ActualTests.Cisco.640-802.Exam.Q.and.A.08.15.08-DDU (1130589), страница 38
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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 310:Part of the Certkiller network is shown below:Based on the information shown above, what is the most efficient summarizationthat Router Certkiller 1 can use to advertise its networks to Router Certkiller 2?A.
172.1.4.0/24172.1.5.0./24172.1.5.0/24172.1.6.0/24172.1.7.0/24B. 172.1.0.0/22C. 172.1.0.0/21D. 172.1.4.0/24172.1.4.128/25172.1.5.0/24172.1.6.0/24172.1.7.0/24E. 172.1.4.0/22F. None of the aboveAnswer: EActualtests.com - The Power of Knowing640-802QUESTION 311:You need subnet a Certkiller network segment. How many subnetworks and hostsare available per subnet if you apply a /28 mask to the 210.10.2.0 class C network?A. 30 networks and 6 hosts.B. 6 networks and 30 hosts.C. 8 networks and 32 hosts.D.
32 networks and 18 hosts.E. 16 networks and 14 hosts.F. None of the aboveAnswer: EExplanation:A 28 bit subnet mask (11111111.11111111.11111111.11110000) applied to a class Cnetwork uses a 4 bits for networks, and leaves 4 bits for hosts. Using the 2n-2 formula,we have 24-2 (or 2x2x2x2-2) which gives us 14 for the number of hosts, and the numberof networks is 24 = 16.Incorrect Answers:A. This would be the result of a /29 (255.255.255.248) network.B. This would be the result of a /27 (255.255.255.224) network.C.
This is not possible, as we must subtract two from the subnets and hosts for thenetwork and broadcast addresses.D. This is not a possible combination of networks and hosts.QUESTION 312:The Certkiller network was assigned the Class C network 199.166.131.0 from theISP. If the administrator at Certkiller were to subnet this class C network using the255.255.255.224 subnet mask, how may hosts will they be able to support on eachsubnet?A. 14B. 16C.
30D. 32E. 62F. 64Answer: CActualtests.com - The Power of Knowing640-802Explanation:The subnet mask 255.255.255.224 is a 27 bit mask(11111111.11111111.11111111.11100000). It uses 3 bits from the last octet for thenetwork ID, leaving 5 bits for host addresses. We can calculate the number of hostssupported by this subnet by using the 2n-2 formula where n represents the number of hostbits. In this case it will be 5. 25-2 gives us 30.Incorrect Answers:A. Subnet mask 255.255.255.240 will give us 14 host addresses.B. Subnet mask 255.255.255.240 will give us a total of 16 addresses.
However, we muststill subtract two addresses (the network address and the broadcast address) to determinethe maximum number of hosts the subnet will support.D. Subnet mask 255.255.255.224 will give us a total of 32 addresses. However, we muststill subtract two addresses (the network address and the broadcast address) to determinethe maximum number of hosts the subnet will support.E. Subnet mask 255.255.255.192 will give us 62 host addresses.F.
Subnet mask 255.255.255.192 will give us a total of 64 addresses. However, we muststill subtract two addresses (the network address and the broadcast address) to determinethe maximum number of hosts the subnet will support.QUESTION 313:What is the subnet for the host IP address 172.16.210.0/22?A.
172.16.42.0B. 172.16.107.0C. 172.16.208.0D. 172.16.252.0E. 172.16.254.0F. None of the aboveAnswer: CExplanation:This question is much easier then it appears when you convert it to binary and do theBoolean operation as shown below:IP address 172.16.210.0 = 10101100.00010000.11010010.00000000/22 mask = 11111111.11111111.11111100.00000000AND result = 11111111.11111111.11010000.00000000AND in decimal= 172 . 16 . 208 . 0QUESTION 314:What is the subnet for the host IP address 201.100.5.68/28?Actualtests.com - The Power of Knowing640-802A. 201.100.5.0B. 201.100.5.32C. 201.100.5.64D. 201.100.5.65E. 201.100.5.31F.
201.100.5.1Answer: CExplanation:This question is much easier then it appears when you convert it to binary and do theBoolean operation as shown below:IP address 201.100.5.68 = 11001001.01100100.00000101.01000100/28 mask = 11111111.11111111.11111111.11000000AND result = 11001001.01100100.00000101.01000000AND in decimal= 200 . 100 . 5 . 64QUESTION 315:Your network uses the172.12.0.0 class B address.
You need to support 459 hosts persubnet, while accommodating the maximum number of subnets. Which mask wouldyou use?A. 255.255.0.0.B. 255.255.128.0.C. 255.255.224.0.D. 255.255.254.0.E. None of the aboveAnswer: DExplanation:To obtain 459 hosts the number of host bits will be 9. This can support a maximum of510 hosts. To keep 9 bits for hosts means the last bit in the 3rd octet will be 0. This gives255.255.254.0 as the subnet mask.QUESTION 316:Using a subnet mask of 255.255.255.224, which of the IP addresses below can youassign to the hosts on this subnet? (Select all that apply)A.
16.23.118.63B. 87.45.16.159Actualtests.com - The Power of Knowing640-802C. 92.11.178.93D. 134.178.18.56E. 192.168.16.87F. 217.168.166.192Answer: C, D, EExplanation:Since the subnet mask is 255.255.255.224, the number of network hosts that is availableis 30. Every network boundary will be a multiple of 32.
