresseson the 192.216.32.32 subnet?A. Certkiller 2(config-if)# ip address 192.216.32.63 255.255.255.248B. Certkiller 2(config-if)# ip address 192.216.32.38 255.255.255.240C.
Certkiller 2(config-if)# ip address 192.216.32.39 255.255.255.248D. Certkiller 2(config-if)# ip address 192.216.32.63 255.255.255.248 no shutE. Certkiller 2(config-if)# ip address 192.216.32.39 255.255.255.248 no shutF. Certkiller 2(config-if)# ip address 192.216.32.38 255.255.255.248Answer: FExplanation:F is the correct answer, as the last usable IP address on this subnet is 192.216.32.38. Thesubnet mask for a /29 is 255.255.255.248Mask/29 11111111.11111111.11111111.11111000 255.255.255.248Subnet 11000000.11011000.00100000.00100000 192.216.32.32Actualtests.com - The Power of Knowing640-802Broadcast 11000000.11011000.00100000.00100111 192.216.32.39Address range = 192.216.32.33 - 192.216.32.38QUESTION 278:The Certkiller Network is displayed as follows:What is a valid possible IP address configuration for Host A?A.
IP 192.168.100.31 255.255.255.240 default-gateway 192.168.100.18B. IP 192.168.100.30 255.255.255.240 default-gateway 172.16.1.1C. IP 192.168.100.20 255.255.255.240 default-gateway 192.168.100.17D. IP 192.168.100.21 255.255.255.248 default-gateway 192.168.100.17E. IP 192.168.100.19 255.255.255.248 default-gateway 172.16.1.1Answer: CExplanation:The network mask for a /28 is 255.255.255.240. The default gateway is always the IPaddress of the router on the local subnet, and the valid IP range for this network is192.168.100.17 - 192.168.100.30.
Choice C is the only one that meets all of these.Incorrect Answers:A. The IP address 192.168.100.31 is the broadcast address. It cannot be used for the host.B. The default gateway should be the fist exit point for the network that the host is on. Inthis case it should be the router interface address 192.168.100.17.Actualtests.com - The Power of Knowing640-802D.
The network uses a 28 bit subnet mask (11111111.11111111.11111111.11110000).This equates to 255.255.255.240, not 255.255.255.248.E. The network uses a 28 bit subnet mask (11111111.11111111.11111111.11110000).This equates to 255.255.255.240, not 255.255.255.248. Also, the default gateway shouldbe the fist exit point for the network that the host is on. In this case it should be the routerinterface address 192.168.100.17.QUESTION 279:A simple Certkiller network is shown below:Based on the information above, which of the following would be a valid IP addressof the PC?A.
192.168.5.55B. 192.168.5.47C. 192.168.5.40D. 192.168.5.32E. 192.168.5.14Answer: CExplanation:The network uses a 28bit subnet (255.255.255.240). This means that 4 bits are used forthe networks and 4 bits for the hosts. This allows for 14 networks and 14 hosts (2n-2).The last bit used to make 240 is the 4th bit (16) therefore the first network will be192.168.5.16.
The network will have 16 addresses (but remember that the first address isthe network address and the last address is the broadcast address). In other words, thenetworks will be in increments of 16 beginning at 192.168.5.16/28. The router interfaceE0 has the IP address 192.168.5.33. Therefore it is on the 2nd network (192.169.5.32/28).The host must also be on this network. Valid IP addresses for hosts on this network are:192.168.5.33-192.168.5.46.Incorrect Answers:Actualtests.com - The Power of Knowing640-802A.
192.168.5.55 is on network 192.168.5.48. It is not on the same network as the routerinterface.B. This is the broadcast address.D. This is the network address.E. This is not a valid address for a 28 bit subnet mask. The first network address shouldbe 192.168.5.16.QUESTION 280:An 802.1Q trunk is configured between a Certkiller switch and router CK1 as shownbelow:Which of the following are valid configuration values for the host shown in thegraphic? (Choose three)A. host A IP address: 192.1.1.65B. host A subnet mask: 255.255.255.224C.
host B IP address: 192.1.1.125D. host B default gateway: 192.1.1.65E. host C IP address: 192.1.1.166F. host C subnet mask: 255.255.255.224Answer: C, D, FExplanation:Host B resides on port 3, which is configured for VLAN 1. As shown in theconfiguration, the default gateway for VLAN is the IP address associated with the FastEthernet 0/1.1 sub-interface. Valid IP hosts for the VLAN 1 subnet is192.1.1.65-192.1.1.126.Incorrect Answers:A. The 192.1.1.65 IP address is already assigned to the router.B. Host A is in VLAN 1, so the subnet mask should be 255.255.255.192E.
Host C belongs to VLAN 10, and this IP address is not in the 192.1.1.128/27 subnet.Actualtests.com - The Power of Knowing640-802QUESTION 281:Which of the following addresses can be assigned to a host when using a subnetmask of 255.255.254.0? (Select three)A. 113.10.4.0B. 186.54.3.0C.
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