子网掩码试题.docx

子网掩码试题 子网掩码练习题(2009-04-03 00:02:57)标签：子网掩码 it 分类：计算机相关 转载于网络1. 网段地址154.27.0.0的网络，若不做子网划分，能支持（d ）台主机 216 - 2 = 65534A254 B1024 C65,536 D655342 某公司申请到一个C类IP地址，但要连接9个的子公司，最大的一个子公司有12台计算机，每个子公司在一个网段中，则子网掩码应设为（ d）。A255.254.255.240 B255.255.255.192C255.255.255.128 D255.255.255.2403 下列哪项是合法的IP主机地址？（d ） A127.2.3.5 B1.255.15.255/22 00000001.11111111.000011 11.11111111 /22C255.23.200.9 D192.240.150.255/23 4 C类地址最大可能子网位数是（a ） 192.240.1001011 0.11111111 00000000.000000 00A6 B8 C12 D145与10.110.12.29 mask 255.255.255.224属于同一网段的主机IP地址是（acd ） 10.01101110.00001100.000 11101A10.110.12.0 B10.110.12.32 001 00000C10.110.12.31 D10.110.12.30 000 111116 一个子网网段地址为2.0.0.0掩码为255.255.224.0网络，他一个有效子网网段地址是（b ）A2.1.16.0 B2.2.160.0 255.111111111.111 00000.000000002. 00000001.000 10000.00000000C2.3.48.0 D2.4.172.0 2. 00000010.101 00000.000000002. 00000011.001 10000.000000002. 00000100.101 01100.000000007 一个子网网段地址为5.96.0.0掩码为255.224.0.0网络，它允许的最大主机地址是（a）A5.127.255.254 B5.128.255.254 5.011 00000.00000000.00000000 5.01111111.11111111.11111110255.111 00000.00000000.00000000. C5.126.255.254 D5.129.255.2558 在一个子网掩码为255.255.248.0的网络中，（d）是合法的网络地址 11111111.11111111.11111 000.00000000A150.150.42.0 B150.150.15.0 00101 010.00000000C150.150.7.0 D150.150.96.0 00001 111.00000000 00000 111.0000000001100 000.00000000 9 如果C类子网的掩码为255.255.255.224，则包含的子网位数、子网数目、每个子网中主机数目正确的是（b）A2，2，62 B3，8，30 11100000C3，14，30 D5，8，610 IP地址为172.168.120.1/20 ，则子网ID、子网掩码、子网个数分别为( a)172.168. 1111 0000.0000 255.255.240.0A172.168.112.0、255.255.240.0 、16 172.168. 0111 0000.0000000B172.168.108.0、255.255.240.0、16C1 72.168.96.0、255.240.0.0、256E、172.168.96.0、255.240.0.0、1611 IP地址为12668240，子网掩码为25519200，求该网段的广播地址（c）A126.68.24.255 B126.64.255.255 255.11000000.0.0 126.01000000.00000000.00000000C126.127.255.255 D126.255.255.255 126.01000100. 126.64.0.0 12下列可以分配给主机的IP地址（e）A131.107.256.80/17 B231.222.0.11/27 C198.121.252.0/23 D126.3.255.255/15 E172.168.126.255/2213一IP为202.196.200.173，掩码为255.255.255.192，求其子网网络ID以及该IP所在子网的广播地址（）A202.196.200.172、202.196.200.255B202.196.200.128、202.196.200.191C202.196.200.0、202.196.200.255D202.196.200.160、202.196.200.19214 一个B类IP地址最多可用（ ）位来划分子网A8 B14 C16 D2215给定IP地址167.77.88.99和掩码255.255.255.240，它的子网号是什么？它的广播地址是什么？（）A167.77.88.96、167.77.88.111 B167.77.88.64、167.77.88.111C167.77.88.92、167.77.88.192 D167.77.88.96、167.77.88.255网工天天一练（12.20）（子网划分专项练习!）1. 192.168.1.0/24 使用掩码255.255.255.240 划分子网，其可用子网数为（），每个子网内可用主机地址数为（）A. 14 14B. 16 14C. 254 6D. 14 62答案：2. 子网掩码为255.255.0.0 ，下列哪个 IP 地址不在同一网段中（）A. 172.25.15.201B. 172.25.16.15C. 172.16.25.16D. 172.25.201.15答案：3. B 类地址子网掩码为 255.255.255.248 ，则每个子网内可用主机地址数为（）A. 10B. 8C. 6D. 4答案：4. 对于C 类 IP 地址，子网掩码为 255.255.255.248 ，则能提供子网数为（）A. 16B. 32C. 30D. 128答案：5. 三个网段 192.168.1.0/24 ， 192.168.2.0/24 ， 192.168.3.0/24 能够汇聚成下面哪个网段（）A. 192.168.1.0/22B. 192.168.2.0/22C. 192.168.3.0/22D. 192.168.0.0/22答案：6.IP 地址219.25.23.56 的缺省子网掩码有几位？A.8B.16C.24D.32ANSWER:7.某公司申请到一个C 类IP 地址，但要连接6 个的子公司，最大的一个子公司有26 台计算机，每个子公司在一个网段中，则子网掩码应设为？