This means that every subnetwill be a multiple (0, 32, 64, 96, 128, 160, 192, 224) and the broadcast address for eachof these subnets will be one less this number (31, 63, 95, 127, 159, 191, 223). Therefore,any IP address that does not end in one of these numbers will be a valid host IP address.C. Valid Host in subnetwork 2 (92.11.178.64 to 92.11.178.95)D. Valid Host in subnetwork 1 (134.178.18.32 to 134.178.18.63)E. Valid Host in subnetwork 2 (192.168.16.64 to 192.168.16.95)Incorrect Answers:A.
This will be the broadcast address for the 16.23.118.32/27 network.B. This will be the broadcast address for the 87.45.16.128/27 networkF. This will be the network address for the 217.168.166.192/27 network.QUESTION 317:Your ISP has assigned you the following IP address and subnet mask:IP address: 199.141.27.0Subnet mask: 255.255.255.240Which of the following addresses can be allocated to hosts on the resulting subnet?(Select all that apply)A.
199.141.27.2B. 199.141.27.175C. 199.141.27.13D. 199.141.27.11E. 199.141.27.208F. 199.141.27.112Answer: A, C, DExplanation:IP address = 11001000.10001101.00011011.00000000 = 199.141.27.0Subnet mask = 11111111.11111111.11111111.11110000 = 255.255.255.240Subnet # = 11001000.10001101.00011011.00000000 = 199.141.27.0Broadcast = 11001000.10001101.00011011.00001111 = 199.141.27.15The valid IP address range = 199.141.27.1 - 199.141.27.14Actualtests.com - The Power of Knowing640-802QUESTION 318:The IP network 210.106.14.0 is subnetted using a /24 mask. How many usablenetworks and host addresses can be obtained from this?A.
1 network with 254 hostsB. 4 networks with 128 hostsC. 2 networks with 24 hostsD. 6 networks with 64 hostsE. 8 networks with 36 hostsAnswer: AExplanation:A subnet with 24 bits on would be 255.255.255.0. Since this is a class C network, thissubnet can have only 1 network and 254 usable hosts.QUESTION 319:Given that you have a class B IP address network range, which of the subnet masksbelow will allow for 100 subnets with 500 usable host addresses per subnet?A. 255.255.0.0B. 255.255.224.0C.
255.255.254.0D. 255.255.255.0E. 255.255.255.224Answer: CExplanation:Using the 2n-2 formula for host addresses, 29-2 = 510 host address, so a 9-bit subnetmask will provide the required number of host addresses. If these 9 bits are used for thehosts in a class B network, then the remaining 7 bits are used for the number of networks.Again using the 2n-2 formula, we have 2n-2 = 126 networks that are available.Incorrect Answers:A. This will provide for only 1 network with 216-2 = 65534 hostsB. This will provide for 6 networks with 8190 host addresses.D. This will provide 254 networks and 254 hosts.E.
This will provide 2046 different networks, but each network will have only 30 hosts.Actualtests.com - The Power of Knowing640-802QUESTION 320:You have a class C network, and you need to design it for 5 usable subnets with eachsubnet handling a minimum of 18 hosts each. Which of the following network masksshould you use?A. 225.225.224.0.B. 225.225.240.0.C. 225.225.255.0.D. 255.255.255.224E. 225.225.255.240Answer: DExplanation:The default subnet mask for class C network is 255.255.255.0. If one has to create 5subnets, then 3 bits are required. With 3 bits we can create 8 subnets. The remaining 5bits are used for Hosts.
One can create 30 hosts using 5 bits in host field. This matcheswith the requirement.Incorrect Answers:A, B: This is an illegal subnet mask for a class C network, as the third octet can not bedivided when using a class C network.C. This is the default subnet mask for a class C network. It provides for one network,with 254 usable host IP addresses.E. This subnet mask will provide for 14 separate networks with 14 hosts each.
This doesnot meet the requirement of a minimum of 18 hosts.QUESTION 321:The 213.115.77.0 network was subnetted using a /28 subnet mask. How many usablesubnets and host addresses per subnet were created as a result of this?A. 2 networks with 62 hostsB. 6 networks with 30 hostsC.
16 networks and 16 hostsD. 62 networks and 2 hostsE. 14 networks and 14 hostsF. None of the aboveAnswer: FExplanation:A class C subnet with a 28 bit mask requires 4 bits for the network address, leaving 4 bitsfor host addresses. Using the 2n-2 formula (24-2 in this case) we have 14 host addressesActualtests.com - The Power of Knowing640-802and 16 network addresses.Incorrect Answers:A. This would be the result of a /26 network maskB. This would be the result of a /27 network maskC. Remember we need to always subtract two for the network and broadcast addresses,so this answer is incorrect.D. This would be the result of a /30 network mask.QUESTION 322:The 201.145.32.0 network is subnetted using a /26 mask. How many networks andIP hosts per network exists using this subnet mask?A.
4 networks and 64 hostsB. 64 networks and 4 hostsC. 4 networks and 62 hostsD. 62 networks and 2 hostsE. 6 network and 30 hostsAnswer: CExplanation:A class C network with a 26 bit mask requires 2 bits for the network address, leaving 6bits for host addresses. Using the 2n-2 formula (22 for the network and 26-2for hosts) wehave 4 network addresses and 62 host addresses.Incorrect Answers:A, B: This is not a possible combination. No network mask will provide for 64 usablehosts, because we must always subtract 2 for the network and broadcast address.D. This would be the result of a /30 mask.E.
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