A.255.255.255.0B.255.255.255.128C.255.255.255.192D.255.255.255.224答案8.一台IP 地址为10.110.9.113/21 主机在启动时发出的广播IP 是？A.10.110.9.255B.10.110.15.255C.10.110.255.255D.10.255.255.255ANSWER:9.规划一个C 类网，需要将网络分为9 个子网，每个子网最多15 台主机，下列哪个是合适的子网掩码？A.255.255.224.0B.255.255.255.224C.255.255.255.240D.没有合适的子网掩码答案10.与10.110.12.29 mask 255.255.255.224 属于同一网段的主机IP 地址是？A.10.110.12.0B.10.110.12.30C.10.110.12.31D.10.110.12.32ANSWER:11.IP 地址190.233.27.13/16 的网络部分地址是？A.190.0.0.0B.190 .233.0.0C.190.233.27.0D.190.233.27.1ANSWER:12.没有任何子网划分的IP 地址125.3.54.56 的网段地址是？A.125.0.0.0B.125.3.0.0C.125.3.54.0D.125.3.54.32ANSWER:13.一个子网网段地址为2.0.0.0 掩码为255.255.224.0 网络，他一个有效子网网段地址是？A.2.1.16.0B.2.2.32.0C.2.3.48.0D.2.4.172.0ANSWER:14.一个子网网段地址为5.32.0.0 掩码为255.224.0.0 网络，它允许的最大主机地址是？A.5.32.254.254B.5.32.255.254C.5.63.255.254D.5.63.255.255ANSWER:15.在一个子网掩码为255.255.240.0 的网络中，哪些是合法的网段地址？A.150.150.0.0B.150.150.0.8C.150.150.8.0D.150.150.16.0ANSWER:16.如果C 类子网的掩码为255.255.255.224，则包含的子网位数.子网数目. 每个子网中主机数目正确的是？A.2，2，62B.3，6，30C.4，14，14D.5，30，6ANSWER:17. 网络地址 :172.16.0.0 ，如果采用子网掩码255.255.192.0 ，那么以下说法正确的是（）A. 划分了 2 个有效子网；B. 划分了 4 个有效子网；C. 其中一个子网的广播地址为： 172.16.191.255 ；D. 其中一个子网的广播地址为： 172.16.128.255 。答案：18. 关于主机地址 192.168.19.125 （子网掩码： 255.255.255.248 ），以下说法正确的是（）A. 子网地址为： 192.168.19.120 ；B. 子网地址为： 192.168.19.121 ；C. 广播地址为： 192.168.19.127 ；D. 广播地址为： 192.168.19.128 。答案：19. 一个 C 类地址： 192.168.5.0 ，进行子网规划，要求每个子网有10 台主机，使用哪个子网掩码划分最合理（）A. 使用子网掩码255.255.255.192 ；B. 使用子网掩码255.255.255.224 ；C. 使用子网掩码255.255.255.240 ；D. 使用子网掩码255.255.255.252 。答案：20. 网络地址 192.168.1.0/24 ，选择子网掩码为255.255.255.224 ，以下说法正确的是（）A. 划分了 4 个有效子网；B. 划分了 6 个有效子网；C. 每个子网的有效主机数是30 个；D. 每个子网的有效主机数是31 个；E. 每个子网的有效主机数是32 个。答案：21. IP 地址:192.168.12.72 ，子网掩码为:255.255.255.192，该地址所在网段的网络地址和广播地址为（）A. 192.168.12.32 ， 192.168.12.127 ；B. 192.168.0.0 ， 255.255.255.255 ；C. 192.168.12.43 ， 255.255.255.128 ；D. 192.168.12.64 ， 192.168.12.127 。答案：22. 172.16.10.32/24 代表的是（）A. 网络地址；B. 主机地址；C. 组播地址；D. 广播地址。答案：23. 一个子网网段地址为10.32.0.0 掩码为 255.224.0.0 的网络，它允许的最大主机地址是（）A. 10.32.254.254 ；B. 10.32.255.254 ；C. 10.63.255.254 ；D. 10.63.255.255 。答案：子网划分专项练习1. What valid host range is the IP address 172.16.10.22 255.255.255.240 a part of?A. 172.16.10.20 through 172.16.10.22B. 172.16.10.1 through 172.16.10.255C. 17 2.16.10.16 through 172.16.10.23D. 172.16.10.17 through 172.16.10.31E. 172.16.10.17 through 172.16.10.302. What is the broadcast address of the subnet address 172.16.8.159 255.255.255.192?A. 172.16.255.255B. 172.16.8.127C. 172.16.8.191D. 172.16.8.2553. What is the broadcast address of the subnet address 192.168.10.33 255.255.255.248?A. 192.168.10.40B. 192.168.10.255C. 192.168.255.255D. 192.168.10.394. If you wanted to have 12 subnets with a Class C network ID, which subnet mask would you use?A. 255.255.255.252B. 255.255.255.248C. 255.255.255.240D. 255.255.255.2555. If you need to have a Class B network address subnetted into exactly 510 subnets, what subnetmask would you assign?A. 255.255.255.252B. 255.255.255.128C. 255.255.255.0D. 255.255.255.1926. If you are using a Class C network ID with two subnets and need 31 hosts per network, whichof the following masks should you use?A. 255.255.255.0B. 255.255.255.192C. 255.255.255.224D. 255.255.255.2487. How many subnets and hosts can you get from the network 192.168.254.0/26?A. 4 networks with 64 hostsB. 2 networks with 62 hostsC. 254 networks with 254 hostsD. 1 network with 254 hosts8. You have the network 172.16.10.0/24. How many subnets and hosts are available?A. 1 subnet with 10 hostsB. 1 subnet with 254 hostsC. 192 subnets with 10 hostsD. 254 subnets with 254 hosts9. What mask would you assign to the network ID of 172.16.0.0 if you needed about 100 subnetswith about 500 hosts each?A. 255.255.255.0B. 255.255.254.0C. 255.255.252.0D. 255.255.0.011You have a Class C 192.168.10.0/28 network. How many usable subnets and hosts do you have?A. 16 subnets, 16 hostsB. 14 subnets, 14 hostsC. 30 subnets, 6 hostsD. 62 subnets, 2 hosts13. You have the network 192.168.10.0/24. How many subnets and hosts are available?A. 1 subnet with 10 hostsB. 1 subnet with 254 hostsC. 192 subnets with 10 hostsD. 254 subnets with 254 hosts14. You have a 255.255.255.240 mask. Which two of the following are valid host IDs?A. 192.168.10.210B. 192.168.10.32C. 192.168.10.94D. 192.168.10.112E. 192.168.10.12715. You have a Class B network ID and need about 450 IP addresses per subnet. What is the bestmask for this network?A. 255.255.240.0B. 255.255.248.0C. 255.255.254.0D. 255.255.255.017. Which of the following is a valid host for network 192.168.10.32/28?A. 192.168.10.39B. 192.168.10.47C. 192.168.10.14D. 192.168.10.5419. What is the valid host range that host 192.168.10.22/30 is a part of?A. 192.168.10.0B. 192.168.10.16C. 192.168.10.20D. 192.168.0.0答案：Answers to Review Questions1. E. This is a Class B network address with 12 bits of subnetting8 in the third octet and 4in the fourth octet. The subnet in the third octet is 10, and the subnets in the fourth octetare 256 240 = 16, 32, 48, etc. Since the fourth octet is using 22, the host is in the 16 subnet, and since the next subnet is 32, the broadcast address for the 16 subnet is 31. The valid hostrange is the numbers in between, or 1730.2. C. This is a Class B network address with 10 bits of subnetting8 in the third octet and 2in the fourth octet. The subnet in the third octet is 8, and the subnets in the fourth octet are256 192 = 64, and 128. However, as long as all the subnet bits in the third octet are not allon at once, the subnets in the fourth octet really can be 0, 64, 128 and 192. This means that thehost is in the 128 subnet and since the next subnet is 192, our broadcast address is 8.191.3. D. This is a Class C network address with 5 bits of subnetting. The valid subnets are 256 248= 8, 16, 24, 32, 40, etc. Since the host ID is 33, we are in the 32 subnet. The next subnet is 40,so our broadcast address is 39.4. C. Take a look at the answers and see which subnet mask will give you what you need for subnetting.252 gives you 62 subnets, 248 gives you 30 subnets, 240 gives you 14 subnets, and 255is invalid. Only the third option (240) gives you what you need.5. B. If you use the mask 255.255.255.0, that only gives you 8 subnet bits, or 254 subnets. You aregoing to have to use 1 subnet bit from the fourth octet, or 255.255.255.128. This is 9 subnet bits(29 2 = 510).6. B. To answer this, you must be able to determine which Class C mask provides how many hosts andsubnets. The 255.255.255.0 mask provides one network with 254 hosts. The 255.255.255.192 providestwo subnets each with 62 hosts. The 255.255.255.224 provides 6 subnets, each with 30 hosts,and the 255.255.255.248 mask provides 30 subnets, each with 6 hosts.7. B. The Class C mask of 255.255.255.192 provides two subnets (four if you are using subnetzerowhich Cisco is not!), each with 62 hosts.8. B. The third octet is used for all subnets, and the fourth octet is used only for hosts. 8 bits forsubnetting, 8 bits for hosts. However, a subnet is already listed, so you have one subnet with 254hosts. If the question stated 172.16.0.0/24, then the answer would be 254 subnets each with254 hosts.9. B. This one takes some thought. 255.255.255.0 would give you 254 hosts each with 254 subnets.Doesnt work for this question. 255.255.254.0 would provide 126 subnets, each with 510hosts; the second option looks good. 255.255.252.0 is 62 subnets, each with 1022 hosts. So255.255.254.0 is the best answer.11B. To answer this, you must know that /28 is 255.255.255.240. 256 240 = 16. You subtract2 from this number for all subnet bits and host bits on/off, so the answer is 14 subnets with 14hosts each.13. B. This is a Class C network using a default mask. This provides a simple single network with254 hosts.14. A, C. To answer this, just do 256 240 = 16. Keep adding 16 together until you reach the subnetmask value. 16 + 16 = 32. Well keep adding 16 until we have all our subnets: 16, 32, 48, 64, 80,96 , 112, 128, 144, 160, 176, 192, 208, 224. The broadcast addresses for each subnet are thenumbers right before the next subnet and are not valid hosts. The answers are 210 and 94.15. C. Start with 255.255.255.0. This provides 254 subnets, each with 254 hosts. Move the subnetbits right if you need more subnets; move them left if you need more hosts. Since we need morehosts, were going to take away subnet bits. The next mask then is 255.255.254.0, which provides7 subnet bits and 9 host bits, or 126 subnets, each with 510 hosts.17. A. You have to know that /28 is 255.255.255.240. 256 240 = 16, 32, 48, etc. The host ID of34 is in the 32 subnet; the next subnet is 48, so the broadcast address is 47. The valid host rangeis 3346, so answer A is correct.19. C. This is as easy as they get. A /30 is a 255.255.255.252 mask. 256 252 = 4, 8, 12, 16, 20, 24. Thishost is in the 20 subnet, the broadcast address is 23 and the valid host range is 21 and 22.IP地址和子网掩码习题1.下列什么有效的主机排列是IP地址172.16.10.22 255.255.255.240的一部分?A.172.16.10.20 到 172.16.10.22B.172.16.